From 3e47c5c4af85420ea75ef8e6b08d972e74c31137 Mon Sep 17 00:00:00 2001
From: wieerwill <robert.jeutter@gmx.de>
Date: Tue, 15 Feb 2022 18:02:55 +0100
Subject: [PATCH] Listen umgewandelt und formatiert

---
 Kryptographie.pdf |  Bin 526830 -> 545932 bytes
 Kryptographie.tex | 1560 +++++++++++++++++++++++++++------------------
 2 files changed, 950 insertions(+), 610 deletions(-)

diff --git a/Kryptographie.pdf b/Kryptographie.pdf
index 6fcc23a0d6f0ab719d98dc42e133fc00d89f8be0..3c042e79ece61227211748d3d37c3b67e72f80c6 100644
GIT binary patch
delta 367798
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diff --git a/Kryptographie.tex b/Kryptographie.tex
index 1cfcb8f..f3b9b2d 100644
--- a/Kryptographie.tex
+++ b/Kryptographie.tex
@@ -136,27 +136,27 @@
 
     Aufgaben von Kryptosystemen
     \begin{description*}
-    \item[Konzelation] Geheimhaltung/Vertraulichkeit/Zugriffsschutz(kein Unberechtigter kann Nachrichteninhalt mithören oder mitlesen
-    \item[Integrität/Fälschungsschutz] stelle sicher, dass Nachrichten auf dem Übertragungsweg nicht manipuliert worden sind
-    \item[Authentizität/Signaturen] Garantiere Absenderidentität; Bob kann kontrollieren, dass Nachricht vom behaupteten Absender Alice kommt
-    \item[Nichtabstreitbarkeit] Bob kann gegenüber Dritten beweisen,dass die Nachricht in der empfangenen Form vom behaupteten Absender Alice kam
-    \end{description*} 
+        \item[Konzelation] Geheimhaltung/Vertraulichkeit/Zugriffsschutz(kein Unberechtigter kann Nachrichteninhalt mithören oder mitlesen
+        \item[Integrität/Fälschungsschutz] stelle sicher, dass Nachrichten auf dem Übertragungsweg nicht manipuliert worden sind
+        \item[Authentizität/Signaturen] Garantiere Absenderidentität; Bob kann kontrollieren, dass Nachricht vom behaupteten Absender Alice kommt
+        \item[Nichtabstreitbarkeit] Bob kann gegenüber Dritten beweisen,dass die Nachricht in der empfangenen Form vom behaupteten Absender Alice kam
+    \end{description*}
 
-    \note{Message Authentication Code (MAC)}{ 
-    Funktion, die aus einer Nachricht x einen Code $mac=MAC(x)$ berechnet. Diese Funktion ist ein Geheimnis von legitimen Sendern von Nachrichten an Bob. Insbesondere kann Eva bei gegebener Nachricht x' keinen korrekten MAC für x' berechnen. Bob verfügt über ein Prüfverfahren, das es ihm erlaubt, ein empfangenes Paar $(x,mac)$ darauf zu testen, ob der zweite Teil der zu x gehörende MAC-Wert ist. Wenn Alice die einzige Instanz ist, die die geheime Funktion MAC kennt, dann kann Bob sogar überprüfen, ob Alice tatsächlich die Absenderin ist.
+    \note{Message Authentication Code (MAC)}{
+        Funktion, die aus einer Nachricht x einen Code $mac=MAC(x)$ berechnet. Diese Funktion ist ein Geheimnis von legitimen Sendern von Nachrichten an Bob. Insbesondere kann Eva bei gegebener Nachricht x' keinen korrekten MAC für x' berechnen. Bob verfügt über ein Prüfverfahren, das es ihm erlaubt, ein empfangenes Paar $(x,mac)$ darauf zu testen, ob der zweite Teil der zu x gehörende MAC-Wert ist. Wenn Alice die einzige Instanz ist, die die geheime Funktion MAC kennt, dann kann Bob sogar überprüfen, ob Alice tatsächlich die Absenderin ist.
     }
 
     \note{Cäsar-Chiffre}{
-    Cäsar ließ Texte verschlüsseln, indem man nimmt immer den Buchstaben, der im Alphabet drei Positionen ,,weiter rechts'' steht, mit ,,wrap around'' am Ende. Nachteil ist offensichtlich: Wer den Trick kannte, konnte jede Nachricht mitlesen. Es gibt einen Schlüssel
+        Cäsar ließ Texte verschlüsseln, indem man nimmt immer den Buchstaben, der im Alphabet drei Positionen ,,weiter rechts'' steht, mit ,,wrap around'' am Ende. Nachteil ist offensichtlich: Wer den Trick kannte, konnte jede Nachricht mitlesen. Es gibt einen Schlüssel
     }
 
     \note{Verschiebechiffre}{
-    Eine einfache ,,Verbesserung'' der Cäsar-Chiffre: Verschiebe zyklisch um eine andere Anzahl k von Buchstaben als nur 3. Um hier die Verschlüsselung und die Entschlüsselung durchzuführen, musste man als ,,Schlüssel''  nur das Bild von A kennen, alternativ die Verschiebeweite k als Zahl. Es gibt dann 21 Schlüssel.
+        Eine einfache ,,Verbesserung'' der Cäsar-Chiffre: Verschiebe zyklisch um eine andere Anzahl k von Buchstaben als nur 3. Um hier die Verschlüsselung und die Entschlüsselung durchzuführen, musste man als ,,Schlüssel''  nur das Bild von A kennen, alternativ die Verschiebeweite k als Zahl. Es gibt dann 21 Schlüssel.
     }
 
     \note{Substitutionschiffre}{
-    Sie sagt, dass das Bild eines Buchstabens ein ganz beliebiger anderer Buchstabe sein soll. Dabei müssen natürlich verschiedene Buchstaben auf verschiedene Buchstaben abgebildet werden. Es er gibt sich eine Chiffre, die durch eine Tabelle mit ganz beliebiger Buchstabenanordnung gegeben ist.
-    Wenn man hier ver- und entschlüsseln möchte, muss man die gesamte zweite Tabellenzeile kennen. Diese kann hier also als ,,Schlüssel''  dienen. Es gibt $21!\approx 5,11* 10^{19}$ viele verschiedene Schlüssel.
+        Sie sagt, dass das Bild eines Buchstabens ein ganz beliebiger anderer Buchstabe sein soll. Dabei müssen natürlich verschiedene Buchstaben auf verschiedene Buchstaben abgebildet werden. Es er gibt sich eine Chiffre, die durch eine Tabelle mit ganz beliebiger Buchstabenanordnung gegeben ist.
+        Wenn man hier ver- und entschlüsseln möchte, muss man die gesamte zweite Tabellenzeile kennen. Diese kann hier also als ,,Schlüssel''  dienen. Es gibt $21!\approx 5,11* 10^{19}$ viele verschiedene Schlüssel.
     }
 
     \note{Vigenère-Verschlüsselung (16.Jhd)}{
@@ -165,57 +165,77 @@
 
     kryptographische Verfahren mit Schlüsseln
     \begin{description*}
-    \item[Symmetrisch (Private-Key)] Es gibt einen geheimen Schlüssel $k$, den beide der kommunizierenden Parteien kennen müssen. Bei Konzelationssystemen bedeutet dies etwa, dass das Verschlüsselungsverfahren und das Entschlüsselungsverfahren beide diesen Schlüssel $k=k'$ benutzen. Bsp. AES (Advanced Encryption Standard).
-    \item[Asymmetrisch (Public-Key)] Nur eine Seite hat einen geheimen Schlüssel $k'$, die andere Seite benutzt einen ,,öffentlichen'' Schlüssel $k\not =k'$. Bsp. RSA
+        \item[Symmetrisch (Private-Key)] Es gibt einen geheimen Schlüssel $k$, den beide der kommunizierenden Parteien kennen müssen. Bei Konzelationssystemen bedeutet dies etwa, dass das Verschlüsselungsverfahren und das Entschlüsselungsverfahren beide diesen Schlüssel $k=k'$ benutzen. Bsp. AES (Advanced Encryption Standard).
+        \item[Asymmetrisch (Public-Key)] Nur eine Seite hat einen geheimen Schlüssel $k'$, die andere Seite benutzt einen ,,öffentlichen'' Schlüssel $k\not =k'$. Bsp. RSA
     \end{description*}
 
     \note{Das Kerckhoffs-Prinzip (1883)}{
-    besagt,dass man davon ausgehen muss, dass Eva die Struktur des Verschlüsselungsverfahrens kennt und die Sicherheit nur von der Geheimhaltung des Schlüssels abhängen darf.
-    \begin{enumerate*}
-    \item Geheimhaltung eines Verfahrens ist schwer sicherzustellen
-    \item Verfahren sind aufwendig zu entwickeln. Ist das geheime Verfahren einmal bekannt, soist es nutzlos. (Mehrfach passiert: Enigma, GSM-Verfahren ,Stromchiffre RC4)
-    \item Allgemein bekannte Verfahren können von mehr Experten auf ,,Sicherheit'' geprüft werden. Findet niemand einen erfolgreichen Angriff, so kann man eher auf Sicherheit des Verfahrens vertrauen
-    \item Nur offen gelegte Verfahren können standardisiert werden und weite Verbreitung finden (DES, AES)
-    \end{enumerate*}
+        besagt,dass man davon ausgehen muss, dass Eva die Struktur des Verschlüsselungsverfahrens kennt und die Sicherheit nur von der Geheimhaltung des Schlüssels abhängen darf.
+        \begin{enumerate*}
+            \item Geheimhaltung eines Verfahrens ist schwer sicherzustellen
+            \item Verfahren sind aufwendig zu entwickeln. Ist das geheime Verfahren einmal bekannt, soist es nutzlos. (Mehrfach passiert: Enigma, GSM-Verfahren ,Stromchiffre RC4)
+            \item Allgemein bekannte Verfahren können von mehr Experten auf ,,Sicherheit'' geprüft werden. Findet niemand einen erfolgreichen Angriff, so kann man eher auf Sicherheit des Verfahrens vertrauen
+            \item Nur offen gelegte Verfahren können standardisiert werden und weite Verbreitung finden (DES, AES)
+        \end{enumerate*}
     }
 
     \section{Symmetrische Verschlüsselung, Sicherheitsmodelle}
     In diesem Teil beschäftigen wir uns ausschließlich mit symmetrischen Konzelationsverfahren, bei denen also Alice und Bob sich auf einen geheimen Schlüssel geeinigt haben.
 
     Mögliche Kommunikationsszenarien:
-    - Alice will nur einmal eine Nachricht (mit bekannter maximaler Länge) an Bob schicken.
-    - Alice will mehrere Nachrichten mit bekannter maximaler Länge schicken.
-    - Alice will beliebig viele Nachrichten beliebiger Länge schicken.
+    \begin{itemize*}
+        \item Alice will nur einmal eine Nachricht (mit bekannter maximaler Länge) an Bob schicken.
+        \item Alice will mehrere Nachrichten mit bekannter maximaler Länge schicken.
+        \item Alice will beliebig viele Nachrichten beliebiger Länge schicken.
+    \end{itemize*}
 
     Angriffsszenarien/Bedrohungsszenarien:
     Das Kerkhoffs-Prinzip impliziert, dass Eva das Ver- und das Entschlüsselungsverfahren kennt (nur den Schlüssel nicht). Folgende Möglichkeiten kann sie weiterhin haben:
-    1. Nur Mithören: Nur-Chiffretext-Angriff (ciphertext-onlyattack,COA).
-    2. Mithören + Eva sind einige Paare von Klartext und Chiffretext bekannt: Angriff mit bekannten Klartexten(known-plaintextattack, KPA).
-    - Beispiele: Einige Klartext-Chiffretext-Paare sind aus Versehen oder absichtlich bekannt geworden, Eva hat einige Chiffretexte mit großem Aufwand entschlüsselt, Eva war früher mit der Verschlüsselung beauftragt (ohne den Schlüssel zu kennen).
-    3. Mithören + Eva kann einige von ihr gewählte Klartexte verschlüsseln: Angriff mit Klartextwahl (chosen-plaintext attack, CPA).
-    - Beispiele: Eva war früher mit der Verschlüsselung beauftragt (ohne den Schlüssel zu kennen)
-    - CPA ist immer möglich bei asymmetrischer Verschlüsselung,die wir aber erst später betrachten.
-    4. Mithören + Eva kann einige von ihr gewählte Chiffretexte entschlüsseln: Angriff mit Chiffretextwahl (chosen-cyphertext attack, CCA).
-    - Beispiele: Verschiedene Authentisierungsverfahren verlangen, dass die zu prüfende Partei einen Chiffretext entschlüsselt und den Klartext zurücksendet; Eva war früher mit der Entschlüsselung beauftragt (ohne den Schlüssel zu kennen).
-    5. Eva hat Möglichkeiten 3. + 4.
+    \begin{enumerate*}
+        \item Nur Mithören: Nur-Chiffretext-Angriff (ciphertext-onlyattack,COA).
+        \item  Mithören + Eva sind einige Paare von Klartext und Chiffretext bekannt: Angriff mit bekannten Klartexten(known-plaintextattack, KPA).
+        \begin{itemize*}
+            \item Beispiele: Einige Klartext-Chiffretext-Paare sind aus Versehen oder absichtlich bekannt geworden, Eva hat einige Chiffretexte mit großem Aufwand entschlüsselt, Eva war früher mit der Verschlüsselung beauftragt (ohne den Schlüssel zu kennen).
+        \end{itemize*}
+        \item  Mithören + Eva kann einige von ihr gewählte Klartexte verschlüsseln: Angriff mit Klartextwahl (chosen-plaintext attack, CPA).
+        \begin{itemize*}
+            \item Beispiele: Eva war früher mit der Verschlüsselung beauftragt (ohne den Schlüssel zu kennen)
+            \item CPA ist immer möglich bei asymmetrischer Verschlüsselung,die wir aber erst später betrachten.
+        \end{itemize*}
+        \item Mithören + Eva kann einige von ihr gewählte Chiffretexte entschlüsseln: Angriff mit Chiffretextwahl (chosen-cyphertext attack, CCA).
+        \begin{itemize*}
+            \item Beispiele: Verschiedene Authentisierungsverfahren verlangen, dass die zu prüfende Partei einen Chiffretext entschlüsselt und den Klartext zurücksendet; Eva war früher mit der Entschlüsselung beauftragt (ohne den Schlüssel zu kennen).
+        \end{itemize*}
+        \item  Eva hat Möglichkeiten 3. + 4.
+    \end{enumerate*}
 
     Wesentlich sind auch noch die Fähigkeiten von Eva. Einige Beispiele:
-    1. Unbegrenzte Rechenkapazitäten. Eva soll keine Information über den Klartext erhalten, egal wieviel sie rechnet (,,informationstheoretische Sicherheit'').
-    2. Konkrete maximale Anzahl an Rechenoperationen, z.B. $2^{60}$. ,,Konkrete Sicherheit'': Mit diesem Aufwand erfährt Eva ,,(fast) nichts'' über Klartexte.
-    3. Begrenzter Speicher (z.B. 1000 TB). Analog zu 2.
-    4. Im Design des Verschlüsselungsverfahrens gibt es einen Stellhebel, einen ,,Sicherheitsparameter''. (Beispiel: Schlüssellänge, Rundenzahl bei DES und AES.) Je nach Leistungsfähigkeit von Eva kann man durch entsprechende Wahl dieses Parameters die Sicherheit des Systems an eine gegebene (geschätzte) Rechenzeitschranke anpassen.
-    5. Man betrachtet ganze Familien von Verschlüsselungsverfahren, für immer längere Klar-und Chiffretexte Typischerweise werden Verschlüsselung und Entschlüsselung von Polynomialzeitalgorithmen geleistet. Wenn asymptotisch, also für wachsende Textlänge, der Rechenzeitaufwand für Eva zum Brechen des Systems schneller als polynomiell wächst, kann man sagen, dass sie für genügend lange Texte keine Chance mehr hat,das System erfolgreich zu brechen. (,,Asymptotische Sicherheit'')
+    \begin{enumerate*}
+        \item Unbegrenzte Rechenkapazitäten. Eva soll keine Information über den Klartext erhalten, egal wieviel sie rechnet (,,informationstheoretische Sicherheit'').
+        \item  Konkrete maximale Anzahl an Rechenoperationen, z.B. $2^{60}$. ,,Konkrete Sicherheit'': Mit diesem Aufwand erfährt Eva ,,(fast) nichts'' über Klartexte.
+        \item  Begrenzter Speicher (z.B. 1000 TB). Analog zu 2.
+        \item Im Design des Verschlüsselungsverfahrens gibt es einen Stellhebel, einen ,,Sicherheitsparameter''. (Beispiel: Schlüssellänge, Rundenzahl bei DES und AES.) Je nach Leistungsfähigkeit von Eva kann man durch entsprechende Wahl dieses Parameters die Sicherheit des Systems an eine gegebene (geschätzte) Rechenzeitschranke anpassen.
+        \item  Man betrachtet ganze Familien von Verschlüsselungsverfahren, für immer längere Klar-und Chiffretexte Typischerweise werden Verschlüsselung und Entschlüsselung von Polynomialzeitalgorithmen geleistet. Wenn asymptotisch, also für wachsende Textlänge, der Rechenzeitaufwand für Eva zum Brechen des Systems schneller als polynomiell wächst, kann man sagen, dass sie für genügend lange Texte keine Chance mehr hat,das System erfolgreich zu brechen. (,,Asymptotische Sicherheit'')
+    \end{enumerate*}
 
     Wir untersuchen in diesem ersten Teil drei verschiedene Szenarien, jeweils symmetrische Konzelationssysteme,mit steigender Komplexität. Alice und Bob haben sich auf einen Schlüssel geeinigt.
 
-    1. Einmalige Verschlüsselung: Ein einzelner Klartext $x$ vorher bekannter Länge wird übertragen, Eva hört mit (COA).
-    - Unvermeidlich: Triviale Information, z.B. der Sachverhalt,dass eine Nachricht übertragenwurde.
-    - Was vermieden werden soll: Eva erhält nicht-triviale Information, z.B. dass der Klartext $x$ ist oder dass der Klartext aller Wahrscheinlichkeit nach weder $x_1$ noch $x_2$ ist.
-    - Gegenstand der Steganographie sind Verfahren,Nachrichten so zu übertragen, dass noch nicht einmal die Existenz der Nachricht entdeckt werden kann.
-    2. Frische Verschlüsselung: Mehrere Klartexte vorher bekannter Länge werden übertragen, Eva hört mit,kann sich einige Klartexte verschlüsseln lassen (CPA).
-    - Triviale Information: z.B.Anzahl der Nachrichten oder Klartext, falls Eva sich zufälligerweise vorher den ,,richtigen'' Klartext hat verschlüsseln lassen.
-    3. Uneingeschränkte symmetrische Verschlüsselung: Mehrere Klartexte verschiedener Länge, Eva hört mit, kann sich einige Klartexte verschlüsseln lassen (CPA).
-    - Triviale Information: Analog zur frischen Verschlüsselung.
+    \begin{enumerate*}
+        \item Einmalige Verschlüsselung: Ein einzelner Klartext $x$ vorher bekannter Länge wird übertragen, Eva hört mit (COA).
+        \begin{itemize*}
+            \item Unvermeidlich: Triviale Information, z.B. der Sachverhalt,dass eine Nachricht übertragenwurde.
+            \item Was vermieden werden soll: Eva erhält nicht-triviale Information, z.B. dass der Klartext $x$ ist oder dass der Klartext aller Wahrscheinlichkeit nach weder $x_1$ noch $x_2$ ist.
+            \item Gegenstand der Steganographie sind Verfahren,Nachrichten so zu übertragen, dass noch nicht einmal die Existenz der Nachricht entdeckt werden kann.
+        \end{itemize*}
+        \item  Frische Verschlüsselung: Mehrere Klartexte vorher bekannter Länge werden übertragen, Eva hört mit,kann sich einige Klartexte verschlüsseln lassen (CPA).
+        \begin{itemize*}
+            \item Triviale Information: z.B.Anzahl der Nachrichten oder Klartext, falls Eva sich zufälligerweise vorher den ,,richtigen'' Klartext hat verschlüsseln lassen.
+        \end{itemize*}
+        \item  Uneingeschränkte symmetrische Verschlüsselung: Mehrere Klartexte verschiedener Länge, Eva hört mit, kann sich einige Klartexte verschlüsseln lassen (CPA).
+        \begin{itemize*}
+            \item Triviale Information: Analog zur frischen Verschlüsselung.
+        \end{itemize*}
+    \end{enumerate*}
 
     \subsection{Einmalige symmetrische Verschlüsselung und klassische Verfahren}
     Wir diskutieren hier eine einführende, einfache Situation, für symmetrische Konzelationssysteme und Sicherheitsmodelle. In einer Fallstudie betrachten wir Methoden zum ,,Brechen'' eines klassischen Kryptosystems.
@@ -228,13 +248,17 @@
     Fragen: Wie soll man vorgehen, damit das verwendete Verfahren als ,,sicher'' gelten kann? Was soll ,,sicher'' überhaupt bedeuten? Wie kann man ,,Sicherheit'' beweisen? Was sind die Risiken von Varianten (mehrere Nachrichten,längere Nachrichten usw.)?
 
     **Definition 1.1** Ein Kryptosystem ist ein Tupel $S=(X,K,Y,e,d)$, wobei
-    - X und K nicht leere endliche Mengen sind [Klartexte bzw. Schlüssel],
-    - Y eine Menge ist [Chiffretexte], und
-    - $e:X\times K\rightarrow Y$ und $d:Y\times K\rightarrow X$ Funktionen sind [Verschlüsselungsfunktion bzw. Entschlüsselungsfunktion],
+    \begin{itemize*}
+        \item X und K nicht leere endliche Mengen sind [Klartexte bzw. Schlüssel],
+        \item Y eine Menge ist [Chiffretexte], und
+        \item $e:X\times K\rightarrow Y$ und $d:Y\times K\rightarrow X$ Funktionen sind [Verschlüsselungsfunktion bzw. Entschlüsselungsfunktion],
+    \end{itemize*}
 
     so dass Folgendes gilt:
-    1. $\forall x\in X\forall k\in K:d(e(x,k),k) =x$ (Dechiffrierbedingung)
-    2. $\forall y\in Y\exists x\in X,k\in K:y=e(x,k)$ (Surjektivität)
+    \begin{enumerate*}
+        \item $\forall x\in X\forall k\in K:d(e(x,k),k) =x$ (Dechiffrierbedingung)
+        \item $\forall y\in Y\exists x\in X,k\in K:y=e(x,k)$ (Surjektivität)
+    \end{enumerate*}
 
     Bemerkung: Surjektivität kann immer hergestellt werden, indem man $Y$ auf das Bild $Bi(e) =e(X\times K)$ einschränkt. Die Forderung ist für unsere Analysen aber bequem.
 
@@ -298,8 +322,10 @@
     Beispiel: $x=1011001,k=1101010$. Dann ist $y=e(x,k)=1011001\oplus_7 1101010 = 0110011$. Zur Kontrolle: $d(y,k) = 0110011\oplus_7 1101010 = 1011001 =x$.
 
     Wir kontrollieren dass das Vernam-System tatsächlich ein Kryptosystem ist.
-    1. Für $x\in X$ und $k\in K$ gelten $d(e(x,k),k)=(x\oplus_l k)\oplus_l k=x\oplus_l(k\oplus_l k) =x\oplus_l 0^l=x$, d.h. die Dechiffrierbedingung ist erfüllt.
-    2. Für $y\in Y$ gilt $e(y,0^l) =y$ und $y\in X,0^l\in K$. Also gilt Surjektivität.
+    \begin{enumerate*}
+        \item Für $x\in X$ und $k\in K$ gelten $d(e(x,k),k)=(x\oplus_l k)\oplus_l k=x\oplus_l(k\oplus_l k) =x\oplus_l 0^l=x$, d.h. die Dechiffrierbedingung ist erfüllt.
+        \item  Für $y\in Y$ gilt $e(y,0^l) =y$ und $y\in X,0^l\in K$. Also gilt Surjektivität.
+    \end{enumerate*}
 
     Wann soll ein Kryptosystemals sicher betrachtet werden?
 
@@ -314,12 +340,16 @@
     **Definition 1.7** Ein Kryptosystem $S=(X,K,Y,e,d)$ heißt possibilistisch sicher,wenn $\forall y\in Y\forall x\in X\exists k\in K:e(x,k)=y$.
 
     Bemerkung
-    1. Sei $S=(X,K,Y,e,d)$ Kryptosystem. Dann sind äquivalent:
-    - $S$ ist possibilistisch sicher.
-    - $\forall x\in X:e(x,K)=\{e(x,k)|k\in K\}=Y$.
-    2. Für $n\geq 2$ ist das Kryptosystem aus Beispiel 1.2 nicht possibilistisch sicher,denn $A_1$ kann nicht Chiffretext zu $a_2$ sein.
-    3. Das Kryptosystem aus Beispiel 1.3 ist nicht possibilistisch sicher, denn $C$ kann nicht Chiffretext zu $0$ sein.
-    4. Das Vernam-Kryptosystem der Länge $l$ ist possibilistisch sicher: Seien $x\in X$ und $y\in Y$. Setze $k=x\oplus_l y$.Dann gilt $e(x,k)=x\oplus_l(x\oplus_l y)=(x\oplus_l x)\oplus_l y= 0^l\oplus_l y=y$.
+    \begin{enumerate*}
+        \item Sei $S=(X,K,Y,e,d)$ Kryptosystem. Dann sind äquivalent:
+        \begin{itemize*}
+            \item $S$ ist possibilistisch sicher.
+            \item $\forall x\in X:e(x,K)=\{e(x,k)|k\in K\}=Y$.
+        \end{itemize*}
+        \item  Für $n\geq 2$ ist das Kryptosystem aus Beispiel 1.2 nicht possibilistisch sicher,denn $A_1$ kann nicht Chiffretext zu $a_2$ sein.
+        \item  Das Kryptosystem aus Beispiel 1.3 ist nicht possibilistisch sicher, denn $C$ kann nicht Chiffretext zu $0$ sein.
+        \item  Das Vernam-Kryptosystem der Länge $l$ ist possibilistisch sicher: Seien $x\in X$ und $y\in Y$. Setze $k=x\oplus_l y$.Dann gilt $e(x,k)=x\oplus_l(x\oplus_l y)=(x\oplus_l x)\oplus_l y= 0^l\oplus_l y=y$.
+    \end{enumerate*}
 
     In der Einführung wurde die Verschiebechiffre betrachtet, bei  der Buchstaben des alten lateinischen Alphabets auf  Chiffrebuchstaben abgebildet wurden, indem man das Bild von $A$ angab und jeder andere Buchstabe um dieselbe Distanz verschoben wurde. Auch die allgemeineren Substitutionschiffren wurden erwähnt, bei der man für jeden Buchstaben $x$ einen beliebigen Bildbuchstaben $\pi(x)$ angibt, auf injektive Weise. Beispiel für eine Substitutionschiffre:
 
@@ -372,18 +402,24 @@
     Wir fassen unsere Grundbegriffe etwas allgemeiner insofern, als wir auch verschiedene Wahrscheinlichkeiten für Elementarereignisse $a\in\Omega$ zulassen und es erlaubt ist,dass $\Omega$ abzählbar unendlich ist. Wir beschränken uns aber auf den Fall endlicher oder abzählbarer Wahrscheinlichkeitsräume, sogenannter diskreter W-Räume.
 
     **Definition**: Ein (diskreter) Wahrscheinlichkeitsraum ist ein Paar $(\Omega,Pr)$, wobei
-    - $\Omega$ eine nichtleere endliche oder abzählbar unendliche Menge und
-    - $Pr:P(\Omega)\rightarrow[0,1]$ eine Abbildung ($P(\Omega)=\{A|A\subseteq\Omega\}$ ist die Potenzmenge)
+    \begin{itemize*}
+        \item $\Omega$ eine nichtleere endliche oder abzählbar unendliche Menge und
+        \item $Pr:P(\Omega)\rightarrow[0,1]$ eine Abbildung ($P(\Omega)=\{A|A\subseteq\Omega\}$ ist die Potenzmenge)
+    \end{itemize*}
 
     ist, sodass Folgendes gilt:
-    1. $Pr(\Omega) = 1$
-    2. für alle $A\subseteq\Omega$ gilt $Pr(A)=1-Pr(A)$, für $A=\Omega\backslash A$
-    3. für alle $A_1,A_2,...\in P(\Omega)$ gilt, falls die Mengen $A_i$ paarweise disjunkt sind: $Pr(\bigcup A_i)=\sum_{i\geq i}^{\infty} Pr(A_i)$ ( ,,$\sigma$-Additivität'' )
+    \begin{enumerate*}
+        \item $Pr(\Omega) = 1$
+        \item  für alle $A\subseteq\Omega$ gilt $Pr(A)=1-Pr(A)$, für $A=\Omega\backslash A$
+        \item für alle $A_1,A_2,...\in P(\Omega)$ gilt, falls die Mengen $A_i$ paarweise disjunkt sind: $Pr(\bigcup A_i)=\sum_{i\geq i}^{\infty} Pr(A_i)$ ( ,,$\sigma$-Additivität'' )
+    \end{enumerate*}
 
     Man nennt
-    - die Elemente von $\Omega$ Ergebnisse oder Elementarereignisse,
-    - die Elemente von $P(\Omega)$ (also die Teilmengen von $\Omega$) Ereignisse und
-    - $Pr$ die Wahrscheinlichkeitsverteilung
+    \begin{itemize*}
+        \item die Elemente von $\Omega$ Ergebnisse oder Elementarereignisse,
+        \item die Elemente von $P(\Omega)$ (also die Teilmengen von $\Omega$) Ereignisse und
+        \item $Pr$ die Wahrscheinlichkeitsverteilung
+    \end{itemize*}
 
     des Wahrscheinlichkeitsraums $(\Omega,Pr)$. Für $A\in P(\Omega)$ heißt $Pr(A)$ die Wahrscheinlichkeit von $A$.
 
@@ -404,8 +440,10 @@
     Achtung: die bedingte Wahrscheinlichkeit $Pr(A|B)$ ist nur definiert, wenn $Pr(B)> 0$ gilt.
 
     Lemma 1.15 Sei $(\Omega,Pr)$ ein Wahrscheinlichkeitsraum.
-    1. (,,Formel von der totalen Wahrscheinlichkeit'') Seien $B_1,...,B_t$ disjunkte Ereignisse mit $Pr(B_1\cup...\cup B_t)=1$. Dann gilt $Pr(A)=\sum_{1\leq s\leq t} Pr(A|B_s)Pr(B_s)$.
-    2. Seien $A,B,C$ Ereignisse mit $Pr(B\cap C),Pr(C\backslash B)>0$. Dann gilt $Pr(A|C)=Pr(A\cap B | C) + Pr(A\backslash B|C)= Pr(A|B\cap C)Pr(B|C) + Pr(A|C\backslash B)Pr(\bar{B}|C)$.
+    \begin{enumerate*}
+        \item (,,Formel von der totalen Wahrscheinlichkeit'') Seien $B_1,...,B_t$ disjunkte Ereignisse mit $Pr(B_1\cup...\cup B_t)=1$. Dann gilt $Pr(A)=\sum_{1\leq s\leq t} Pr(A|B_s)Pr(B_s)$.
+        \item Seien $A,B,C$ Ereignisse mit $Pr(B\cap C),Pr(C\backslash B)>0$. Dann gilt $Pr(A|C)=Pr(A\cap B | C) + Pr(A\backslash B|C)= Pr(A|B\cap C)Pr(B|C) + Pr(A|C\backslash B)Pr(\bar{B}|C)$.
+    \end{enumerate*}
 
     Beispiel: In dem Würfel-Wahrscheinlichkeitsraum mit $\Omega=\{1,...,6\}$ und der uniformen Verteilung betrachten wir die Ereignisse $A=\{3,6\}$ (durch 3 teilbare Augenzahl) und $B=\{2,4,6\}$ (gerade Augenzahl). Wir haben: $Pr(A\cap B) = Pr(\{6\})=\frac{1}{6}=\frac{1}{3}*\frac{1}{2}=Pr(A)*Pr(B)$.
 
@@ -421,9 +459,11 @@
     Zufallsvariablen mit $R\subseteq R$ heißen reelle Zufallsvariable.
 
     Beispiel 1.18 Zu $\Omega=\{1,2,...,N\}^q$ (q,$N\geq 1$) betrachten wir den Wahrscheinlichkeitsraum $(\Omega,Pr)$ mit der Gleichverteilung $Pr$. Beispiele für Zufallsvariablen sind:
-    - $R=\mathbb{N}$ und $X:\Omega\rightarrow R,(a_1,...,a_q)\rightarrow a_5$ (eine Projektion, definiert für $q\geq 5$)
-    - $R=\{-1,0,1\}$ und $Y_{ij}((a_1,...,a_q))=\begin{cases} -1\quad\text{falls } a_i< a_j\\ 0\quad\text{falls} a_i=a_j, \text{für } 1\leq i < j\leq n \\ 1\quad\text{falls } a_i> a_j\end{cases}$
-    - $R=\mathbb{N}$ und $Z:\Omega\rightarrow R,(a_1 ,...,a_q)\rightarrow\sum_{1\leq i\leq q} a_i$
+    \begin{itemize*}
+        \item $R=\mathbb{N}$ und $X:\Omega\rightarrow R,(a_1,...,a_q)\rightarrow a_5$ (eine Projektion, definiert für $q\geq 5$)
+        \item $R=\{-1,0,1\}$ und $Y_{ij}((a_1,...,a_q))=\begin{cases} -1\quad\text{falls } a_i< a_j\\ 0\quad\text{falls} a_i=a_j, \text{für } 1\leq i < j\leq n \\ 1\quad\text{falls } a_i> a_j\end{cases}$
+        \item $R=\mathbb{N}$ und $Z:\Omega\rightarrow R,(a_1 ,...,a_q)\rightarrow\sum_{1\leq i\leq q} a_i$
+    \end{itemize*}
 
     Sei $X:\Omega\rightarrow R$ eine Zufallsvariable. Für $S\subseteq R$ setze $Pr^X(S):= Pr(X^{-1}(S))=Pr(\{a\in\Omega|X(a)\in S\})$. Dann ist $(R,Pr^X)$ ein Wahrscheinlichkeitsraum. Dieser heißt der von $X$ auf $R$ induzierte Wahrscheinlichkeitsraum. $Pr^X$ heißt auch die Verteilung von $X$.
 
@@ -442,10 +482,12 @@
     Man beachte, dass die Verteilung $Pr_K$ ,,Teil des Kryptosystems'' ist, also der Kontrolle von Alice und Bob unterliegt, während $Pr_X$ ,,Teil der Anwendung'' oder ,,Teil der Realität'' ist, also von den Teilnehmern normalerweise nicht beeinflusst werden kann. Die Verteilung $Pr_X$ braucht beim Entwurf des Kryptosystems nicht einmal bekannt zu sein. (Alice und Bob sollten ihr Kryptosystem ohne Kenntnis von $Pr_X$ planen können. Die Annahme, dass Eva $Pr_X$ kennt, ist eine worst-case-Annahme, sie muss in der Realität nicht unbedingt erfüllt sein.)
 
     **Definition 1.19** Ein Kryptosystem mit Schlüsselverteilung (KSV) ist ein 6-Tupel $V=(X,K,Y,e,d,Pr_K)$, wobei
-    - $S=(X,K,Y,e,d)$ ein Kryptosystem (das zugrundeliegende Kryptosystem) ist und
-    - $Pr_K:K\rightarrow (0,1]$ eine Wahrscheinlichkeitsfunktion (die Schlüsselverteilung) ist.
-    - Für $V=(X,K,Y,e,d,Pr_K)$ schreiben wir auch $S[Pr_K]$.
-    - Achtung: Die Definition verlangt $Pr_K(k)\in (0,1]$, also $Pr_K(k)> 0$ für alle $k\in K$. (Hat man Schlüssel mit Wahrscheinlichkeit 0, kann man sie aus K einfach weg lassen.)
+    \begin{itemize*}
+        \item $S=(X,K,Y,e,d)$ ein Kryptosystem (das zugrundeliegende Kryptosystem) ist und
+        \item $Pr_K:K\rightarrow (0,1]$ eine Wahrscheinlichkeitsfunktion (die Schlüsselverteilung) ist.
+        \item Für $V=(X,K,Y,e,d,Pr_K)$ schreiben wir auch $S[Pr_K]$.
+        \item Achtung: Die Definition verlangt $Pr_K(k)\in (0,1]$, also $Pr_K(k)> 0$ für alle $k\in K$. (Hat man Schlüssel mit Wahrscheinlichkeit 0, kann man sie aus K einfach weg lassen.)
+    \end{itemize*}
 
     Sei weiter $Pr_X:X\rightarrow [0,1]$ eine Wahrscheinlichkeitsfunktion auf der Menge der Klartexte. Das heißt: $\sum_{x\in X}Pr_X(x)=1$. Diese Wahrscheinlichkeitsfunktion definiert natürlich eine W-Verteilung auf X, die wir wieder $Pr_X$ nennen. (Achtung: Es kann Klartextexte mit $Pr(x)=0$ geben. Solche Klartexte heißen passiv, die anderen, mit $Pr_X(x)>0$, aktiv.) Wir definieren die gemeinsame Wahrscheinlichkeitsfunktion $Pr:X\times K\rightarrow [0,1]$ durch $Pr((x,k)):=Pr_X(x)*Pr_K(k)$.
     Dies definiert einen Wahrscheinlichkeitsraum auf $X\times K$, für den $Pr(X'\times K')=Pr_X(X')*Pr_K(K')$, für alle $X'\subseteq X,K'\subseteq K$ gilt. Durch diese Definition wird die Annahme modelliert, dass der Schlüssel k unabhängig vom Klartext durch ein von $Pr_K$ gesteuertes Zufallsexperiment gewählt wird.
@@ -462,12 +504,16 @@
 
     Der Chiffretext y ist dann eine Zufallsvariable auf diesem Wahrscheinlichkeitsraum: $X_3((x,k)):=e(x,k)$.
     Auch die beiden Komponenten $x$ und $k$ werden als Zufallsvariable betrachtet (Projektionen):
-    - $X_1:X\times K\rightarrow X,(x,k) \rightarrow x$
-    - $X_2:X\times K\rightarrow K,(x,k) \rightarrow k$
+    \begin{itemize*}
+        \item $X_1:X\times K\rightarrow X,(x,k) \rightarrow x$
+        \item $X_2:X\times K\rightarrow K,(x,k) \rightarrow k$
+    \end{itemize*}
 
     Wir beobachten einige einfache Zusammenhänge, für $x_0\in X,k_0\in K,y_0\in Y$:
-    - $Pr(x_0):=Pr(X_1=x_0)=Pr(\{x_0\}\times K) = Pr_X(x_0)*Pr_K(K) = Pr_X(x_0)$.
-    - $Pr(k_0):=Pr(X_2=k_0)=Pr(X\times\{k_0\})=Pr_X(X)*Pr_K(k_0)=Pr_K(k_0)$
+    \begin{itemize*}
+        \item $Pr(x_0):=Pr(X_1=x_0)=Pr(\{x_0\}\times K) = Pr_X(x_0)*Pr_K(K) = Pr_X(x_0)$.
+        \item $Pr(k_0):=Pr(X_2=k_0)=Pr(X\times\{k_0\})=Pr_X(X)*Pr_K(k_0)=Pr_K(k_0)$
+    \end{itemize*}
 
     (Man erhält also die ursprünglichen Wahrscheinlichkeiten für Klartexte und Schlüssel zurück. Dies ist eine einfache Grundeigenschaft von Produkträumen.)
 
@@ -484,8 +530,10 @@
     (In Beispiel 1.20 gilt $Pr(c|A)=0,15/0,25=0,6$.) Die letzte Formel ist nur für $y_0$ mit $Pr(y_0)>0$ definiert.
 
     **Definition 1.21** Sei $V=(X,K,Y,e,d,Pr_K)$ ein Kryptosystem mit Schlüsselverteilung.
-    1. Sei $Pr_X$ eine Wahrscheinlichkeitsfunktion auf den Klartexten. Dann heißt $V$ informationstheoretisch sicher bezüglich $Pr_X$, wenn für alle $x\in X,y\in Y$ mit $Pr(y)>0$ gilt: $Pr(x) = Pr(x|y)$.
-    2. Das KSV $V$ heißt informationstheoretisch sicher, wenn es bezüglich jeder beliebigen Klartextverteilung $Pr_X$ informationstheoretisch sicher ist.
+    \begin{enumerate*}
+        \item Sei $Pr_X$ eine Wahrscheinlichkeitsfunktion auf den Klartexten. Dann heißt $V$ informationstheoretisch sicher bezüglich $Pr_X$, wenn für alle $x\in X,y\in Y$ mit $Pr(y)>0$ gilt: $Pr(x) = Pr(x|y)$.
+        \item  Das KSV $V$ heißt informationstheoretisch sicher, wenn es bezüglich jeder beliebigen Klartextverteilung $Pr_X$ informationstheoretisch sicher ist.
+    \end{enumerate*}
 
     Bemerkungen: Hinter Definition 1. steckt die folgende Vorstellung: Eva kennt (im schlimmsten Fall) die Wahrscheinlichkeitsfunktion $Pr_X$. Das System gilt als sicher, wenn sich durch Abfangen eines Chiffretextes y aus Evas Sicht die Wahrscheinlichkeiten der einzelnen Klartexte x nicht ändern. Die Bedingung $Pr(y)>0$ in 1. ist nötig, damit $Pr(x|y)$ definiert ist. Sie bedeutet aber keine Einschränkung, da Chiffretexte $y$ mit $Pr(y)=0$ nie vorkommen, also auch nicht abgefangen werden können. Das Konzept in 2. ist relevant, weil man beim Entwurf eines Kryptosystems meistens die Klartextverteilung nicht oder nicht genau kennt.
 
@@ -519,20 +567,28 @@
     Damit folgt $Pr(x)=\frac{Pr(x,y)}{Pr(y)}= Pr(x|y)$, wie bei der informationstheoretischen Sicherheit verlangt.
 
     Bemerkung 1.25
-    1. Der Beweis und damit das Vernamsystem kommt mit jeder beliebigen Klartextverteilung zurecht.
-    2. Im KSV V wird die Gleichverteilung $Pr_K$ auf den Schlüsseln benutzt.
+    \begin{enumerate*}
+        \item Der Beweis und damit das Vernamsystem kommt mit jeder beliebigen Klartextverteilung zurecht.
+        \item  Im KSV V wird die Gleichverteilung $Pr_K$ auf den Schlüsseln benutzt.
+    \end{enumerate*}
 
     Wir wollen nun überlegen, dass diese beiden Sachverhalte nicht zufällig sind.  Es wird sich herausstellen, dass informationstheoretische Sicherheit inbestimmten  Fällen (nämlich wenn y und K möglichst ,,sparsam'' gebaut sind) Gleichverteilung auf den Schlüsseln erzwingt, und dass die informationstheoretische Sicherheit eines KSV nichts mit den konkreten Wahrscheinlichkeiten der Klartextverteilung $Pr_X$ zu tun hat, sondern nur die Menge $\{x\in X|Pr_X(x)> 0\}$ der ''aktiven'' Klartexte relevant ist.
 
     Lemma 1.26 Sei $V=(X,K,Y,e,d,Pr_K)$ ein KSV. Sei $V$ informationstheoretisch sicher bezüglich einer Klartextverteilung $Pr_X$ mit $Pr(x)>0$ für alle $x\in X$. Dann gilt:
-    1. $Pr(y)>0$ für alle $y\in Y$, und $S=(X,K,Y,e,d)$ ist possibilistisch sicher.
-    2. Gilt zusätzlich $|X|=|Y|=|K|$, so gilt $Pr_K(k)=\frac{1}{|K|}$ für alle $k\in K$.
+    \begin{enumerate*}
+        \item $Pr(y)>0$ für alle $y\in Y$, und $S=(X,K,Y,e,d)$ ist possibilistisch sicher.
+        \item  Gilt zusätzlich $|X|=|Y|=|K|$, so gilt $Pr_K(k)=\frac{1}{|K|}$ für alle $k\in K$.
+    \end{enumerate*}
 
     Beweis:
-    1. Sei $y\in Y$ beliebig. Nach Definition 1.1(2) gibt es $x_0\in X$ und $k_0\in K$ mit $e(x_0,k_0)=y$. Da $Pr_X(x_0)>0$ (nach Vor.) und $Pr_K(k_0)>0$ (nach Def 1.19),erhalten wir $Pr(y)\geq Pr_X(x_0)Pr_K(k_0)>0$. Sei nun zusätzlich auch $x\in X$ beliebig. Dann gilt: $\sum_{k\in K:e(x,k)=y} Pr(x)Pr(k)= Pr(x,y)= Pr(x|y)Pr(y)=^* Pr(x)Pr(y)> 0$. ((*) gilt, da V informationstheoretisch sicher bzgl. $Pr_X$ ist.) Also existiert $k\in K$ mit $e(x,k)=y$. Da $x$ und $y$ beliebig waren, ist S possibilistisch sicher.
-    2. Nun nehmen wir zusätzlich $|X|=|Y|=|K|$ an. Wir beobachten zuerst zwei Dinge:
-    1. Für jedes $x\in X$ ist die Abbildung $K\ni k \rightarrow e(x,k)\in Y$  bijektiv. (Dass diese Abbildung surjektiv ist, ist eine Umformulierung der possibilistischen Sicherheit, die nach 1. gegeben ist. Aus Surjektivität folgt Bijektivität, wegen $|K|=|Y|$.)
-    2. Für jedes $k\in K$ ist die Abbildung $X\ni x \rightarrow e(x,k)\in Y$ bijektiv. (Dass die Abbildung injektiv ist, folgt aus der Dechiffrierbedingung. Aus Injektivität folgt Bijektivität, wegen $|X|=|Y|$.)
+    \begin{enumerate*}
+        \item Sei $y\in Y$ beliebig. Nach Definition 1.1(2) gibt es $x_0\in X$ und $k_0\in K$ mit $e(x_0,k_0)=y$. Da $Pr_X(x_0)>0$ (nach Vor.) und $Pr_K(k_0)>0$ (nach Def 1.19),erhalten wir $Pr(y)\geq Pr_X(x_0)Pr_K(k_0)>0$. Sei nun zusätzlich auch $x\in X$ beliebig. Dann gilt: $\sum_{k\in K:e(x,k)=y} Pr(x)Pr(k)= Pr(x,y)= Pr(x|y)Pr(y)=^* Pr(x)Pr(y)> 0$. ((*) gilt, da V informationstheoretisch sicher bzgl. $Pr_X$ ist.) Also existiert $k\in K$ mit $e(x,k)=y$. Da $x$ und $y$ beliebig waren, ist S possibilistisch sicher.
+        \item  Nun nehmen wir zusätzlich $|X|=|Y|=|K|$ an. Wir beobachten zuerst zwei Dinge:
+        \begin{enumerate*}
+            \item Für jedes $x\in X$ ist die Abbildung $K\ni k \rightarrow e(x,k)\in Y$  bijektiv. (Dass diese Abbildung surjektiv ist, ist eine Umformulierung der possibilistischen Sicherheit, die nach 1. gegeben ist. Aus Surjektivität folgt Bijektivität, wegen $|K|=|Y|$.)
+            \item  Für jedes $k\in K$ ist die Abbildung $X\ni x \rightarrow e(x,k)\in Y$ bijektiv. (Dass die Abbildung injektiv ist, folgt aus der Dechiffrierbedingung. Aus Injektivität folgt Bijektivität, wegen $|X|=|Y|$.)
+        \end{enumerate*}
+    \end{enumerate*}
 
     Nun seien $k_1,k_2\in K$ beliebig. Unser Ziel ist zu zeigen, dass $Pr(k_1)=Pr(k_2)$ gilt. (Dann ist gezeigt,dass $Pr_K$ die uniforme Verteilung ist.) Wähle $x\in X$ beliebig und setze $y:=e(x,k_1)$. Beachte, dass es wegen 1. keinen Schlüssel $k\not=k_1$ mit $y=e(x,k)$ gibt. Wegen 2. gibt es ein $x'\in X$ mit $e(x',k_2)=y$. Auch hier gibt es kein $k'\not=k_2$ mit $e(x',k')=y$. Es gilt also: $Pr(x)Pr(k_1)=\sum_{k\in K:e(x,k)=y} Pr(x)Pr(k) = Pr(x,y) = Pr(x|y)Pr(y) =^* Pr(x)Pr(y)$, und daher $Pr(k_1)=Pr(y)$, wegen $Pr(x)>0$. (* gilt, weil $V$ informationstheoretisch sicher ist.) Analog gilt $Pr(x')Pr(k_2)=Pr(x')Pr(y)$, und daher $Pr(k_2)=Pr(y)$. Es folgt $Pr(k_1)=Pr(k_2)$, wie gewünscht.
     Teil 2. dieses Lemmas hat eine Umkehrung.
@@ -547,8 +603,10 @@
     Aus den beiden Lemmas erhalten wir den folgenden Satz, der die informationstheoretisch sicheren KSVs für den Fall $|X|=|Y|=|K|$ vollständig beschreibt.
 
     Satz 1.28 Sei $V= (X,K,Y,e,d,Pr_K)$ ein KSV mit $|X|=|Y|=|K|$. Dann sind äquivalent:
-    1. $V$ ist informationstheoretisch sicher.
-    2. $(X,K,Y,e,d)$ ist possibilistisch sicher und $Pr_K(k)=\frac{1}{|K|}$ für alle $k\in K$.
+    \begin{enumerate*}
+        \item $V$ ist informationstheoretisch sicher.
+        \item  $(X,K,Y,e,d)$ ist possibilistisch sicher und $Pr_K(k)=\frac{1}{|K|}$ für alle $k\in K$.
+    \end{enumerate*}
 
     Beweis:  ,,$(a)\Rightarrow (b)$'': Wenn V informationstheoretisch sicher ist, dann auch bezüglich einer Klartextverteilung $Pr_X$, in der alle Klartexte aktiv sind. Lemma 1.26 liefert 2. ,,$(b)\Rightarrow (a)$'': Lemma 1.27.
 
@@ -569,34 +627,42 @@
 
     Technisch hilfreich sind die folgenden Größen, die nur von der Verschlüsselungsfunktion und der Schlüsselverteilung abhängen (nicht von irgendeiner Klartextverteilung): $P^x(y):=\sum_{k\in K, e(x,k)=y} Pr(k)$, für $x\in X,y\in Y$ (1.1).
     Man beobachtet sofort die folgenden Gleichungen, die aus der Unabhängigkeit der Verteilungen $Pr_X$ und $Pr_K$ folgen:
-    - Für alle $x\in X:Pr(x,y) = Pr(x)*P^x(y)$. (1.2)
-    - Wenn $Pr(x)> 0$:$Pr(y|x) = \frac{Pr(x,y)}{Pr(x)}=P^x(y)$. (1.3)ümgekehrt wie bei der Definition der informationstheoretischen Sicherheit stellt man sich hier vor, dass ein Klartext x gegeben ist und man fragt nach der resultierenden Verteilung auf den Chiffretexten.
+    \begin{itemize*}
+        \item Für alle $x\in X:Pr(x,y) = Pr(x)*P^x(y)$. (1.2)
+        \item Wenn $Pr(x)> 0$:$Pr(y|x) = \frac{Pr(x,y)}{Pr(x)}=P^x(y)$. (1.3)ümgekehrt wie bei der Definition der informationstheoretischen Sicherheit stellt man sich hier vor, dass ein Klartext x gegeben ist und man fragt nach der resultierenden Verteilung auf den Chiffretexten.
+    \end{itemize*}
     Das nächste Lemma besagt, dass man die Wahrscheinlichkeiten aktiver Klartexte beliebig ändern kann (auch auf 0, also sie weglassen), ohne dass eine bestehende informationstheoretische Sicherheit zerstört wird.
 
     Lemma 1.30 Sei $V=(X,K,Y,e,d,Pr_K)$ KSV und seien $Pr_X$ und $Pr'_X$ Klartextverteilungen mit $Pr'_X(x)>0\Rightarrow Pr_X(x)>0$. Dann gilt: Ist $V$ informationstheoretisch sicher bzgl. $Pr_X$, so ist V informationstheoretisch sicher bzgl. $Pr'_X$.
 
     Beweis: Sei $V$ informationstheoretisch sicher bzgl. $Pr_X$. Wir haben es jetzt mit zwei Wahrscheinlichkeitsräumen zu tun, einem zu $Pr_X$ und $Pr_K$ (bezeichnet mit $(X\times K,Pr)$) und einem zu $Pr'_X$ und $Pr_K$ (bezeichnet mit $(X\times K,Pr')$).
     Wir zeigen nacheinander vier Aussagen.
-    1. $Pr_X(x)> 0 \Rightarrow P^x(y) = Pr(y|x) = Pr(y)$ für alle $y\in Y$. (Die Verteilungen $Pr^X(*)=Pr(*|x)$ auf den Chiffretexten sind für alle (Pr-)aktiven Klartexte x gleich und sind auch gleich der globalen Verteilung auf den Chiffretexten.) Beweis hierzu: Sei $Pr(x)>0$. Dann gilt $P^x(y)=Pr(y|x)$, siehe (1.3). Wenn $Pr(y)=0$ gilt, folgt auch $Pr(y|x)=0$. Sei also $Pr(y)>0$. Dann gilt: $Pr(y|x) =\frac{Pr(x,y)}{Pr(x)}=\frac{Pr(x|y)Pr(y)}{Pr(x)}=^* \frac{Pr(x)Pr(y)}{Pr(x)}= Pr(y)$. (* gilt, weil V informationstheoretisch sicher bzgl. $Pr_X$ ist.)
-    2. $Pr'_X(x)> 0 \Rightarrow Pr'(y|x) = Pr(y)$ für alle $y\in Y$. Beweis hierzu: Aus $Pr'(x)>0$ folgt $Pr(x)>0$, nach Voraussetzung. Wir wenden (1.3) für $Pr'$ und für $Pr$ an und erhalten für alle $y\in Y$: $Pr'(y|x)=P^x(y)=Pr(y|x)=^a Pr(y)$.
-    3. $Pr'(y)=Pr(y)$ für alle $y\in Y$. Beweis hierzu: Mit Lemma 1.15(a) (Formel von der totalen Wahrscheinlichkeit): $Pr'(y)=\sum_{x\in X: Pr'(x)> 0} Pr'(y|x)Pr'(x)=^b \sum_{x\in X: Pr'(x)> 0} Pr(y)Pr'(x) = Pr(y)$.
-    4. $Pr'(x)=Pr'(x|y)$ für alle $x\in X,y\in Y$ mit $Pr'(y)>0$. (D.h.: V ist bzgl. $Pr'_X$ informationstheoretisch sicher.) Beweis hierzu: Wenn $Pr'(x)=0$ gilt, dann folgt $Pr'(x|y)=0=Pr'(x)$. Sei nun $Pr'(x)>0$. Dann: $Pr'(x|y)=\frac{Pr'(x,y)}{Pr'(y)}=\frac{Pr'(y|x)Pr'(x)}{Pr'(y)}=^{b,c} \frac{Pr(y)Pr'(x)}{Pr(y)} = Pr'(x)$.
+    \begin{enumerate*}
+        \item $Pr_X(x)> 0 \Rightarrow P^x(y) = Pr(y|x) = Pr(y)$ für alle $y\in Y$. (Die Verteilungen $Pr^X(*)=Pr(*|x)$ auf den Chiffretexten sind für alle (Pr-)aktiven Klartexte x gleich und sind auch gleich der globalen Verteilung auf den Chiffretexten.) Beweis hierzu: Sei $Pr(x)>0$. Dann gilt $P^x(y)=Pr(y|x)$, siehe (1.3). Wenn $Pr(y)=0$ gilt, folgt auch $Pr(y|x)=0$. Sei also $Pr(y)>0$. Dann gilt: $Pr(y|x) =\frac{Pr(x,y)}{Pr(x)}=\frac{Pr(x|y)Pr(y)}{Pr(x)}=^* \frac{Pr(x)Pr(y)}{Pr(x)}= Pr(y)$. (* gilt, weil V informationstheoretisch sicher bzgl. $Pr_X$ ist.)
+        \item  $Pr'_X(x)> 0 \Rightarrow Pr'(y|x) = Pr(y)$ für alle $y\in Y$. Beweis hierzu: Aus $Pr'(x)>0$ folgt $Pr(x)>0$, nach Voraussetzung. Wir wenden (1.3) für $Pr'$ und für $Pr$ an und erhalten für alle $y\in Y$: $Pr'(y|x)=P^x(y)=Pr(y|x)=^a Pr(y)$.
+        \item  $Pr'(y)=Pr(y)$ für alle $y\in Y$. Beweis hierzu: Mit Lemma 1.15(a) (Formel von der totalen Wahrscheinlichkeit): $Pr'(y)=\sum_{x\in X: Pr'(x)> 0} Pr'(y|x)Pr'(x)=^b \sum_{x\in X: Pr'(x)> 0} Pr(y)Pr'(x) = Pr(y)$.
+        \item  $Pr'(x)=Pr'(x|y)$ für alle $x\in X,y\in Y$ mit $Pr'(y)>0$. (D.h.: V ist bzgl. $Pr'_X$ informationstheoretisch sicher.) Beweis hierzu: Wenn $Pr'(x)=0$ gilt, dann folgt $Pr'(x|y)=0=Pr'(x)$. Sei nun $Pr'(x)>0$. Dann: $Pr'(x|y)=\frac{Pr'(x,y)}{Pr'(y)}=\frac{Pr'(y|x)Pr'(x)}{Pr'(y)}=^{b,c} \frac{Pr(y)Pr'(x)}{Pr(y)} = Pr'(x)$.
+    \end{enumerate*}
 
     Satz 1.31 Sei $V=(X,K,Y,e,d,Pr_K)$ KSV und sei $Pr_X$ eine Klartextverteilung. Dann sind äquivalent:
-    1. V ist informationstheoretisch sicher für $Pr_X$.
-    2. Für jedes $x\in X$ und jedes $y\in Y$ gilt: $Pr(x,y)=Pr(x)Pr(y)$ (das Eintreten von x und das Eintreten von y sind unabhängig).
-    3. Für alle $x\in X$ mit $Pr(x)>0$ und alle $y\in Y$ gilt $Pr(y)=Pr(y|x)$ (andere Formulierung der Unabhängigkeit).
-    4. Für alle $x,x'\in X$ mit $Pr(x),Pr(x')>0$ und alle $y\in Y$ gilt $P^x(y)=P^{x'}(y)$.
+    \begin{enumerate*}
+        \item V ist informationstheoretisch sicher für $Pr_X$.
+        \item Für jedes $x\in X$ und jedes $y\in Y$ gilt: $Pr(x,y)=Pr(x)Pr(y)$ (das Eintreten von x und das Eintreten von y sind unabhängig).
+        \item  Für alle $x\in X$ mit $Pr(x)>0$ und alle $y\in Y$ gilt $Pr(y)=Pr(y|x)$ (andere Formulierung der Unabhängigkeit).
+        \item  Für alle $x,x'\in X$ mit $Pr(x),Pr(x')>0$ und alle $y\in Y$ gilt $P^x(y)=P^{x'}(y)$.
+    \end{enumerate*}
 
     Bemerkung: Bedingung 1. fragt nach der Situation bei gegebenem Chiffretext y mit $Pr(y)>0$. Bedingung 2. ist die wahrscheinlichkeitstheoretisch klarste Charakterisierung von informationstheoretischer Sicherheit, ohne bedingte Wahrscheinlichkeiten zu verwenden. Bedingungen 3. und 4. machen deutlich, dass es nur auf das Verhalten des Kryptosystems (mit seiner Verteilung $Pr_K$) auf den aktiven Klartexten ankommt, nicht auf die Klartextverteilung. Sie sagen auch, worauf genau es ankommt: Für jeden beliebigen aktiven Buchstaben ist die von $e(x,*)$ und der Schlüsselverteilung erzeugte Verteilung auf den Chiffretexten gleich, und zwar gleich der absoluten Verteilung auf den Chiffretexten. Informationstheoretische Sicherheit von $V$ (also für alle Klartextverteilungen) heißt also, dass alle Funktionen $P^x:Y\rightarrow [0,1]$, für $x\in X$, gleich sind (weil man als $Pr_X$ eine Verteilung wählen kann, bei der alle Klartexte aktiv sind, zum Beispiel die Gleichverteilung).
 
     Beweis:
-    - ,,$1.\Rightarrow 2.$'': Wenn $Pr(y)=0$, gilt $Pr(x,y)=0=Pr(x)Pr(y)$. Sei jetzt $Pr(y)>0$. Dann gilt $Pr(x,y)=Pr(y)Pr(x|y) = Pr(y)Pr(x)$, nach 1.
-    - ,,$2.\Rightarrow 3.$'': Wegen 2. gilt $Pr(y)Pr(x)=Pr(x,y)$. Andererseits ist $Pr(y|x)Pr(x)=Pr(x,y)$, also folgt 3. durch Kürzen mit $Pr(x)>0$.
-    - ,,$3.\Rightarrow 4.$'': Verwende (1.3) für $x$ und $x'$ und benutze 3.
-    - ,,$4.\Rightarrow 1.$'': (Dies ist natürlich der entscheidende und schwierigste Beweisschritt!) Nach Voraussetzung 4. gibt es für jedes $y\in Y$ ein $p_y$ mit $P^x(y)=p_y$ für alle aktiven $x\in X$.
-    - Nach Lemma 1.15.1 (Formel von der totalen Wahrscheinlichkeit) gilt dann für jedes y: $Pr(y)=\sum_{x\in X:Pr(x)>0} Pr(y|x)*Pr(x) = \sum_{x\in X: Pr(x)>0} P^x(y)*Pr(x) = \sum_{x\in X:Pr(x)>0} p_y*Pr(x) =p_y$.
-    - Sei nun $y\in Y$ mit $Pr(y)>0$, und sei $x\in X$. Wenn $Pr(x)=0$ gilt, folgt auch $Pr(x|y)=0$. Wenn $x$ aktiv ist, dann gilt $Pr(x|y)=\frac{Pr(x,y)}{Pr(y)}=\frac{Pr(y|x)Pr(x)}{p_y}=\frac{P^x(y)Pr(x)}{p_y}=Pr(x)$, wie gewünscht.
+    \begin{itemize*}
+        \item ,,$1.\Rightarrow 2.$'': Wenn $Pr(y)=0$, gilt $Pr(x,y)=0=Pr(x)Pr(y)$. Sei jetzt $Pr(y)>0$. Dann gilt $Pr(x,y)=Pr(y)Pr(x|y) = Pr(y)Pr(x)$, nach 1.
+        \item ,,$2.\Rightarrow 3.$'': Wegen 2. gilt $Pr(y)Pr(x)=Pr(x,y)$. Andererseits ist $Pr(y|x)Pr(x)=Pr(x,y)$, also folgt 3. durch Kürzen mit $Pr(x)>0$.
+        \item ,,$3.\Rightarrow 4.$'': Verwende (1.3) für $x$ und $x'$ und benutze 3.
+        \item ,,$4.\Rightarrow 1.$'': (Dies ist natürlich der entscheidende und schwierigste Beweisschritt!) Nach Voraussetzung 4. gibt es für jedes $y\in Y$ ein $p_y$ mit $P^x(y)=p_y$ für alle aktiven $x\in X$.
+        \item Nach Lemma 1.15.1 (Formel von der totalen Wahrscheinlichkeit) gilt dann für jedes y: $Pr(y)=\sum_{x\in X:Pr(x)>0} Pr(y|x)*Pr(x) = \sum_{x\in X: Pr(x)>0} P^x(y)*Pr(x) = \sum_{x\in X:Pr(x)>0} p_y*Pr(x) =p_y$.
+        \item Sei nun $y\in Y$ mit $Pr(y)>0$, und sei $x\in X$. Wenn $Pr(x)=0$ gilt, folgt auch $Pr(x|y)=0$. Wenn $x$ aktiv ist, dann gilt $Pr(x|y)=\frac{Pr(x,y)}{Pr(y)}=\frac{Pr(y|x)Pr(x)}{p_y}=\frac{P^x(y)Pr(x)}{p_y}=Pr(x)$, wie gewünscht.
+    \end{itemize*}
 
     Beispiel 1.32 Wir geben noch ein Beispiel für ein informationstheoretisch sicheres Kryptosystem mit $|X|=4,|Y|=6$ und $|K|=8$ an. Die Klartextverteilung ist irrelevant. Sei $X=\{a,b,c,d\},K=\{k_0,...,k_7\},Y=\{A,B,C,D,E,F\}$, und $e$ durch die folgende Tabelle gegeben. (Sie entsteht durch Zusammensetzen zweier informationstheoretisch sicherer Kryptosysteme mit jeweils vier Schlüsseln und vier Chiffretexten.)
     | e                    | a   | b   | c   | d   |
@@ -626,22 +692,26 @@
     Die einfachste Methode ist folgende Version der Cäsar-Chiffre: Wähle einen Schlüssel k aus $K=\{0,1,...,25\}=\{A,...,Z\}$ zufällig. Um ,,Texte'' (d.h. Wörterüber $\mathbb{Z}_n$) zu verschlüsseln, wird S buchstabenweise angewandt: Aus $x_0 x_1...x_{l-1}$ wird $e(x_0,k)e(x_1,k)...e(x_{l-1},k)$.
 
     Diese Methode ist allerdings sehr leicht zu brechen, sogar ,,von Hand'', also ohne massiven Einsatz von Computern. Es gibt mindestens die folgenden naheliegenden Möglichkeiten, einen gegebenen Chiffretext $y_0...y_{l-1}$, der aus einem natürlichsprachigen Text entstanden ist, zu entschlüsseln:
-    1. probiere die 26 möglichen Schlüssel aus, oder
-    2. zähle, welche Buchstaben am häufigsten im Chiffretext vorkommen und teste die Hypothese, dass einer von diesen für ,,e'' steht.
+    \begin{enumerate*}
+        \item probiere die 26 möglichen Schlüssel aus, oder
+        \item  zähle, welche Buchstaben am häufigsten im Chiffretext vorkommen und teste die Hypothese, dass einer von diesen für ,,e'' steht.
+    \end{enumerate*}
 
     Betrachte beispielsweise den Chiffretext $RYFWAVSVNPLVOULHUZAYLUNBUN$.
-    - Zählen liefert folgende Häufigkeiten für die häufigsten Buchstaben: $U:4,L:3,N:3,V:3$.
-    - Vermutung: Einer dieser Buchstaben entspricht dem ,,e''.
-    - Der Schlüssel $k$ mit $e(e,k)=U$ ist $k=Q$. Entschlüsselung mit $Q$ liefert das Wort $bipgkfcfxzvfyevrejkivexlex$, das nicht sehr sinnvoll erscheint.
-    - Der Schlüssel $k$ mit $e(e,k)=L$ ist $k=H$. Entschlüsselung mit $H$ liefert kryptologie ohne anstrengung, und wir sind fertig.
-    - Als Basis für solche Entschlüsselungsansätze benutzt(e) man Häufigkeitstabellen für Buchstaben, wie die folgende (Angaben in Prozent):
-    | Englisch  | Deutsch   | Italienisch |
-    | --------- | --------- | ----------- |
-    | E,e 12,31 | E,e 18,46 | E,e 11,79   |
-    | T,t 9,59  | N,n 11,42 | A,a 11,74   |
-    | A,a 8,05  | I,i 8,02  | I,i 11,28   |
-    | O,o 7,94  | R,r 7,14  | O,o 9,83    |
-    - Dass das ,,e'' im Deutschen deutlich häufiger als im Englischen ist, liegt auch daran, dass bei der Umschreibung der Umlaute ä,ö und ü als ae, oe, ue jeweils ein ,,e'' entsteht.)
+    \begin{itemize*}
+        \item Zählen liefert folgende Häufigkeiten für die häufigsten Buchstaben: $U:4,L:3,N:3,V:3$.
+        \item Vermutung: Einer dieser Buchstaben entspricht dem ,,e''.
+        \item Der Schlüssel $k$ mit $e(e,k)=U$ ist $k=Q$. Entschlüsselung mit $Q$ liefert das Wort $bipgkfcfxzvfyevrejkivexlex$, das nicht sehr sinnvoll erscheint.
+        \item Der Schlüssel $k$ mit $e(e,k)=L$ ist $k=H$. Entschlüsselung mit $H$ liefert kryptologie ohne anstrengung, und wir sind fertig.
+        \item Als Basis für solche Entschlüsselungsansätze benutzt(e) man Häufigkeitstabellen für Buchstaben, wie die folgende (Angaben in Prozent):
+        | Englisch  | Deutsch   | Italienisch |
+        | --------- | --------- | ----------- |
+        | E,e 12,31 | E,e 18,46 | E,e 11,79   |
+        | T,t 9,59  | N,n 11,42 | A,a 11,74   |
+        | A,a 8,05  | I,i 8,02  | I,i 11,28   |
+        | O,o 7,94  | R,r 7,14  | O,o 9,83    |
+        \item Dass das ,,e'' im Deutschen deutlich häufiger als im Englischen ist, liegt auch daran, dass bei der Umschreibung der Umlaute ä,ö und ü als ae, oe, ue jeweils ein ,,e'' entsteht.)
+    \end{itemize*}
 
     Man kann auch die Häufigkeiten von ,,Digrammen'' (zwei Buchstaben, z.B. ng) oder ,,Trigrammen''  (drei Buchstaben, z.B. ung oder eit) heranziehen, auch um unterschiedliche Sprachen zu unterscheiden.
 
@@ -685,21 +755,23 @@
     | VYVFG BVIIO VWLEW DBXMS SFEJG FHFVJ PLWZS FCRVU FMXVZ MNIRI |
     | GAESS HYPFS TNLRH UYR                                       |
 
-    - $y^0 =EFFBUTFSSMJFPOFTMOPJUFXFTFTUESPHMBVFFJUAVOOLFEVTEJIJBUUPVTSJVBVDSFPFFMGHTU$.
-    - Buchstaben in $y^0$ mit Häufigkeiten $>1:AB(4)DE(4)F(14)GH(2)IJ(5)LM(3)O(4)P(5)S(5)T(7)U(7)V(6)X$
-    - Mögliches Bild von ,,e'': F. Schlüsselbuchstabe wäre: B
-    - $y^1 =YWHHCNLXSIHXMZCNCVNAUOOMWMYUCVCYUNHLNALYECUUMYYIUYLBLHAYLCZBYVWBFHLCMNAYNY$.
-    - Buchstaben in $y^1:A(4)B(3)C(8)EFH(6)I(2)L(7)M(5)N(7)O(2)SU(5)V(3)W(2)X(2)Y(19)Z(2)$
-    - Mögliches Bild von ,,e'':Y. Schlüsselbuchstabe wäre: U
-    - $y^2 =RLSMSEJMTKZMMSREGYEOYIVKLWOIIYQRWXHMEOYLEQXQAHVVWAIVFMIVHIXVVILXEFWRXIEPLR$.
-    - Buchstaben in $y^2:A(2)E(7)F(2)GH(3)I(8)JK(2)L(5)M(6)O(2)PQ(3)R(5)S(3)TV(7)W(4)X(5)Y(4)Z$
-    - Mögliche Bilder von ,,e'':I,V. Schlüsselbuchstaben wären: E,R
-    - $y^3 =YJIJEELVKZVXVIVRYVVVJSURCVIIZVUGKVWYVVEEKDSVLRXXJREVVTJKVUVVFIEMJVZVVRSFR$.
-    - Buchstaben in $y^3 :CDE(6)F(2)GI(5)J(6)K(4)L(2)MR(6)S(3)TU(3)V(21)WX(3)Y(3)Z(3)$
-    - Mögliches Bild von ,,e'':V. Schlüsselbuchstabe wäre: R
-    - $y^4 =CHUOGRVYCSBKWAFHSFHWUHSPIZSSVFCOVBIFHWRRSCSBFTSHGSRFWVHKBIBPGOWSGJSUZISSH$.
-    - Buchstaben in $y^4 :AB(5)C(4)F(6)G(4)H(8)I(4)JK(2)O(3)P(2)R(4)S(13)TU(3)V(4)W(5)YZ(2)$
-    - Mögliches Bild von ,,e'':S. Schlüsselbuchstabe wäre: O
+    \begin{itemize*}
+        \item $y^0 =EFFBUTFSSMJFPOFTMOPJUFXFTFTUESPHMBVFFJUAVOOLFEVTEJIJBUUPVTSJVBVDSFPFFMGHTU$.
+        \item Buchstaben in $y^0$ mit Häufigkeiten $>1:AB(4)DE(4)F(14)GH(2)IJ(5)LM(3)O(4)P(5)S(5)T(7)U(7)V(6)X$
+        \item Mögliches Bild von ,,e'': F. Schlüsselbuchstabe wäre: B
+        \item $y^1 =YWHHCNLXSIHXMZCNCVNAUOOMWMYUCVCYUNHLNALYECUUMYYIUYLBLHAYLCZBYVWBFHLCMNAYNY$.
+        \item Buchstaben in $y^1:A(4)B(3)C(8)EFH(6)I(2)L(7)M(5)N(7)O(2)SU(5)V(3)W(2)X(2)Y(19)Z(2)$
+        \item Mögliches Bild von ,,e'':Y. Schlüsselbuchstabe wäre: U
+        \item $y^2 =RLSMSEJMTKZMMSREGYEOYIVKLWOIIYQRWXHMEOYLEQXQAHVVWAIVFMIVHIXVVILXEFWRXIEPLR$.
+        \item Buchstaben in $y^2:A(2)E(7)F(2)GH(3)I(8)JK(2)L(5)M(6)O(2)PQ(3)R(5)S(3)TV(7)W(4)X(5)Y(4)Z$
+        \item Mögliche Bilder von ,,e'':I,V. Schlüsselbuchstaben wären: E,R
+        \item $y^3 =YJIJEELVKZVXVIVRYVVVJSURCVIIZVUGKVWYVVEEKDSVLRXXJREVVTJKVUVVFIEMJVZVVRSFR$.
+        \item Buchstaben in $y^3 :CDE(6)F(2)GI(5)J(6)K(4)L(2)MR(6)S(3)TU(3)V(21)WX(3)Y(3)Z(3)$
+        \item Mögliches Bild von ,,e'':V. Schlüsselbuchstabe wäre: R
+        \item $y^4 =CHUOGRVYCSBKWAFHSFHWUHSPIZSSVFCOVBIFHWRRSCSBFTSHGSRFWVHKBIBPGOWSGJSUZISSH$.
+        \item Buchstaben in $y^4 :AB(5)C(4)F(6)G(4)H(8)I(4)JK(2)O(3)P(2)R(4)S(13)TU(3)V(4)W(5)YZ(2)$
+        \item Mögliches Bild von ,,e'':S. Schlüsselbuchstabe wäre: O
+    \end{itemize*}
 
     Man versucht Schlüssel BURRO und erhält keinen sinnvollen Text. Mit BUERO ergibt sich:
     denho echst enorg anisa tions stand erfuh rdiek rypto logie
@@ -774,17 +846,23 @@
     Algorithmen bilden das Herzstück jeder nichttrivialen Anwendung von Computern. Daher sollte jede Informatikerin und jeder Informatiker Kenntnisse über die wesentlichen algorithmischen Werkzeuge haben: über Strukturen, die es erlauben, Daten effizient zu organisieren und aufzufinden, über häufig benutzte Algorithmen und über die Standardtechniken, mit denen man algorithmische Probleme modellieren, verstehen und lösen kann.
 
     Bemerkungen:
-    - (i) Der Test funktioniert nur gut, wenn die Schlüssellänge s gering im Verhältnis zur Chiffretextlänge l ist.
-    - (ii) Um ihn anwenden zu können, muss die Klartextsprache bekannt sein.
-    - (iii) Der Test kann auch in der viel allgemeineren Situation benutzt werden, in der Schlüssel nicht s Verschiebungen, sondern s beliebige Substitutionschiffren auf $X$ bestimmen (z.B. $X=Y$ und Schlüssel ist Tupel($\pi_0,...,\pi_{s-1}$) von Permutationen von $X$).
+    \begin{itemize*}
+        \item (i) Der Test funktioniert nur gut, wenn die Schlüssellänge s gering im Verhältnis zur Chiffretextlänge l ist.
+        \item (ii) Um ihn anwenden zu können, muss die Klartextsprache bekannt sein.
+        \item (iii) Der Test kann auch in der viel allgemeineren Situation benutzt werden, in der Schlüssel nicht s Verschiebungen, sondern s beliebige Substitutionschiffren auf $X$ bestimmen (z.B. $X=Y$ und Schlüssel ist Tupel($\pi_0,...,\pi_{s-1}$) von Permutationen von $X$).
+    \end{itemize*}
 
     Was passiert im Extremfall $s=l$?
-    - Grundsätzlich hat man dann ein informationstheoretisch sicheres one-time pad vor sich...
-    - ... aber nur dann, wenn die Schlüssel gleichverteilt gewählt werden.  Wenn der Schlüssel selbst ein deutscher Text ist (z.B. ein Textstück aus einem Buch), so weist der Chiffretext wieder statistische Merkmale auf, die zum Brechen ausgenutzt werden können. (Beispiel: Wenn Schlüssel und Klartext beides deutsche Texte sind, werden ca. $7,6\%$ der Buchstaben mit sich selbst verschlüsselt, d.h. Chiffretextbuchstabe$= 2 *$ Klartextbuchstabe modulo 26.)
+    \begin{itemize*}
+        \item Grundsätzlich hat man dann ein informationstheoretisch sicheres one-time pad vor sich...
+        \item ... aber nur dann, wenn die Schlüssel gleichverteilt gewählt werden.  Wenn der Schlüssel selbst ein deutscher Text ist (z.B. ein Textstück aus einem Buch), so weist der Chiffretext wieder statistische Merkmale auf, die zum Brechen ausgenutzt werden können. (Beispiel: Wenn Schlüssel und Klartext beides deutsche Texte sind, werden ca. $7,6\%$ der Buchstaben mit sich selbst verschlüsselt, d.h. Chiffretextbuchstabe$= 2 *$ Klartextbuchstabe modulo 26.)
+    \end{itemize*}
 
     Effektive Verfahren der Schlüsselverlängerung (die aber keine informationstheoretische Sicherheit bringen):
-    - Autokey-Vigenère: Schlüssel k, Klartext m. Dann wird klassische Vigenère-Chiffre mit Schlüssel km auf m angewendet.
-    - Pseudozufallszahlen: Geheimer Schlüssel ist seed eines (Pseudo-)Zufallszahlengenerators, mit dem eine lange Schlüsselfolge $k_0,...,k_{l-1}$ erzeugt wird.
+    \begin{itemize*}
+        \item Autokey-Vigenère: Schlüssel k, Klartext m. Dann wird klassische Vigenère-Chiffre mit Schlüssel km auf m angewendet.
+        \item Pseudozufallszahlen: Geheimer Schlüssel ist seed eines (Pseudo-)Zufallszahlengenerators, mit dem eine lange Schlüsselfolge $k_0,...,k_{l-1}$ erzeugt wird.
+    \end{itemize*}
 
     \subsubsection{Koinzidenzindex und Friedman-Methode}
     Wir betrachten noch eine andere interessante Methode zur Abschätzung der Schlüssellänge, die bei der Verwendung einer Vigenère-Chiffre oder anderen Substitutionschiffren mit fester Schlüssellänge s helfen können, diese zu ermitteln. Die Methode beruht darauf, dass die Buchstabenhäufigkeiten (zu einer gegebenen Sprache) fest stehen und sich bei der Verschlüsselung mit einer einfachen Substitutionschiffre nicht ändert. Ebenso ändert sich nicht die Wahrscheinlichkeit, bei der zufälligen Wahl eines Buchstabenpaars zwei identische Buchstaben zu erhalten. Die Methode stammt von William F. Friedman (1891, 1969), einem amerikanischen Kryptographen.
@@ -852,25 +930,33 @@
     Sehr oft werden wir annehmen, dass $\beta$ selbst invers ist, dass also ${\beta}^{-1}=\beta$ ist (oder ${\beta}({\beta}(i))=i$ für $0\leq i<l$). Dann erhält man $x^{\beta}$ aus x, indem man $x(i)$ an die Stelle ${\beta}(i)$ des neuen Vektors schreibt.
 
     Definition 2.5 Ein Substitutions-Permutations-Netzwerk (SPN) ist ein Tupel $N=(m,n,r,s,S,\beta,\kappa)$ wobei
-    - die positiven ganzen Zahlen $m,n,r$ und $s$ die Wortanzahl $m$, die Wortlängen, die Rundenzahl r und die Schlüssellänge s angeben,
-    - $S:\{0,1\}^n\rightarrow\{0,1\}^n$ eine bijektive Funktion ist (,,Substitution'',die S-Box),
-    - ${\beta}:\{0,...,mn-1\}\rightarrow\{0,...,mn-1\}$ eine selbstinverse Permutation (die Bitpermutation) ist und
-    - $\kappa :\{0,1\}s\times\{0,...,r\}\rightarrow\{0,1\}^{mn}$ die Rundenschlüsselfunktion ist.
+    \begin{itemize*}
+        \item die positiven ganzen Zahlen $m,n,r$ und $s$ die Wortanzahl $m$, die Wortlängen, die Rundenzahl r und die Schlüssellänge s angeben,
+        \item $S:\{0,1\}^n\rightarrow\{0,1\}^n$ eine bijektive Funktion ist (,,Substitution'',die S-Box),
+        \item ${\beta}:\{0,...,mn-1\}\rightarrow\{0,...,mn-1\}$ eine selbstinverse Permutation (die Bitpermutation) ist und
+        \item $\kappa :\{0,1\}s\times\{0,...,r\}\rightarrow\{0,1\}^{mn}$ die Rundenschlüsselfunktion ist.
+    \end{itemize*}
 
     Das zu N gehörende Substitutions-Permutations-Kryptosystem (SPKS) ist das mn-Block-Kryptosystem $B(N)=(\{0,1\}^{mn},\{0,1\}^s,\{0,1\}^{mn},e,d)$, das durch folgende Operationen beschrieben ist:
 
     Chiffrierung: $e(x,k)$ wird für $x\in\{0,1\}^{mn}$ und $k\in\{0,1\}^s$ wie folgt berechnet:
-    1. Initialisierung (,,Weißschritt''): Berechne $u=x\oplus_{mn} \kappa (k,0)$.
-    2. Verschlüsselung in Runden: für $i=1,...,r-1$ berechne
-    1. $v(j)=S(u(j))$ für $0\leq j<m$ (jedes Wort einzeln durch die S-Box)
-    2. $w=v^{\beta}$ (Bitpermutation auf dem Gesamtwort der Länge mn)
-    3. $u=w\oplus_{mn} \kappa (k,i)$ (XOR mit dem Rundenschlüssel)
-    3. Schlussrunde: $v(j)=S(u(j))$ für $0\leq j<m$
-    4. Ausgabe: $y=v\oplus \kappa (k,r)$ (wie gewöhnliche Runde, ohne Bitpermutation)
+    \begin{enumerate*}
+        \item Initialisierung (,,Weißschritt''): Berechne $u=x\oplus_{mn} \kappa (k,0)$.
+        \item Verschlüsselung in Runden: für $i=1,...,r-1$ berechne
+        \begin{enumerate*}
+            \item $v(j)=S(u(j))$ für $0\leq j<m$ (jedes Wort einzeln durch die S-Box)
+            \item  $w=v^{\beta}$ (Bitpermutation auf dem Gesamtwort der Länge mn)
+            \item  $u=w\oplus_{mn} \kappa (k,i)$ (XOR mit dem Rundenschlüssel)
+            \item Schlussrunde: $v(j)=S(u(j))$ für $0\leq j<m$
+            \item  Ausgabe: $y=v\oplus \kappa (k,r)$ (wie gewöhnliche Runde, ohne Bitpermutation)
+        \end{enumerate*}
+    \end{enumerate*}
 
     Dechiffrierung: $d(y,k)$ wird aus y und k nach demselben Verfahren berechnet, wobei jedoch
-    1. die S-Box (die eine Permutation ist) durch ihre Inverse $S^{-1}$ ersetzt wird und
-    2. die Rundenschlüsselfunktion $\kappa$ ersetzt wird durch die Funktion $\kappa':\{0,1\}^s\times\{0,...,r\}\rightarrow\{0,1\}^{mn}$, $(k,i)\rightarrow\begin{cases} \kappa (k,r-i)\quad\text{ für } i\in\{0,r\}\\ \kappa(k,r-i)^{\beta} \quad\text{ für } 0<i<r \end{cases}$
+    \begin{enumerate*}
+        \item die S-Box (die eine Permutation ist) durch ihre Inverse $S^{-1}$ ersetzt wird und
+        \item  die Rundenschlüsselfunktion $\kappa$ ersetzt wird durch die Funktion $\kappa':\{0,1\}^s\times\{0,...,r\}\rightarrow\{0,1\}^{mn}$, $(k,i)\rightarrow\begin{cases} \kappa (k,r-i)\quad\text{ für } i\in\{0,r\}\\ \kappa(k,r-i)^{\beta} \quad\text{ für } 0<i<r \end{cases}$
+    \end{enumerate*}
 
     Vorteil der Struktur: Man kann dieselbe Hardware für Verschlüsselung und Entschlüsselung benutzen. Bei der Entschlüsselung läuft der Vorgang rückwärts ab, wobei die Runden anders gruppiert sind. Anstelle von S verwendet man $S^{-1}$. Da $\beta$ selbst invers ist, kann man für die Permutation auch bei der Dekodierung $\beta$ verwenden. Da Permutation und Schlüsselanwendung in den Runden vertauscht sind, muss man auf die Rundenschlüssel der ,,inneren'' Runden bei der Dekodierung auch noch $\beta$ anwenden.
 
@@ -882,11 +968,15 @@
     | S(b) | 0101 | 0100 | 1101 | 0001 | 0011 | 1100 | 1011 | 1000 | 1010 | 0010 | 0110 | 1111 | 1001 | 1110 | 0000 | 0111 |
 
     Für $x= 0000 1111 0000$ und $k=0000 0001 0010 0011 0100 0101$ ergeben sich nacheinander
-    1. $\kappa (k,0) = 0000 0001 0010$ und damit $u= 0000 1110 0010$
-    2.
-    - $i=1$ $v=0101 0000 1101$, $w=0011 0101 0100$, $\kappa (k,1) = 0001 0010 0011$ und $u=0010 0111 0111$
-    - $i=2$ $v=1101 1000 1000$, $w=1010 1100 0100$, $\kappa (k,2) = 0010 0011 0100$ und damit $u= 1000 1111 0000$
-    3. $v=1010 0111 0101$, $\kappa (k,3) = 0011 0100 0101$ und damit $y=1001 0011 0000$
+    \begin{enumerate*}
+        \item $\kappa (k,0) = 0000 0001 0010$ und damit $u= 0000 1110 0010$
+        \item
+        \begin{itemize*}
+            \item $i=1$ $v=0101 0000 1101$, $w=0011 0101 0100$, $\kappa (k,1) = 0001 0010 0011$ und $u=0010 0111 0111$
+            \item $i=2$ $v=1101 1000 1000$, $w=1010 1100 0100$, $\kappa (k,2) = 0010 0011 0100$ und damit $u= 1000 1111 0000$
+        \end{itemize*}
+        \item  $v=1010 0111 0101$, $\kappa (k,3) = 0011 0100 0101$ und damit $y=1001 0011 0000$
+    \end{enumerate*}
 
 
     \subsubsection{Einschub: Endliche Körper}
@@ -990,8 +1080,10 @@
     Weil die üblichen Rechenregeln in der Struktur $(F^k,\oplus,\odot,0,1)$ mit den angegebenen Operationen leicht nachzuweisen sind, bedeutet dies, dass wir einen Körper mit $|F|^k$ Elementen erhalten haben. Wir nennen ihn $F[X]/(f)$, für das gewählte irreduzible Polynom $f$ vom Grad $k$. Wenn $|F|^k=q$ ist, nennen wir diesen Körper $GF(q)$. Im Fall $F=\mathbb{Z}_2$ erhalten wir so Körper $GF(2^k)$, deren zugrunde liegende Menge einfach $\{0,1\}^k$ ist, die Menge der Bitstrings der Länge k.
 
     Bemerkung: In der Algebra zeigt man folgende Fakten über endliche Körper:
-    - Es gibt einen endlichen Körper $GF(q)$ mit $q$ Elementen genau dann wenn $q=p^r$ für eine Primzahl p und einen Exponenten $r\geq 1$.
-    - Es ist gleichgültig, auf welchem Konstruktionsweg man zu einem Körper mit $q$ Elementen gelangt: alle diese Körper sind isomorph, also strukturell identisch. (Insbesondere ist der Körper $GF(2^k)$ immer ,,der gleiche'' , ganz egal welches irreduzible Polynom f man benutzt.) Die Tabellen für die Operationen können eventuell unterschiedlich aussehen, aber nur aufgrund von Umbenennungen von Objekten.
+    \begin{itemize*}
+        \item Es gibt einen endlichen Körper $GF(q)$ mit $q$ Elementen genau dann wenn $q=p^r$ für eine Primzahl p und einen Exponenten $r\geq 1$.
+        \item Es ist gleichgültig, auf welchem Konstruktionsweg man zu einem Körper mit $q$ Elementen gelangt: alle diese Körper sind isomorph, also strukturell identisch. (Insbesondere ist der Körper $GF(2^k)$ immer ,,der gleiche'' , ganz egal welches irreduzible Polynom f man benutzt.) Die Tabellen für die Operationen können eventuell unterschiedlich aussehen, aber nur aufgrund von Umbenennungen von Objekten.
+    \end{itemize*}
 
     Ab hier schreiben wir $+$ für die Körperaddition und $*$ für die Körpermultiplikation.
 
@@ -1047,19 +1139,23 @@
     Aus jeder solchen Matrix ergibt sich durch spaltenweises Auslesen wieder ein Bitstring $A_{00} A_{10} A_{20} A_{30} A_{01} A_{11} A_{21} A_{31} A_{02} A_{12} A_{22} A_{32} A_{03} A_{13} A_{23} A_{33}$.
 
     Abkürzungen für Zeilen und Spalten einer Matrix A:
-    - Zeile $i: row_i(A)=(A_{i0},A_{i1},A_{i2} ,A_{i3})$, für $i=0,1,2,3$.
-    - Spalte $j:col_j(A)=\begin{pmatrix} A_{0j}\\ A_{1j}\\ A_{2j}\\ A_{3j} \end{pmatrix}$, für $j=0,1,2,3$.
+    \begin{itemize*}
+        \item Zeile $i: row_i(A)=(A_{i0},A_{i1},A_{i2} ,A_{i3})$, für $i=0,1,2,3$.
+        \item Spalte $j:col_j(A)=\begin{pmatrix} A_{0j}\\ A_{1j}\\ A_{2j}\\ A_{3j} \end{pmatrix}$, für $j=0,1,2,3$.
+    \end{itemize*}
 
     Elementare Operationen auf Matrizen:
-    - Für Matrizen A und B in $GF(2^8)^{4\times 4}$ bedeutet $A\oplus B$ die komponentenweise Anwendung der Addition im Körper $GF(2^8)$. (Diese ist wiederum $\oplus_8$, die bitweise XOR-Operation auf Bytes.)
-    - Zyklischer Linksshift von Zeile i um $h=1,2,3$ Positionen:
-    - $rotLeft_1((A_{i0}, A_{i1}, A_{i2}, A_{i3})) =(A_{i1}, A_{i2}, A_{i3}, A_{i0})$ usw.
-    - Beispiel: $rotLeft_2((A1,06,4B,EF)) = (4B,EF,A1,06)$.
-    - In AES wird Zeile i um i Positionen nach links rotiert. Die Abbildung ist also folgende: $\begin{pmatrix} A_{00}& A_{01}& A_ {02}& A_{03}\\ A_{10}& A_{11}& A_{12}& A_{13}\\ A_{20}& A_{21}& A_{22}& A_{23}\\ A_{30}& A_{31}& A_{32}& A_{33}\end{pmatrix} \rightarrow \begin{pmatrix} A_{00}& A_{01}& A_{02}& A_{03}\\ A_{11}& A_{12}& A_{13}& A_{10}\\ A_{22}& A_{23}& A_{20}& A_{21}\\ A_{33}& A_{30}& A_{31}& A_{32}\end{pmatrix}$
-    - Diese Zeilenrotation ist eine Bitpermutation. Ihre Inverse besteht offenbar darin, Zeile i um i Positionen nach rechts zu rotieren. (Sie ist also nicht selbst invers)
-    - Zyklischer Shift von Spalte $j$ um $h=1,2,3$ Positionen nach oben: $rotUp_1((A_{0j},A_{1j},A_{2j},A_{3j})^T) = (A_{1j},A_{2j},A_{3j},A_{0j})^T$ usw. (Benötigt bei der Rundenschlüsselerzeugung.)
-    -  ,,Lineare Spaltendurchmischung''
-    - $col_j(A) \rightarrow M*col_j(A)$, für $0\geq j\geq 3$, für die feste Matrix $M=\begin{pmatrix} 02& 03& 01& 01\\ 01& 02& 03& 01\\ 01& 01& 02& 03\\ 03& 01& 01& 02\end{pmatrix}\in GF(2^8)^{4\times 4}$.
+    \begin{itemize*}
+        \item Für Matrizen A und B in $GF(2^8)^{4\times 4}$ bedeutet $A\oplus B$ die komponentenweise Anwendung der Addition im Körper $GF(2^8)$. (Diese ist wiederum $\oplus_8$, die bitweise XOR-Operation auf Bytes.)
+        \item Zyklischer Linksshift von Zeile i um $h=1,2,3$ Positionen:
+        \item $rotLeft_1((A_{i0}, A_{i1}, A_{i2}, A_{i3})) =(A_{i1}, A_{i2}, A_{i3}, A_{i0})$ usw.
+        \item Beispiel: $rotLeft_2((A1,06,4B,EF)) = (4B,EF,A1,06)$.
+        \item In AES wird Zeile i um i Positionen nach links rotiert. Die Abbildung ist also folgende: $\begin{pmatrix} A_{00}& A_{01}& A_ {02}& A_{03}\\ A_{10}& A_{11}& A_{12}& A_{13}\\ A_{20}& A_{21}& A_{22}& A_{23}\\ A_{30}& A_{31}& A_{32}& A_{33}\end{pmatrix} \rightarrow \begin{pmatrix} A_{00}& A_{01}& A_{02}& A_{03}\\ A_{11}& A_{12}& A_{13}& A_{10}\\ A_{22}& A_{23}& A_{20}& A_{21}\\ A_{33}& A_{30}& A_{31}& A_{32}\end{pmatrix}$
+        \item Diese Zeilenrotation ist eine Bitpermutation. Ihre Inverse besteht offenbar darin, Zeile i um i Positionen nach rechts zu rotieren. (Sie ist also nicht selbst invers)
+        \item Zyklischer Shift von Spalte $j$ um $h=1,2,3$ Positionen nach oben: $rotUp_1((A_{0j},A_{1j},A_{2j},A_{3j})^T) = (A_{1j},A_{2j},A_{3j},A_{0j})^T$ usw. (Benötigt bei der Rundenschlüsselerzeugung.)
+        \item  ,,Lineare Spaltendurchmischung''
+        \item $col_j(A) \rightarrow M*col_j(A)$, für $0\geq j\geq 3$, für die feste Matrix $M=\begin{pmatrix} 02& 03& 01& 01\\ 01& 02& 03& 01\\ 01& 01& 02& 03\\ 03& 01& 01& 02\end{pmatrix}\in GF(2^8)^{4\times 4}$.
+    \end{itemize*}
 
     Achtung: Gerechnet wird in $GF(2^8)$!
 
@@ -1067,32 +1163,58 @@
     In AES werden einmal alle vier Spalten auf diese Weise transformiert. Dies kann man auch zu einer Matrixmultiplikation zusammenfassen: $A \rightarrow M*A$.
 
     Diese Operation ,,vermischt'' die Einträge, ist aber keine Bitpermutation mehr, da Bits nicht nur vertauscht werden, sondern mit Körperoperationen verrechnet. Allerdings ist M invertierbar, sodass man die Operation auch wieder rückgängig machen kann, indem man mit $M^{-1}=\begin{pmatrix} 0E& 0B& 0D& 09\\ 09& 0E& 0B& 0D\\ 0D& 09& 0E& 0B\\ 0B& 0D& 09& 0E\end{pmatrix}$ multipliziert.
-    - Die S-Box von AES ist eine feste bijektive Abbildung $\{0,1\}^8\rightarrow {0,1}^8$. Der mathematische Hintergrund der S-Box wird weiter unten diskutiert.
-    - Rundenschlüssel: Aus Schlüssel $k\in\{0,1\}^{128}$ und Rundennummer r wird Rundenschlüssel $\kappa (k,r)\in\{0,1\}^{128}=(\{0,1\}^8)^{4\times 4}$ berechnet. Details hierzu folgen.
+    \begin{itemize*}
+        \item Die S-Box von AES ist eine feste bijektive Abbildung $\{0,1\}^8\rightarrow {0,1}^8$. Der mathematische Hintergrund der S-Box wird weiter unten diskutiert.
+        \item Rundenschlüssel: Aus Schlüssel $k\in\{0,1\}^{128}$ und Rundennummer r wird Rundenschlüssel $\kappa (k,r)\in\{0,1\}^{128}=(\{0,1\}^8)^{4\times 4}$ berechnet. Details hierzu folgen.
+    \end{itemize*}
 
     \paragraph{AES-Chiffrieralgorithmus}
-    - X:128-Bit-Klartext,
-    - k:128-Bit-Schlüssel)
-    - Umarrangiert: $X\in GF(2^8)^{4\times 4}$: Klartextmatrix; $k\in\{0,1\}^{128}$: Schlüssel.
-    1. Initialschritt (,,Weißschritt''):
-    - $U\leftarrow X\oplus \kappa (k,0)$ // addiere Rundenschlüssel für Runde $r=0$, als 128-Bit-String.
-    2. Für $r=1,...,9$ tue
-    1. Substitution: Anwendung der S-Box auf jedes Byte in U
-    - $V_{ij}\leftarrow S(U_{ij})$, für $0\geq i,j\geq 3$;
-    2. Zeilenrotation: i-te Zeile um i Positionen nach links
-    - $row_i(W)\leftarrow rotLeft_i(row_i(V))$, für $0\geq i\geq 3$;
-    3. Lineare Spalten durchmischung: M wie oben beschrieben
-    - $Z\leftarrow M*W$, für $0\geq i\geq 3$;
-    4. Schlüsseladdition: Rundenschlüssel für Runde r
-    - $U\leftarrow Z\oplus \kappa (k,r)$ // komponentenweise Addition
-    3. Verkürzte Schlussrunde, $r=10$, keine Spaltendurchmischung:
-    1. Substitution
-    - $V_{ij}\leftarrow S(U_{ij})$, für $0\geq i,j\geq 3$;
-    2. Zeilenrotation
-    - $row_i(Z)\leftarrow rotLeft_i(row_i(V))$, für $0\geq i\geq 3$;
-    3. Schlüsseladdition
-    - $Y\leftarrow Z\oplus \kappa (k,10)$
-    4. Ausgabe: $Y\in GF(2^8)^{4\times 4}$, geschrieben als 128-Bit-Wort.
+    \begin{itemize*}
+        \item X:128-Bit-Klartext,
+        \item k:128-Bit-Schlüssel)
+        \item Umarrangiert: $X\in GF(2^8)^{4\times 4}$: Klartextmatrix; $k\in\{0,1\}^{128}$: Schlüssel.
+        \begin{enumerate*}
+            \item Initialschritt (,,Weißschritt''):
+            \begin{itemize*}
+                \item $U\leftarrow X\oplus \kappa (k,0)$ // addiere Rundenschlüssel für Runde $r=0$, als 128-Bit-String.
+            \end{itemize*}
+            \item  Für $r=1,...,9$ tue
+            \begin{enumerate*}
+                \item Substitution: Anwendung der S-Box auf jedes Byte in U
+                \begin{itemize*}
+                    \item $V_{ij}\leftarrow S(U_{ij})$, für $0\geq i,j\geq 3$;
+                \end{itemize*}
+                \item  Zeilenrotation: i-te Zeile um i Positionen nach links
+                \begin{itemize*}
+                    \item $row_i(W)\leftarrow rotLeft_i(row_i(V))$, für $0\geq i\geq 3$;
+                \end{itemize*}
+                \item  Lineare Spalten durchmischung: M wie oben beschrieben
+                \begin{itemize*}
+                    \item $Z\leftarrow M*W$, für $0\geq i\geq 3$;
+                \end{itemize*}
+                \item  Schlüsseladdition: Rundenschlüssel für Runde r
+                \begin{itemize*}
+                    \item $U\leftarrow Z\oplus \kappa (k,r)$ // komponentenweise Addition
+                \end{itemize*}
+            \end{enumerate*}
+            \item  Verkürzte Schlussrunde, $r=10$, keine Spaltendurchmischung:
+            \begin{enumerate*}
+                \item Substitution
+                \begin{itemize*}
+                    \item $V_{ij}\leftarrow S(U_{ij})$, für $0\geq i,j\geq 3$;
+                \end{itemize*}
+                \item  Zeilenrotation
+                \begin{itemize*}
+                    \item $row_i(Z)\leftarrow rotLeft_i(row_i(V))$, für $0\geq i\geq 3$;
+                \end{itemize*}
+                \item  Schlüsseladdition
+                \begin{itemize*}
+                    \item $Y\leftarrow Z\oplus \kappa (k,10)$
+                \end{itemize*}
+            \end{enumerate*}
+            \item  Ausgabe: $Y\in GF(2^8)^{4\times 4}$, geschrieben als 128-Bit-Wort.
+        \end{enumerate*}
+    \end{itemize*}
 
     Abbildung 3: Die S-Box für AES. $S(z_1 z_0)$ für Hexziffern $z_1,z_0$ steht in Zeile $z_1$, Spalte $z_0$. Beispiel: $S(A4)=49$.
     |     | 0   | 1   | 2   | 3   | 4   | 5   | 6   | 7   | 8   | 9   | A   | B   | C   | D   | E   | F   |
@@ -1132,16 +1254,20 @@
     $K_0 =\kappa (k,0)\leftarrow k$ //128-Bit-Schlüssel als $4\times 4$-Matrix
 
     Für $r= 1,..., 10$ führe Runde r aus.
-    - Input: Rundenschlüssel $K_{r-1}$ als $4\times 4$-Matrix
-    - Output: Rundenschlüssel $K_r$ als $4\times 4$-Matrix
-    - Berechnung von Spalte 0:
-    1. Rotation (analog zur Zeilenrotation): $U\leftarrow rotUp_1(col_3(K_{r-1}))$ //Spalte 3 zyklisch eine Position nach oben
-    2. Substitution: $V_i\leftarrow S(U_i)$, für $0\geq i\geq 3$;
-    3. Konstantenaddition: $V_0\leftarrow V_0+02^{r-1}$; //Potenz in $GF(2^8)$;
-    4. Addition auf Spalte 0: $col_0(K_r)\leftarrow col_0(K_{r-1})+V$; //Vektoraddition;
-    - Berechnung von Spalten 1 bis 3:
-    - Iterative Addition: für $i=1,2,3: col_i(K_r)\leftarrow col_i(K_{r-1})+col_{i-1}(K_r)$; //Vektoraddition;
-    - Ergebnis: $K_r=\kappa (k,r)$ für $r=0,1,...10$.
+    \begin{itemize*}
+        \item Input: Rundenschlüssel $K_{r-1}$ als $4\times 4$-Matrix
+        \item Output: Rundenschlüssel $K_r$ als $4\times 4$-Matrix
+        \item Berechnung von Spalte 0:
+        \begin{enumerate*}
+            \item Rotation (analog zur Zeilenrotation): $U\leftarrow rotUp_1(col_3(K_{r-1}))$ //Spalte 3 zyklisch eine Position nach oben
+            \item  Substitution: $V_i\leftarrow S(U_i)$, für $0\geq i\geq 3$;
+            \item  Konstantenaddition: $V_0\leftarrow V_0+02^{r-1}$; //Potenz in $GF(2^8)$;
+            \item Addition auf Spalte 0: $col_0(K_r)\leftarrow col_0(K_{r-1})+V$; //Vektoraddition;
+        \end{enumerate*}
+        \item Berechnung von Spalten 1 bis 3:
+        \item Iterative Addition: für $i=1,2,3: col_i(K_r)\leftarrow col_i(K_{r-1})+col_{i-1}(K_r)$; //Vektoraddition;
+        \item Ergebnis: $K_r=\kappa (k,r)$ für $r=0,1,...10$.
+    \end{itemize*}
 
     Entschlüsselung:
     Man geht nach dem schon diskutierten grundsätzlichen Ansatz bei Substitutions-Permutations-Kryptosystemen vor. Erst werden alle Rundenschlüssel berechnet. Diese werden in umgekehrter Reihenfolge benutzt. Für die Substitution wird stets die inverse S-Box $S^{-1}$ benutzt. Rotationen werden in umgekehrter Richtung ausgeführt. Die ,,lineare Spaltendurchmischung'' wird mit der zu $M$ inversen Matrix $M^{-1}$ ausgeführt. Ansonsten ist der Ablauf völlig analog der Verschlüsselung.
@@ -1160,16 +1286,20 @@
     Sei $B=(\{0,1\}^l,\{0,1\}^s,\{0,1\}^l,e,d)$ ein Block-Kryptosystem und seien $x_i,y_i\in\{0,1\}^l$ für $1\geq i\geq t$ paarweise verschiedene Klar- bzw. Chiffretexte. Ein Schlüssel $k\in\{0,1\}^s$ ist konsistent mit den Paaren $(x_i,y_i)$, wenn $e(x_i,k)=y_i$ für alle $1\geq i\geq t$ gilt. Für realistische (d.h. praktisch relevante) SPKS gibt es im Fall $t\approx 5$ höchstens einen konsistenten Schlüssel. (Man denke an AES. $t=5$ bedeutet, dass $5*128$ Bit Information über $k$ vorliegen, eventuell redundant, aber $k$ ist nur $128$ Bit lang.)
 
     Wir können also für Eva in Szenario 2 den folgenden Angriff nach dem CPA-Muster planen:
-    1. Lasse fünf Klartexte $x_1,...,x_5$ zu $y_1,...,y_5$ verschlüsseln.
-    2. ,,Schaue nach'', welcher Schlüssel $k$ mit den Paaren $(x_1,y_1),...,(x_5,y_5)$ konsistent ist (wenigstens einer ist es und es gibt höchstens einen).
-    3. Höre Chiffretext $y$ ab und berechne $x=d(y,k)$ aus $y$.üm in Schritt 2. ,,nachsehen'' zu können, benötigt Eva eine Tabelle mit allen möglichen Chiffretextkombinationen für die fünf Klartexte $x_1,...,x_5$. Dafür gibt es $\geq |K|= 2^s$ Möglichkeiten, nämlich für jeden Schlüssel eine. Die Hashtabelle oder der Suchbaum (oder die entsprechende Struktur in einem Turingmaschinen-Programm) hat Größe $2^s$. Wenn der Programmtext oder die Turingmaschine binäre Suche (oder einen binären Suchbaum) benutzt, dann führt dieser ,,Angriff '' in Zeit $O(log(|K|)) =O(log(2^s)) =O(s)$ zum Klartext $x$.
+    \begin{enumerate*}
+        \item Lasse fünf Klartexte $x_1,...,x_5$ zu $y_1,...,y_5$ verschlüsseln.
+        \item  ,,Schaue nach'', welcher Schlüssel $k$ mit den Paaren $(x_1,y_1),...,(x_5,y_5)$ konsistent ist (wenigstens einer ist es und es gibt höchstens einen).
+        \item Höre Chiffretext $y$ ab und berechne $x=d(y,k)$ aus $y$.üm in Schritt 2. ,,nachsehen'' zu können, benötigt Eva eine Tabelle mit allen möglichen Chiffretextkombinationen für die fünf Klartexte $x_1,...,x_5$. Dafür gibt es $\geq |K|= 2^s$ Möglichkeiten, nämlich für jeden Schlüssel eine. Die Hashtabelle oder der Suchbaum (oder die entsprechende Struktur in einem Turingmaschinen-Programm) hat Größe $2^s$. Wenn der Programmtext oder die Turingmaschine binäre Suche (oder einen binären Suchbaum) benutzt, dann führt dieser ,,Angriff '' in Zeit $O(log(|K|)) =O(log(2^s)) =O(s)$ zum Klartext $x$.
+    \end{enumerate*}
 
     Dieses Vorgehen ist natürlich unrealistisch, denn niemand kann ein so großes Programm schreiben oder speichern, wenn etwa (wie bei AES) die Zahl der Schlüssel $2^{128}>10^{38}$ beträgt. Um diesen Trick auszuschließen, werden wir den Ressourcenverbrauch eines Algorithmus als Summe von Laufzeit und Länge des Programmcodes (=Größe der Transitionstabelle der Turingmaschine) messen. Die Programmgröße ist eine untere Schranke für die Zeit, die Eva zur Erstellung ihres Programms benötigt. Unser Ressourcenmaß erfasst also die Zeit für die Erstellung und die Zeit für die Ausführung des Algorithmus.ünsere Algorithmen werden oft mit Kryptosystemen einer festen Blocklänge $l$ und einer fixen Anzahl von Klartext-/Chiffretext-Paaren arbeiten. Damit sind nur endlich viele Eingaben möglich, der Ressourcenverbrauch kann also nicht asymptotisch (,,bei großen Eingaben'') angegeben werden. Daher betrachten wir zunächst Algorithmen mit konstanter Laufzeit. Das führt dazu, dass alle eventuell vorhandenen Schleifen nur konstant oft durchlaufen werden. Damit können wir aber auf Schleifenanweisungen vollständig verzichten. Die resultierenden Programme heißen ,,Straight-Line-Programme''. Man kann sie sich in Pseudocode geschrieben vorstellen (ohne ,,while'' und ,,for'' und ohne Rückwärtssprünge), oder als Turingmaschinenprogramme mit der Eigenschaft, dass nie eine Programmzeile zweimal ausgeführt wird.
 
     \paragraph{Randomisierung bei Straight-Line-Programmen}
     In unseren Algorithmen wollen wir zusätzlich die Anweisung ,,$y\leftarrow flip(M)$'' erlauben, wobei $M$ eine endliche Menge ist. Die Idee dabei ist, dass der Variable $y$ ein zufälliges Element von $M$ (bzgl. der Gleichverteilung auf $M$) zugewiesen wird. Damit berechnen unsere Algorithmen keine Funktionen, sondern zufällige Werte, die durch eine Wahrscheinlichkeitsverteilung gesteuert werden.üm den Wahrscheinlichkeitsraum definieren zu können, machen wir die folgenden Annahmen (die keine Einschränkung darstellen):
-    1. Bei nacheinander ausgeführten Kommandos der Form $y\leftarrow flip(M)$, mit identischem oder unterschiedlichem $M$, werden immer neue, unabhängige Zufallsexperimente ausgeführt.
-    2. Wie eben diskutiert, enthalten unsere Algorithmen keine Schleifen. Wir können daher jedes Ergebnis eines Zufallsexperiments in einer eigenen Variable speichern. Zusätzlich können wir die flip-Kommandos aus dem gesamten Programm, auch den bedingten Anweisungen, herausziehen und sie alle (vorab, auf Vorrat) ausführen. Damit können wir annehmen: Wenn die Anweisung $y\leftarrow flip(M)$ im Programmtext vorkommt, dann wird sie in jedem Durchlauf des Algorithmus (unabhängig von der Eingabe und den Ergebnissen der flip-Anweisungen) genau einmal ausgeführt.
+    \begin{enumerate*}
+        \item Bei nacheinander ausgeführten Kommandos der Form $y\leftarrow flip(M)$, mit identischem oder unterschiedlichem $M$, werden immer neue, unabhängige Zufallsexperimente ausgeführt.
+        \item Wie eben diskutiert, enthalten unsere Algorithmen keine Schleifen. Wir können daher jedes Ergebnis eines Zufallsexperiments in einer eigenen Variable speichern. Zusätzlich können wir die flip-Kommandos aus dem gesamten Programm, auch den bedingten Anweisungen, herausziehen und sie alle (vorab, auf Vorrat) ausführen. Damit können wir annehmen: Wenn die Anweisung $y\leftarrow flip(M)$ im Programmtext vorkommt, dann wird sie in jedem Durchlauf des Algorithmus (unabhängig von der Eingabe und den Ergebnissen der flip-Anweisungen) genau einmal ausgeführt.
+    \end{enumerate*}
 
     Abkürzung: $flip(l)$ für $flip(\{0,1\}^l)$ (l-facher Münzwurf) und $flip()$ für $flip(1)$, also $flip(\{0,1\})$ (einfacher Münzwurf).
 
@@ -1178,20 +1308,24 @@
     $I$ sei die Menge der Eingaben, $Z$ sei die Menge der Ausgaben.
 
     Ist $x\in I$ eine Eingabe, so erhalten wir für jedes $m=(m_1,...,m_r)\in M$ genau eine Ausgabe $A^m(x)\in\mathbb{Z}$, indem wir in $A$ die Anweisung $y_i\leftarrow flip(M_i)$ durch $y_i\leftarrow m_i$ ersetzen. Auf diese Weise erhalten wir
-    - Für jedes $m\in M$ eine Funktion $A^m:I\rightarrow Z,x \rightarrow A^m(x)$, und
-    - für jedes $x\in I$ eine Zufallsgröße $A(x):M\rightarrow Z,m\rightarrow A^m(x)$.
+    \begin{itemize*}
+        \item Für jedes $m\in M$ eine Funktion $A^m:I\rightarrow Z,x \rightarrow A^m(x)$, und
+        \item für jedes $x\in I$ eine Zufallsgröße $A(x):M\rightarrow Z,m\rightarrow A^m(x)$.
+    \end{itemize*}
 
     Beispiel 2.8 Betrachte den folgenden Algorithmus:
-    - $A(x:\{0,1\}^4):\{0,1\}$        // nach dem Doppelpunkt: Typ der Eingabe bzw. Ausgabe
-    - $b_0 \leftarrow flip()$
-    - $b_1 \leftarrow flip()$
-    - $b_2 \leftarrow flip()$
-    - $b_3 \leftarrow flip()$
-    - $if b_0 = 1 \text{ then return } x(0)$
-    - $if b_1 = 1 \text{ then return } x(1)$
-    - $if b_2 = 1 \text{ then return } x(2)$
-    - $if b_3 = 1 \text{ then return } x(3)$
-    - return 1
+    \begin{itemize*}
+        \item $A(x:\{0,1\}^4):\{0,1\}$        // nach dem Doppelpunkt: Typ der Eingabe bzw. Ausgabe
+        \item $b_0 \leftarrow flip()$
+        \item $b_1 \leftarrow flip()$
+        \item $b_2 \leftarrow flip()$
+        \item $b_3 \leftarrow flip()$
+        \item $if b_0 = 1 \text{ then return } x(0)$
+        \item $if b_1 = 1 \text{ then return } x(1)$
+        \item $if b_2 = 1 \text{ then return } x(2)$
+        \item $if b_3 = 1 \text{ then return } x(3)$
+        \item return 1
+    \end{itemize*}
 
     Dann ist $M=\{0,1\}^4$, und es gilt: $A^{0110}(1101)=1$ und $A^{0010}(1101)=0$. Kompakt: Wenn $b_0 b_1 b_2 b_3$ mit $i$ Nullen und einer $1$ (an Position $i$) beginnt, dann ist die Ausgabe $x(i)$, für $i=0,1,2,3$. Ist $b_0 b_1 b_2 b_3=0000$, so ist die Ausgabe 1. Also gilt: $Pr(A(1101)=0) = Pr(\{w\in\{0,1\}^4 |w_0=w_1=0, w_2=1\}) =(\frac{1}{2})^3 =\frac{1}{8}$ und $Pr(b_1=1) =Pr(\{w\in\{0,1\}^4 |w(1)=1\})=\frac{1}{2}$
 
@@ -1201,16 +1335,20 @@
 
     \paragraph{Prozeduren als Parameter}
     Wir werden Algorithmen $A$ betrachten, die als Eingabe eine Prozedur $B$ (z.B. die Verschlüsselungsfunktion einer Blockchiffre mit fest eingesetzem Schlüssel) erhalten. Diese Prozedur darf nur aufgerufen werden, sie wird nicht als Text in den Rumpf von $A$ eingefügt. Sie könnte aber wiederum zufallsgesteuert sein. Um den Wahrscheinlichkeitsraum von $A$ mit einem solchen Funktionsparameter zu bestimmen, müssen folgende Informationen gegeben sein:
-    - $B$,
-    - die Anzahl der Aufrufe von $B$ in $A$,
-    - welche Aufrufe $y\leftarrow flip(M)$ in $B$ vorkommen.
+    \begin{itemize*}
+        \item $B$,
+        \item die Anzahl der Aufrufe von $B$ in $A$,
+        \item welche Aufrufe $y\leftarrow flip(M)$ in $B$ vorkommen.
+    \end{itemize*}
 
     Wir behandeln dann die Variablen $y$ in verschiedenen Aufrufen von $B$ genau wie die aus einem gewöhnlichen randomisierten Programm (Umbenennen der Variablen, Herausziehen der Zufallsexperimente an den Anfang). In dem resultierenden Wahrscheinlichkeitsraum sind dann die Zufallsexperimente in den verschiedenen Aufrufen von $B$ und die in Teilen von $A$ außerhalb von $B$ unabhängig.
 
     \subsubsection{Sicherheit von Block-Kryptosystemen}
     Wir betrachten hier l-Block-Kryptosysteme $B=(X,K,Y,e,d)$ mit $X=Y=\{0,1\}^l$ und $K\subseteq\{0,1\}^s$ für ein $s$. $e$ ist die Verschlüsselungsfunktion und $d$ die Entschlüsselungsfunktion. Wir erinnern uns:
-    1. Im Szenario 2 (chosen-plaintext attack,CPA) kann die Angreiferin Eva sich eine ,,geringe Zahl'' von Klartexten verschlüsseln lassen, also hat sie eine Liste von Klartext-Chiffretext-Paaren: $(x_1,y_1),...,(x_r,y_r)$. Nun kann jedenfalls nur ein ,,neuer'' Klartext $x$, also ein $x\in X\{x_1,...,x_r\}$, geheim übertragen werden. Die possibilistische Sicherheit verlangt genau, dass dies möglich ist: Wenn $y\in Y\{y_1,...,y_r\}$ ein neuer gegebener Chiffretext ist, dann gibt es für jeden Klartext $x\in X\{x_1,...,x_r\}$ einen Schlüssel $k$, der $x_i$ zu $y_i$ chiffriert, $1 \geq i\geq r$, und $x$ zu $y$ chiffriert.
-    2. Wenn dabei $r$ beliebig groß sein darf, können nach Prop.2.3 nur Substitutionskryptosysteme in diesem Sinne sicher sein. Da sie $|X|!$ Schlüssel haben müssen und daher immense Schlüssellänge (mindestens $|X|log|X|-O(|X|)$ Bits) erfordern, kommen sie in der Praxis nicht in Frage.
+    \begin{enumerate*}
+        \item Im Szenario 2 (chosen-plaintext attack,CPA) kann die Angreiferin Eva sich eine ,,geringe Zahl'' von Klartexten verschlüsseln lassen, also hat sie eine Liste von Klartext-Chiffretext-Paaren: $(x_1,y_1),...,(x_r,y_r)$. Nun kann jedenfalls nur ein ,,neuer'' Klartext $x$, also ein $x\in X\{x_1,...,x_r\}$, geheim übertragen werden. Die possibilistische Sicherheit verlangt genau, dass dies möglich ist: Wenn $y\in Y\{y_1,...,y_r\}$ ein neuer gegebener Chiffretext ist, dann gibt es für jeden Klartext $x\in X\{x_1,...,x_r\}$ einen Schlüssel $k$, der $x_i$ zu $y_i$ chiffriert, $1 \geq i\geq r$, und $x$ zu $y$ chiffriert.
+        \item  Wenn dabei $r$ beliebig groß sein darf, können nach Prop.2.3 nur Substitutionskryptosysteme in diesem Sinne sicher sein. Da sie $|X|!$ Schlüssel haben müssen und daher immense Schlüssellänge (mindestens $|X|log|X|-O(|X|)$ Bits) erfordern, kommen sie in der Praxis nicht in Frage.
+    \end{enumerate*}
 
     Idee eines neuen Sicherheitsbegriffs (für Block-Kryptosysteme): Gegeben sei eine Angreiferin Eva mit beschränkten Berechnungsressourcen. Wir fragen, wie sehr aus Evas Sicht das $l$-Block-Kryptosystem $B=(\{0,1\}^l,K,\{0,1\}^l,e,d)$ dem Substitutionskryptosystem $S'=(\{0,1\}^l,P\{0,1\}^l,\{0,1\}^l,e',d')$ (siehe Def.1.9) ähnelt. Ist es ihr mit ,,signifikanter Erfolgswahrscheinlichkeit'' möglich, aufgrund der ihr vorliegenden Information und im Rahmen ihrer Ressourcen, zu unterscheiden, welches der beiden Systeme verwendet wird? Wenn dies nicht der Fall ist, kann das Kryptosystem $B$ als sicher gelten, wie die folgende Überlegung zeigt.
     Wenn Eva aufgrund der ihr vorliegenden Information (2.4) nicht mit nennenswerter Erfolgswahrscheinlichkeit unterscheiden kann, ob sie es mit der Chiffre $e(.,k)$ (für ein zufälliges $k\in K$) oder mit $e'(.,\pi)$ (für ein zufälliges $\pi\in P_{\{0,1\}^l}$) zu tun hat, dann hat sie aus der ihr vorliegenden Information keine nennenswerte Information über die konkrete Chiffree $(.,k)$ gelernt. Insbesondere kann sich  ihre Information über die Klartextverteilung nicht (wesentlich) ändern, wenn ihr ein neuer Chiffretext $y$ vorgelegt wird, da bei $S'=(\{0,1\}^l,P\{0,1\}^l,\{0,1\}^l,e',d')$ keine solche Änderung eintritt.
@@ -1224,21 +1362,29 @@
     Am liebsten hätte man folgendes Verhalten, für ein gegebenes Block-Kryptosystem $B$: Programm $U$ sollte 1 liefern, wenn $F$ eine Chiffre $e(.,k)$ zu $B$ ist, und $0$, wenn $F=\pi$ für eine Permutation $\pi\in P\{0,1\}^l$ ist, die keine $B$-Chiffre ist. Das gewünschte Verhalten wird sich bei $U$ natürlich niemals mit absoluter Sicherheit, sondern nur mit mehr oder weniger großer Wahrscheinlichkeit einstellen, je nach Situation.
 
     Beispiel 2.10 Als Beispiel betrachten wir das Vernam-Kryptosystem $B=B_{Vernam}$, siehe Beispiel 1.6. Wir definieren einen l-Unterscheider $U=U_{Vernam}$, der als Parameter eine Funktion $F:\{0,1\}^l\rightarrow\{0,1\}^l$ erhält und Folgendes tut:
-    1. $k\leftarrow F(0^l)$
-    2. $y\leftarrow F(1^l)$
-    3. falls $1^l\oplus_l k=y$, dann gib $1$ aus, sonst $0$.
+    \begin{enumerate*}
+        \item $k\leftarrow F(0^l)$
+        \item  $y\leftarrow F(1^l)$
+        \item  falls $1^l\oplus_l k=y$, dann gib $1$ aus, sonst $0$.
+    \end{enumerate*}
 
     Dieser Unterscheider benutzt keine Zufallsexperimente, obwohl es ihm erlaubt wäre. Man sieht leicht Folgendes: Wenn $F(.) =e(.,k)$ für die Vernam-Chiffre mit beliebigem Schlüssel $k$, liefert $U_{Vernam}$ immer das Ergebnis 1. Wenn hingegen $F$ eine zufällige Permutation $\pi$ von $\{0,1\}^l$ ist, dann ist die Wahrscheinlichkeit, dass $F(1^l)=1^l\oplus_l F(0^l)$ gilt, genau $\frac{1}{2^{l-1}}$, also wird die Ausgabe $1$ nur mit sehr kleiner Wahrscheinlichkeit ausgegeben (wenn $l$ nicht ganz klein ist).
 
     Wir definieren nun ein Spiel (,,game''), mit dem ein beliebiges Block-Kryptosystem $B$ und ein beliebiger Unterscheider $U$ darauf getestet werden, ob $B$ gegenüber $U$ ,,anfällig'' ist oder nicht. (Das Spiel ist ein Gedankenexperiment, es ist nicht algorithmisch.) Die Idee ist folgende: Man entscheidet mit einem Münzwurf (Zufallsbit b), ob $U$ für seine Untersuchungen als $F(.)$ eine zufällige Chiffre $e(.,k)$ von $B$ (,,Realwelt'') oder eine zufällige Permutation $\pi$ von $\{0,1\}^l$ (,,Idealwelt'') erhalten soll. Dann rechnet $U$ mit $F$ als Orakel und gibt dann seine Meinung ab, ob er sich in der Realwelt oder in der Idealwelt befindet. U ,,gewinnt'', wenn diese Meinung zutrifft.
 
     Definition 2.11 Sei $B=(\{0,1\}^l,K,\{0,1\}^l,e,d)$ ein l-Block-Kryptosystem, und sei $U$ ein Unterscheider. Das zugehörige Experiment (oder Spiel) ist der folgende Algorithmus:
-    - $GBU():\{0,1\}$ // kein Argument, Ausgabe ist ein Bit
-    1. $b\leftarrow flip(\{0,1\})$
-    - if $b=1$ (,,Realwelt'') then $k\leftarrow flip(K);F\leftarrow e(.,k)$ // Zufallschiffre von B
-    - if $b=0$ (,,Idealwelt'') then $F\leftarrow flip(P\{0,1\}^l)$ // Zufallspermutation
-    2. $b'\leftarrow U(F)$  // Der l-Unterscheider versucht herauszubekommen, ob $b=0$ oder $b=1$ gilt.
-    3. if $b=b'$ then return 1 else return 0.     // 1 heißt, dass $U$ das richtige Ergebnis hat.
+    \begin{itemize*}
+        \item $GBU():\{0,1\}$ // kein Argument, Ausgabe ist ein Bit
+        \begin{enumerate*}
+            \item $b\leftarrow flip(\{0,1\})$
+            \begin{itemize*}
+                \item if $b=1$ (,,Realwelt'') then $k\leftarrow flip(K);F\leftarrow e(.,k)$ // Zufallschiffre von B
+                \item if $b=0$ (,,Idealwelt'') then $F\leftarrow flip(P\{0,1\}^l)$ // Zufallspermutation
+            \end{itemize*}
+            \item  $b'\leftarrow U(F)$  // Der l-Unterscheider versucht herauszubekommen, ob $b=0$ oder $b=1$ gilt.
+            \item  if $b=b'$ then return 1 else return 0.     // 1 heißt, dass $U$ das richtige Ergebnis hat.
+        \end{enumerate*}
+    \end{itemize*}
 
     Das verkürzte Experiment/Spiel $S^B_U$ (,,short'') gibt im 3.Schritt einfach $b'$ aus.
 
@@ -1249,9 +1395,11 @@
     Beispiel 2.12 Für einen l-Bit-String $z$ sei $p(z)=\oplus_{1 \geq i\geq l} z_i$ seine Parität. Nehmen wir an, das Chiffrierverfahren von $N$ ist unvorsichtig geplant und zwar so, dass $p(e(x,k)) =p(x)$ ist für beliebige $x\in X$ und $k\in K$. Bei der Chiffrierung ändert sich also die Parität nicht. (Die Parität ist sicher ,,nichttriviale Information'', auch wenn sie vielleicht nicht unmittelbar nützlich ist.)
 
     $U_{Paritaet}(F)$:
-    1. Wähle Klartexte $x_1,...,x_q$ mit $p(x_1)=...=p(x_q) = 0$.
-    2. $y_r\leftarrow F(x_r)$, für $r=1,...,q$.
-    3. falls $p(y_1)=...=p(y_q)=0$, dann gib 1 aus, sonst 0.
+    \begin{enumerate*}
+        \item Wähle Klartexte $x_1,...,x_q$ mit $p(x_1)=...=p(x_q) = 0$.
+        \item  $y_r\leftarrow F(x_r)$, für $r=1,...,q$.
+        \item  falls $p(y_1)=...=p(y_q)=0$, dann gib 1 aus, sonst 0.
+    \end{enumerate*}
 
     Wenn wir in der  ,,Realwelt''  sind $(b=1)$, also $F$ eine Chiffre $e(.,k)$ ist, dann ist die Antwort von $U$ immer ,,1'', also korrekt. Wenn wir in der ,,Idealwelt'' sind $(b=0)$, also $F$ eine zufällige Permutation ist, dann ist die Antwort nur mit Wahrscheinlichkeit $\frac{2^{l-1} (2^{l-1} -1)(2^{l-1}-2)...(2^{l-1} -q+ 1)}{2^l (2^l-1)(2^l-2)...(2^l-q+ 1)} \geq \frac{1}{2^q}$ falsch. (Alle Werte $F(x_1),...,F(x_q)$ müssen zufällig zu Chiffretexten geführt haben, die Parität 0 haben, von denen es $2^{l-1}$ viele gibt.) Es ergibt sich $Pr(G^B_U= 1)\geq\frac{1}{2} (1+1-2^{-q}) =1-2^{-(q+1)}$. Mit der effizienten Berechenbarkeit der ,,nichttrivialen Information'' hat man also einen Unterscheider gefunden, der $Pr(G^B_U=1)$ ,, groß'' macht.
 
@@ -1261,19 +1409,23 @@
     Definition 2.13 Sei $U$ ein l-Unterscheider und $B$ ein l-Block-KS. Der Vorteil von $U$ bzgl. B ist $adv(U,B):= 2(Pr(G^B_U=1)-\frac{1}{2})$-
 
     Klar ist:
-    - Für jeden l-Unterscheider $U$ und jedes l-Block-KS $B$ gilt $-1\geq adv(U,B)\geq 1$.
-    - Werte $adv(U,B)<0$ sind uninteressant. (Wenn man einen Unterscheider $U$ mit $adv(U,B)<0$ hat, sollte man in $U$ die Ausgaben $0$ und $1$ vertauschen und erhält einen Unterscheider mit positivem Vorteil.)
-    - Für den trivialen l-Unterscheider $U_{trivial}$ gilt $adv(U,B) = 0$.üm den Vorteil eines Unterscheiders auszurechnen, sind seine ,,Erfolgswahrscheinlichkeit'' und seine ,,Misserfolgswahrscheinlichkeit'' hilfreiche Werte: Der Erfolg von U bzgl. B ist (für das verkürzte Spiel $S=S_U^B$) $suc(U,B) := Pr(S\langle b= 1\rangle = 1)$, d.h. die Wahrscheinlichkeit, dass U die Verwendung des l-Block-KS B richtig erkennt. Der Misserfolg ist $fail(U,B) := fail(U) := Pr(S\langle b= 0\rangle = 1)$, d.h. die Wahrscheinlichkeit, dass U die Verwendung des idealen Substitutionskryptosystems nicht erkennt. Man kann in der Notation für ,,fail'' das ,,B'' auch weglassen, da im Fall $b=0$ das Kryptosystem B überhaupt keine Rolle spielt.
+    \begin{itemize*}
+        \item Für jeden l-Unterscheider $U$ und jedes l-Block-KS $B$ gilt $-1\geq adv(U,B)\geq 1$.
+        \item Werte $adv(U,B)<0$ sind uninteressant. (Wenn man einen Unterscheider $U$ mit $adv(U,B)<0$ hat, sollte man in $U$ die Ausgaben $0$ und $1$ vertauschen und erhält einen Unterscheider mit positivem Vorteil.)
+        \item Für den trivialen l-Unterscheider $U_{trivial}$ gilt $adv(U,B) = 0$.üm den Vorteil eines Unterscheiders auszurechnen, sind seine ,,Erfolgswahrscheinlichkeit'' und seine ,,Misserfolgswahrscheinlichkeit'' hilfreiche Werte: Der Erfolg von U bzgl. B ist (für das verkürzte Spiel $S=S_U^B$) $suc(U,B) := Pr(S\langle b= 1\rangle = 1)$, d.h. die Wahrscheinlichkeit, dass U die Verwendung des l-Block-KS B richtig erkennt. Der Misserfolg ist $fail(U,B) := fail(U) := Pr(S\langle b= 0\rangle = 1)$, d.h. die Wahrscheinlichkeit, dass U die Verwendung des idealen Substitutionskryptosystems nicht erkennt. Man kann in der Notation für ,,fail'' das ,,B'' auch weglassen, da im Fall $b=0$ das Kryptosystem B überhaupt keine Rolle spielt.
+    \end{itemize*}
 
     Lemma 2.14 $adv(U,B) = suc(U,B)-fail(U)$.
 
     Beweis: $Pr(G^B_U= 1) = Pr(S_U^B=b)$
-    - $= Pr(S_U^B= 1,b= 1) + Pr(S_U^B= 0,b= 0)$
-    - $= Pr(S_U^B= 1|b= 1)* Pr(b= 1) + Pr(S^B_U= 0|b= 0)*Pr(b= 0)$
-    - $=\frac{1}{2}( Pr(S_U^B\langle b= 1\rangle = 1) + Pr(S_U^B\langle b= 0\rangle = 0))$
-    - $=\frac{1}{2}( Pr(S_U^B\langle b= 1\rangle = 1) + (1-Pr(S_U^B\langle b= 0\rangle = 1)))$
-    - $=\frac{1}{2}( (suc(U,B) + (1-fail(U)))$
-    - $=\frac{1}{2}(suc(U,B)-fail(U)) + \frac{1}{2}$
+    \begin{itemize*}
+        \item $= Pr(S_U^B= 1,b= 1) + Pr(S_U^B= 0,b= 0)$
+        \item $= Pr(S_U^B= 1|b= 1)* Pr(b= 1) + Pr(S^B_U= 0|b= 0)*Pr(b= 0)$
+        \item $=\frac{1}{2}( Pr(S_U^B\langle b= 1\rangle = 1) + Pr(S_U^B\langle b= 0\rangle = 0))$
+        \item $=\frac{1}{2}( Pr(S_U^B\langle b= 1\rangle = 1) + (1-Pr(S_U^B\langle b= 0\rangle = 1)))$
+        \item $=\frac{1}{2}( (suc(U,B) + (1-fail(U)))$
+        \item $=\frac{1}{2}(suc(U,B)-fail(U)) + \frac{1}{2}$
+    \end{itemize*}
 
     Durch Umstellen ergibt sich die Behauptung.ünterscheider mit Werten $adv(U,B)$ substanziell über 0 können als interessant (oder möglicherweise gefährlich) für B gelten. Wir müssen noch die beschränkten Ressourcen des Unterscheiders ins Spiel bringen.
 
@@ -1314,34 +1466,42 @@
     Szenarium 3: Alice möchte Bob mehrere Klartexte beliebiger Länge schicken. Sie verwendet dafür immer denselben Schlüssel. Eva hört die Chiffretexte mit und kann sich einige wenige Klartexte mit dem verwendeten Schlüssel verschlüsseln lassen.
 
     Erweiterungen im Vergleich zu vorher:
-    1. Beliebig lange Texte sind erlaubt.
-    2. Mehrfaches Senden desselben Textes ist möglich; Eva sollte dies aber nicht erkennen.
+    \begin{enumerate*}
+        \item Beliebig lange Texte sind erlaubt.
+        \item  Mehrfaches Senden desselben Textes ist möglich; Eva sollte dies aber nicht erkennen.
+    \end{enumerate*}
 
     Wir müssen die bisher benutzten Konzepte anpassen, um auch das mehrfache Senden derselben Nachricht zu erfassen. Ein grundlegender Ansatz, um mit identischen Botschaften umzugehen, ist Folgendes: Alices Verschlüsselungsalgorithmus ist randomisiert, liefert also zufallsabhängig unterschiedliche Chiffretexte für ein und denselben Klartext. Allerdings ist normalerweise die Anzahl der Zufallsexperimente fest und nicht von der Klartextlänge abhängig, sodass wir wie vorher davon ausgehen können, dass nur ein Experiment am Anfang ausgeführt wird. Auch der Sicherheitsbegriff und die Rechenzeitbeschränkungen müssen verändert werden.
 
     Klartexte und Chiffretexte sind nun endliche Folgen von Bitvektoren (,,Blöcken'') der festen Länge $l$. Die Menge aller dieser Folgen bezeichnen wir mit $(\{0,1\}^l)^*$ oder kürzer mit $\{0,1\}^{l*}$. Es gibt also unendlich viele Klartexte und Chiffretexte. Die Menge der Schlüssel heißt $K$. Der Verschlüsselungsalgorithmus $E$ ist randomisiert und transformiert einen Klartext $x$ und einen Schlüssel $k$ in einen Chiffretext. Der Entschlüsselungsalgorithmus $D$ ist deterministisch und transformiert einen Chiffretext $y$ und einen Schlüssel $k$ in einen Klartext. Sicher muss man den Algorithmen eine Rechenzeit zugestehen, die von der Länge der zu verarbeitenden Texte abhängt. Als effizient werden Algorithmen angesehen, die eine polynomielle Rechenzeit haben. Es muss eine verallgemeinerte Dechiffrierbedingung gelten, die die Randomisierung der Verschlüsselung berücksichtigt.
 
     Definition 3.1 Ein symmetrisches $l$-Kryptoschema ist ein Tupel $S= (K,E,D)$, wobei
-    - $K\subseteq\{0,1\}^s$ eine endliche Menge ist (für ein $s\in\mathbb{N}$),
-    - $E(x:\{0,1\}^{l*},k:K) :\{0,1\}^{l*}$ ein randomisierter Algorithmus und
-    - $D(y:\{0,1\}^{l*},k:K) :\{0,1\}^{l*}$ ein deterministischer Algorithmus
-    sind, so dass gilt:
-    - Die Laufzeiten von $E$ und $D$ sind polynomiell beschränkt in der Länge von $x$ bzw. $y$.
-    - Für jedes $x\in\{0,1\}^{l*},k\in K$ und jedes $m\in M_1\times...\times M_r$ (die Ausgänge der flip-Anweisungen in $E$) gilt: $D(E^m(x,k),k)=x$ (Dechiffrierbedingung).
+    \begin{itemize*}
+        \item $K\subseteq\{0,1\}^s$ eine endliche Menge ist (für ein $s\in\mathbb{N}$),
+        \item $E(x:\{0,1\}^{l*},k:K) :\{0,1\}^{l*}$ ein randomisierter Algorithmus und
+        \item $D(y:\{0,1\}^{l*},k:K) :\{0,1\}^{l*}$ ein deterministischer Algorithmus
+        sind, so dass gilt:
+        \item Die Laufzeiten von $E$ und $D$ sind polynomiell beschränkt in der Länge von $x$ bzw. $y$.
+        \item Für jedes $x\in\{0,1\}^{l*},k\in K$ und jedes $m\in M_1\times...\times M_r$ (die Ausgänge der flip-Anweisungen in $E$) gilt: $D(E^m(x,k),k)=x$ (Dechiffrierbedingung).
+    \end{itemize*}
 
     Die Elemente von
-    - $K$ heißen ,,Schlüssel''
-    - $\{0,1\}^{l*}$ heißen ,,Klartexte'' bzw. ,,Chiffretexte'', je nachdem, welche Rolle sie gerade spielen.
+    \begin{itemize*}
+        \item $K$ heißen ,,Schlüssel''
+        \item $\{0,1\}^{l*}$ heißen ,,Klartexte'' bzw. ,,Chiffretexte'', je nachdem, welche Rolle sie gerade spielen.
+    \end{itemize*}
 
     $E$ ist der Chiffrieralgorithmus, $D$ der Dechiffrieralgorithmus.
 
     Bemerkungen:
-    - Zentral: Die Nachrichtenlänge ist unbestimmt.
-    - Wir werden immer davon ausgehen, dass der Schlüssel $k$ uniform zufällig aus $K$ gewählt wurde und beiden Parteien bekannt ist. Die Angreiferin Eva kennt $k$ natürlich nicht.
-    - Etwas allgemeiner ist es, wenn man den Schlüssel nicht uniform zufällig aus einer Menge wählt, sondern von einem randomisierten Schlüsselerzeugungsalgorithmus $G(s:integer):\{0,1\}^*$ generieren lässt. Konzeptuell besteht zwischen den beiden Situationen aber kein großer Unterschied.
-    - Jeder Klar-und jeder Chiffretext ist ein Bitvektor der Länge $l*h$ für ein $h\in\mathbb{N}$. (Um auf ein Vielfaches der Blocklänge zu kommen, muss man die Texte notfalls mit Nullen auffüllen.)
-    - Um einen Klartext $x$ zu verschicken, wird der Algorithmus $E$ mit einem neuen, uniform zufällig gewählten Element $m$ abgearbeitet und es wird $E^m(x,k)$ als Chiffretext verschickt. Insbesondere entsteht bei wiederholter Verschlüsselung eines Textes $x$ (mit sehr großer Wahrscheinlichkeit) jedes mal ein anderer Chiffretext.
-    - In der Literatur finden sich auch Kryptoschemen, bei denen auch die Dechiffrierung randomisiert ist. Wir betrachten dies hier nicht.
+    \begin{itemize*}
+        \item Zentral: Die Nachrichtenlänge ist unbestimmt.
+        \item Wir werden immer davon ausgehen, dass der Schlüssel $k$ uniform zufällig aus $K$ gewählt wurde und beiden Parteien bekannt ist. Die Angreiferin Eva kennt $k$ natürlich nicht.
+        \item Etwas allgemeiner ist es, wenn man den Schlüssel nicht uniform zufällig aus einer Menge wählt, sondern von einem randomisierten Schlüsselerzeugungsalgorithmus $G(s:integer):\{0,1\}^*$ generieren lässt. Konzeptuell besteht zwischen den beiden Situationen aber kein großer Unterschied.
+        \item Jeder Klar-und jeder Chiffretext ist ein Bitvektor der Länge $l*h$ für ein $h\in\mathbb{N}$. (Um auf ein Vielfaches der Blocklänge zu kommen, muss man die Texte notfalls mit Nullen auffüllen.)
+        \item Um einen Klartext $x$ zu verschicken, wird der Algorithmus $E$ mit einem neuen, uniform zufällig gewählten Element $m$ abgearbeitet und es wird $E^m(x,k)$ als Chiffretext verschickt. Insbesondere entsteht bei wiederholter Verschlüsselung eines Textes $x$ (mit sehr großer Wahrscheinlichkeit) jedes mal ein anderer Chiffretext.
+        \item In der Literatur finden sich auch Kryptoschemen, bei denen auch die Dechiffrierung randomisiert ist. Wir betrachten dies hier nicht.
+    \end{itemize*}
 
     Der Standardansatz zur Konstruktion eines Kryptoschemas besteht darin, von einer Block-chiffre wie in Kapitel 2 auszugehen und sie zu einem Kryptoschema auszubauen. Dies wird im Folgenden beschrieben.
 
@@ -1352,14 +1512,16 @@
     Dies ist die nächstliegende Methode. Ein Schlüssel ist ein Schlüssel $k$ von $B$. Man verschlüsselt einfach die einzelnen Blöcke von $x$ mit $B$, jedes mal mit demselben Schlüssel $k$.
 
     Definition: Das zu $B$ gehörende $l$-ECB-Kryptoschema $S=ECB(B)=(KB,E,D)$ ist gegeben durch die folgenden Algorithmen:
-    - $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$
-    - zerlege $x$ in Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
-    - für $0\geq i < m$ setze $y_i\leftarrow e_B(x_i,k)$;
-    - gib $y=y_0 ...y_{m-1}$ zurück.
-    - $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$
-    - zerlege $y$ in Blöcke der Länge $l:y=y_0 y_1 ...y_{m-1}$;
-    - für $0\geq i < m$ setze $x_i\leftarrow d_B(y_i,k)$;
-    - gib $x=x_0 ...x_{m-1}$ zurück.
+    \begin{itemize*}
+        \item $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$
+        \item zerlege $x$ in Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
+        \item für $0\geq i < m$ setze $y_i\leftarrow e_B(x_i,k)$;
+        \item gib $y=y_0 ...y_{m-1}$ zurück.
+        \item $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$
+        \item zerlege $y$ in Blöcke der Länge $l:y=y_0 y_1 ...y_{m-1}$;
+        \item für $0\geq i < m$ setze $x_i\leftarrow d_B(y_i,k)$;
+        \item gib $x=x_0 ...x_{m-1}$ zurück.
+    \end{itemize*}
 
     Die Verschlüsselung verzichtet auf die Option, Randomisierung zu verwenden. (Sie hat den großen Vorteil, parallel ausführbar zu sein.) Es ist klar, dass die Dechiffrierbedingung erfüllt ist. Jedoch hat dieses Kryptoschema ein ziemlich offensichtliches Problem, nämlich,
     dass ein Block $x\in\{0,1\}^l$ immer gleich verschlüsselt wird, Eva also ganz leicht nicht-triviale Informationen aus dem Chiffretext erhalten kann. Zum Beispiel kann sie sofort sehen, ob der Klartext die Form $x=x_1 x_1$, mit $x_1\in\{0,1\}^l$, hat oder nicht.
@@ -1372,16 +1534,18 @@
     Diese Betriebsart weicht dem zentralen Problem von ECB aus, indem man die Blöcke in Runden $i=0, 1 ,...,m-1$ nacheinander verschlüsselt und das Ergebnis einer Runde zur Modifikation des Klartextblocks der nächsten Runde benutzt. Konkret: Es wird nicht $x_i$ mit $B$ verschlüsselt, sondern $x_i\oplus_l y_{i-1}$ (bitweises XOR). Man benötigt dann einen Anfangsvektor $y_{-1}$ für die erste Runde. Dieser ist Teil des Schlüssels des Kryptoschemas (nicht von B), ein Schlüssel des Schemas ist also ein Paar $(k,v)$ mit $k\in K_B$ und $v\in\{0,1\}^l$.
 
     Definition: Das zu $B$ gehörende $l$-CBC-Kryptoschema $S=CBC(B)=(KB\times\{0,1\}^l,E,D)$ ist durch die folgenden Algorithmen gegeben:
-    - $E(x:\{0,1\}^{l*},(k,v) :KB\times\{0,1\}^l) :\{0,1\}^{l*}$
-    - zerlege $x$ in Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
-    - $y_{-1} \leftarrow v$;
-    - für $i= 0,...,m-1$ nacheinander: $y_i\leftarrow e_B(x_i\oplus_l y_{i-1},k)$;
-    - gib $y=y_0 ...y_{m-1}$ zurück.
-    - $E(x:\{0,1\}^{l*},(k,v) :K_B\times\{0,1\}^l) :\{0,1\}^{l*}$
-    - zerlege $y$ in Blöcke der Länge $l:y=y_0 y_1 ...y_{m-1}$
-    - $y_{-1} \leftarrow v$;
-    - für $i=0,...,m-1$ nacheinander: $x_i\leftarrow d_B(y_i,k)\oplus_l y_{i-1}$;
-    - gib $x=x_0 ...x_{m-1}$ zurück.
+    \begin{itemize*}
+        \item $E(x:\{0,1\}^{l*},(k,v) :KB\times\{0,1\}^l) :\{0,1\}^{l*}$
+        \item zerlege $x$ in Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
+        \item $y_{-1} \leftarrow v$;
+        \item für $i= 0,...,m-1$ nacheinander: $y_i\leftarrow e_B(x_i\oplus_l y_{i-1},k)$;
+        \item gib $y=y_0 ...y_{m-1}$ zurück.
+        \item $E(x:\{0,1\}^{l*},(k,v) :K_B\times\{0,1\}^l) :\{0,1\}^{l*}$
+        \item zerlege $y$ in Blöcke der Länge $l:y=y_0 y_1 ...y_{m-1}$
+        \item $y_{-1} \leftarrow v$;
+        \item für $i=0,...,m-1$ nacheinander: $x_i\leftarrow d_B(y_i,k)\oplus_l y_{i-1}$;
+        \item gib $x=x_0 ...x_{m-1}$ zurück.
+    \end{itemize*}
 
     Der Vektor $v$ wird Initialisierungsvektor genannt. Man versteht recht gut, was beim Chiffrieren passiert, wenn man sich das Bild auf Seite 104 im Buch von Küsters/Wilke ansieht. Beim Dechiffrieren geht man den umgekehrten Weg: Entschlüssele einen Block $y_i$ mittels B, dann addiere $y_{i-1}$, um den Klartextblock $x_i$ zu erhalten. Es ist klar, dass die Dechiffrierbedingung erfüllt ist.
 
@@ -1395,26 +1559,32 @@
     Um das Problem der identischen Verschlüsselung identischer Klartexte zu beseitigen, muss in die Verschlüsselung eine Zufalls komponente eingebaut werden. Beispielsweise kann man dazu CBC leicht modifizieren. Der Initialisierungsvektor $y_{-1}=v\in\{0,1\}^l$  ist nicht mehr Teil des Schlüssels, sondern wird vom Verschlüsselungsalgorithmus einfach zufällig gewählt, und zwar für jeden Klartext immer aufs Neue. Damit der Empfänger entschlüsseln kann, benötigt er $v$. Daher wird $y_{-1}$ als Zusatzkomponente dem Chiffretext vorangestellt. Damit ist der Chiffretext um einen Block länger als der Klartext, und Eva kennt auch $v=y_{-1}$.
 
     Definition: Das zu $B$ gehörende l-R-CBC-Kryptoschema $S=R-CBC(B) = (K_B,E,D)$ ist gegeben durch die folgenden Algorithmen:
-    - $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$
-    - setze $y_{-1}= flip(\{0,1\}^l)$;
-    - für $i=0,...,m-1$ nacheinander: $y_i\leftarrow e_B(x_i\oplus_l y_{i-1} ,k)$;
-    - gib $y=y_{-1} y_0 ...y_{m-1}$ zurück. //Länge: $m+1$ Blöcke
-    - $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$
-    - für $i=0,...,m-1$ nacheinander: $x_i\leftarrow d_B(y_i,k)\oplus_l y_{i-1}$;
-    - gib $x=x_0 ...x_{m-1}$ zurück. //Länge: m Blöcke
+    \begin{itemize*}
+        \item $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$
+        \item setze $y_{-1}= flip(\{0,1\}^l)$;
+        \item für $i=0,...,m-1$ nacheinander: $y_i\leftarrow e_B(x_i\oplus_l y_{i-1} ,k)$;
+        \item gib $y=y_{-1} y_0 ...y_{m-1}$ zurück. //Länge: $m+1$ Blöcke
+        \item $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$
+        \item für $i=0,...,m-1$ nacheinander: $x_i\leftarrow d_B(y_i,k)\oplus_l y_{i-1}$;
+        \item gib $x=x_0 ...x_{m-1}$ zurück. //Länge: m Blöcke
+    \end{itemize*}
 
     Es gibt zwei Unterschiede zu CBC:
-    1. Für jede Verschlüsselung eines Klartextes wird ein neuer zufälliger Initialisierungsvektor verwendet. Dadurch wird ein Klartext $x$ bei mehrfachem Auftreten mit hoher Wahrscheinlichkeit immer verschieden verschlüsselt.
-    2. Der Initialisierungsvektor ist nicht Teil des geheimen Schlüssels, sondern ist dem Angreifer bekannt, da er Teil des Chiffretextes ist. (Intuitiv würde man vielleicht sagen, dass dies ,,die Sicherheit verringert'' .)
+    \begin{enumerate*}
+        \item Für jede Verschlüsselung eines Klartextes wird ein neuer zufälliger Initialisierungsvektor verwendet. Dadurch wird ein Klartext $x$ bei mehrfachem Auftreten mit hoher Wahrscheinlichkeit immer verschieden verschlüsselt.
+        \item  Der Initialisierungsvektor ist nicht Teil des geheimen Schlüssels, sondern ist dem Angreifer bekannt, da er Teil des Chiffretextes ist. (Intuitiv würde man vielleicht sagen, dass dies ,,die Sicherheit verringert'' .)
+    \end{enumerate*}
 
     Tatsächlich und etwas unerwartet kann man nach einer sorgfältigen Formulierung eines Sicherheitsbegriffs für Kryptoschemen beweisen, dass die Betriebsart R-CBC zu ,,sicheren'' Verfahren führt, wenn die zugrundeliegende Blockchiffre ,,sicher'' ist. (Der Beweis ist
     aufwendig.)
 
     Warnung, als Beispiel für harmlos erscheinende Modifikationen, die Kryptoschemen unsicher machen:
-    1. Man könnte meinen, dass es genügt, bei jeder Verschlüsselung einen neuen Initialisierungsvektor zu benutzen, also zum Beispiel nacheinander $v,v+1,v+2,...$. Dies führt jedoch zu einem unsicheren Kryptoschema.
-    2. Um Kommunikation zu sparen, könnte man auf die Idee kommen, dass Alice und Bob sich von einer Kommunikationsrunde zur nächsten den letzten Chiffretextblock merken und ihn bei der nächsten Runde als Initialisierungsvektor benutzen. Dieses Verfahren heißt ,,chained CBC'' und wurde in SSL3.0 und TLS1.0 verwendet. Es stellte sich heraus, dass dieses Verfahren mit einem Angriff mit gewähltem Klartext erfolgreich attackiert werden kann!
+    \begin{enumerate*}
+        \item Man könnte meinen, dass es genügt, bei jeder Verschlüsselung einen neuen Initialisierungsvektor zu benutzen, also zum Beispiel nacheinander $v,v+1,v+2,...$. Dies führt jedoch zu einem unsicheren Kryptoschema.
+        \item  Um Kommunikation zu sparen, könnte man auf die Idee kommen, dass Alice und Bob sich von einer Kommunikationsrunde zur nächsten den letzten Chiffretextblock merken und ihn bei der nächsten Runde als Initialisierungsvektor benutzen. Dieses Verfahren heißt ,,chained CBC'' und wurde in SSL3.0 und TLS1.0 verwendet. Es stellte sich heraus, dass dieses Verfahren mit einem Angriff mit gewähltem Klartext erfolgreich attackiert werden kann!
+    \end{enumerate*}
 
     \subsubsection{OFB-Betriebsart( ,,Output Feed Back mode'' )}
     Wir kommen nun zu zwei Betriebsarten, die einen ganz anderen Ansatz für die Verschlüsselung der einzelnen Blöcke von $x$ verfolgen. Es wird dazu nämlich nicht $B$ mit Schlüssel $k$ benutzt, sondern der Mechanismus des Vernam-Systems (One-Time-Pads, siehe Beispiel 1.6) :$y_i=x_i\oplus_l k_i$, wobei $k_i\in\{0,1\}^l$ ein ,,Rundenschlüssel'' für Block $x_i$ ist. Das Kryptosystem $B$ wird nur dafür benutzt, diese Rundenschlüssel herzustellen, die bei Verschlüsselung und bei Entschlüsselung identisch sind. Die Dechiffrierbedingung folgt dann daraus, dass das Vernam-System korrekt dechiffriert. Der Ansatz führt dazu, dass die Entschlüsselungsfunktiond $B$ des Block-Kryptosystems gar nicht benötigt wird.
@@ -1422,16 +1592,18 @@
     Zuerst betrachten wir die Betriebsart ,,Output Feedback''. Dabei wird ein zufälliger Startvektor $v$ aus $\{0,1\}^l$ gewählt. Man setzt $k_{-1}=v$, und konstruiert die Rundenschlüssel $k_0,...,k_{m-1}$ dadurch, dass man iteriert den letzten Rundenschlüssel durch die Verschlüsselungsfunktion von $B$ schickt: $k_i=e_B(k_{i-1}, k)$, für $i=0,1,...,m-1$. (Der Name ''Output Feedback''rührt daher, dass das Ergebnis einer Verschlüsselung durch $e_B$ wieder als Input des nächsten Aufrufs von $e_B$ benutzt wird.) Der Empfänger benötigt $v$, um seinerseits die Rundenschlüssel zu berechnen; daher wird $v$ als $y_{-1}$ dem Chiffretext vorangestellt, wie beim R-CBC-Modus.
 
     Definition: Das zu $B$ gehörende l-OFB-Kryptoschema $S=(K_B,E,D) =OFB(B) =(K_B,E,D)$ ist gegeben durch die folgenden Algorithmen:
-    - $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
-    - $k_{-1} \leftarrow y_{-1} \leftarrow flip(\{0,1\}^l)$;
-    - für $i=0,...,m-1$ nacheinander: $k_i\leftarrow e_B(k_{i-1},k)$ und $y_i\leftarrow x_i\oplus_l k_i$;
-    - gib $y=y_{-1} y_0 ...y_{m-1}$ zurück.
-    - $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$;
-    - $k_{-1} \leftarrow y_{-1}$;
-    - für $i=0,...,m-1$ nacheinander: $k_i\leftarrow e_B(k_{i-1} ,k)$ und $x_i\leftarrow y_i\oplus k_i$;
-    - gib $x=x_0 ...x_{m-1}$ zurück.
+    \begin{itemize*}
+        \item $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
+        \item $k_{-1} \leftarrow y_{-1} \leftarrow flip(\{0,1\}^l)$;
+        \item für $i=0,...,m-1$ nacheinander: $k_i\leftarrow e_B(k_{i-1},k)$ und $y_i\leftarrow x_i\oplus_l k_i$;
+        \item gib $y=y_{-1} y_0 ...y_{m-1}$ zurück.
+        \item $D(y:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$;
+        \item $k_{-1} \leftarrow y_{-1}$;
+        \item für $i=0,...,m-1$ nacheinander: $k_i\leftarrow e_B(k_{i-1} ,k)$ und $x_i\leftarrow y_i\oplus k_i$;
+        \item gib $x=x_0 ...x_{m-1}$ zurück.
+    \end{itemize*}
 
     Dieses Verfahren hat einen interessanten Vorteil. Oft werden die Blöcke des Chiffretextes beim Empfänger nacheinander eintreffen. Die Hauptarbeit, nämlich die iterierte Verschlüsselung mit $e_B$ zur Ermittlung der Rundenschlüssel $k_i$, kann unabhängig von der Verfügbarkeit der Klartextblöcke erfolgen, sobald $y_{-1}$ eingetroffen ist.
 
@@ -1441,23 +1613,27 @@
     Dies ist die zweite Betriebsart, bei der das Kryptosystem $B$ nur zur Herstellung von m ,,Rundenschlüsseln'' benutzt wird, mit denen dann die Blöcke per $\oplus_l$ verschlüsselt werden. Anstatt iteriert zu verschlüsseln, was bei OFB eine sequentielle Verarbeitung erzwingt, werden hier die mit $B$ zu verschlüsselnden Strings anders bestimmt. Man fasst $\{0,1\}^l$ als äquivalent zur Zahlenmenge $\{0,1,...,2^{l-1}\}$ auf, interpretiert einen $l$-Bit-String also als Block oder als Zahl, wie es passt. In dieser Menge wählt man  eine Zufallszahl $r$. Man ,,zählt'' von $r$ ausgehend nach oben und berechnet die Rundenschlüssel $k_0,...,k_{m-1}$ durch Verschlüsseln von $r,r+1,...,r+m-1$ (modulo $2^l$ gerechnet) mittelse $B(.,k)$. Rundenschlüssel $k_i$ ist also $e_B((r+i) mod\ 2^l,k)$, und Chiffretextblock $y_i$ ist $k_i\oplus_l x_i$. Interessant ist, dass hier die Berechnung der Rundenschlüssel und die Ver- bzw. Entschlüsselung der Blöcke parallel erfolgen kann, also sehr schnell, falls mehrere Prozessoren zur Verfügung stehen.
 
     Definition: Das zu $B$ gehörende l-R-CTR-Kryptoschema $S=R-CTR(B) = (K_B,E,D)$ ist gegeben durch die folgenden Algorithmen:
-    - $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
-    - $r\leftarrow flip(\{0,..., 2^l-1\})$;
-    - für $0\geq i < m:y_i\leftarrow e_B((r+i) mod\ 2^l,k)\oplus_l x_i$;
-    - gib $y=r y_0 ...y_{m-1}$ zurück.
-    - $D(y: (\{0,1\}^l)+,k:K_B) :\{0,1\}^{l*}$;
-    - zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$;
-    - $r\leftarrow y_{-1}$;
-    - für $0\geq i < m:x_i\leftarrow e_B((r+i) mod\ 2^l,k)\oplus_l y_i$;
-    - gib $x=x_0 ...x_{m-1}$ zurück.
+    \begin{itemize*}
+        \item $E(x:\{0,1\}^{l*},k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $x$ in $m$ Blöcke der Länge $l:x=x_0 x_1 ...x_{m-1}$;
+        \item $r\leftarrow flip(\{0,..., 2^l-1\})$;
+        \item für $0\geq i < m:y_i\leftarrow e_B((r+i) mod\ 2^l,k)\oplus_l x_i$;
+        \item gib $y=r y_0 ...y_{m-1}$ zurück.
+        \item $D(y: (\{0,1\}^l)+,k:K_B) :\{0,1\}^{l*}$;
+        \item zerlege $y$ in $m+1$ Blöcke der Länge $l:y=y_{-1} y_0 y_1 ...y_{m-1}$;
+        \item $r\leftarrow y_{-1}$;
+        \item für $0\geq i < m:x_i\leftarrow e_B((r+i) mod\ 2^l,k)\oplus_l y_i$;
+        \item gib $x=x_0 ...x_{m-1}$ zurück.
+    \end{itemize*}
 
     Es ist offensichtlich, dass die Dechiffrierbedingung erfüllt ist.
 
     Bemerkungen:
-    - Wie bei R-CBC und OFB wird hier ein zufälliger Initialisierungswert $r$ verwendet, der als Teil des Chiffretextes dem Angreifer bekannt ist.
-    - Wie bei OFB wird die Entschlüsselungsfunktion $d_B$ gar nicht verwendet, man kann also anstelle der Verschlüsselungsfunktion $e_B$ eine beliebige Funktion $e_B:\{0,1\}^l\times\{0,1\}^l\rightarrow\{0,1\}^l$ benutzen, bei der die ,,Chiffren'' $e_B(.,k)$ nicht einmal injektiv sein müssen.
-    - Man kann dieses Kryptoschema auch wie folgt verstehen: Zu einem gegebenen Klartext $x\in\{0,1\}^{lm}$ wird aus einem zufälligen Initialwert $r$ ein langer Bitstring $k'=E_B(r,k) E_B((r+1) mod\ 2^l,k)... E_B((r+m-1) mod\ 2^l,k)$ berechnet und der Klartext $x$ dann mittels Vernamsystem und diesem Schlüssel verschlüsselt. Der Empfänger erhält $r$ und den Chiffretext, kann also ebenfalls $k'$ b erechnen und damit entschlüsseln. Ist $B$ ein sicheres Block-Kryptosystem, so kann ein Angreifer aus $r$ den Vernam-Schlüssel $k'$ nicht so einfach berechnen, da er $k$ nicht kennt. Die R-CTR-Betriebsart liefert also intuitiv einen hohen Grad an Sicherheit.
+    \begin{itemize*}
+        \item Wie bei R-CBC und OFB wird hier ein zufälliger Initialisierungswert $r$ verwendet, der als Teil des Chiffretextes dem Angreifer bekannt ist.
+        \item Wie bei OFB wird die Entschlüsselungsfunktion $d_B$ gar nicht verwendet, man kann also anstelle der Verschlüsselungsfunktion $e_B$ eine beliebige Funktion $e_B:\{0,1\}^l\times\{0,1\}^l\rightarrow\{0,1\}^l$ benutzen, bei der die ,,Chiffren'' $e_B(.,k)$ nicht einmal injektiv sein müssen.
+        \item Man kann dieses Kryptoschema auch wie folgt verstehen: Zu einem gegebenen Klartext $x\in\{0,1\}^{lm}$ wird aus einem zufälligen Initialwert $r$ ein langer Bitstring $k'=E_B(r,k) E_B((r+1) mod\ 2^l,k)... E_B((r+m-1) mod\ 2^l,k)$ berechnet und der Klartext $x$ dann mittels Vernamsystem und diesem Schlüssel verschlüsselt. Der Empfänger erhält $r$ und den Chiffretext, kann also ebenfalls $k'$ b erechnen und damit entschlüsseln. Ist $B$ ein sicheres Block-Kryptosystem, so kann ein Angreifer aus $r$ den Vernam-Schlüssel $k'$ nicht so einfach berechnen, da er $k$ nicht kennt. Die R-CTR-Betriebsart liefert also intuitiv einen hohen Grad an Sicherheit.
+    \end{itemize*}
 
     \subsection{Sicherheit von symmetrischen Kryptoschemen}
     Wir werden hier ein Sicherheitsmodell definieren, das es gestattet, Aussagen wie die folgende zu formulieren (und zu beweisen): Wenn B ein ,,sicheres'' l-Block-Kryptosystem ist (bzgl. einer Reihe von Parametern), und das Kryptoschema $S$ wird aus $B$ konstruiert, indem man einen geeigneten Betriebsmodus verwendet, dann ist $S$ ebenfalls ,,sicher'' (bzgl. einer variierten Reihe von Parametern). Ziel ist dabei, Betriebsmodi zu identifizieren, die keine unnötigen neuen Unsicherheitskomponenten ins Spiel bringen, die nicht im Block-KS
@@ -1468,11 +1644,13 @@
     Wir skizzieren zunächst eine Idee für ein Sicherheitsmodell, das die Fähigkeit, in so einer Situation ,,Information zu ermitteln'', formalisiert. Eva behauptet, sie könne ,,aus $y$ Information über $x$ ermitteln, die über die Länge von $x$ hinausgeht''. Um das zu überprüfen, stellt ihr ein ,,Herausforderer'' Charlie folgende Aufgabe: Eva darf sich zunächst eine Reihe von Klartexten ihrer Wahl verschlüsseln lassen. Dann wählt sie selbst zwei verschiedene, gleichlange Klartexte $z_0$ und $z_1$.  Charlie verschlüsselt einen von diesen; der Chiffretext ist $y$. Eva bekommt $y$. Sie soll herausfinden, ob $y$ von $z_0$ oder von $z_1$ kommt. Für diese Entscheidung darf sie sich weitere Klartexte verschlüsseln lassen und weiter rechnen. Wir betrachten ein Kryptoschema als unsicher, wenn Eva eine signifikant von ,,purem Raten'' abweichende Erfolgswahrscheinlichkeit hat, sie also mit guten Chancen unterscheiden kann, ob $z_0$ oder $z_1$ zu $y$ verschlüsselt wurde.
 
     Wie in Abschnitt 2.4 formulieren wir den Vorgang wieder als Spiel. Gegeben ist also $S=(K,E,D)$. Akteure sind Eva, hier ,,Angreifer'' (engl.: adversary) genannt, und Charlie (,,Herausforderer'' ,engl.: challenger). Die Parameter, mit denen wir die Erfolgschancen von Eva unter Ressourcenbeschränkungen messen, sind der ,,Vorteil'' in Analogie zu Definition 2.13, die Rechenzeit (inklusive Speicherplatz, wie vorher), Anzahl der Orakelaufrufe und Anzahl der bei der Verschlüsselung bearbeiteten Blöcke.
-    - Charlie wählt zufällig einen Schlüssel $k$ aus $K$ und legt damit die Chiffre $H=E(.,k)$ fest, die Klartexte aus $\{0,1\}^{l*}$ in Chiffretexte aus $\{0,1\}^{l*}$ transformiert.
-    - Eva wählt einige Klartexte und lässt sie sich von Charlie mit $H$ verschlüsseln.
-    - Eva wählt zwei Klartexte $z_0$ und $z_1$ gleicher Länge und gibt sie an Charlie.
-    - Verdeckt vor Eva: Charlie wirft eine Münze, um zufällig einen der beiden Klartexte zu wählen. Er verschlüsselt ihn mit $H$, das Ergebnis ist $y$. Charlie gibt $y$ an Eva.
-    - Eva kann sich weitere Klartexte verschlüsseln lassen und (mit beschränkten Ressourcen) rechnen und muss schließlich sagen (raten?), ob Charlie $z_0$ oder $z_1$ zu $y$ verschlüsselt hat.
+    \begin{itemize*}
+        \item Charlie wählt zufällig einen Schlüssel $k$ aus $K$ und legt damit die Chiffre $H=E(.,k)$ fest, die Klartexte aus $\{0,1\}^{l*}$ in Chiffretexte aus $\{0,1\}^{l*}$ transformiert.
+        \item Eva wählt einige Klartexte und lässt sie sich von Charlie mit $H$ verschlüsseln.
+        \item Eva wählt zwei Klartexte $z_0$ und $z_1$ gleicher Länge und gibt sie an Charlie.
+        \item Verdeckt vor Eva: Charlie wirft eine Münze, um zufällig einen der beiden Klartexte zu wählen. Er verschlüsselt ihn mit $H$, das Ergebnis ist $y$. Charlie gibt $y$ an Eva.
+        \item Eva kann sich weitere Klartexte verschlüsseln lassen und (mit beschränkten Ressourcen) rechnen und muss schließlich sagen (raten?), ob Charlie $z_0$ oder $z_1$ zu $y$ verschlüsselt hat.
+    \end{itemize*}
 
     Wenn die Wahrscheinlichkeit, dass Eva richtig antwortet, weit von zufälligem Raten abweicht, wollen wir das Kryptoschema als unsicher ansehen.
 
@@ -1481,8 +1659,10 @@
     Wir beschreiben den Part von Eva in diesem Spiel als Algorithmenpaar. Der erste Algorithmus $AF$ (der ,,Finder'', ,,find'' ) ist für das Erzeugen von $z_0$ und $z_1$ zuständig. Als Argument erhält dieser Teil eine Chiffre $H$, die er im Sinne eines Orakels benutzen kann. Außerdem werden Aufzeichnungen über die angeforderten Verschlüsselungen und ihre Ergebnisse gemacht. Diese Aufzeichnungen sind als Element $v$ einer endlichen Menge $V$ kodiert. Der zweite Algorithmus $AG$ (der ,,Rater'' , ,,guess'' ) ist dafür zuständig, herauszufinden, ob $z_0$ oder $z_1$ verschlüsselt wurde. Dieser Algorithmus bekommt $H$ als Orakel, die Aufzeichnungen $v$ von $AF$ und den Chiffretext $y$ als Input.
 
     Definition 3.2 Ein l-Angreifer $A$ ist ein Paar von randomisierten Algorithmen
-    - $AF(H(\{0,1\}^{l*}) :\{0,1\}^{l*}) : (\{0,1\}^l\times\{0,1\}^l)^+\times V$
-    - $AG(v:V,H(\{0,1\}^{l*}) :\{0,1\}^{l*},y:\{0,1\}^{l*}) :\{0,1\}$
+    \begin{itemize*}
+        \item $AF(H(\{0,1\}^{l*}) :\{0,1\}^{l*}) : (\{0,1\}^l\times\{0,1\}^l)^+\times V$
+        \item $AG(v:V,H(\{0,1\}^{l*}) :\{0,1\}^{l*},y:\{0,1\}^{l*}) :\{0,1\}$
+    \end{itemize*}
 
     Hierbei ist $H$ ein randomisierter Algorithmus (und nicht als Funktion zu verstehen).
 
@@ -1490,11 +1670,13 @@
     Im zweiten Schritt verwendet der ,,Rater'' $AG$ die Notizen $v$, die Chiffre $H=E(.,k)$ und die ,,Probe'' $y$, um zu bestimmen, ob $z_0$ oder $z_1$ verschlüsselt wurde.
 
     Definition 3.3 Sei $S=(K,E,D)$ ein symmetrisches Kryptoschema, und sei $A=(AF,AG)$ ein l-Angreifer. Das zugehörige Experiment (oder Spiel) ist der folgende Algorithmus $G^S_A:\{0,1\}:$
-    1. $k\leftarrow flip(K)$, $H\leftarrow E(.,k)$ (In diesem Schritt wählt Charlie zufällig eine Chiffre des Kryptoschemas $S$.)
-    2. $(z_0, z_1 ,v)\leftarrow AF(H)$ (In dieser Phase berechnet der Finder ein Paar von Klartexten gleicher Länge, von denen er annimmt,ihre Chiffretexte unterscheiden zu können.)
-    3. $b\leftarrow flip(\{0,1\})$ und $y\leftarrow E(z_b,k)$ (In dieser Phase wählt Charlie zufällig einen der beiden Klartexte und verschlüsselt ihn zu $y$.)
-    4. $b'\leftarrow AG(v,H,y)$ (In dieser Phase versucht der Rater herauszubekommen, ob $z_0$ oder $z_1$ verschlüsselt wurde.)
-    5. falls $b=b'$, so gib $1$ zurück, sonst $0$. (Charlies Auswertung: Hat $AG$ recht oder nicht?)
+    \begin{enumerate*}
+        \item $k\leftarrow flip(K)$, $H\leftarrow E(.,k)$ (In diesem Schritt wählt Charlie zufällig eine Chiffre des Kryptoschemas $S$.)
+        \item  $(z_0, z_1 ,v)\leftarrow AF(H)$ (In dieser Phase berechnet der Finder ein Paar von Klartexten gleicher Länge, von denen er annimmt,ihre Chiffretexte unterscheiden zu können.)
+        \item  $b\leftarrow flip(\{0,1\})$ und $y\leftarrow E(z_b,k)$ (In dieser Phase wählt Charlie zufällig einen der beiden Klartexte und verschlüsselt ihn zu $y$.)
+        \item  $b'\leftarrow AG(v,H,y)$ (In dieser Phase versucht der Rater herauszubekommen, ob $z_0$ oder $z_1$ verschlüsselt wurde.)
+        \item  falls $b=b'$, so gib $1$ zurück, sonst $0$. (Charlies Auswertung: Hat $AG$ recht oder nicht?)
+    \end{enumerate*}
 
     Das verkürzte Experiment oder Spiel $S^S_A$ gibt im 5.Schritt einfach $b'$ aus.
 
@@ -1503,18 +1685,22 @@
     Wenn ein Angreifer $A$ mit ,,nicht zu großem Rechenaufwand'' einen Vorteil erzielen kann, der deutlich über $0$ liegt, wird man das Kryptoschema als unsicher einstufen.
 
     Beispiel 3.4 Ziel ist es, die ECB-Betriebsart anzugreifen, d.h. sei $S=ECB(B)$ für ein l-Block-Kryptosystem $B$. Wir wollen zeigen, dass es einen l-Angreifer $A$ mit $Pr(G^S_A= 1) = 1$, also $adv(A,S) = 1$, gibt. Dieser ist wie folgt aufgebaut.
-    - l-Angreifer $A$ mit $V=\{1\}$ ($V=\{1\}$ bedeutet, dass stets $v=1$ gilt, dass die Aufzeichnung $v$ also keinerlei Information übermittelt.)
-    - $AF(H)$ arbeitet wie folgt: $z_0\leftarrow 0^{2l}; z_1\leftarrow 0^l 1^l;$ Ausgabe: $(z_0,z_1 ,1)$
-    - $AG(v,H,y)$ tut Folgendes: falls $y=y_1y_1$ für ein $y_1\in\{0,1\}^l$, gib $0$ aus, sonst $1$.
+    \begin{itemize*}
+        \item l-Angreifer $A$ mit $V=\{1\}$ ($V=\{1\}$ bedeutet, dass stets $v=1$ gilt, dass die Aufzeichnung $v$ also keinerlei Information übermittelt.)
+        \item $AF(H)$ arbeitet wie folgt: $z_0\leftarrow 0^{2l}; z_1\leftarrow 0^l 1^l;$ Ausgabe: $(z_0,z_1 ,1)$
+        \item $AG(v,H,y)$ tut Folgendes: falls $y=y_1y_1$ für ein $y_1\in\{0,1\}^l$, gib $0$ aus, sonst $1$.
+    \end{itemize*}
 
     Im Ablauf des Spiels $G^S_A$ wird der Rater $AG$ mit $y=E(0^{2l},k)=e_B(0^l,k)e_B(0^l,k)$ oder mit $y=E(0^l 1^l,k)=e_B(0^l,k)e_B(1^l,k)$ gestartet. Im ersten Fall ist $y=y_1y_1$ für ein $y_1\in\{0,1\}^l$, im zweiten Fall ist dies nicht so, wegen der Dechiffrierbedingung. Daher gilt $Pr(G^S_A= 1) = 1$, d.h. $adv(A,S) = 1$.
 
     Die Ressourcen, die $A$ benötigt, sind sehr klein: Zwei Aufrufe des Verschlüsselungsverfahrens $e_B$ des Block-Kryptosystems. Wir können schließen, dass es einen effizienten Angreifer $A$ gibt, dem das Sicherheitsmodell Vorteil 1 gibt. Damit gilt das Kryptoschema $ECB(B)$ als komplett unsicher.
 
     Beispiel 3.5 Ziel ist es, die CBC-Betriebsart anzugreifen, d.h. es sei $S=CBC(B)$ für ein Block-Kryptosystem $B$. Das Problem mit dieser Betriebsart ist, dass ein Klartext bei Wiederholung identisch verschlüsselt wird. Um dies auszunutzen, verwenden wir den folgenden l-Angreifer $A$ mit $V=\{0,1\}^l$, der zwei verschiedene Klartexte benutzt, die nur einen Block enthalten:
-    - $AF(H)$ arbeitet wie folgt: $z_0\leftarrow 0^l;v\leftarrow H(z_0);z_1\leftarrow 1^l;$ Ausgabe: $(z_0,z_1,v)$
-    - ($A$ merkt sich den Chiffretext zu $x=0^l$.)
-    - $AG(v,H,y)$ tut Folgendes: falls $v=y$, so gib $0$ aus, sonst $1$.
+    \begin{itemize*}
+        \item $AF(H)$ arbeitet wie folgt: $z_0\leftarrow 0^l;v\leftarrow H(z_0);z_1\leftarrow 1^l;$ Ausgabe: $(z_0,z_1,v)$
+        \item ($A$ merkt sich den Chiffretext zu $x=0^l$.)
+        \item $AG(v,H,y)$ tut Folgendes: falls $v=y$, so gib $0$ aus, sonst $1$.
+    \end{itemize*}
 
     Im Ablauf des Spiels $G^S_A$ wird der Rater $AG$ mit $E(0^l,k)$ oder mit $E(1^l,k)$ gestartet. Wegen $e_B(0^l,k)\not=e_B(1^l,k)$ (wegen der Dechiffrierbedingung) gilt also $Pr(G^S_A=1)=1$, d.h. $adv(A,S)=1$.
 
@@ -1562,8 +1748,10 @@
 
     \subsection{Fakten aus der Zahlentheorie und grundlegende Algorithmen}
     Unsere Zahlenbereiche:
-    - $\mathbb{N}=\{ 0 , 1 , 2 , 3 ,...\}$,
-    - $\mathbb{Z}=\{...,- 2 ,- 1 , 0 , 1 , 2 , 3 ,...\}$
+    \begin{itemize*}
+        \item $\mathbb{N}=\{ 0 , 1 , 2 , 3 ,...\}$,
+        \item $\mathbb{Z}=\{...,- 2 ,- 1 , 0 , 1 , 2 , 3 ,...\}$
+    \end{itemize*}
 
     Wir stellen uns die Zahlen immer als zu einer passenden Basis $b$ dargestellt vor: Binärdarstellung, (Oktaldarstellung,) Dezimaldarstellung, Hexadezimaldarstellung, Darstellung zur Basis 256 (eine Ziffer ist ein Byte) oder $2^{32}$ oder $2^{64}$ (eine Ziffer ist ein 32- bzw. 64-Bit-Wort, passend für die Darstellung in einem Rechner).
 
@@ -1587,44 +1775,56 @@
     Beobachtungen: Die Teilbarkeitsrelation $|$ ist reflexiv und transitiv. Es handelt sich damit um eine ,,Präordnung'' (oft auch ,,Quasiordnung'' genannt). Zahlen $x$ und $-x$ können von ihr nicht unterschieden werden: Es gilt $x|y\Leftrightarrow -x|y$ und $y|x\Leftrightarrow y|-x$, und weiter $x|-x$ und $-x|x$. Die Präordnung ist also nicht antisymmetrisch. Sie ist auch nicht total, weil manche Elemente nicht verglichen werden können: $4\not|9$ und $9\not|4$. Aus $0|x$ folgt $x=0$; für jede ganze Zahl $y$ gilt $y|0$; also ist in dieser Präordnung $0$ das eindeutig bestimmte größte Element. Für jede ganze Zahl $x$ gilt: $1|x$ und $-1|x$, also sind $1$ und $-1$ kleinste Elemente. Wenn $m\geq 1$ ist, ist $m|x$ gleichbedeutend mit $x\ mod\ m= 0$.
 
     Fakt 4.2 Teilbarkeit: Für beliebige $x,y,z\in\mathbb{Z}$ gilt:
-    1. Aus $x|y$ und $x|z$ folgt $x|uy+vz$ für alle $u,v\in\mathbb{Z}$.
-    2. Aus $x|y$ folgt $ux|uy$ für alle $u\in\mathbb{Z}$.
-    3. Aus $x|y$ und $y|z$ folgt $x|z$ (Transitivität).
-    4. Aus $x|y$ und $y\not= 0$ folgt $0<|x|\leq |y|$.
-    5. Aus $x|y$ und $y|x$ folgt $|x|=|y|$. Wenn zudem $x,y\geq 0$ gilt, folgt $x=y$.
+    \begin{enumerate*}
+        \item Aus $x|y$ und $x|z$ folgt $x|uy+vz$ für alle $u,v\in\mathbb{Z}$.
+        \item  Aus $x|y$ folgt $ux|uy$ für alle $u\in\mathbb{Z}$.
+        \item  Aus $x|y$ und $y|z$ folgt $x|z$ (Transitivität).
+        \item  Aus $x|y$ und $y\not= 0$ folgt $0<|x|\leq |y|$.
+        \item  Aus $x|y$ und $y|x$ folgt $|x|=|y|$. Wenn zudem $x,y\geq 0$ gilt, folgt $x=y$.
+    \end{enumerate*}
 
     ![Abbildung 1](Assets/Kryptographie-teilbarkeitsbeziehung.png)
     Einige Zahlen und ihre Teilbarkeitsbeziehungen. Beziehungen, die aus der Transitivität folgen, sind nicht eingetragen. Man erkennt $1$ und $-1$ als kleinste Elemente und $0$ als größtes Element der Teilbarkeitsbeziehung als Präordnung. Die Elemente in der Ebene unmittelbar über $\{1,-1\}$ sind die Primzahlen, positiv und negativ, also Zahlen $x\not=\pm 1$, die durch keine Zahl außer $\pm x$ und $\pm 1$ teilbar sind.
     Der Beweis ist eine einfache Übung.
 
     Definition 4.3 Größter gemeinsamer Teiler:
-    1. Für $x,y\in\mathbb{Z}$ heißt $t\in\mathbb{Z}$ ein gemeinsamer Teiler von $x$ und $y$, wenn $t|x$ und $t|y$ gilt. (Bemerkung: 1 ist stets gemeinsamer Teiler von $x$ und $y$.)
-    2. Für $x,y\in\mathbb{Z}$ sei $ggT(x, y)$, der größte gemeinsame Teiler von $x$ und $y$, die (eindeutig bestimmte) nichtnegative Zahl $d$ mit:
-    - $d$ ist gemeinsamer Teiler von $x$ und $y$;
-    - jeder gemeinsame Teiler von $x$ und $y$ ist Teiler von $d$.
-    3. $x,y\in\mathbb{Z}$ heißen teilerfremd, wenn $ggT(x,y)=1$ gilt, d.h. wenn sie nicht beide $0$ sind und keine Zahl $>1$ beide teilt.
+    \begin{enumerate*}
+        \item Für $x,y\in\mathbb{Z}$ heißt $t\in\mathbb{Z}$ ein gemeinsamer Teiler von $x$ und $y$, wenn $t|x$ und $t|y$ gilt. (Bemerkung: 1 ist stets gemeinsamer Teiler von $x$ und $y$.)
+        \item  Für $x,y\in\mathbb{Z}$ sei $ggT(x, y)$, der größte gemeinsame Teiler von $x$ und $y$, die (eindeutig bestimmte) nichtnegative Zahl $d$ mit:
+        \begin{itemize*}
+            \item $d$ ist gemeinsamer Teiler von $x$ und $y$;
+            \item jeder gemeinsame Teiler von $x$ und $y$ ist Teiler von $d$.
+        \end{itemize*}
+        \item  $x,y\in\mathbb{Z}$ heißen teilerfremd, wenn $ggT(x,y)=1$ gilt, d.h. wenn sie nicht beide $0$ sind und keine Zahl $>1$ beide teilt.
+    \end{enumerate*}
 
     Bei Definition 4.3.2 stellt sich die Frage nach Existenz und Eindeutigkeit von $ggT(x,y)$. Wir beweisen diese Eigenschaften im Anhang.
 
     Wir bemerken, dass $ggT(0,0) = 0$ gilt. (Sei $d = ggT(0,0)$. Weil $0|0$, folgt mit Def. 4.3.2 $0|d$ und damit $d=0$.) Wenn $x\not= 0$ oder $y\not= 0$ gilt, kann es keinen gemeinsamen Teiler geben, der größer als $max\{|x|,|y|\}$ ist, und der größte gemeinsame Teiler ist auch größtmöglich im Sinn der gewöhnlichen Ordnung auf $\mathbb{Z}$. Weil das Vorzeichen für die Teilbarkeit irrelevant ist, gilt stets $ggT(x,y) = ggT(|x|,|y|)$, und man kann sich immer auf den Fall nichtnegativer Argumente beschränken. Weiter gilt $$ggT(x,y) = ggT(x+uy,y) \quad\quad(4.1)$$, für beliebige $x,y,u\in\mathbb{Z}$. (Wenn $d$ gemeinsamer Teiler von $x$ und $y$ ist, dann teilt $d$ auch $x+uy$. Wenn $d$ gemeinsamer Teiler von $z=x+uy$ und $y$ ist, dann teilt $d$ auch $z-uy =x$. Also haben die Paare $(x,y)$ und $(x+uy,y)$ dieselbe Menge gemeinsamer Teiler, und es folgt $ggT(x,y) = ggT(x+uy,y)$.)
 
     Es gibt einen effizienten Algorithmus zur Ermittlung des größten gemeinsamen Teilers. Er beruht auf den Gleichungen
-    1. $ggT(x,y) = ggT(|x|,|y|)$ für alle $x,y\in\mathbb{Z}$,
-    2. $ggT(x,y) = ggT(y,x)$ für alle $x,y\in\mathbb{Z}$,
-    3. $ggT(a,0) =a$ für $a\geq 0$,
-    4. $ggT(a,b) = ggT(b,a\ mod\ b)$ für $a\geq b >0$.
+    \begin{enumerate*}
+        \item $ggT(x,y) = ggT(|x|,|y|)$ für alle $x,y\in\mathbb{Z}$,
+        \item  $ggT(x,y) = ggT(y,x)$ für alle $x,y\in\mathbb{Z}$,
+        \item  $ggT(a,0) =a$ für $a\geq 0$,
+        \item  $ggT(a,b) = ggT(b,a\ mod\ b)$ für $a\geq b >0$.
+    \end{enumerate*}
 
     (1. gilt, weil Teilbarkeit das Vorzeichen ignoriert. 2. ist trivial. 3. folgt daraus, dass jede Zahl Teiler von $0$ ist. 4. folgt aus 4.1 und 2., weil $a\ mod\ b=a-qb$ mit $q=ba/bc$ gilt.)
 
     Wir setzen die Beobachtung in ein iteratives Verfahren um.
 
     Algorithmus 4.1 Euklidischer Algorithmus:
-    - Input: Zwei ganze Zahlen $x$ und $y$.
-    - Methode:
-    1. $a,b:integer;a\leftarrow |x|;b\leftarrow |y|;$
-    2. $while\ b> 0\ repeat$
-    3. $(a,b)\leftarrow (b,amodb);$ // simultane Zuweisung
-    4. return $a$.
+    \begin{itemize*}
+        \item Input: Zwei ganze Zahlen $x$ und $y$.
+        \item Methode:
+        \begin{enumerate*}
+            \item $a,b:integer;a\leftarrow |x|;b\leftarrow |y|;$
+            \item  $while\ b> 0\ repeat$
+            \item  $(a,b)\leftarrow (b,amodb);$ // simultane Zuweisung
+            \item  return $a$.
+        \end{enumerate*}
+    \end{itemize*}
 
     Die eigentliche Rechnung findet in der while-Schleife statt. In dieser Schleife wird immer ein Zahlenpaar durch ein anderes ersetzt, das dieselben gemeinsamen Teiler hat wie $x$ und $y$. Wenn der Algorithmus terminiert, weil der Inhalt $b$ von $b$ Null geworden ist, kann man den Inhalt von $a$ ausgeben.
 
@@ -1647,37 +1847,49 @@
     Wenn $|x|<|y|$, hat der erste Schleifendurchlauf nur den Effekt, die beiden Zahlen zu vertauschen. Wir ignorieren diesen trivialen Schleifendurchlauf. Wir betrachten die Zahlen $a$ in $a$ und $b$ in $b$. Es gilt stets $a>b$, und $b$ nimmt in jeder Runde strikt ab, also terminiert der Algorithmus. Um einzusehen, dass er sogar sehr schnell terminiert, bemerken wir Folgendes. Betrachte den Beginn eines Schleifendurchlaufs. Der Inhalt von $a$ sei $a$, der Inhalt von $b$ sei $b$, mit $a\geq b >0$. Nach einem Schleifendurchlauf enthält $a$ den Wert $a'=b$ und $b$ den Wert $b'=a\ mod\ b$. Falls $b'=0$, endet der Algorithmus. Sonst wird noch ein Durchlauf ausgeführt, an dessen Ende $a$ den Wert $b'=a\ mod\ b$ enthält. Wir behaupten: $b'<\frac{1}{2} a$. Um dies zu beweisen, betrachten wir zwei Fälle: Wenn $b>\frac{1}{2} a$ ist, gilt $b'=a\ mod\ b=a-b<\frac{1}{2} a$. Wenn $b\leq\frac{1}{2} a$ ist, gilt $b'=a\ mod\ b < b\leq\frac{1}{2} a$. - Also wird der Wert in $a$ in jeweils zwei Durchläufen mindestens halbiert. Nach dem ersten Schleifendurchlauf enthält $a$ den Wert $min\{x,y\}$. Daraus ergibt sich Teil 1. der folgenden Aussage.
 
     Fakt 4.5
-    1. Die Schleife in Zeilen $2-3$ wird höchstens $O(log(min\{x,y\}))$-mal ausgeführt.
-    2. Die gesamte Anzahl von Ziffernoperationen für den Euklidischen Algorithmus ist $O((log\ x)(log\ y))$.
+    \begin{enumerate*}
+        \item Die Schleife in Zeilen $2-3$ wird höchstens $O(log(min\{x,y\}))$-mal ausgeführt.
+        \item  Die gesamte Anzahl von Ziffernoperationen für den Euklidischen Algorithmus ist $O((log\ x)(log\ y))$.
+    \end{enumerate*}
 
     Man beachte, dass $\lceil log(x+1)\rceil\approx log\ x$ die Anzahl der Bits in der Binärdarstellung von $x$ ist. Damit hat der Euklidische Algorithmus bis auf einen konstanten Faktor denselben Aufwand wie die Multiplikation von $x$ und $y$, wenn man die Schulmethode benutzt. (Der Beweis der Schranke in 2. benötigt eine Rechnung, die die Längen der beteiligten Zahlen genauer verfolgt.)
 
     Beispiel:
-    1. $21$ und $25$ sind teilerfremd. Es gilt $31*21 + (-26)*25 = 651-650 = 1$.
-    2. Auch $-21$ und $25$ sind teilerfremd. Aus 1. folgt sofort $(-31)*(-21) + (-26)*25 =651 -650 = 1$.
-    3. Es gilt $ggT(21,35) = 7$, und $2* 35 - 3 *21 = 7$.
+    \begin{enumerate*}
+        \item $21$ und $25$ sind teilerfremd. Es gilt $31*21 + (-26)*25 = 651-650 = 1$.
+        \item  Auch $-21$ und $25$ sind teilerfremd. Aus 1. folgt sofort $(-31)*(-21) + (-26)*25 =651 -650 = 1$.
+        \item  Es gilt $ggT(21,35) = 7$, und $2* 35 - 3 *21 = 7$.
+    \end{enumerate*}
 
     Die folgende sehr nützliche Aussage verallgemeinert diese Beobachtung:
 
     Lemma 4.6... von Bezout
-    1. Wenn $x,y\in\mathbb{Z}$ teilerfremd sind, gibt es $s,t\in\mathbb{Z}$ mit $sx+ty= 1$.
-    2. Für $x,y\in\mathbb{Z}$ gibt es $s,t\in\mathbb{Z}$ mit $sx+ty= ggT(x,y)$.
+    \begin{enumerate*}
+        \item Wenn $x,y\in\mathbb{Z}$ teilerfremd sind, gibt es $s,t\in\mathbb{Z}$ mit $sx+ty= 1$.
+        \item  Für $x,y\in\mathbb{Z}$ gibt es $s,t\in\mathbb{Z}$ mit $sx+ty= ggT(x,y)$.
+    \end{enumerate*}
 
     Wir geben einen Algorithmus an, der zu $x$ und $y$ die Werte $s$ und $t$ (sehr effizient) berechnet. Damit ist die Frage der Existenz natürlich gleich mit erledigt. Vorab bemerken wir noch, dass es eine Art Umkehrung von 1. gibt: Wenn $sx+ty= 1$ für ganze Zahlen $s$ und $t$ gilt, dann sind $x$ und $y$ teilerfremd. (Beweis: Alle gemeinsamen Teiler von $x$ und $y$ teilen auch $1$, sind also $1$ oder $-1$. Daraus folgt $ggT(x,y) = 1$.)
 
     Für den Algorithmus können wir o.B.d.A. annehmen, dass $x,y\geq 0$ gilt. Die Umrechnung für negative Inputs ist offensichtlich.
 
     Algorithmus 4.2 Erweiterter Euklidischer Algorithmus:
-    - Eingabe: Natürliche Zahlen $x$ und $y$.
-    - Methode:
-    1. $a,b,sa,ta,sb,tb,q:integer;$
-    2. $a\leftarrow x; b\leftarrow y;$
-    3. $sa\leftarrow 1; ta\leftarrow 0; sb\leftarrow 0; tb\leftarrow 1;$
-    4. while $b> 0$ repeat
-    1. $q\leftarrow a\ div\ b$;
-    2. $(a,b)\leftarrow (b,a-q*b)$;
-    3. $(sa,ta,sb,tb)\leftarrow (sb,tb,sa-q*sb,ta-q*tb)$;
-    5. return$(a,sa,ta)$;
+    \begin{itemize*}
+        \item Eingabe: Natürliche Zahlen $x$ und $y$.
+        \item Methode:
+        \begin{enumerate*}
+            \item $a,b,sa,ta,sb,tb,q:integer;$
+            \item  $a\leftarrow x; b\leftarrow y;$
+            \item  $sa\leftarrow 1; ta\leftarrow 0; sb\leftarrow 0; tb\leftarrow 1;$
+            \item  while $b> 0$ repeat
+            \begin{enumerate*}
+                \item $q\leftarrow a\ div\ b$;
+                \item  $(a,b)\leftarrow (b,a-q*b)$;
+                \item  $(sa,ta,sb,tb)\leftarrow (sb,tb,sa-q*sb,ta-q*tb)$;
+            \end{enumerate*}
+            \item  return$(a,sa,ta)$;
+        \end{enumerate*}
+    \end{itemize*}
 
     Genau wie im ursprünglichen Euklidischen Algorithmus findet die eigentliche Arbeit in der while-Schleife (Zeilen 4 - 7 ) statt.
 
@@ -1705,9 +1917,11 @@
     gilt. - Allgemein gilt:
 
     Fakt 4.7: Wenn Algorithmus 4.2 auf Eingabe $(x,y)$ mit $x,y\geq 0$ gestartet wird, dann gilt:
-    1. Für die Ausgabe $(d,s,t)$ gilt $d= ggT(x,y) =sx+ty$.
-    2. Die Anzahl der Schleifendurchläufe ist dieselbe wie beim gewöhnlichen Euklidischen Algorithmus.
-    3. Die Anzahl von Ziffernoperationen für Algorithmus 4.2 ist $O((log\ x)(log\ y))$.
+    \begin{enumerate*}
+        \item Für die Ausgabe $(d,s,t)$ gilt $d= ggT(x,y) =sx+ty$.
+        \item  Die Anzahl der Schleifendurchläufe ist dieselbe wie beim gewöhnlichen Euklidischen Algorithmus.
+        \item  Die Anzahl von Ziffernoperationen für Algorithmus 4.2 ist $O((log\ x)(log\ y))$.
+    \end{enumerate*}
 
     Wir notieren noch eine wichtige Folgerung aus dem Lemma von Bezout. Die Aussage ist aus der Schule bekannt: Wenn eine Zahl z.B. durch $3$ und durch $5$ teilbar ist, dann ist sie auch durch 15 teilbar. Dort benutzt man die Primzahlzerlegung zur Begründung. Diese ist aber gar nicht nötig.
 
@@ -1721,8 +1935,10 @@
     Man sagt: ,,$x$ ist kongruent zu $y$ modulo $m$.'' In der Mathematik sieht man auch oft die kompaktere Notation $x\equiv y(m)$ oder $x\equiv_m y$. Es besteht eine enge Beziehung zwischen dieser Relation und der Division mit Rest.
 
     Fakt 4.10:
-    1. $x\equiv y(mod\ m)$ gilt genau dann wenn $x\ mod\ m=y\ mod\ m$ gilt.
-    2. Die zweistellige Relation $*\equiv *(mod\ m)$ ist eine Äquivalenzrelation, sie ist also reflexiv, transitiv und symmetrisch.
+    \begin{enumerate*}
+        \item $x\equiv y(mod\ m)$ gilt genau dann wenn $x\ mod\ m=y\ mod\ m$ gilt.
+        \item  Die zweistellige Relation $*\equiv *(mod\ m)$ ist eine Äquivalenzrelation, sie ist also reflexiv, transitiv und symmetrisch.
+    \end{enumerate*}
 
     Beispiel für 1.: $29\ mod\ 12 = 53\ mod\ 12 = 5$ und $53-29 = 24$ ist durch $12$ teilbar.
 
@@ -1731,13 +1947,17 @@
     Die Kongruenzrelation $* \equiv  *(mod\ m)$ führt (wie jede Äquivalenzrelation) zu einer Zerlegung der Grundmenge $\mathbb{Z}$ in Äquivalenzklassen (die hier ,,Restklassen'' heißen): $[x]_m= [x] =\{y\in\mathbb{Z}|x\equiv y(mod\ m)\}=\{y\in\mathbb{Z}|x\ mod\ m=y\ mod\ m\}$. Wir definieren: $m\mathbb{Z}:=\{...,-3m,-2m,-m,0,m,2m,3m,...\}$ und $x+A:=\{x+y|y\in A\}$, für $A\supseteq Z$.
 
     Beispiel: Für $m=3$ gibt es die drei Restklassen
-    - $[0] = [0]_3 =\{...,-6,-3,0,3,6,...\}= 0 + 3\mathbb{Z}$,
-    - $[1] = [1]_3 =\{...,-5,-2,1,4,7,...\}= 1 + 3\mathbb{Z}$,
-    - $[2] = [2]_3 =\{...,-4,-1,2,5,8,...\}= 2 + 3\mathbb{Z}$.
+    \begin{itemize*}
+        \item $[0] = [0]_3 =\{...,-6,-3,0,3,6,...\}= 0 + 3\mathbb{Z}$,
+        \item $[1] = [1]_3 =\{...,-5,-2,1,4,7,...\}= 1 + 3\mathbb{Z}$,
+        \item $[2] = [2]_3 =\{...,-4,-1,2,5,8,...\}= 2 + 3\mathbb{Z}$.
+    \end{itemize*}
 
     Mit den Restklassen kann man dann wieder rechnen: Addition und Multiplikation lassen sich wie folgt definieren.
-    - $[x]_m+ [y]_m := [x+y]_m$,
-    - $[x]_m*[y]_m := [x*y]_m$
+    \begin{itemize*}
+        \item $[x]_m+ [y]_m := [x+y]_m$,
+        \item $[x]_m*[y]_m := [x*y]_m$
+    \end{itemize*}
 
     Beispielsweise gelten für $m=3$ die Gleichheiten $[4] + [5] = [9] = [0]$ und $[4]*[5] =[20] = [2]$.
 
@@ -1768,14 +1988,16 @@
     Man beachte noch, dass $\lfloor y\backslash 2\rfloor=\begin{cases} y/2\quad\quad\text{ für gerade y},\\ (y-1)/2\quad\quad\text{ für ungerade y}\end{cases}$. Diese Formeln führen unmittelbar zu folgender rekursiver Prozedur.
 
     Algorithmus 4.3 Schnelle modulare Exponentiation, rekursiv
-    - function $modexp(x,y,m)$
-    - Eingabe: Ganze Zahlen $x,y\geq 0$, $m\geq 1$, mit $0\leq x < m$.
-    - Methode:
-    - if $y= 0$ then return $1$;
-    - if $y= 1$ then return $x$;
-    - $z\leftarrow modexp((x*x) mod\ m,\lfloor y/2\rfloor,m);$ // rekursiver Aufruf
-    - if $y$ ist ungerade then $z\leftarrow (z*x) mod\ m$
-    - return $z$.
+    \begin{itemize*}
+        \item function $modexp(x,y,m)$
+        \item Eingabe: Ganze Zahlen $x,y\geq 0$, $m\geq 1$, mit $0\leq x < m$.
+        \item Methode:
+        \item if $y= 0$ then return $1$;
+        \item if $y= 1$ then return $x$;
+        \item $z\leftarrow modexp((x*x) mod\ m,\lfloor y/2\rfloor,m);$ // rekursiver Aufruf
+        \item if $y$ ist ungerade then $z\leftarrow (z*x) mod\ m$
+        \item return $z$.
+    \end{itemize*}
 
     Man erkennt sofort, dass in jeder Rekursionsebene die Bitanzahl des Exponenten $y$ um 1 sinkt, dass also die Anzahl der Rekursionsebenen etwa $log\ y$ beträgt. In jeder Rekursionsstufe ist eine oder sind zwei Multiplikationen modulo m auszuführen, was $O((log\ m)^2)$ Ziffernoperationen erfordert (Schulmethode).
     Beispiel: Wir berechnen $13^{43} mod\ 19$.
@@ -1793,12 +2015,14 @@
     Lemma 4.14: Sei $x<m$. Die Berechnung von $x^y\ mod\ m$ benötigt $O(log\ y)$ Multiplikationen und Divisionen modulo m von Zahlen aus $\{0,...,m^2-1\}$, und damit $O((log\ y)(log\ m)^2)$ Ziffernoperationen.
 
     Bemerkung: Man kann denselben Algorithmus in einem beliebigen Monoid $(M,\circ,e)$ ($M\not=\varnothing$ ist eine Menge,$\circ:M\times M \rightarrow M$ ist assoziative Operation mit neutralem Element $e\in M$) benutzen. Monoide bilden zum Beispiel:
-    - $(\mathbb{Z}_m ,*_m,1)$, wo $*_m$ die Multiplikation modulo m ist;
-    - $(\mathbb{N},+,0)$: die natürlichen Zahlen mit der Addition (neutral: $0$);
-    - $(\mathbb{N},*,1)$: die natürlichen Zahlen mit der Multiplikation (neutral: $1$);
-    - quadratische Matrizen über einem Ring mit $1$, mit Matrixmultiplikation (neutral: Einheitsmatrix);
-    - die Menge $\sum^*$ aller Wörter über einem Alphabet $\sum$, mit der Konkatenation (neutral: das leere Wort);
-    - jede Gruppe $(G,\circ,e)$ (G ist die Grundmenge, $\circ$ die Operation und $e$ das neutrale Element.)
+    \begin{itemize*}
+        \item $(\mathbb{Z}_m ,*_m,1)$, wo $*_m$ die Multiplikation modulo m ist;
+        \item $(\mathbb{N},+,0)$: die natürlichen Zahlen mit der Addition (neutral: $0$);
+        \item $(\mathbb{N},*,1)$: die natürlichen Zahlen mit der Multiplikation (neutral: $1$);
+        \item quadratische Matrizen über einem Ring mit $1$, mit Matrixmultiplikation (neutral: Einheitsmatrix);
+        \item die Menge $\sum^*$ aller Wörter über einem Alphabet $\sum$, mit der Konkatenation (neutral: das leere Wort);
+        \item jede Gruppe $(G,\circ,e)$ (G ist die Grundmenge, $\circ$ die Operation und $e$ das neutrale Element.)
+    \end{itemize*}
 
     Wenn man nur eine assoziative Operation und kein neutrales Element hat, funktioniert der Algorithmus für Exponenten $y\geq 1$.
 
@@ -1817,8 +2041,10 @@
     Fakt 4.16: Für jedes $m\geq 2$ und jedes $x\in\mathbb{Z}$ gilt: Es gibt ein $y$ mit $xy\ mod\ m=1$ genau dann wenn $ggT(x,m)=1$.
 
     Beweis:
-    - ,,$\Rightarrow$'': Es sei $xy\ mod\ m= 1$. Das heißt: Es gibt ein $q\in\mathbb{Z}$ mit $xy-qm=1$. Dann teilt jeder gemeinsame Faktor von $x$ und $m$ auch $1$, also sind $x$ und $m$ teilerfremd.
-    - ,,$\Leftarrow$'': $x$ und $m$ seien teilerfremd. Nach dem Lemma von Bezout (Lemma 4.6.1) gibt es $s,t\in\mathbb{Z}$ mit $sx+tm=1$. Setze $y:=s\ mod\ m$. Dann gilt: $(x*y) mod\ m= (x*(s\ mod\ m)) mod\ m=sx\ mod\ m= 1$.
+    \begin{itemize*}
+        \item ,,$\Rightarrow$'': Es sei $xy\ mod\ m= 1$. Das heißt: Es gibt ein $q\in\mathbb{Z}$ mit $xy-qm=1$. Dann teilt jeder gemeinsame Faktor von $x$ und $m$ auch $1$, also sind $x$ und $m$ teilerfremd.
+        \item ,,$\Leftarrow$'': $x$ und $m$ seien teilerfremd. Nach dem Lemma von Bezout (Lemma 4.6.1) gibt es $s,t\in\mathbb{Z}$ mit $sx+tm=1$. Setze $y:=s\ mod\ m$. Dann gilt: $(x*y) mod\ m= (x*(s\ mod\ m)) mod\ m=sx\ mod\ m= 1$.
+    \end{itemize*}
 
     Beispiel: $x=22$ und $m=15$ sind teilerfremd. Mit dem erweiterten Euklidischen Algorithmus findet man $s=-2$ und $t=3$, so dass $sx+tm=(-2)*22 + 3*15 =-44 + 45 = 1$ gilt. Setze $y=(-2)\ mod\ 15 = 13$. Man kontrolliert: $22*13 =286=19 *15 + 1\equiv 1\ mod\ 15$.
 
@@ -1843,15 +2069,19 @@
     Eine ganze Zahl $p\geq 1$ heißt Primzahl, wenn $p$ genau zwei positive Teiler hat, nämlich $1$ und $p$. Die Folge der Primzahlen beginnt mit $2, 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 ,....$
 
     Fakt 4.19 (*): Für jedes $m\geq 2$ sind folgende Aussagen äquivalent:
-    1. $m$ ist eine Primzahl.
-    2. $\mathbb{Z}^*_m=\{ 1 ,...,m-1\}$.
-    3. $\mathbb{Z}_m$ ist ein Körper.
+    \begin{enumerate*}
+        \item $m$ ist eine Primzahl.
+        \item  $\mathbb{Z}^*_m=\{ 1 ,...,m-1\}$.
+        \item  $\mathbb{Z}_m$ ist ein Körper.
+    \end{enumerate*}
 
     Der Beweis erfolgt durch einen Ringschluss.
-    - ,,1. $\Rightarrow$ 2.'': Sei $m$ Primzahl. Dann kann für kein Element $x\in\{1 ,...,m-1\}$ die Beziehung $ggT(x,m)>1$ gelten, weil sonst die Zahl $ggT(x,m)$ ein Teiler von $m$ strikt zwischen $1$ und $m$ wäre.
-    - ,,2. $\Rightarrow$ 3.'': Wenn $\mathbb{Z}^*_m=\{1 ,...,m-1\}$ gilt, hat nach Fakt 4.16 jedes Element von $\mathbb{Z}_m -\{0\}$ ein multiplikatives Inverses. Das ist genau die Eigenschaft, die dem Ring $\mathbb{Z}_m$ zum Körper fehlt.
-    - ,,3. $\Rightarrow$ 1.'': Das beweisen wir durch Kontraposition. Sei also 1. falsch, d.h. sei $m$ keine Primzahl. Dann gibt es ein $x\in\{2,...,m-1\}$, das Teiler von $m$ ist. Insbesondere ist $ggT(x,m) =x >1$.
-    - Mit Fakt 4.16 folgt, dass $x$ kein multiplikatives Inverses modulo $m$ hat, also ist $\mathbb{Z}_m$ kein Körper, d.h. 3. ist falsch.
+    \begin{itemize*}
+        \item ,,1. $\Rightarrow$ 2.'': Sei $m$ Primzahl. Dann kann für kein Element $x\in\{1 ,...,m-1\}$ die Beziehung $ggT(x,m)>1$ gelten, weil sonst die Zahl $ggT(x,m)$ ein Teiler von $m$ strikt zwischen $1$ und $m$ wäre.
+        \item ,,2. $\Rightarrow$ 3.'': Wenn $\mathbb{Z}^*_m=\{1 ,...,m-1\}$ gilt, hat nach Fakt 4.16 jedes Element von $\mathbb{Z}_m -\{0\}$ ein multiplikatives Inverses. Das ist genau die Eigenschaft, die dem Ring $\mathbb{Z}_m$ zum Körper fehlt.
+        \item ,,3. $\Rightarrow$ 1.'': Das beweisen wir durch Kontraposition. Sei also 1. falsch, d.h. sei $m$ keine Primzahl. Dann gibt es ein $x\in\{2,...,m-1\}$, das Teiler von $m$ ist. Insbesondere ist $ggT(x,m) =x >1$.
+        \item Mit Fakt 4.16 folgt, dass $x$ kein multiplikatives Inverses modulo $m$ hat, also ist $\mathbb{Z}_m$ kein Körper, d.h. 3. ist falsch.
+    \end{itemize*}
 
     Beispiel: $m=13$. Wir geben für jedes $x\in\mathbb{Z}^*_{13}$ das Inverse $y$ sowie das Produkt $x*y$ an (das natürlich bei der Division durch $13$ Rest $1$ lassen muss).
     | x     | 1   | 2   | 3   | 4   | 5   | 6   | 7   | 8   | 9   | 10  | 11  | 12  |
@@ -1883,9 +2113,11 @@
     Der Chinesische Restsatz sagt im wesentlichen, dass eine solche strukturelle Entsprechung zwischen den Resten modulo $mn$ und Paaren von Resten modulo $m$ bzw. $n$ immer gilt, wenn $m$ und $n$ teilerfremd sind.
 
     Fakt 4.21 Chinesischer Restsatz (*): $m$ und $n$ seien teilerfremd. Dann ist die Abbildung $\Phi:\mathbb{Z}_{mn} \owns x \rightarrow (x\ mod\ m, x\ mod\ n)\in\mathbb{Z}_m\times\mathbb{Z}_n$ bijektiv. Weiterhin: Wenn $\Phi(x)=(x_1,x_2)$ und $\Phi(y)=(y_1,y_2)$, dann gilt:
-    1. $\Phi(x+_{mn} y) = (x_1 +_m y_1 , x_2 +_n y_2)$
-    2. $\Phi(x*_{mn} y) = (x_1 *_m y_1 , x_2 *_n y_2)$
-    3. $\Phi(1) = (1,1)$
+    \begin{enumerate*}
+        \item $\Phi(x+_{mn} y) = (x_1 +_m y_1 , x_2 +_n y_2)$
+        \item $\Phi(x*_{mn} y) = (x_1 *_m y_1 , x_2 *_n y_2)$
+        \item  $\Phi(1) = (1,1)$
+    \end{enumerate*}
 
     (Dabei bezeichnen $+_j$ und $*_j$ die Addition und die Multiplikation modulo $j$.)
 
@@ -1927,8 +2159,10 @@
     Jede positive ganze Zahl $x$ ist durch 1 und durch $x$ teilbar.
 
     Definition 4.26:
-    1. Eine Zahl $p\geq 1$ heißt Primzahl, wenn $p$ genau zwei positive Teiler hat. Diese Teiler sind dann $1$ und $p$. (Die Zahl 1 hat nur einen positiven Teiler, nämlich 1. Also ist 1 keine Primzahl.)
-    2. Eine Zahl $x\geq 1$ heißt zusammengesetzt, wenn sie einen Teiler $y$ mit $1<y < x$ besitzt. (Die Zahl 1 besitzt keinen Teiler $y$ mit $1<y<1$. Also ist 1 nicht zusammengesetzt.)
+    \begin{enumerate*}
+        \item Eine Zahl $p\geq 1$ heißt Primzahl, wenn $p$ genau zwei positive Teiler hat. Diese Teiler sind dann $1$ und $p$. (Die Zahl 1 hat nur einen positiven Teiler, nämlich 1. Also ist 1 keine Primzahl.)
+        \item  Eine Zahl $x\geq 1$ heißt zusammengesetzt, wenn sie einen Teiler $y$ mit $1<y < x$ besitzt. (Die Zahl 1 besitzt keinen Teiler $y$ mit $1<y<1$. Also ist 1 nicht zusammengesetzt.)
+    \end{enumerate*}
 
     Bemerkung: Ein Blick auf die Teilbarkeitsbeziehung zeigt, welche besondere Rolle $1,-1, 0$ und die Primzahlen spielen. Die Zahlen $1$ und $-1$ bilden das Minimum in der Teilbarkeitsrelation, die Primzahlen sitzen unmittelbar darüber, und die zusammengesetzten Elemente liegen strikt über den Primzahlen. Die $0$, als Maximum in der Ordnung, sitzt strikt über allen zusammengesetzten Zahlen.
 
@@ -1987,8 +2221,10 @@
         In diesem Abschnitt lernen wir einen randomisierten Algorithmus kennen, der es erlaubt, zu einer gegebenen Zahl $N$ zu entscheiden, ob $N$ eine Primzahl ist oder nicht.
 
         Ein idealer Primzahltest sieht so aus:
-        - Eingabe: Eine natürliche Zahl $N\geq 3$.
-        - Ausgabe: $0$, falls $N$ eine Primzahl ist; $1$, falls $N$ zusammengesetzt ist.
+        \begin{itemize*}
+            \item Eingabe: Eine natürliche Zahl $N\geq 3$.
+            \item Ausgabe: $0$, falls $N$ eine Primzahl ist; $1$, falls $N$ zusammengesetzt ist.
+        \end{itemize*}
 
         Wozu braucht man Primzahltests? Zunächst ist die Frage ,,Ist $N$ eine Primzahl?'' eine grundlegende mathematisch interessante Fragestellung. Spätestens mit dem Siegeszug des RSA-Kryptosystems hat sich die Situation jedoch dahin entwickelt, dass man Algorithmen benötigt, die immer wieder neue vielziffrige Primzahlen (etwa mit 1000 oder 1500 Bits bzw. 301 oder 452 Dezimalziffern) bereitstellen können. Den Kern dieser Primzahlerzeugungs-Verfahren bildet ein Verfahren, das eine gegebene Zahl $N$ darauf testet, ob sie prim ist. Ein naiver Primzahltest (,,versuchsweise Division''), der dem brute-force-Paradigma folgt, findet durch direkte Division der Zahl $N$ durch $2,3,4,...,\lfloor\sqrt{N}\rfloor$ heraus, ob $N$ einen nichttrivialen Teiler hat. Man kann dieses Verfahren durch einige Tricks beschleunigen, aber die Rechenzeit wächst dennoch mit $O(\sqrt{N})$. Dies macht es für Zahlen mit mehr als $40$ Dezimalstellen praktisch undurchführbar, von Zahlen mit mehr als $100$ Dezimalstellen ganz zu schweigen. (Achtung: Damit wird nichts über den Zeitaufwand bei anderen Faktorisierungsalgorithmen gesagt. Es gibt andere, sehr fortgeschrittene Faktorisierungsalgorithmen, die bei entsprechendem Zeitaufwand und mit sehr leistungsstarken Rechnern auch noch mit $200$-stelligen Zahlen zurechtkommen. Für Information zu früheren und aktuelleren Faktorisierungserfolgen siehe z.B. \href{http://en.wikipedia.org/wiki/RSA_numbers}{Wikipedia RSA Numbers}.
 
@@ -2030,10 +2266,14 @@
         | F-Zeugen in $\mathbb{Z}^*_{91}$ :                | 2, 5, 6, 8, 11, 15, 18, 19, 20, 24, 31, 32, 33, 34, 37, 41, 44, 45, 46, 47, 50, 54, 57, 58, 59, 60, 67, 71, 72, 73, 76, 80, 83, 85, 86, 89 |
 
         Algorithmus 4.4: Fermat-Test
-        - Eingabe: Ungerade Zahl $N\geq 3$
-        - Methode:
-        1. Wähle $a$ zufällig aus $\{1,...,N-1\}$
-        2. if $a^{N-1}\ mod\ N\not= 1$ then return $1$ else return $0$
+        \begin{itemize*}
+            \item Eingabe: Ungerade Zahl $N\geq 3$
+            \item Methode:
+            \begin{enumerate*}
+                \item Wähle $a$ zufällig aus $\{1,...,N-1\}$
+                \item if $a^{N-1}\ mod\ N\not= 1$ then return $1$ else return $0$
+            \end{enumerate*}
+        \end{itemize*}
 
         Die Laufzeitanalyse liegt auf der Hand: Der teuerste Teil ist die Berechnung der Potenz $a^{N-1}\ mod\ N$ durch schnelle Exponentiation, die nach den Ergebnissen von Lemma 4.14 $O(log\ N)$ arithmetische Operationen und $O((log\ N)^3)$ Ziffernoperationen benötigt. Weiter ist es klar, dass der Algorithmus einen F-Zeugen gefunden hat, wenn er ,,1'' ausgibt, dass in diesem Fall also $N$ zusammengesetzt sein muss. Umgekehrt ausgedrückt: Wenn $N$ eine Primzahl ist, gibt der Fermat-Test garantiert ,,0'' aus. Für $N=91$ wird das falsche Ergebnis $0$ ausgegeben, wenn als $a$ einer der 36 F-Lügner gewählt wird. Die Wahrscheinlichkeit hierfür ist $\frac{36}{90} =\frac{2}{5} = 0,4$.
 
@@ -2046,12 +2286,18 @@
         Eine Fehlerwahrscheinlichkeit in der Nähe von $\frac{1}{2}$ ist natürlich viel zu groß. Wir verringern die Fehlerschranke durch wiederholte Ausführung des Fermat-Tests.
 
         Algorithmus 4.5: Iterierter Fermat-Test
-        - Eingabe: Ungerade Zahl $N\geq 3$, eine Zahl $l\geq 1$
-        - Methode:
-        1. repeat $l$ times
-        1. $a\leftarrow$ ein zufälliges Element von $\{1 ,...,N-1\}$
-        2. if $a^{N-1}\ mod\ N\not= 1$ then return $1$
-        2. return 0
+        \begin{itemize*}
+            \item Eingabe: Ungerade Zahl $N\geq 3$, eine Zahl $l\geq 1$
+            \item Methode:
+            \begin{enumerate*}
+                \item repeat $l$ times
+                \begin{itemize*}
+                    \item $a\leftarrow$ ein zufälliges Element von $\{1 ,...,N-1\}$
+                    \item if $a^{N-1}\ mod\ N\not= 1$ then return $1$
+                \end{itemize*}
+                \item return 0
+            \end{enumerate*}
+        \end{itemize*}
 
         Wenn die Ausgabe $1$ ist, hat der Algorithmus einen F-Zeugen für $N$ gefunden, also ist $N$ zusammengesetzt. D.h.: Wenn $N$ eine Primzahl ist, ist die Ausgabe $0$. Andererseits: Wenn $N$ zusammengesetzt ist, und es mindestens einen F-Zeugen $b\in\mathbb{Z}^*_N$ gibt, dann ist nach Satz 4.36 die Wahrscheinlichkeit für die falsche Ausgabe ,,0'' höchstens $(\frac{1}{2})^l= 2^{-l}$. Indem wir $l$ genügend groß wählen, können wir die Fehlerwahrscheinlichkeit so klein wie gewünscht einstellen.
 
@@ -2118,14 +2364,16 @@
         Wie kann die Folge $b_0,...,b_k$ überhaupt aussehen? Wenn $b_i=1$ oder $b_i=N-1$ gilt, dann sind $b_{i+1},...,b_k$ alle gleich $1$. Daher beginnt die Folge $b_0,...,b_k$ mit einer (eventuell leeren) Folge von Elementen $\not\in\{1,N-1\}$ und endet mit einer (eventuell leeren) Folge von Einsen. Diese beiden Teile können von einem Eintrag $N-1$ getrennt sein oder nicht.
 
         Die möglichen Muster sind in Tabelle 5 angegeben. Dabei steht ,,*'' für ein Element $\not\in\{1,N-1\}$. Wir unterscheiden vier Fälle (mit Unterfällen).
-        - Fall 1: $b_0 = 1$. - Dann ist $b_1 =...=b_k= 1$.
-        - Fall 2: Die Folge endet mit $b_k= 1$, und vor der ersten $1$ in der Folge steht eine $N-1$.
-        - Fall 2a: $b_0 =N-1$
-        - Fall 2b: $b_i=N-1$ für ein $i\in\{1,...,k-1\}$
-        - Fall 3: Die Folge endet mit $b_k\not= 1$. - Dann ist $N$ zusammengesetzt, weil $a$ ein F-Zeuge für $N$ ist. In $b_0,...,b_{k-1}$ können $1$ und $N-1$ nicht vorkommen.
-        - Fall 4: Die Folge endet mit $b_k=1$, aber $b_0\not= 1$ und in $b_0,...,b_{k-1}$ kommt $N-1$ nicht vor. - Dann gibt es ein $i\in\{1,...,k\}$ mit $b_{i-1}\not\in\{1,N-1\}$ und $b_i=1$. Das heißt, dass $b_{i-1}$ eine nichttriviale Quadratwurzel der $1$ ist. Also ist $N$ zusammengesetzt.
-        - Fall 4a: $i<k$
-        - Fall 4b: $i=k$
+        \begin{itemize*}
+            \item Fall 1: $b_0 = 1$. - Dann ist $b_1 =...=b_k= 1$.
+            \item Fall 2: Die Folge endet mit $b_k= 1$, und vor der ersten $1$ in der Folge steht eine $N-1$.
+            \item Fall 2a: $b_0 =N-1$
+            \item Fall 2b: $b_i=N-1$ für ein $i\in\{1,...,k-1\}$
+            \item Fall 3: Die Folge endet mit $b_k\not= 1$. - Dann ist $N$ zusammengesetzt, weil $a$ ein F-Zeuge für $N$ ist. In $b_0,...,b_{k-1}$ können $1$ und $N-1$ nicht vorkommen.
+            \item Fall 4: Die Folge endet mit $b_k=1$, aber $b_0\not= 1$ und in $b_0,...,b_{k-1}$ kommt $N-1$ nicht vor. - Dann gibt es ein $i\in\{1,...,k\}$ mit $b_{i-1}\not\in\{1,N-1\}$ und $b_i=1$. Das heißt, dass $b_{i-1}$ eine nichttriviale Quadratwurzel der $1$ ist. Also ist $N$ zusammengesetzt.
+            \item Fall 4a: $i<k$
+            \item Fall 4b: $i=k$
+        \end{itemize*}
 
         In Tabelle 4 findet man konkrete Beispiele für das Eintreten aller Fälle für die zusammengesetzte Zahl $N=325$.
 
@@ -2150,30 +2398,38 @@
         Aus der obigen Diskussion der möglichen Fälle folgt sofort:
 
         Lemma 4.41: Sei $N\geq 3$ ungerade.
-        1. $a$ ist F-Zeuge für $N \Rightarrow a$ ist MR-Zeuge für $N$ (Fall 3).
-        2. Wenn $N$ eine Primzahl ist, dann gilt (4.4) für alle $a<N$.
+        \begin{enumerate*}
+            \item $a$ ist F-Zeuge für $N \Rightarrow a$ ist MR-Zeuge für $N$ (Fall 3).
+            \item Wenn $N$ eine Primzahl ist, dann gilt (4.4) für alle $a<N$.
+        \end{enumerate*}
 
         Im Jahr 1976 stellte Gary M. Miller einen deterministischen Algorithmus vor, der auf der Idee beruhte, die Folge $b_0,...,b_k$ zu betrachten. Er testete die kleinsten $O((log\ N)^2)$ Elemente $a\in\{2,3,...,N-1\}$ auf die Eigenschaft, MR-Zeugen zu sein. Der Algorithmus hat polynomielle Laufzeit und liefert das korrekte Ergebnis, wenn die ,,erweiterte Riemannsche Vermutung (ERH)'' stimmt, eine berühmte Vermutung aus der Zahlentheorie, die bis heute unbewiesen ist. Um 1980 schlugen Michael Rabin und (unabhängig) Louis Monier vor, Millers Algorithmus so zu modifizieren, dass die zu testende Zahl $a$ zufällig gewählt wird, und bewiesen, dass dieser Algorithmus konstante Fehlerwahrscheinlichkeit hat. Der Algorithmus heißt heute allgemein der Miller-Rabin-Test.
 
         Der Test selbst ist leicht und sehr effizient durchführbar: Man wählt $a$ zufällig und berechnet zunächst $b_0=a^u\ mod\ N$ und testet, ob $b_0\in\{1,N-1\}$ ist. Falls ja, trifft Fall 1 bzw. Fall 2a zu, die Ausgabe ist $0$. Falls nein, berechnet man durch iteriertes Quadrieren nacheinander die Werte $b_1,b_2,...,$ bis
-        - entweder der Wert $N-1$ auftaucht (Fall 2b, Ausgabe 0)
-        - oder der Wert $1$ auftaucht (Fall 4a, $a$ ist MR-Zeuge, Ausgabe 1)
-        - oder $b_1,...,b_{k-1}$ berechnet worden sind, ohne dass Werte $N-1$ oder $1$ vorgekommen sind (Fall 3, $a$ ist F-Zeuge, oder 4b, $a$ ist MR-Zeuge, Ausgabe 1).
+        \begin{itemize*}
+            \item entweder der Wert $N-1$ auftaucht (Fall 2b, Ausgabe 0)
+            \item oder der Wert $1$ auftaucht (Fall 4a, $a$ ist MR-Zeuge, Ausgabe 1)
+            \item oder $b_1,...,b_{k-1}$ berechnet worden sind, ohne dass Werte $N-1$ oder $1$ vorgekommen sind (Fall 3, $a$ ist F-Zeuge, oder 4b, $a$ ist MR-Zeuge, Ausgabe 1).
+        \end{itemize*}
 
         In Pseudocode sieht das Verfahren dann folgendermaßen aus. Im Unterschied zur bisherigen Beschreibung wird nur eine Variable $b$ benutzt, die nacheinander die Werte $b_0,b_1,...$ aufnimmt.
 
         Algorithmus 4.6 Der Miller-Rabin-Primzahltest
-        - Eingabe: Eine ungerade Zahl $N\geq 3$
-        - Methode:
-        - Bestimme $???$ ungerade und $k\geq 1$ mit $N-1 =u*2^k$
-        - wähle zufällig ein $a$ aus $\{1 ,...,N-1\}$
-        - $b \leftarrow a^u\ mod\ N$ // mit schnellem Potenzieren
-        - if $b\in\{1,N-1\}$ then return $0$; // Fall 1 oder 2a
-        - for $j$ from $1$ to $k-1$ do //,,wiederhole (k-1)-mal''
-        - $b\leftarrow b^2\ mod\ N$
-        - if $b=N-1$ then return $0$ // Fall 2b
-        - if $b=1$ then return $1$ // Fall 4a
-        - return $1$ // Fall 3 oder 4b
+        \begin{itemize*}
+            \item Eingabe: Eine ungerade Zahl $N\geq 3$
+            \item Methode:
+            \begin{itemize*}
+                \item Bestimme $???$ ungerade und $k\geq 1$ mit $N-1 =u*2^k$
+                \item wähle zufällig ein $a$ aus $\{1 ,...,N-1\}$
+                \item $b \leftarrow a^u\ mod\ N$ // mit schnellem Potenzieren
+                \item if $b\in\{1,N-1\}$ then return $0$; // Fall 1 oder 2a
+                \item for $j$ from $1$ to $k-1$ do //,,wiederhole (k-1)-mal''
+                \item $b\leftarrow b^2\ mod\ N$
+                \item if $b=N-1$ then return $0$ // Fall 2b
+                \item if $b=1$ then return $1$ // Fall 4a
+                \item return $1$ // Fall 3 oder 4b
+            \end{itemize*}
+        \end{itemize*}
 
         Wie steht es mit der Rechenzeit des Algorithmus? Man benötigt höchstens $log\ N$ Divisionen durch 2, um $u$ und $k$ zu finden (Zeile 1). Wenn man benutzt, dass $N$ normalerweise in Binärdarstellung gegeben sein wird, ist die Sache noch einfacher: $k$ ist die Zahl der Nullen, mit der die Binärdarstellung von $N-1$ aufhört, $u$ ist die Zahl, die aus der Binärdarstellung von $N$ durch Weglassen dieser letzten Nullen entsteht. Die Berechnung von $a^u\ mod\ N$ mit schneller Exponentiation in Zeile 3 benötigt $O(log\ N)$ arithmetische Operationen und $O((log\ N)^3)$ Ziffernoperationen (mit der Schulmethode für Multiplikation und Division). Die Schleife in Zeilen 5-8 wird höchstens $(k-1)$-mal durchlaufen; offenbar ist $k\leq log\ N$. In jedem Durchlauf ist die Multiplikation modulo N die teuerste Operation. Insgesamt benutzt der Algorithmus $O(log\ N)$ arithmetische Operationen auf Zahlen, die kleiner als $N^2$ sind, und $O((log\ N)^3)$ Ziffernoperationen. Wenn man die Rechenzeit mit der des Fermat-Tests vergleicht, stellt man fest, dass die Rechenwege etwas unterschiedlich sind, aber dieselbe Anzahl von Multiplikationen modulo N anfallen. Nur werden beim Miller-Rabin-Test ,,unterwegs'' noch einige Zwischenergebnisse darauf getestet, ob sie gleich $1$ oder gleich $N-1$ sind. Dies verursacht aber kaum Mehraufwand.
         Nun wenden wir uns dem Ein-/Ausgabeverhalten des Miller-Rabin-Tests zu.
@@ -2215,38 +2471,48 @@
         Wir werden zeigen, dass diese Menge $B_N$ die gewünschten Eigenschaften hat.
 
         Lemma 4.43:
-        1. $L^{MR}_N \supseteq B_N$
-        2. $B_N \not\subseteq \mathbb{Z}^*_N$
-        3. $|B_N|\leq \frac{1}{2} | \mathbb{Z}^*_N|$, also $Pr_{a\in\{1,...,N-1\}}$ ($a$ ist MR-Lügner) $<\frac{1}{2}$
+        \begin{enumerate*}
+            \item  $L^{MR}_N \supseteq B_N$
+            \item  $B_N \not\subseteq \mathbb{Z}^*_N$
+            \item  $|B_N|\leq \frac{1}{2} | \mathbb{Z}^*_N|$, also $Pr_{a\in\{1,...,N-1\}}$ ($a$ ist MR-Lügner) $<\frac{1}{2}$
+        \end{enumerate*}
 
         Beweis: Teil 1. ist nur die Überprüfung, dass die Definition technisch sinnvoll ist. Der eigentlich interessante Teil ist 2.; Teil 3. ist dann Routine.
-        1. Sei $a$ ein beliebiger MR-Lügner.
-        - Fall 1: $a^u\ mod\ N = 1$. - Dann ist auch $a^{u*2^{i_0}} mod\ N = 1$, und daher gilt $a\in B_N$. (Die Zeile zu $a$ in Abb. 6 sieht aus wie die von $a_1=1$.)
-        - Fall2: $a^{u*2^i} mod\ N=N-1$ für ein $i<k$ - Dann ist $0\leq i\leq i_0$ nach der Definition von $i_0$. Wenn $i=i_0$ ist, haben wir direkt $a\in B_N$ (Beispiel in Tab. 6: $a_3$ oder $a_{\#}$); wenn $i<i_0$ ist, dann gilt $a^{u*2^{i_0}} mod\ N=(a^{u*2^{i}} mod\ N)^{2^{i_{0} -i}} mod\ N= (-1)^{2^{i_{0} -i}} mod\ N= 1$, also ebenfalls $a\in B_N$ (Beispiel in Tab. 6: $a_2$ oder $a_4$).
-        2. Dass $B_N\supseteq \mathbb{Z}^*_N$ gilt, folgt aus der Definition. Wir müssen also nur zeigen, dass $\mathbb{Z}^*_N - B_N\not=\varnothing$ gilt. Wir benutzen die in Lemma 4.38 beobachtete Eigenschaft von Carmichael-Zahlen, keine Primzahlpotenz zu sein. Nach diesem Lemma können wir $N=N_1 *N_2$ schreiben, für teilerfremde ungerade Zahlen $N_1,N_2\geq 3$. Im Folgenden werden die Eigenschaften aus dem Chinesischen Restsatz (Fakt 4.21) benutzen.
-        - Die grobe Idee der Konstruktion ist folgende: Wir haben das Element $1$ mit $1^{u*2^{i_0}} mod\ N= 1$ und das Element $a_{\#}$ mit $a^{u*2^{i_0}} \equiv -1 (mod\ N)$. Aus diesen beiden Elementen ,,basteln'' wir mit Hilfe des Chinesischen Restsatzes ein $b\in\mathbb{Z}^*_N$, das sich modulo $N_1$ wie $1$ und modulo $N_2$ wie $a_{\#}$ verhält. Es wird sich zeigen, dass $b^{u*2^{i_0}} mod\ N \not\in\{1 ,N-1\}$ gilt, also $b\not\in B_N$ ist.
-        - Wir führen diese Idee jetzt formal durch. Sei $x_2=a_{\#}\ mod\ N_2$. Nach dem Chinesischen Restsatz (Fakt 4.21) gibt es eine eindeutig bestimmte Zahl $b\in\{0,1,...,N- 1\}$, die $b\equiv 1(mod\ N_1)$ und $b\equiv x_2(mod\ N_2)$ (4.6) erfüllt. (Es folgt $b\equiv a_{\#}(mod\ N_2)$.) Wir zeigen, dass $b$ in $\mathbb{Z}^*_N-B_N$ liegt.
-        - Wir notieren, dass für beliebige $x,y\in\mathbb{Z}$ gilt: $x\equiv x'(mod\ N)\Rightarrow x\equiv x'(mod\ N_i)$, für $i=1,2$. (4.7)
-        - (Wenn $x-x'$ durch $N$ teilbar ist, dann auch durch $N_1$ und $N_2$.)
-        - Beh. 1: $b\in\mathbb{Z}^*_N$.
-        - Bew.: Offensichtlich gilt $b^{N-1} \equiv 1^{N-1} \equiv 1 (mod\ N_1)$. Weiter haben wir, modulo $N_2$ gerechnet: $b^{N-1} \equiv a^{N-1}_{\#}\equiv ((a^{N-1}_{\#}) mod\ N)\equiv 1 (mod\ N_2)$.
-        - Wegen der Eindeutigkeitsaussage im Chinesischen Restsatz folgt daraus $b^{N-1} \equiv 1(mod\ N)$. Nach Lemma 4.35 folgt $b\in\mathbb{Z}^*_N$.
-        - Beh. 2: $b\not\in B_N$.
-        - Bew.: Indirekt. Annahme: $b\in B_N$, d.h. $b^{u*2^{i_0}} \equiv 1 (mod\ N)$ oder $b^{u*2^{i_0}} \equiv -1 (mod\ N)$.
-        - 1. Fall: $b^{u*2^{i_0}} \equiv 1 (mod\ N)$. - Mit (4.7) folgt $b^{u*2^{i_0}} \equiv 1 (mod\ N_2)$. Andererseits gilt $b^{u*2^{i_0}} \equiv x^{u*2^{i_0}}_2 \equiv a^{u*2^{i_0}}_{\#}\equiv (a^{u*2^{i_0}}_{\#}mod\ N)\equiv N-1 \equiv -1 (mod\ N_2)$, also $2\equiv 0 (mod\ N_2)$, ein Widerspruch, weil $N_2 \geq 3$ ist.
-        - 2. Fall: $b^{u*2^{i_0}} \equiv  -1 (mod\ N)$. - Mit (4.7) folgt $b^{u*2^{i_0}} \equiv -1(mod\ N_1)$. Andererseits gilt $b^{u*2^{i_0}} \equiv 1^{u*2^{i_0}} \equiv 1 (mod\ N_1)$, also $2\equiv 0 (mod\ N_1)$, ebenfalls ein Widerspruch.
-        3. Wir verwenden die in 2. konstruierte Zahl $b\in\mathbb{Z}^*_N-B_N$. Wie im Beweis von Satz 4.36 betrachtet man die injektive Funktion $g_b:B_N \ni a\rightarrow ba\ mod\ N \in\mathbb{Z}^*_N$. Es gilt $g_b(a)\not\in B_N$ für jedes $a\in B_N$, denn $g_b(a)^{u*2^{i_0}} mod\ N= (b^{u*2^{i_0}} mod\ N)(a^{u*2^{i_0}}) mod\ N\in\{b^{u*2^{i_0}} mod\ N, (N-b^{u*2^{i_0}}) mod\ N\}$, und die letzte Menge ist disjunkt zu $\{1 ,N-1\}$. Man erhält sofort $|B_N|\leq \frac{1}{2} | \mathbb{Z}^*_N|$.
+        \begin{enumerate*}
+            \item Sei $a$ ein beliebiger MR-Lügner.
+            \begin{itemize*}
+                \item Fall 1: $a^u\ mod\ N = 1$. - Dann ist auch $a^{u*2^{i_0}} mod\ N = 1$, und daher gilt $a\in B_N$. (Die Zeile zu $a$ in Abb. 6 sieht aus wie die von $a_1=1$.)
+                \item Fall2: $a^{u*2^i} mod\ N=N-1$ für ein $i<k$ - Dann ist $0\leq i\leq i_0$ nach der Definition von $i_0$. Wenn $i=i_0$ ist, haben wir direkt $a\in B_N$ (Beispiel in Tab. 6: $a_3$ oder $a_{\#}$); wenn $i<i_0$ ist, dann gilt $a^{u*2^{i_0}} mod\ N=(a^{u*2^{i}} mod\ N)^{2^{i_{0} -i}} mod\ N= (-1)^{2^{i_{0} -i}} mod\ N= 1$, also ebenfalls $a\in B_N$ (Beispiel in Tab. 6: $a_2$ oder $a_4$).
+            \end{itemize*}
+            \item Dass $B_N\supseteq \mathbb{Z}^*_N$ gilt, folgt aus der Definition. Wir müssen also nur zeigen, dass $\mathbb{Z}^*_N - B_N\not=\varnothing$ gilt. Wir benutzen die in Lemma 4.38 beobachtete Eigenschaft von Carmichael-Zahlen, keine Primzahlpotenz zu sein. Nach diesem Lemma können wir $N=N_1 *N_2$ schreiben, für teilerfremde ungerade Zahlen $N_1,N_2\geq 3$. Im Folgenden werden die Eigenschaften aus dem Chinesischen Restsatz (Fakt 4.21) benutzen.
+            \begin{itemize*}
+                \item Die grobe Idee der Konstruktion ist folgende: Wir haben das Element $1$ mit $1^{u*2^{i_0}} mod\ N= 1$ und das Element $a_{\#}$ mit $a^{u*2^{i_0}} \equiv -1 (mod\ N)$. Aus diesen beiden Elementen ,,basteln'' wir mit Hilfe des Chinesischen Restsatzes ein $b\in\mathbb{Z}^*_N$, das sich modulo $N_1$ wie $1$ und modulo $N_2$ wie $a_{\#}$ verhält. Es wird sich zeigen, dass $b^{u*2^{i_0}} mod\ N \not\in\{1 ,N-1\}$ gilt, also $b\not\in B_N$ ist.
+                \item Wir führen diese Idee jetzt formal durch. Sei $x_2=a_{\#}\ mod\ N_2$. Nach dem Chinesischen Restsatz (Fakt 4.21) gibt es eine eindeutig bestimmte Zahl $b\in\{0,1,...,N- 1\}$, die $b\equiv 1(mod\ N_1)$ und $b\equiv x_2(mod\ N_2)$ (4.6) erfüllt. (Es folgt $b\equiv a_{\#}(mod\ N_2)$.) Wir zeigen, dass $b$ in $\mathbb{Z}^*_N-B_N$ liegt.
+                \item Wir notieren, dass für beliebige $x,y\in\mathbb{Z}$ gilt: $x\equiv x'(mod\ N)\Rightarrow x\equiv x'(mod\ N_i)$, für $i=1,2$. (4.7)
+                \item (Wenn $x-x'$ durch $N$ teilbar ist, dann auch durch $N_1$ und $N_2$.)
+                \item Beh. 1: $b\in\mathbb{Z}^*_N$.
+                \item Bew.: Offensichtlich gilt $b^{N-1} \equiv 1^{N-1} \equiv 1 (mod\ N_1)$. Weiter haben wir, modulo $N_2$ gerechnet: $b^{N-1} \equiv a^{N-1}_{\#}\equiv ((a^{N-1}_{\#}) mod\ N)\equiv 1 (mod\ N_2)$.
+                \item Wegen der Eindeutigkeitsaussage im Chinesischen Restsatz folgt daraus $b^{N-1} \equiv 1(mod\ N)$. Nach Lemma 4.35 folgt $b\in\mathbb{Z}^*_N$.
+                \item Beh. 2: $b\not\in B_N$.
+                \item Bew.: Indirekt. Annahme: $b\in B_N$, d.h. $b^{u*2^{i_0}} \equiv 1 (mod\ N)$ oder $b^{u*2^{i_0}} \equiv -1 (mod\ N)$.
+                \item 1. Fall: $b^{u*2^{i_0}} \equiv 1 (mod\ N)$. - Mit (4.7) folgt $b^{u*2^{i_0}} \equiv 1 (mod\ N_2)$. Andererseits gilt $b^{u*2^{i_0}} \equiv x^{u*2^{i_0}}_2 \equiv a^{u*2^{i_0}}_{\#}\equiv (a^{u*2^{i_0}}_{\#}mod\ N)\equiv N-1 \equiv -1 (mod\ N_2)$, also $2\equiv 0 (mod\ N_2)$, ein Widerspruch, weil $N_2 \geq 3$ ist.
+                \item 2. Fall: $b^{u*2^{i_0}} \equiv  -1 (mod\ N)$. - Mit (4.7) folgt $b^{u*2^{i_0}} \equiv -1(mod\ N_1)$. Andererseits gilt $b^{u*2^{i_0}} \equiv 1^{u*2^{i_0}} \equiv 1 (mod\ N_1)$, also $2\equiv 0 (mod\ N_1)$, ebenfalls ein Widerspruch.
+            \end{itemize*}
+            \item Wir verwenden die in 2. konstruierte Zahl $b\in\mathbb{Z}^*_N-B_N$. Wie im Beweis von Satz 4.36 betrachtet man die injektive Funktion $g_b:B_N \ni a\rightarrow ba\ mod\ N \in\mathbb{Z}^*_N$. Es gilt $g_b(a)\not\in B_N$ für jedes $a\in B_N$, denn $g_b(a)^{u*2^{i_0}} mod\ N= (b^{u*2^{i_0}} mod\ N)(a^{u*2^{i_0}}) mod\ N\in\{b^{u*2^{i_0}} mod\ N, (N-b^{u*2^{i_0}}) mod\ N\}$, und die letzte Menge ist disjunkt zu $\{1 ,N-1\}$. Man erhält sofort $|B_N|\leq \frac{1}{2} | \mathbb{Z}^*_N|$.
+        \end{enumerate*}
 
         Ab hier wieder prüfungsrelevant.
 
         Wir haben also eine Schranke von $\frac{1}{2}$ für die Irrtumswahrscheinlichkeit im Miller-Rabin-Algorithmus bewiesen. Eine etwas kompliziertere Analyse zeigt, dass sogar die Fehlerschranke $\frac{1}{4}$ gilt; man kann auch zeigen, dass es zusammengesetzte Zahlen $N$ gibt (zum Beispiel $703 = 19*37$), bei denen eine Fehlerwahrscheinlichkeit von fast $\frac{1}{4}$ tatsächlich auftritt. Durchl-fache Wiederholung, ebenso wie in Algorithmus 4.5, kann man die Fehlerschranke auf $4^{-l}$ reduzieren. Wir kürzen den Miller-Rabin-Test mit ,,MiRa-Test'' ab. Man beachte, dass bei jedem neuen Aufruf eine neue Zufallszahlagewählt wird.
 
         Algorithmus 4.7 Iterierter MR-Test
-        - Eingabe: Ungerade Zahl $N\geq 3$, eine Zahl $l\geq 1$.
-        - Methode:
-        - repeat l times
-        - if MiRa-Test(N) = 1 then return 1;
-        - return 0;
+        \begin{itemize*}
+            \item Eingabe: Ungerade Zahl $N\geq 3$, eine Zahl $l\geq 1$.
+            \item Methode:
+            \item repeat l times
+            \item if MiRa-Test(N) = 1 then return 1;
+            \item return 0;
+        \end{itemize*}
 
         Diesen Test wollen wir kurz IterMiRa(N,l) nennen. - Wir erhalten zusammenfassend:
 
@@ -2263,37 +2529,45 @@
         Ein naheliegender Ansatz zur Erzeugung einer zufälligen n-Bit-Primzahl ist dann folgender: Man wählt wiederholt eine (ungerade) Zahl $N$ aus $[2^{n-1}, 2^n)$ zufällig und wendet auf sie den (iterierten) Miller-Rabin-Test an. Dies wiederholt man, bis eine Zahl gefunden wurde, die die Ausgabe $0$ liefert.
 
         Algorithmus 4.8: GetPrime - Randomisierte Primzahlerzeugung
-        - Eingabe: Bitanzahl $n\geq 9$, Zahl $l\geq 1$ // Zuverlässigkeitsparameter
-        - Methode:
-        - repeat
-        - $N\leftarrow$ zufällige ungerade Zahl in $[2^{n-1}, 2^n)$;
-        - until IterMiRa(N,$l$) = 0; // Algorithmus 4.7
-        - return $N$.
+        \begin{itemize*}
+            \item Eingabe: Bitanzahl $n\geq 9$, Zahl $l\geq 1$ // Zuverlässigkeitsparameter
+            \item Methode:
+            \item repeat
+            \item $N\leftarrow$ zufällige ungerade Zahl in $[2^{n-1}, 2^n)$;
+            \item until IterMiRa(N,$l$) = 0; // Algorithmus 4.7
+            \item return $N$.
+        \end{itemize*}
 
         Die Ausgabe ist korrekt, wenn der Algorithmus eine Primzahl zurückgibt, ein Fehler tritt auf, wenn eine zusammengesetzte Zahl zurückgegeben wird. Auf den ersten Blick könnte man meinen (aufgrund von Proposition 4.44), dass die Fehlerwahrscheinlichkeit höchstens $1/4^l$ ist - eben die Fehlerwahrscheinlichkeit des iterierten MiRa-Tests. Wir werden sehen, dass wir auf der Basis der Analyse des Miller-Rabin-Tests nur ein etwas schwächeres Ergebnis bekommen. (Tatsächlich ist die Fehlerwahrscheinlichkeit durch $1/4^l$ beschränkt, für (von $l$ abhängig) genügend großen. Dies kann man aber nur durch fortgeschrittene zahlentheoretische Untersuchungen über die erwartete Wahrscheinlichkeit, dass eine zufällige ungerade zusammengesetzte Zahl den $l$-fach iterierten MiRa-Test übersteht, beweisen [P. Beauchemin, G. Brassard, C. Crèpeau, C. Goutier, C. Pomerance: The generation of random numbers that are probably prime.J. Cryptology 1 (1): 53-64 (1988)].)
 
         Wir definieren: $Prim_n:=\{p\in [2^{n-1}, 2^n)| \text{p ist Primzahl}\}$.
 
         Satz 4.45: Bei der Anwendung von Algorithmus 4.8 auf $n\geq 9$ gilt:
-        1. Wenn das Ergebnis eine Primzahl ist, hat jede Primzahl in $[2^{n-1}, 2^n)$ dieselbe Wahrscheinlichkeit, als Ergebnis zu erscheinen.
-        2. $Pr(GetPrime(n,l)\not\in Prim_n)\leq \frac{5n}{12*4^l}=O(\frac{n}{4^l})$. ( Nicht $1/4^l$, wie naiv vermutet.)
-        3. Die erwartete Rundenzahl ist $O(n)$, der erwartete Rechenaufwand ist $O(n^2)$ arithmetische Operationen und $O(n^4)$ Ziffernoperationen.
+        \begin{enumerate*}
+            \item Wenn das Ergebnis eine Primzahl ist, hat jede Primzahl in $[2^{n-1}, 2^n)$ dieselbe Wahrscheinlichkeit, als Ergebnis zu erscheinen.
+            \item  $Pr(GetPrime(n,l)\not\in Prim_n)\leq \frac{5n}{12*4^l}=O(\frac{n}{4^l})$. ( Nicht $1/4^l$, wie naiv vermutet.)
+            \item  Die erwartete Rundenzahl ist $O(n)$, der erwartete Rechenaufwand ist $O(n^2)$ arithmetische Operationen und $O(n^4)$ Ziffernoperationen.
+        \end{enumerate*}
 
         Beweis:
-        1. Eine Primzahl wird als Resultat genau dann geliefert, wenn keine zusammengesetzte Zahl fälschlicherweise vom Miller-Rabin-Test akzeptiert wird, bevor (in Zeile 2 ) die erste echte Primzahl gewählt wird. Jede der Primzahlen in $[2^{n-1}, 2^n)$ hat dieselbe Wahrscheinlichkeit, diese erste gewählte Primzahl zu sein.
-        2. $Pr(GetPrime(n,l)\not\in Prim_n)= Pr(\exists i\geq 1 :\text{ in Runden j=1,...,i-1 wird eine zusammengesetzte Zahl N gewählt und erkannt}\wedge\text{in Runde i wird zusammengesetzte Zahl N gewählt und der iterierte MiRa-Test auf N liefert 0})$ $\leq\sum_{i\geq 1} (1 -\frac{|Prim_n|}{2^{n-2}})^{i-1}*\frac{1}{4^l} = \frac{2^{n-2}}{|Prim_n|} *\frac{1}{4^l} \leq \frac{2^{n-2}}{\frac{6}{5} *2^{n-1} /n}*\frac{1}{4^l} = \frac{5n}{12 * 4^l}$. Wir haben benutzt, dass es in $[2^{n-1}, 2^n)$ genau $2^{n-2}$ ungerade Zahlen gibt.
-        3. Da man in jeder Runde mit Wahrscheinlichkeit mindestens $\frac{|Prim_n|}{2^{n-2}}$ eine Primzahl wählt, ist die erwartete Rundenzahl nicht größer als $\frac{2^{n-2}}{|Prim_n|} \leq \frac{5n}{12}$.
+        \begin{enumerate*}
+            \item  Eine Primzahl wird als Resultat genau dann geliefert, wenn keine zusammengesetzte Zahl fälschlicherweise vom Miller-Rabin-Test akzeptiert wird, bevor (in Zeile 2 ) die erste echte Primzahl gewählt wird. Jede der Primzahlen in $[2^{n-1}, 2^n)$ hat dieselbe Wahrscheinlichkeit, diese erste gewählte Primzahl zu sein.
+            \item  $Pr(GetPrime(n,l)\not\in Prim_n)= Pr(\exists i\geq 1 :\text{ in Runden j=1,...,i-1 wird eine zusammengesetzte Zahl N gewählt und erkannt}\wedge\text{in Runde i wird zusammengesetzte Zahl N gewählt und der iterierte MiRa-Test auf N liefert 0})$ $\leq\sum_{i\geq 1} (1 -\frac{|Prim_n|}{2^{n-2}})^{i-1}*\frac{1}{4^l} = \frac{2^{n-2}}{|Prim_n|} *\frac{1}{4^l} \leq \frac{2^{n-2}}{\frac{6}{5} *2^{n-1} /n}*\frac{1}{4^l} = \frac{5n}{12 * 4^l}$. Wir haben benutzt, dass es in $[2^{n-1}, 2^n)$ genau $2^{n-2}$ ungerade Zahlen gibt.
+            \item  Da man in jeder Runde mit Wahrscheinlichkeit mindestens $\frac{|Prim_n|}{2^{n-2}}$ eine Primzahl wählt, ist die erwartete Rundenzahl nicht größer als $\frac{2^{n-2}}{|Prim_n|} \leq \frac{5n}{12}$.
+        \end{enumerate*}
 
         Bemerkung: Bei der Erzeugung zufälliger n-Bit-Primzahlen für kryptographische Zwecke wird man aus Effizienzgründen nicht unseren Algorithmus anwenden, der $\Theta (n^2)$ Multiplikationen benötigt, sondern eine Kombination zweier verschiedener Primzahltests (z. B. Miller-Rabin und ,,Lucas Strong Probable Prime Test'') mit sehr wenigen Iterationen und einem Test auf Teilbarkeit durch sehr kleine Primteiler. Dies erfordert nur $O(n)$ Multiplikationen. Die Dichte der zusammengesetzten Zahlen, die von einem solchen Test nicht erkannt werden, ist als sehr gering einzuschätzen. Über eine interessante experimentelle Untersuchung hierzu berichtet die kurze Notiz http://people.csail.mit.edu/rivest/Rivest-FindingFourMillionLargeRandomPrimes.ps von Ron Rivest (bekannt von ,,RSA'' und von ,,Cormen, Leiserson, Rivest und Stein'').
 
         \subsection{Beweise und Bemerkungen zu Kapitel 4}
         Beweis der Existenz und der Eindeutigkeit des größten gemeinsamen Teilers $ggT(x,y)$ zweier Zahlen (Definition 4.3.2):
-        - *Eindeutigkeit*: Wenn in 2. $d$ und $d'$ beide (i) und (ii) erfüllen und nichtnegativ sind, dann folgt $d|d'$ und $d'|d$, also $d=d'$ nach Fakt 4.2.5.
-        - *Existenz*: Weil $t$ gemeinsamer Teiler von $x$ und $y$ ist genau dann wenn $t$ gemeinsamer Teiler von $a=|x|$ und $b=|y|$ ist, und weil offenbar das Vertauschen von $x$ und $y$ nichts ändert, können wir uns auf den Fall $x=a\geq y=b\geq 0$ beschränken. Wir zeigen durch Induktion über $b=min\{a,b\}$, dass $ggT(a,b)$ existiert.
-        - *Induktionsanfang*: $b=0$.
-        - 1. Fall: $a= 0$. Dann ist jede Zahl $t$ ein gemeinsamer Teiler von $a$ und $b$. Wir wählen $d=0$. Dann gilt (i), weil $0$ Teiler von $0$ ist, und (ii), weil jede Zahl $t$ die Zahl $0$ teilt. ($0$ ist ,,größter'' gemeinsamer Teiler von 0 und 0 im Sinn der Quasiordnung ,,Teilbarkeit''. Hier ist 0 das größte Element überhaupt.)
-        - 2. Fall: $a >0$. Wir wählen $d=a$. Dann gilt (i), weil $a$ Teiler von $a$ und von 0 ist, und es gilt (ii), weil jeder gemeinsame Teiler von $a$ und 0 auf jeden Fall Teiler von $a$ ist.
-        - *Induktionsschritt*: $b>0$. Setze $q:=a\ div\ b$ und $r:=a-qb=a\ mod\ b$ und $(a',b'):=(b,r)$. Dann ist $b'=r < b=a'$. Nun haben $a$ und $b$ genau dieselben gemeinsamen Teiler wie $a'$ und $b'$. (Aus $t|a$ und $t|b$ folgt $t|(a-qb)$, also $t|a'$, und aus $t|a'$ und $t|b'$ folgt $t|r+qb$, also $t|a$.) Nach I.V. existiert $d= ggT(a',b')$, und dieses $d$ ist dann auch größter gemeinsamer Teiler von $a$ und $b$.
+        \begin{itemize*}
+            \item *Eindeutigkeit*: Wenn in 2. $d$ und $d'$ beide (i) und (ii) erfüllen und nichtnegativ sind, dann folgt $d|d'$ und $d'|d$, also $d=d'$ nach Fakt 4.2.5.
+            \item  *Existenz*: Weil $t$ gemeinsamer Teiler von $x$ und $y$ ist genau dann wenn $t$ gemeinsamer Teiler von $a=|x|$ und $b=|y|$ ist, und weil offenbar das Vertauschen von $x$ und $y$ nichts ändert, können wir uns auf den Fall $x=a\geq y=b\geq 0$ beschränken. Wir zeigen durch Induktion über $b=min\{a,b\}$, dass $ggT(a,b)$ existiert.
+            \item  *Induktionsanfang*: $b=0$.
+            \item  1. Fall: $a= 0$. Dann ist jede Zahl $t$ ein gemeinsamer Teiler von $a$ und $b$. Wir wählen $d=0$. Dann gilt (i), weil $0$ Teiler von $0$ ist, und (ii), weil jede Zahl $t$ die Zahl $0$ teilt. ($0$ ist ,,größter'' gemeinsamer Teiler von 0 und 0 im Sinn der Quasiordnung ,,Teilbarkeit''. Hier ist 0 das größte Element überhaupt.)
+            \item  2. Fall: $a >0$. Wir wählen $d=a$. Dann gilt (i), weil $a$ Teiler von $a$ und von 0 ist, und es gilt (ii), weil jeder gemeinsame Teiler von $a$ und 0 auf jeden Fall Teiler von $a$ ist.
+            \item  *Induktionsschritt*: $b>0$. Setze $q:=a\ div\ b$ und $r:=a-qb=a\ mod\ b$ und $(a',b'):=(b,r)$. Dann ist $b'=r < b=a'$. Nun haben $a$ und $b$ genau dieselben gemeinsamen Teiler wie $a'$ und $b'$. (Aus $t|a$ und $t|b$ folgt $t|(a-qb)$, also $t|a'$, und aus $t|a'$ und $t|b'$ folgt $t|r+qb$, also $t|a$.) Nach I.V. existiert $d= ggT(a',b')$, und dieses $d$ ist dann auch größter gemeinsamer Teiler von $a$ und $b$.
+        \end{itemize*}
 
 
         \section{Asymmetrische Verschlüsselung: RSA \& Co.}
@@ -2307,12 +2581,14 @@
         Die Menge $X$ der möglichen Nachrichten kann als Teilmenge von $\{0,1\}^*$ aufgefasst werden. Wir möchten vermeiden, dass sich Alice und Bob vor der Kommunikation auf einen gemeinsamen Schlüssel einigen müssen. Stattdessen wird ein Schlüsselpaar $(k,\hat{k})$ verwendet. Die erste Komponente $k$ heißt der öffentliche Schlüssel von Bob und wird von Bob allen zugänglich gemacht, etwa über seine Webseite oder als Anhang von E-Mails. Die zweite Komponente $\hat{k}$ heißt der private Schlüssel von Bob und ist nur ihm bekannt. Die Menge der Schlüsselpaare $(k,\hat{k})$, die zusammengehören, nennen wir die Schlüsselmenge $K$.
 
         Definition 5.1: Ein Public-Key-Kryptosystem $(X,Y,K,E,D)$ hat 5 Komponenten:
-        - Klartextmenge $X$ (endlich),
-        - Chiffretextmenge $Y$ (endlich),
-        - Schlüsselmenge $K$, wobei $K\supseteq K_{pub} \times K_{priv}$ für Mengen $K_{pub}$ und $K_{priv}$,
-        - Verschlüsselungsfunktion $E:X\times K_{pub} \rightarrow Y$,
-        - Entschlüsselungsfunktion $D:Y\times K_{priv} \rightarrow X$,
-        - wobei die folgende Dechiffrierbedingung gilt: $D(E(x,k), \hat{k}) =x$, für alle $x\in X,(k,\hat{k})\in K$.
+        \begin{itemize*}
+            \item Klartextmenge $X$ (endlich),
+            \item  Chiffretextmenge $Y$ (endlich),
+            \item  Schlüsselmenge $K$, wobei $K\supseteq K_{pub} \times K_{priv}$ für Mengen $K_{pub}$ und $K_{priv}$,
+            \item  Verschlüsselungsfunktion $E:X\times K_{pub} \rightarrow Y$,
+            \item  Entschlüsselungsfunktion $D:Y\times K_{priv} \rightarrow X$,
+            \item  wobei die folgende Dechiffrierbedingung gilt: $D(E(x,k), \hat{k}) =x$, für alle $x\in X,(k,\hat{k})\in K$.
+        \end{itemize*}
 
         Um ein solches System benutzen zu können, benötigt man noch ein Verfahren, um Schlüsselpaare $(k,\hat{k})$ zu erzeugen. Dieses Verfahren heißt $G$, ist randomisiert und wird von Bob angestoßen, dem auch das Ergebnis mitgeteilt wird. Dies ist notwendig, da die Ausgabe von $G$ den geheimen Schlüsselteil $\hat{k}$ enthält. Nach Erzeugung des Paars $(k,\hat{k})$ gibt Bob $k$ bekannt und speichert $\hat{k}$ geheim ab. Wenn Alice Bob eine Nachricht $x\in X$ schicken will, berechnet sie $y=E(x,k)$ und schickt $y$. Wenn Bob einen Chiffretext $y$ empfängt, berechnet er $z=D(y,\hat{k})$. Nach der Dechiffrierbedingung ist dies wieder $x$.
 
@@ -2333,20 +2609,26 @@
         Hierzu kann man im Prinzip wie in Abschnitt 4.7 beschrieben vorgehen. Nun wird das Produkt $N=pq$ berechnet. Die Zahl $N$ hat $l$ oder $l-1$ Bits. Weiter wird $\varphi(N) = (p-1)(q-1)$ berechnet. Es wird eine Zahl $e\in\{3,...,\varphi(N)-1\}$ mit $ggT(e,\varphi(N)) = 1$ gewählt. (Überprüfung mittels des erweiterten Euklidischen Algorithmus (Algorithmus 4.2), Rechenzeit $O(l^3)$.) Dann wird das multiplikative Inverse $d<\varphi(N) modulo\ \varphi(N)$ von $e$ bestimmt, so dass also $ed\ mod\ \varphi(N) = 1$ gilt. (Man beachte, dass nur ungerade $e$ in Frage kommen, weil $\varphi(N)$ gerade ist. Man weiß, dass die Anzahl der geeigneten Werte $e$ mindestens $\frac{\varphi(N)}{log(\varphi(N))}$ ist, so dass die erwartete Anzahl von Versuchen $O(log(\varphi(N)))=O(logN)$ ist.)
 
         Bob erhält aus seiner Rechnung $N$, $e$ und $d$. Aus diesen wird das Schlüsselpaar $(k,\hat{k})$ gebildet:
-        - Der öffentliche Schlüssel $k$ ist das Paar $(N,e)$. Dieser wird bekanntgegeben.
-        - Der geheime Schlüssel $\hat{k}$ ist $(N,d)$. (Natürlich ist nur der Teil $d$ wirklich geheim.)
+        \begin{itemize*}
+            \item Der öffentliche Schlüssel $k$ ist das Paar $(N,e)$. Dieser wird bekanntgegeben.
+            \item  Der geheime Schlüssel $\hat{k}$ ist $(N,d)$. (Natürlich ist nur der Teil $d$ wirklich geheim.)
+        \end{itemize*}
 
         Der erwartete Berechnungsaufwand für die Schlüsselerzeugung ist $O((log\ N)^4) =O(l^4)$, weil dies die Kosten der Primzahlerzeugung sind (siehe Abschnitt 4.7); die Kosten für das Finden von $e$ sind geringer.
 
         \subsubsection{Verschlüsselung}
-        - Gegeben: Klartext $x$.
-        - Benötigt: Öffentlicher Schlüssel $k= (N,e)$.
+        \begin{itemize*}
+            \item Gegeben: Klartext $x$.
+            \item  Benötigt: Öffentlicher Schlüssel $k= (N,e)$.
+        \end{itemize*}
 
         Verschlüsselung von $x\in X= [N]: y=E(x,(N,e)) :=x^e\ mod\ N$. (Zu berechnen mit schneller Exponentiation, Rechenzeit $O((log\ N)^3) =O(l^3)$.)
 
         \subsubsection{Entschlüsselung}
-        - Gegeben: Chiffretext $y$.
-        - Benötigt: Privater Schlüssel $\hat{k}= (N,d)$.
+        \begin{itemize*}
+            \item Gegeben: Chiffretext $y$.
+            \item  Benötigt: Privater Schlüssel $\hat{k}= (N,d)$.
+        \end{itemize*}
 
     $z=D(y,(N,d)) :=y^d\ mod\ N$. (Zu berechnen mit schneller Exponentiation, Rechenzeit $O((log\ N)^3) =O(l^3)$.)
 
@@ -2361,19 +2643,23 @@
         Die Behauptung gilt naturlich genauso für $q$: Auch $q$ ist ein Teiler von $x^{ed}-x$. Da $p$ und $q$ teilerfremd sind, folgt aus Fakt 4.8, dass $N=pq$ Teiler von $x^{ed}-x$ ist, wie gewünscht.
 
         Beispiel:
-        - Schlüsselerzeugung: $N= 55 = 5*11$.
-        - $\varphi(N) = 4*10 = 40$.
-        - $e= 3, ggT(3,40) = 1, d=e^{-1}\ mod\ 40 = 27$, weil $3*27 = 81\equiv 1 (mod\ 40)$ gilt.
-        - Also: $k= (55,3)$, $\hat{k}= (55,27)$.
-        - Sei der Klartext $x=9$ gegeben.
-        - Verschlüsselung: $y= 9^3\ mod\ 55 = 729\ mod\ 55 = 14$.
-        - Entschlüsselung: Wir rechnen modulo 55: $y^{27}\equiv (14^2 * 14)^9 \equiv (31*14)^9 \equiv 434^9 \equiv ((-6)^3)^3 \equiv 4^3 \equiv 64 \equiv 9\ (mod\ 55)$.
+        \begin{itemize*}
+            \item Schlüsselerzeugung: $N= 55 = 5*11$.
+            \item  $\varphi(N) = 4*10 = 40$.
+            \item  $e= 3, ggT(3,40) = 1, d=e^{-1}\ mod\ 40 = 27$, weil $3*27 = 81\equiv 1 (mod\ 40)$ gilt.
+            \item  Also: $k= (55,3)$, $\hat{k}= (55,27)$.
+            \item  Sei der Klartext $x=9$ gegeben.
+            \item  Verschlüsselung: $y= 9^3\ mod\ 55 = 729\ mod\ 55 = 14$.
+            \item  Entschlüsselung: Wir rechnen modulo 55: $y^{27}\equiv (14^2 * 14)^9 \equiv (31*14)^9 \equiv 434^9 \equiv ((-6)^3)^3 \equiv 4^3 \equiv 64 \equiv 9\ (mod\ 55)$.
+        \end{itemize*}
 
         RSA, unsichere Version, insgesamt
-        - Bob erzeugt einen Schlüsselsatz $(k, \hat{k}) = ((N,e),(N,d))$.
-        - Er veröffentlicht $(N,e)$ und hält $d$ geheim.
-        - Wenn Alice $(x_1,...,x_r)\in [N]^r$ an Bob schicken will, verschlüsselt sie $y_i:=x^e_i\ mod\ N$, für $i=1,...,r$, und sendet $(y_1,...,y_r)$ an Bob.
-        - Dieser entschlüsselt $z_i:=y^d_i\ mod\ N$, für $i=1,...,r$, und erhält $(z_1,...,z_r) = (x_1,...,x_r)$.
+        \begin{itemize*}
+            \item Bob erzeugt einen Schlüsselsatz $(k, \hat{k}) = ((N,e),(N,d))$.
+            \item  Er veröffentlicht $(N,e)$ und hält $d$ geheim.
+            \item  Wenn Alice $(x_1,...,x_r)\in [N]^r$ an Bob schicken will, verschlüsselt sie $y_i:=x^e_i\ mod\ N$, für $i=1,...,r$, und sendet $(y_1,...,y_r)$ an Bob.
+            \item  Dieser entschlüsselt $z_i:=y^d_i\ mod\ N$, für $i=1,...,r$, und erhält $(z_1,...,z_r) = (x_1,...,x_r)$.
+        \end{itemize*}
 
         \subsection{Asymmetrische Kryptoschemen, Sicherheitsbegriff}
         Wir betrachten die Situation asymmetrisch verschlüsselter Kommunikation allgemein, also für beliebig lange Klartexte, und formulieren ein Sicherheitskonzept. Der Einfachheit halber nehmen wir als Menge $Y$ der Chiffretexte die Menge aller Binärstrings an.
@@ -2382,20 +2668,24 @@
 
         \subsubsection{Asymmetrische Kryptoschemen}
         Definition 5.3: Ein asymmetrisches Kryptoschema ist ein Tupel $S= (X,K,G,E,D)$, wobei
-        - $X,K\supseteq K_{pub} \times K_{priv}$ Mengen,
-        - $G():K_{pub} \times K_{priv}$ ein randomisierter Algorithmus,
-        - $E(x:X,k:K_{pub}):\{0,1\}^*$ ein randomisierter Algorithmus und
-        - $D(y:\{0,1\}^*,k:K_{priv}):\{0,1\}^*$ ein deterministischer Algorithmus sind,
-        - so dass gelten
-        - die (erwartete) Laufzeit von $?$ ist beschränkt durch eine Konstante,
-        - die Laufzeiten von $?$ und $D$ sind polynomiell beschränkt in der Länge von $x$ bzw. $y$,
-        - für jedes $x\in X,k\in K_{pub}$, jede Ausgabe $(k,\hat{k})$ von $?$ und jedes $m\in\{0,1\}^{p(|x|)}$ (die Ausgänge der flip-Anweisungen in $E$, für ein Polynom $p(n)$) gilt: $D(E^m (x,k), \hat{k}) =x$.
+        \begin{itemize*}
+            \item $X,K\supseteq K_{pub} \times K_{priv}$ Mengen,
+            \item  $G():K_{pub} \times K_{priv}$ ein randomisierter Algorithmus,
+            \item  $E(x:X,k:K_{pub}):\{0,1\}^*$ ein randomisierter Algorithmus und
+            \item  $D(y:\{0,1\}^*,k:K_{priv}):\{0,1\}^*$ ein deterministischer Algorithmus sind,
+            \item so dass gelten
+            \item die (erwartete) Laufzeit von $?$ ist beschränkt durch eine Konstante,
+            \item die Laufzeiten von $?$ und $D$ sind polynomiell beschränkt in der Länge von $x$ bzw. $y$,
+            \item für jedes $x\in X,k\in K_{pub}$, jede Ausgabe $(k,\hat{k})$ von $?$ und jedes $m\in\{0,1\}^{p(|x|)}$ (die Ausgänge der flip-Anweisungen in $E$, für ein Polynom $p(n)$) gilt: $D(E^m (x,k), \hat{k}) =x$.
+        \end{itemize*}
 
         Begriffe
-        - Die Elemente von X heißen ,,Klartexte'', die Menge der ,,Chiffretexte'' ist $\{0,1\}^*$.
-        - Die Elemente von $K_{pub}$ heißen ,,Öffentliche Schlüssel'', die von $K_{priv}$ ,,private Schlüssel'', mit $K\supseteq K_{pub} \times K_{priv}$ bezeichnen wir die Menge der Schlüsselpaare, die $G$ möglicherweise ausgibt.
-        - $G$ ist der ,,Schlüsselgenerierungsalgorithmus''. Er hat im Prinzip kein Argument, eventuell gibt es verschiedene Varianten, die von einem ,,Sicherheitsparameter'' $l$ abhängen.
-        - $E$ ist der ,,Verschlüsselungs-'' und $D$ der ,,Entschlüsselungsalgorithmus''.
+        \begin{itemize*}
+            \item Die Elemente von X heißen ,,Klartexte'', die Menge der ,,Chiffretexte'' ist $\{0,1\}^*$.
+            \item Die Elemente von $K_{pub}$ heißen ,,Öffentliche Schlüssel'', die von $K_{priv}$ ,,private Schlüssel'', mit $K\supseteq K_{pub} \times K_{priv}$ bezeichnen wir die Menge der Schlüsselpaare, die $G$ möglicherweise ausgibt.
+            \item $G$ ist der ,,Schlüsselgenerierungsalgorithmus''. Er hat im Prinzip kein Argument, eventuell gibt es verschiedene Varianten, die von einem ,,Sicherheitsparameter'' $l$ abhängen.
+            \item $E$ ist der ,,Verschlüsselungs-'' und $D$ der ,,Entschlüsselungsalgorithmus''.
+        \end{itemize*}
 
         Beispiel: Die Erweiterung des RSA-Schemas auf Strings beliebiger Länge durch Anwendung auf Blöcke der Länge $w\leq log\ N$, wie oben beschrieben, liefert ein Kryptoschema. Die Frage ist, wie gut dieses Kryptoschema ist.
 
@@ -2405,21 +2695,27 @@
         Die Idee ist wie folgt: Der Finder $AF$ bekommt einen öffentlichen Schlüssel $k$. Daraus berechnet er zwei verschiedene Klartexte $(z_0,z_1)$ gleicher Länge und ,,Notizen'' $v\in V$. Danach wird zufällig $z_0$ oder $z_1$ zu $y$ verschlüsselt. Im zweiten Schritt verwendet der Rater $AG$ die Zwischeninformation $v$, den öffentlichen Schlüssel $k$ und die ,,Probe'' $y$, um zu bestimmen, ob $z_0$ oder $z_1$ verschlüsselt wurde.
 
         Definition 5.5: Sei $S=(X,K,G,E,D)$ ein asymmetrisches Kryptoschema und $A=(AF,AG)$ ein Angreifer. Das zugehörige Experiment oder Spiel ist der folgende Algorithmus $G^S_A:\{0,1\}$:
-        1. $(k,\hat{k})\leftarrow G()$ (ein Schlüsselpaar des Kryptoschemas $S$ wird gewählt)
-        2. $(z_0,z_1,v)\leftarrow AF(k)$ (der Finder berechnet ein Paar von Klartexten gleicher Länge, von denen er annimmt, ihre Chiffretexte unterscheiden zu können)
-        3. $b\leftarrow flip()$ und $y\leftarrow E(z_b,k)$ (einer der Klartexte wird zuyverschlüsselt)
-        4. $b'\leftarrow AG(v,k,y)$ (der Rater versucht herauszubekommen, ob $z_0$ oder $z_1$ verschlüsselt wurde)
-        5. falls $b=b'$, so gib $1$ zurück, sonst $0$.
-        - Das verkürzte Experiment oder Spiel $S^S_A$ gibt im 5. Schritt einfach $b'$ aus.
+        \begin{enumerate*}
+            \item  $(k,\hat{k})\leftarrow G()$ (ein Schlüsselpaar des Kryptoschemas $S$ wird gewählt)
+            \item $(z_0,z_1,v)\leftarrow AF(k)$ (der Finder berechnet ein Paar von Klartexten gleicher Länge, von denen er annimmt, ihre Chiffretexte unterscheiden zu können)
+            \item  $b\leftarrow flip()$ und $y\leftarrow E(z_b,k)$ (einer der Klartexte wird zuyverschlüsselt)
+            \item  $b'\leftarrow AG(v,k,y)$ (der Rater versucht herauszubekommen, ob $z_0$ oder $z_1$ verschlüsselt wurde)
+            \item  falls $b=b'$, so gib $1$ zurück, sonst $0$.
+            \begin{itemize*}
+                \item Das verkürzte Experiment oder Spiel $S^S_A$ gibt im 5. Schritt einfach $b'$ aus.
+            \end{itemize*}
+        \end{enumerate*}
 
         Dann ist $Pr(G^S_A= 1)$ die Wahrscheinlichkeit dafür, dass der Angreifer $A$ den korrekten Klartext erkennt. Man kann jetzt wie in Abschnitt 2.4 (Sicherheit von $l$-Blockkryptosystemen) den Vorteil $adv(A,S) = 2 Pr(G^S_A = 1)- 1$, den ,,Erfolg'' $suc(A,S) = Pr(G^S_A\langle b = 1\rangle  = 1)$ und den ,,Misserfolg'' $fail(A,S) =Pr(G^S_A\langle b= 0\rangle = 1)$ definieren. Lemma 2.16 gilt dann wörtlich.
 
         Beispiel 5.6: Sei $S=(X,K,G,E,D)$ ein asymmetrisches Kryptoschema, in dem der Verschlüsselungsalgorithmus $E$ deterministisch ist. Wir betrachten den folgenden Angreifer $A$ mit $V=\{0,1\}^*$. Seien $z_0$ und $z_1$ verschiedene Elemente von $X$:
-        - $AF(k:K_{pub}) : (\{0,1\}\times\{0,1\})^+\times V$
-        - $v\leftarrow E(z_0,k);return(z_0,z_1,v)$
-        - $AG(v:V,k:K_{pub},y:\{0,1\}^*):\{0,1\}$
-        - if $v=y$ then return 0 else return 1.
-        - Im Ablauf des Spiels $G^S_A$ wird der Rater $AG$ also mit $E(z_0,k)$ oder mit $E(z_1,k)$ gestartet. Wegen $E(z_0,k)\not=E(z_1,k)$ gilt $Pr(G^S_A=1)=1$, d.h. $adv(A,S)=1$.
+        \begin{itemize*}
+            \item $AF(k:K_{pub}) : (\{0,1\}\times\{0,1\})^+\times V$
+            \item $v\leftarrow E(z_0,k);return(z_0,z_1,v)$
+            \item $AG(v:V,k:K_{pub},y:\{0,1\}^*):\{0,1\}$
+            \item if $v=y$ then return 0 else return 1.
+            \item Im Ablauf des Spiels $G^S_A$ wird der Rater $AG$ also mit $E(z_0,k)$ oder mit $E(z_1,k)$ gestartet. Wegen $E(z_0,k)\not=E(z_1,k)$ gilt $Pr(G^S_A=1)=1$, d.h. $adv(A,S)=1$.
+        \end{itemize*}
 
         Dieses Beispiel lässt sich verallgemeinern:
 
@@ -2435,14 +2731,18 @@
         Betrachten wir nun konkret das RSA-System mit nur einem Block, also $X=[N]$. Kann Eva aus Kenntnis von $y=x^e\ mod\ N$, $e$ und $N$ irgendeine konkrete Information über $x$ ermitteln?
 
         Definition 5.9:
-        1. Das ,,Legendre-Symbol'' gibt an, ob $a\in\mathbb{Z}$ modulo einer Primzahl $p$ ein Quadrat ist. Für $p>2$ prim und $a\in\mathbb{Z}$ setze $L_p(a) =\begin{cases} 0 \quad\text{ falls } p|a \\ 1 \quad\text{ falls } p\not|a, \exists b:b^2 \equiv a (mod\ p) \\ -1 \quad\text{ falls } p\not|a, \lnot\exists b:b^2 \equiv a (mod\ p)\end{cases}$.
-        2. Das ,,Jacobi-Symbol'' verallgemeinert dies auf Moduli $N$, die nicht notwendig Primzahlen sind. Für $N>2$ ungerade mit Primzahlzerlegung $N=\prod_{1\leq i\leq r} p^{alpha_i}_i$ und $a\in\mathbb{Z}$ setze $J_N(a) =\prod_{1 \leq i\leq r} L_{p_i}(a)^{\alpha_i}$
+        \begin{enumerate*}
+            \item  Das ,,Legendre-Symbol'' gibt an, ob $a\in\mathbb{Z}$ modulo einer Primzahl $p$ ein Quadrat ist. Für $p>2$ prim und $a\in\mathbb{Z}$ setze $L_p(a) =\begin{cases} 0 \quad\text{ falls } p|a \\ 1 \quad\text{ falls } p\not|a, \exists b:b^2 \equiv a (mod\ p) \\ -1 \quad\text{ falls } p\not|a, \lnot\exists b:b^2 \equiv a (mod\ p)\end{cases}$.
+            \item  Das ,,Jacobi-Symbol'' verallgemeinert dies auf Moduli $N$, die nicht notwendig Primzahlen sind. Für $N>2$ ungerade mit Primzahlzerlegung $N=\prod_{1\leq i\leq r} p^{alpha_i}_i$ und $a\in\mathbb{Z}$ setze $J_N(a) =\prod_{1 \leq i\leq r} L_{p_i}(a)^{\alpha_i}$
+        \end{enumerate*}
 
         Beobachtungen und Beispiele:
-        - Weil $4^2\ mod\ 7 = 2$, gilt $L_7(2) = 1$. Weiter gilt $L_7(21) = 0$ und $L_7(5) =-1$, weil $1,2,4$ die einzigen Quadrate modulo $7$ sind.
-        - $J_N(a) = 0$ genau dann, wenn $ggT(a,N)>1$.
-        - Wenn $p\not|a$ und $p$ ist ein Primfaktor, der in $N$ mit geradem Exponenten vorkommt, dann ist der Beitrag von $p$ zu $J_N(a)$ der Faktor 1, spielt also keine Rolle.
-        - Wenn $ggT(a,N) = 1$, dann ist $J_N(a) = (-1)^{\#}\{i\leq r|\text{a ist Nichtquadrat modulo } p_i\}$.
+        \begin{itemize*}
+            \item Weil $4^2\ mod\ 7 = 2$, gilt $L_7(2) = 1$. Weiter gilt $L_7(21) = 0$ und $L_7(5) =-1$, weil $1,2,4$ die einzigen Quadrate modulo $7$ sind.
+            \item $J_N(a) = 0$ genau dann, wenn $ggT(a,N)>1$.
+            \item Wenn $p\not|a$ und $p$ ist ein Primfaktor, der in $N$ mit geradem Exponenten vorkommt, dann ist der Beitrag von $p$ zu $J_N(a)$ der Faktor 1, spielt also keine Rolle.
+            \item Wenn $ggT(a,N) = 1$, dann ist $J_N(a) = (-1)^{\#}\{i\leq r|\text{a ist Nichtquadrat modulo } p_i\}$.
+        \end{itemize*}
 
         Das Legendre-Symbol ist leicht mit schneller modularer Exponentiation zu berechnen.
 
@@ -2494,11 +2794,13 @@
         \subsubsection{Bemerkungen zum Faktorisierungsproblem, allgemein}
         Nach wie vor gibt es keinen Polynomialzeitalgorithmus, der beliebige zusammengesetzte Zahlen $N$ in ihre Primfaktoren zerlegen kann (äquivalent: in polynomieller
         Zeit einen nichttrivialen Faktor von $N$ bestimmen kann). Allerdings gab es in den letzten Jahrzehnten auch gewaltige Fortschritte bei der Entwicklung immer besserer Verfahren.
-        - Pollards $(p-1)$-Methode (klassisch): Führt zu schneller Ermittlung eines Faktors von $N$, wenn für einen Primfaktor $p$ von $N$ gilt, dass $p-1$ nur kleine Primfaktoren hat.
-        - Pollards ${\rho}$-Methode (klassisch): Führt zur Ermittlung eines Faktors von $N$ in Zeit $O(\sqrt{p})$, wo $p$ ein Primfaktor von $N$ ist. Da $N$ immer einen Primfaktor in $O(\sqrt{n})$ hat, ist die Rechenzeit im schlechtesten Fall in $O(\sqrt[4]{n}) =O(2^{(log\ N)/ 4)})$. Schnell zum Ziel kommt man also, wenn $N$ einen kleinen Primfaktor $p$ hat. (Die oben angegebenen Vorschriften zur Wahl von $p$ und $q$ vermeiden diese Situation.)
-        - Quadratisches Sieb (Pomerance, 1981): Rechenzeit $O(e^{sqrt{(ln\ N)(ln\ ln\ N)}})$.
-        - Faktorisierung mit Elliptischen Kurven (Hendrik W. Lenstra, 1987): Rechenzeit $O(e^{(1+o(1)) \sqrt{(ln\ p)(ln\ ln\ p)}})$, wo $p$ der kleinste Primteiler von $N$ ist.
-        - Zahlkörpersieb (um 1990, zwei Varianten): Die schnellsten heute bekannten Faktorisierungsverfahren. Die anfallenden Rechnungen können auf vielen Rechnern parallel durchgeführt werden. Die Gesamtrechenzeit ist beschränkt durch $O(e^{C(ln\ N)^{\frac{1}{3}} (ln\ ln\ N)^{\frac{2}{3}}})$, für eine Konstante $C$. An der Verbesserung der Verfahren wird laufend gearbeitet.
+        \begin{itemize*}
+            \item Pollards $(p-1)$-Methode (klassisch): Führt zu schneller Ermittlung eines Faktors von $N$, wenn für einen Primfaktor $p$ von $N$ gilt, dass $p-1$ nur kleine Primfaktoren hat.
+            \item Pollards ${\rho}$-Methode (klassisch): Führt zur Ermittlung eines Faktors von $N$ in Zeit $O(\sqrt{p})$, wo $p$ ein Primfaktor von $N$ ist. Da $N$ immer einen Primfaktor in $O(\sqrt{n})$ hat, ist die Rechenzeit im schlechtesten Fall in $O(\sqrt[4]{n}) =O(2^{(log\ N)/ 4)})$. Schnell zum Ziel kommt man also, wenn $N$ einen kleinen Primfaktor $p$ hat. (Die oben angegebenen Vorschriften zur Wahl von $p$ und $q$ vermeiden diese Situation.)
+            \item Quadratisches Sieb (Pomerance, 1981): Rechenzeit $O(e^{sqrt{(ln\ N)(ln\ ln\ N)}})$.
+            \item Faktorisierung mit Elliptischen Kurven (Hendrik W. Lenstra, 1987): Rechenzeit $O(e^{(1+o(1)) \sqrt{(ln\ p)(ln\ ln\ p)}})$, wo $p$ der kleinste Primteiler von $N$ ist.
+            \item Zahlkörpersieb (um 1990, zwei Varianten): Die schnellsten heute bekannten Faktorisierungsverfahren. Die anfallenden Rechnungen können auf vielen Rechnern parallel durchgeführt werden. Die Gesamtrechenzeit ist beschränkt durch $O(e^{C(ln\ N)^{\frac{1}{3}} (ln\ ln\ N)^{\frac{2}{3}}})$, für eine Konstante $C$. An der Verbesserung der Verfahren wird laufend gearbeitet.
+        \end{itemize*}
 
         Der Rechenaufwand für die Faktorisierung auch nur 260-stelliger Zahlen in Dezimaldarstellung ist heute noch so immens, dass diese Art von Angriff auf das RSA-System noch nicht als ernsthafte Bedrohung angesehen wird. Das BSI (Bundesamt für Sicherheit in der Informationstechnik) empfiehlt für RSA-Anwendungen Schlüssellängen von 2000 Bits oder etwa 600 Dezimalziffern (bzw. 3000 Bits oder 900 Dezimalziffern, wenn auch Entwicklungen der nächsten Jahre mit einkalkuliert werden sollen.)
         Wenn sich $p$ und $q$ nur geringfügig unterscheiden, lässt sich $N$ effizient faktorisieren. Man sollte also darauf achten, dass $p$ und $q$ in nicht mehr als etwa den ersten 20 Binärstellen übereinstimmen. Bei zufälliger Wahl wird dies mit sehr hoher Wahrscheinlichkeit eintreten.
@@ -2528,10 +2830,12 @@
         Weil $r^2\ mod\ p =((x^2\ mod\ N)^{(p+1)/4})^2\ mod\ p=x^{p+1}\ mod\ p=(x^p *x) mod\ p = x^2\ mod\ p$ (wir haben den kleinen Satz von Fermat in der Version ,,$x^p\equiv x(mod\ p)$ für alle $x$'' benutzt), gilt $r^2-x^2\equiv (r-x)(r+x)\equiv 0 (mod\ p)$. Das heißt, dass entweder $r\equiv x(mod\ p)$ oder $p-r\equiv x(mod\ p)$ gilt.
         Genauso sieht man, dass $s\equiv x(mod\ q)$ oder $q-s\equiv x(mod\ q)$ gilt.
         Mit der konstruktiven Variante des chinesischen Restsatzes (Bemerkung nach Fakt 4.28) können wir nun vier Zahlen $z_1,...,z_4 \in [N]$ berechnen, die die folgenden Kongruenzen erfullen:
-        1. $z_1 \equiv r (mod\ p)$ und $z_1 \equiv s (mod\ q)$
-        2. $z_2 \equiv r (mod\ p)$ und $z_2 \equiv q-s (mod\ q)$
-        3. $z_3 \equiv p-r (mod\ p)$ und $z_3 \equiv s (mod\ q)$
-        4. $z_4 \equiv p-r (mod\ p)$ und $z_4 \equiv q-s (mod\ q)$
+        \begin{itemize*}
+            \item  $z_1 \equiv r (mod\ p)$ und $z_1 \equiv s (mod\ q)$
+            \item  $z_2 \equiv r (mod\ p)$ und $z_2 \equiv q-s (mod\ q)$
+            \item  $z_3 \equiv p-r (mod\ p)$ und $z_3 \equiv s (mod\ q)$
+            \item  $z_4 \equiv p-r (mod\ p)$ und $z_4 \equiv q-s (mod\ q)$
+        \end{itemize*}
 
         Wegen der obigen Überlegung ist $x\in\{z_1,...,z_4\}$. Wir wählen eine dieser vier Möglichkeiten. (Man kann Vorkehrungen treffen, dass ,,sinnvolle'' Blöcke $x$ leicht zu erkennen sind. Beispielsweise könnte man den Block $x$ mit einer bestimmten Bitfolge wie 10000 abschließen. Es ist dann nicht anzunehmen, dass die Binärdarstellung einer der anderen Möglichkeiten zufällig ebenso endet.)
         Welcher Rechenaufwand ist nötig? Für die Verschlüsselung muss nur eine Quadrierung modulo $N$ durchgeführt werden; sie kostet nur Zeit $O((log\ N)^2)$. Die Entschlüsselung erfordert eine Exponentiation modulo $p$ und eine modulo $q$ und mehrere Anwendungen des erweiterten Euklidischen Algorithmus - insgesamt Zeit $O((log\ N)^3)$.
@@ -2539,33 +2843,41 @@
         *Sicherheit*: Wir nehmen an, Eva hätte ein effizientes Verfahren $B$, mit dem sie alle Chiffretexte zum öffentlichen Schlüssel $N$ entschlüsseln kann. Wir zeigen, dass sie dann auch $N$ faktorisieren kann. (Solange man annimmt, dass dies ein schwieriges Problem ist, kann auch die Annahme, dass Eva Verfahren $B$ hat, als unwahrscheinlich gelten.)
         Eva geht so vor: Sie wählt eine Zahl $x$ aus $[N]$ zufällig. Wenn $ggT(x,N)>1$, ist sie fertig, denn dieser größte gemeinsame Teiler ist entweder $p$ oder $q$. Andernfalls berechnet sie $y=x^2\ mod\ N$. Dann wendet sie ihr Entschlüsselungsverfahren an und berechnet ein $z=B(y)$ mit $z^2 \equiv y\ mod\ N$. Dieses $z$ hängt wohlgemerkt nicht von $x$, sondern nur von $y$ ab. Es gilt $x^2\equiv z^2
     (mod\ N)$. Wie oben gesehen gibt es vier Möglichkeiten:
-        1. $x\equiv z (mod\ p)$ und $x\equiv z (mod\ q)$
-        2. $x\equiv z (mod\ p)$ und $x\equiv -z (mod\ q)$
-        3. $x\equiv -z (mod\ p)$ und $x\equiv z (mod\ q)$
-        4. $x\equiv -z (mod\ p)$ und $x\equiv -z (mod\ q)$
+        \begin{enumerate*}
+            \item $x\equiv z (mod\ p)$ und $x\equiv z (mod\ q)$
+            \item  $x\equiv z (mod\ p)$ und $x\equiv -z (mod\ q)$
+            \item  $x\equiv -z (mod\ p)$ und $x\equiv z (mod\ q)$
+            \item  $x\equiv -z (mod\ p)$ und $x\equiv -z (mod\ q)$
+        \end{enumerate*}
 
         Welche dieser Möglichkeiten die richtige ist, hängt vom Zufall ab, der die Auswahl von $x$ steuert. Jede der 4 Quadratwurzeln von $y$ hat dieselbe Wahrscheinlichkeit $1/4$, als $x$ gewählt worden zu sein.
 
-        1. Fall: $x=z$, Misserfolg.
-        2. Fall: $0<|x-z|< N$ und $x-z$ durch $p$ teilbar, woraus $ggT(x-z,N) =p$ folgt: Erfolg!
-        3. Fall: $0<|x-z|< N$ und durch $q$ teilbar, woraus $ggT(x-z,N) =q$ folgt: Erfolg!
-        4. Fall: $x+z=N$, also $x-z\equiv 2x(mod\ N)$. Weil $2x$ teilerfremd zu $N$ ist, ergibt sich $ggT(x-z,N)=1$, Misserfolg.
+        \begin{enumerate*}
+            \item  Fall: $x=z$, Misserfolg.
+            \item  Fall: $0<|x-z|< N$ und $x-z$ durch $p$ teilbar, woraus $ggT(x-z,N) =p$ folgt: Erfolg!
+            \item  Fall: $0<|x-z|< N$ und durch $q$ teilbar, woraus $ggT(x-z,N) =q$ folgt: Erfolg!
+            \item  Fall: $x+z=N$, also $x-z\equiv 2x(mod\ N)$. Weil $2x$ teilerfremd zu $N$ ist, ergibt sich $ggT(x-z,N)=1$, Misserfolg.
+        \end{enumerate*}
 
         Eva muss also nur $ggT(x-z,N)$ berechnen! Damit gelingt es ihr mit Wahrscheinlichkeit $1/2$, die Faktoren von $N$ zu ermitteln. Durch l-fache Wiederholung desselben Experiments lässt sich die Erfolgswahrscheinlichkeit auf $1-\frac{1}{2^l}$ erhöhen.
 
         \subsection{Diskrete Logarithmen und Anwendungen}
         \subsubsection{Diskrete Logarithmen}
         Wir betrachten eine endliche zyklische Gruppe $(G,\circ,e)$ (multiplikativ geschrieben) mit einem erzeugenden Element $g$. Das bedeutet : $G=\{g^0=e,g^1=g,g^2,...,g^{|G|- 1}\}$. Sei $N=|G|$ die Größe (,,Ordnung'') dieser Gruppe. Dann haben wir:
-        - Die Exponentiationsabbildung $exp_g:\{0,1,...,N-1\}\rightarrow G, a\rightarrow g^a$, ist eine Bijektion (sogar ein Gruppenisomorphismus zwischen $(\mathbb{Z}_N,+,0)$ und $(G,\circ,e)$).
-        - Die Umkehrfunktion $log_g:G\rightarrow\{0,1,...,N-1\},g^a\rightarrow a$, heißt der diskrete Logarithmus zur Basis $g$.
-        - Das DL-Problem für $G$ und $g$ bezeichnet die Aufgabe, zu gegebenem $h\in G$ den Exponenten $a=log_g (h) \in\{0,1,...,N-1\}$ zu berechnen, also den Exponenten $a$ mit $g^a=h$.
+        \begin{itemize*}
+            \item Die Exponentiationsabbildung $exp_g:\{0,1,...,N-1\}\rightarrow G, a\rightarrow g^a$, ist eine Bijektion (sogar ein Gruppenisomorphismus zwischen $(\mathbb{Z}_N,+,0)$ und $(G,\circ,e)$).
+            \item Die Umkehrfunktion $log_g:G\rightarrow\{0,1,...,N-1\},g^a\rightarrow a$, heißt der diskrete Logarithmus zur Basis $g$.
+            \item Das DL-Problem für $G$ und $g$ bezeichnet die Aufgabe, zu gegebenem $h\in G$ den Exponenten $a=log_g (h) \in\{0,1,...,N-1\}$ zu berechnen, also den Exponenten $a$ mit $g^a=h$.
+        \end{itemize*}
 
         Für alle kryptographischen Verfahren, die mit zyklischen Gruppen arbeiten, müssen die Gruppenelemente eine explizite Darstellung haben, auf denen die Gruppenoperationen $\circ$ und $^{-1}$ effizient ausführbar sind.
         Dann ist auch die Exponentiation $a\rightarrow h^a$ effizient ausführbar (mit $\leq 2 log\ a$ Gruppenoperationen), mittels einer Variante der schnellen modularen Exponentiation. Das DL-Problem darf jedoch keinen (bekannten) effizienten Algorithmus haben.
         Als Gruppen $G$ kommen u.a. in Frage:
-        - Für Primzahlen $p$: Die multiplikative Gruppe $\mathbb{Z}^*_p$ mit Grundmenge $\{1,...,p-1\}$ und Multiplikation modulo $p$ als Operation. Die Ordnung (d.h. die Größe) dieser Gruppe ist bekanntermaßen $N=p-1$. Siehe die folgende Tabelle. Es ist ein recht einfacher Fakt aus der Zahlentheorie, dass diese Gruppe für jede Primzahlpzyklisch ist, also ein erzeugendes Element $g$ hat. Die Gruppenoperationen sind effizient ausführbar: Die Gruppenoperation $\circ$ ist die Multiplikation modulo $p$; das neutrale Element ist die 1; zu gegebenem $h\in G$ kann man $h^{-1}$ mit dem erweiterten Euklidischen Algorithmus berechnen. Für $p$ mit 2048 Bits oder mehr, wobei $p-1$ einen ,,großen'' Primteiler enthält, gilt das DL-Problem als praktisch nicht lösbar (für ,,allgemeine'' $a$).
-        - Die multiplikative Gruppe eines beliebigen endlichen Körpers, z. B. $GF(2^k)$. Es muss nur sichergestellt sein, dass die Multiplikation im Körper effizient ausführbar ist. Hierfür benötigt man ein irreduzibles Polynom von Grad $k$ und man muss Implementierungen von Polynommultiplikation und -Division haben. Für das Bilden von Inversen kann man den Potenzierungstrick benutzen oder den erweiterten Euklidischen Algorithmus für Polynome. Schließlich muss ein erzeugendes Element der multiplikativen Gruppe $GF(2^k)^*$ bekannt sein. (Die Kardinalität der Gruppe sollte mindestens $2^{2000}$ sein.)
-        - Zyklische Untergruppen von ,,elliptischen Kurven'' über endlichen Körpern. (Hier genügen Körperelemente mit einer Bitlänge von 256 Bits.)
+        \begin{itemize*}
+            \item Für Primzahlen $p$: Die multiplikative Gruppe $\mathbb{Z}^*_p$ mit Grundmenge $\{1,...,p-1\}$ und Multiplikation modulo $p$ als Operation. Die Ordnung (d.h. die Größe) dieser Gruppe ist bekanntermaßen $N=p-1$. Siehe die folgende Tabelle. Es ist ein recht einfacher Fakt aus der Zahlentheorie, dass diese Gruppe für jede Primzahlpzyklisch ist, also ein erzeugendes Element $g$ hat. Die Gruppenoperationen sind effizient ausführbar: Die Gruppenoperation $\circ$ ist die Multiplikation modulo $p$; das neutrale Element ist die 1; zu gegebenem $h\in G$ kann man $h^{-1}$ mit dem erweiterten Euklidischen Algorithmus berechnen. Für $p$ mit 2048 Bits oder mehr, wobei $p-1$ einen ,,großen'' Primteiler enthält, gilt das DL-Problem als praktisch nicht lösbar (für ,,allgemeine'' $a$).
+            \item Die multiplikative Gruppe eines beliebigen endlichen Körpers, z. B. $GF(2^k)$. Es muss nur sichergestellt sein, dass die Multiplikation im Körper effizient ausführbar ist. Hierfür benötigt man ein irreduzibles Polynom von Grad $k$ und man muss Implementierungen von Polynommultiplikation und -Division haben. Für das Bilden von Inversen kann man den Potenzierungstrick benutzen oder den erweiterten Euklidischen Algorithmus für Polynome. Schließlich muss ein erzeugendes Element der multiplikativen Gruppe $GF(2^k)^*$ bekannt sein. (Die Kardinalität der Gruppe sollte mindestens $2^{2000}$ sein.)
+            \item Zyklische Untergruppen von ,,elliptischen Kurven'' über endlichen Körpern. (Hier genügen Körperelemente mit einer Bitlänge von 256 Bits.)
+        \end{itemize*}
 
         Tabelle: Die multiplikative Gruppe $\mathbb{Z}^*_{11}$ für $p=11$ mit erzeugendem Element $g=2$. Die erzeugenden Elemente sind mit $*$ markiert. Sie entsprechen den Potenzen $g^a$ mit $ggT(a,10) = 1$.
 
@@ -2576,11 +2888,15 @@
 
         Für kryptographische Anwendungen ungeeignetist dagegen die bekannteste zyklische Gruppe $(\mathbb{Z}_N,+,0)$. Hier ist die Gruppenoperation die Addition modulo $N$; die Potenz $g^a$ für $a\in\mathbb{Z}$ entspricht dem Element $a*g\ mod\ N$. Die erzeugenden Elemente sind gerade die Elemente von $\mathbb{Z}^*_N$. Wenn $g\in\mathbb{Z}^*_N$ ist, dann besteht das DL-Problem für $g$ in folgendem: Gegeben $h=g^a$ (in der Gruppe gerechnet, das ist also $h=a*g\ mod\ N)$, finde $a$. Dies ist mit dem erweiterten Euklidischen Algorithmus leicht möglich: finde $g-1\ mod\ N$, berechne $h*g^{-1}\ mod\ N$. Das DL-Problem in dieser Gruppe ist also effizient lösbar.
         Wenn eine zyklische Gruppe $G$ mit effizienten Operationen gefunden worden ist, muss man immer noch ein erzeugendes Element finden. Das ist unter Umständen nicht ganz einfach. Wir wissen:
-        - Eine zyklische Gruppe $G=\langle g\rangle $ mit $|G|=N$ hat genau $\varphi(N) = |\mathbb{Z}^*_N|$ viele erzeugende Elemente, nämlich die Elemente $g^a$ mit $a\in\mathbb{Z}^*_N$.
+        \begin{itemize*}
+            \item Eine zyklische Gruppe $G=\langle g\rangle $ mit $|G|=N$ hat genau $\varphi(N) = |\mathbb{Z}^*_N|$ viele erzeugende Elemente, nämlich die Elemente $g^a$ mit $a\in\mathbb{Z}^*_N$.
+        \end{itemize*}
 
         Das bedeutet, dass ein Anteil von $\varphi(N)/N$ der Elemente von $G$ erzeugende Elemente sind. Aus Kapitel 4 wissen wir, dass $\varphi(N)/N=\prod_{p\ ist\ prim, p|N}(1-\frac{1}{p})$ gilt. Man kann zeigen, dass dies $\Omega(1/log\ log\ N)$ ist. Wir können also analog zum Vorgehen bei der Primzahlerzeugung ein erzeugendes Element finden, wenn wir folgende Operationen zur Verfügung haben:
-        1. zufälliges Wählen eines Elements von $G$
-        2. Test, ob $h\in G$ ein erzeugendes Element ist oder nicht
+        \begin{itemize*}
+            \item  zufälliges Wählen eines Elements von $G$
+            \item  Test, ob $h\in G$ ein erzeugendes Element ist oder nicht
+        \end{itemize*}
 
         Jedoch setzen alle bekannten effizienten Verfahren für den Test ,,Ist $h$ erzeugendes Element?'' voraus, dass man die Primfaktoren von $N=|G|$ kennt.
         Für den Fall $G=\mathbb{Z}^*_p$ ist die Situation so: Für große zufällige Primzahlen $p$ (512, 1024 oder 2048 Bits) ist die Primfaktorzerlegung von $N=p-1$ normalerweise nicht leicht zu ermitteln. Ein Ausweg ist, gezielt nach Primzahlen $p$ zu suchen, für die $p-1$ eine übersichtliche Primfaktorzerlegung hat. Besonders angenehm ist die Situation, wenn $p=2q+1$ für eine Primzahl $q$ ist.
@@ -2594,11 +2910,15 @@
         benannt nach W. Diffie und M. E. Hellman, die das Schlüsselaustauschverfahren 1976 vorgeschlagen haben.) Wenn Eva das DL-Problem lösen könnte, wäre dies leicht. Andere Möglichkeiten, das DH-Problem effizient zu lösen, ohne einen effizienten Algorithmus für das DL-Problem, sind prinzipiell denkbar, aber nicht bekannt.
 
         *Protokoll ,,Diffie-Hellman-Schlüsselaustausch''*
-        - Voraussetzung: Alice and Bob kennen $G,|G|$ und $g$.
-        1. Alice wählt $a\in\{2 ,...,|G|- 2\}$ zufällig, und sendet $A=g^a$ an Bob.
-        2. Bob wählt $b\in\{2 ,...,|G|-2\}$ zufällig, und sendet $B=g^b$ an Alice.
-        3. Alice berechnet $B^a= (g^b)^a=g^{ab}=k$.
-        4. Bob berechnet $A^b= (g^a)^b=g^{ab}=k$.
+        \begin{itemize*}
+            \item Voraussetzung: Alice and Bob kennen $G,|G|$ und $g$.
+        \end{itemize*}
+        \begin{enumerate*}
+            \item  Alice wählt $a\in\{2 ,...,|G|- 2\}$ zufällig, und sendet $A=g^a$ an Bob.
+            \item  Bob wählt $b\in\{2 ,...,|G|-2\}$ zufällig, und sendet $B=g^b$ an Alice.
+            \item  Alice berechnet $B^a= (g^b)^a=g^{ab}=k$.
+            \item  Bob berechnet $A^b= (g^a)^b=g^{ab}=k$.
+        \end{enumerate*}
 
         Dabei wird benutzt, dass die Multiplikation der Exponenten $a$ und $b$ kommutativ ist. Alice und Bob kennen nun $k$; Eva kennt nur $A$ und $B$, sowie (nach dem Kerckhoffs-Prinzip) $G$ und $|G|$.
 
@@ -2628,17 +2948,23 @@
         Kommentar: Der Rechenaufwand der Entschlüsselung liegt im Wesentlichen in der schnellen Exponentiation und der Invertierung von $k$.
 
         *Sicherheit*: Man kann sichuberlegen, dass für eine Angreiferin Eva und eine Gruppe G mit erzeugendem Element $g$ folgende Situationen äquivalent sind:
-        1. Eva kann alle mit dem ElGamal-Verfahren bzgl. $G$ und $g$ verschlüsselten Nachrichten effizient entschlüsseln, also aus $B$, $A$ und $y$ die Nachricht $x$ berechnen, die zum Chiffretext $(A,y)$ geführt hat.
-        2. Eva kann das DH-Problem für $G$ lösen.
+        \begin{enumerate*}
+            \item  Eva kann alle mit dem ElGamal-Verfahren bzgl. $G$ und $g$ verschlüsselten Nachrichten effizient entschlüsseln, also aus $B$, $A$ und $y$ die Nachricht $x$ berechnen, die zum Chiffretext $(A,y)$ geführt hat.
+            \item Eva kann das DH-Problem für $G$ lösen.
+        \end{enumerate*}
 
         Wenn Eva diskrete Logarithmen bezüglich $G$ und $g$ berechnen kann, gelten natürlich 1. und 2. Wir beweisen die Äquivalenz.
 
-        - ,,1.$\Rightarrow$2.'': Eva hat $B=g^b$ und $A=g^a$ vorliegen und möchte $k=g^{ab}$ bestimmen. Sie wendet ihr Entschlüsselungsverfahren auf $B$, $A$ und $y=1$ an. Es ergibt sich ein Wert $x$ mit $g^{ab}\circ x=k\circ x=y=1$. Es gilt also $x=k^{-1}$, und Eva kann $k$ durch Invertierung von $x$ in $G$ berechnen.
-        - ,,2.$\Rightarrow$1.'': Eva hat $B=g^b$,$A=g^a$,$y=g^{ab}\circ x$ vorliegen. Weil sie das DH-Problem lösen kann, kann sie $k=g^{ab}$ berechnen und damit natürlich $x=k^{-1}\circ y$ bestimmen.
+        \begin{itemize*}
+            \item ,,1.$\Rightarrow$2.'': Eva hat $B=g^b$ und $A=g^a$ vorliegen und möchte $k=g^{ab}$ bestimmen. Sie wendet ihr Entschlüsselungsverfahren auf $B$, $A$ und $y=1$ an. Es ergibt sich ein Wert $x$ mit $g^{ab}\circ x=k\circ x=y=1$. Es gilt also $x=k^{-1}$, und Eva kann $k$ durch Invertierung von $x$ in $G$ berechnen.
+            \item ,,2.$\Rightarrow$1.'': Eva hat $B=g^b$,$A=g^a$,$y=g^{ab}\circ x$ vorliegen. Weil sie das DH-Problem lösen kann, kann sie $k=g^{ab}$ berechnen und damit natürlich $x=k^{-1}\circ y$ bestimmen.
+        \end{itemize*}
 
         Unterschiede zwischen RSA und ElGamal:
-        - RSA in der puren Form benötigt einen Chiffretext $y=x^e\ mod\ N$, der die gleiche Bitlänge hat wie der Klartext $x$. ElGamal hat einen doppelt so langen Chiffretext $(B,y)$.
-        - ElGamal ist erzwungenermaßen randomisiert. Daher führt die wiederholte Verschlüsselung desselben Klartextes $x$ stets zu unterschiedlichen Chiffretexten, weil der Schlüssel $k$ zufällig ist.
+        \begin{itemize*}
+            \item RSA in der puren Form benötigt einen Chiffretext $y=x^e\ mod\ N$, der die gleiche Bitlänge hat wie der Klartext $x$. ElGamal hat einen doppelt so langen Chiffretext $(B,y)$.
+            \item ElGamal ist erzwungenermaßen randomisiert. Daher führt die wiederholte Verschlüsselung desselben Klartextes $x$ stets zu unterschiedlichen Chiffretexten, weil der Schlüssel $k$ zufällig ist.
+        \end{itemize*}
 
         Allerdings gibt es die Empfehlung, beim Arbeiten mit RSA den Klartext $x$ durch das Anhängen eines nicht ganz kurzen Zufallsstrings zu
         randomisieren. Wenn dieser angehängte Zufallsstring die gleiche Länge wie $x$ hat, ist der Chiffretext genauso lang wie bei ElGamal.
@@ -2657,12 +2983,14 @@
 
         Babystep-Giantstep-Algorithmus von Shanks:
         Wähle $m=\lceil \sqrt{N}\rceil$. Der gesuchte Exponent $a$ hat eine Zerlegung $a=bm+c$, für $0\leq c < m$ und passendes $b\leq a/m < N/m\leq \sqrt{N}$.
-        - Es gilt: $h=g^a=g^{bm+c}=g^{bm} \circ g^c$, also $g^{-c}=h^{-1}\circ g^{bm}$.
-        - Gesucht: $b$ und $c$.
-        - Algorithmus: Berechne alle Potenzen $g^{bm}$, $0\leq b < N/m$, und speichere $h^{-1}\circ g^{bm}$ (als Schlüssel) mit Wert $b$ in einer Hashtabelle $T$, Umfang $2N/m\leq 2\sqrt{N}$. Berechne $g^{-c}$, für $c=0,1,...,m-1$ und suche $g^{-c}$ in $T$. Wenn gefunden, gilt $h^{-1} \circ g^{bm}=g^{-c}$, also $h=g^{bm+c}$.
-        - Rechenzeit: $O(\sqrt{N})$ (erwartet) für Tabellenaufbau und $O(\sqrt{N})$ (erwartet) für die Suche, zusammen $O(\sqrt{N})$.
-        - Platzbedarf: $O(\sqrt{N})$
-        - Wenn $N=2^{200}$, ist $\sqrt{N}= 2^{100}> 10^{30}$. Selbst wenn man nur in dem unwahrscheinlichen Fall $a\leq 2^{160}$ erfolgreich sein möchte, muss man Tabellengröße $2^{80}> 10^{24}$ veranschlagen. Auch dies ist nicht durchführbar.
+        \begin{itemize*}
+            \item Es gilt: $h=g^a=g^{bm+c}=g^{bm} \circ g^c$, also $g^{-c}=h^{-1}\circ g^{bm}$.
+            \item Gesucht: $b$ und $c$.
+            \item Algorithmus: Berechne alle Potenzen $g^{bm}$, $0\leq b < N/m$, und speichere $h^{-1}\circ g^{bm}$ (als Schlüssel) mit Wert $b$ in einer Hashtabelle $T$, Umfang $2N/m\leq 2\sqrt{N}$. Berechne $g^{-c}$, für $c=0,1,...,m-1$ und suche $g^{-c}$ in $T$. Wenn gefunden, gilt $h^{-1} \circ g^{bm}=g^{-c}$, also $h=g^{bm+c}$.
+            \item Rechenzeit: $O(\sqrt{N})$ (erwartet) für Tabellenaufbau und $O(\sqrt{N})$ (erwartet) für die Suche, zusammen $O(\sqrt{N})$.
+            \item Platzbedarf: $O(\sqrt{N})$
+            \item Wenn $N=2^{200}$, ist $\sqrt{N}= 2^{100}> 10^{30}$. Selbst wenn man nur in dem unwahrscheinlichen Fall $a\leq 2^{160}$ erfolgreich sein möchte, muss man Tabellengröße $2^{80}> 10^{24}$ veranschlagen. Auch dies ist nicht durchführbar.
+        \end{itemize*}
 
         Pollards ${\rho}$-Algorithmus für DL
         Idee: Definiere eine Folge $(x_i,a_i,b_i),i= 0,1,2,...,$ in $G\times\mathbb{Z}_N\times\mathbb{Z}_N$, so dass $x_i=F(x_{i-1})$ gilt, für eine als zufällig geltende Funktion $F:G\rightarrow G$. Dann verhalten sich die ersten Komponenten $x_i$ wie zufällige Elemente in $G$, solange noch keine Wiederholung in der Folge $(Z_i)_{i=0, 1 , 2 ,...}$ aufgetreten ist. Nach dem Geburtstagsparadoxon weiß man, dass für eine genügend große Konstante $K$ die Wahrscheinlichkeit, dass $K\sqrt{N}$ zufällig gewählte Elemente von $G$ alle verschieden sind, sehr klein ist. Wir erwarten also nach $O(\sqrt{N})$ Schritten die Wiederholung eines Elements, also $i_0< j_0=O(\sqrt{N})$ mit $x_{i_0}= x_{j_0}$. Danach wiederholt sich die Folge: $x_{i_0 +1}= F(x_{i_0}) =F(x_{j_0}) = x_{j_0 +1}$, usw., also gilt $x_i=x_{i+kl}$ für alle $i\geq i_0$ und alle $k\geq 1$, wenn man $l=j_0 -i_0$ definiert. Man kann das Verhalten der Folge wie folgt zeichnen: $x_0,...,x_{i0}$ als gerade Linie ohne Wiederholung, daran angehängt $x_{i0},...,x_{j0}=x_{i0}$ als Kreis. Dies gibt die Form ,,${\rho}$'', wie beim griechischen Buchstaben ,,rho''. Wenn man es nun noch schafft, aus zwei Indizes $i < j$ mit $x_i=x_j$ den gesuchten Exponenten $a$ mit $h=g^a$ auszurechnen, ist man fertig.
@@ -2675,9 +3003,11 @@
     Falls nun $ggT(c_{2i}-c_i,N) = 1$ ist, können wir mit $a= (b_i-b_{2i})(c_{2i}-c_i)^{-1}\ mod\ N$ den gesuchten Exponenten berechnen.
     Die Rechenzeit ist $O(\sqrt{N})$, wenn man unterstellt, dass die Abbildung $F:x_{i-1}\rightarrow x_i$ rein zufällig ist. (In der Praxis bestätigt sich diese Vorstellung.)
     Weitere Algorithmen für das DL-Problem:
-    - Pohlig-Hellman-Algorithmus. Dieser Algorithmus benötigt die Primfaktorzerlegung von $N=|G|$. Seine Rechenzeit ist $O(\sum_{1 \leq i\leq k} e_i(log|G|+\sqrt{p_i}) + (log\ |G|)^2)$, wenn $|G|$ die Primfaktorzerlegung $p^{e_1}_1... p^{e_k}_k$ hat. Dieser Algorithmus ist also effizient, wenn $|G|$ nur eher kleine Primfaktoren hat. Wenn man also mit $G$ arbeiten will, muss $N=|G|$ mindestens einen ,,großen'' Primfaktor enthalten, damit nicht der Pohlig-Hellman-Algorithmus das DL-Problem effizient löst.
-    - Indexkalkül. Dieser Algorithmus ist nur für die multiplikative Gruppe $GF(q)^*$ in endlichem Körper $GF(q)$ anwendbar.
-    - Zahlkörpersieb. Ebenso nur für $GF(q)^*$ (mit ähnlichen subexponentiellen Rechenzeiten wie bei dem gleichnamigen Algorithmus bei der Faktorisierung).
+    \begin{itemize*}
+        \item Pohlig-Hellman-Algorithmus. Dieser Algorithmus benötigt die Primfaktorzerlegung von $N=|G|$. Seine Rechenzeit ist $O(\sum_{1 \leq i\leq k} e_i(log|G|+\sqrt{p_i}) + (log\ |G|)^2)$, wenn $|G|$ die Primfaktorzerlegung $p^{e_1}_1... p^{e_k}_k$ hat. Dieser Algorithmus ist also effizient, wenn $|G|$ nur eher kleine Primfaktoren hat. Wenn man also mit $G$ arbeiten will, muss $N=|G|$ mindestens einen ,,großen'' Primfaktor enthalten, damit nicht der Pohlig-Hellman-Algorithmus das DL-Problem effizient löst.
+        \item Indexkalkül. Dieser Algorithmus ist nur für die multiplikative Gruppe $GF(q)^*$ in endlichem Körper $GF(q)$ anwendbar.
+        \item Zahlkörpersieb. Ebenso nur für $GF(q)^*$ (mit ähnlichen subexponentiellen Rechenzeiten wie bei dem gleichnamigen Algorithmus bei der Faktorisierung).
+    \end{itemize*}
 
     Letzteres ist eine allgemeine Beobachtung: DL in $GF(q)^*$ scheint nicht viel schwieriger zu sein als das Faktorisierungsproblem für Zahlen in der Größenordnung von $q$.
 
@@ -2712,9 +3042,11 @@
     Weiter definiert man: $P\circ O=O\circ P=P$ für alle Punkte $P$.
 
     Es ergeben sich (mit einigem Rechnen) die folgenden Formeln. Diese werden dann wörtlich auch als Definition für eine Operation $\circ$ in einer elliptischen Kurve über $\mathbb{Z}_p$ benutzt.
-    - $O+O=O,$
-    - $O+ (x,y) = (x,y) +O$ für alle $(x,y)\in\mathbb{Z}^2_p$,
-    - $(x_1,y_1) + (x_2,y_2) =\begin{cases} O,\quad\text{ falls } x_1=x_2 \text{ und } y_1=-y_2 \\ (x_3,y_3),\quad\text{ sonst,}\end{cases}$
+    \begin{itemize*}
+        \item $O+O=O,$
+        \item $O+ (x,y) = (x,y) +O$ für alle $(x,y)\in\mathbb{Z}^2_p$,
+        \item $(x_1,y_1) + (x_2,y_2) =\begin{cases} O,\quad\text{ falls } x_1=x_2 \text{ und } y_1=-y_2 \\ (x_3,y_3),\quad\text{ sonst,}\end{cases}$
+    \end{itemize*}
 
     wobei $(x_3,y_3)$ folgendermaßen berechnet wird: $x_3=\lambda^2-x_1-x_2$, $y_3=\lambda(x_1-x_3)-y_1$ mit $\lambda=\begin{cases} (y_2-y_1)/(x_2-x_1),\quad\text{ falls } (x_1,y_1)\not= (x_2,y_2)\\ (3x^2_1+A)/(2y_1),\quad\text{ falls } (x_1,y_1) = (x_2,y_2)\end{cases}$.
     Der erste Fall bezieht sich auf den dritten Schnittpunkt einer Geraden durch zwei Punkte; der zweite auf den Tangentenfall.
@@ -2748,29 +3080,37 @@
     *Klartextmenge*: $X=\mathbb{Z}^*_p$.
 
     *Chiffretextmenge*: $Y=(\mathbb{Z}_p \times\{0,1\})\times\mathbb{Z}^*_p$.
-    - (Paare aus: (komprimiertes) Element von $G$ und Element von $\mathbb{Z}^*_p$).
-    - Öffentliche Schlüssel: $K_{pub}=G$. Private Schlüssel: $K_{priv}=\mathbb{Z}_N$.
-    - Schlüsselmenge: $K=\{(Q,b)|Q\in K_{pub} = G,b\in K_{priv} =\mathbb{Z}_N,Q=bP\}$.
+    \begin{itemize*}
+        \item (Paare aus: (komprimiertes) Element von $G$ und Element von $\mathbb{Z}^*_p$).
+        \item Öffentliche Schlüssel: $K_{pub}=G$. Private Schlüssel: $K_{priv}=\mathbb{Z}_N$.
+        \item Schlüsselmenge: $K=\{(Q,b)|Q\in K_{pub} = G,b\in K_{priv} =\mathbb{Z}_N,Q=bP\}$.
+    \end{itemize*}
 
     *Schlüsselerzeugung*: (Gegeben sind $E$ [also $p,A,B$], $P$, $N$.)
-    - Wähle $b\in\mathbb{Z}_N$ zufällig und berechne $Q=bP$ (schnelle Exponentiation).
-    - Schlüssel:$(Q,b)$.
-    - Der öffentliche Schlüssel ist $k_{pub}=Q$.
-    - Der private Schlüssel ist $k_{priv}=b$.
+    \begin{itemize*}
+        \item Wähle $b\in\mathbb{Z}_N$ zufällig und berechne $Q=bP$ (schnelle Exponentiation).
+        \item Schlüssel:$(Q,b)$.
+        \item Der öffentliche Schlüssel ist $k_{pub}=Q$.
+        \item Der private Schlüssel ist $k_{priv}=b$.
+    \end{itemize*}
 
     *Verschlüsselungsfunktion* $E:X\times G\rightarrow Y$, als randomisierter Algorithmus.
-    - Gegeben: Klartext $x\in\mathbb{Z}^*_p$.
-    - Öffentlicher Schlüssel $Q\in G$.
-    - Wähle zufällig $a\in\mathbb{Z}_N$ und berechne $(k,y) =aQ$ mit $k\in\mathbb{Z}^*_p$. //(Falls $k=0$, wähle neues $a$.)
-    - Berechne $E^a (x,Q)\leftarrow (Point-Compress(aP),x*k\ mod\ p) =: (y',y'');$ das Paar $(y',y'')\in(\mathbb{Z}_p\times\{0,1\})\times\mathbb{Z}^*_p$ ist der Chiffretext.
+    \begin{itemize*}
+        \item Gegeben: Klartext $x\in\mathbb{Z}^*_p$.
+        \item Öffentlicher Schlüssel $Q\in G$.
+        \item Wähle zufällig $a\in\mathbb{Z}_N$ und berechne $(k,y) =aQ$ mit $k\in\mathbb{Z}^*_p$. //(Falls $k=0$, wähle neues $a$.)
+        \item Berechne $E^a (x,Q)\leftarrow (Point-Compress(aP),x*k\ mod\ p) =: (y',y'');$ das Paar $(y',y'')\in(\mathbb{Z}_p\times\{0,1\})\times\mathbb{Z}^*_p$ ist der Chiffretext.
+    \end{itemize*}
 
     Bemerkung: $k\in\mathbb{Z}^*_p$, die erste Komponente eines Punktes in $G$, wird durch eine Operation in $G$ erstellt, und dann wie beim One-Time-Pad (oder beim Vernam-System) benutzt, wobei diese Verschlüsselung durch Multiplikation in $\mathbb{Z}^*_p$ ausgeführt wird.
 
     *Entschlüsselungsfunktion* $D:Y\times\mathbb{Z}_N \rightarrow X$, als (deterministischer) Algorithmus.
-    - Gegeben: Chiffretext $y=(y',y'')$ mit $y'\in\mathbb{Z}_p\times\{0,1\}$ und $y''\in\mathbb{Z}^*_p$. Privater Schlüssel $b$.
-    - Berechne $(x_1,y_1)\leftarrow Point-Decompress (y')$ //nun gilt $(x_1,y_1) =aP$
-    - $(x_0,y_0)\leftarrow b(x_1,y_1)$ (in $G$, nun gilt $(x_0,y_0) = (ba)P=a(bP) =aQ= (k,y))$, und schließlich in $\mathbb{Z}^*_p$
-    - $D((y',y''),b)\leftarrow y''*(x_0)^{-1}\ mod\ p$.
+    \begin{itemize*}
+        \item Gegeben: Chiffretext $y=(y',y'')$ mit $y'\in\mathbb{Z}_p\times\{0,1\}$ und $y''\in\mathbb{Z}^*_p$. Privater Schlüssel $b$.
+        \item Berechne $(x_1,y_1)\leftarrow Point-Decompress (y')$ //nun gilt $(x_1,y_1) =aP$
+        \item $(x_0,y_0)\leftarrow b(x_1,y_1)$ (in $G$, nun gilt $(x_0,y_0) = (ba)P=a(bP) =aQ= (k,y))$, und schließlich in $\mathbb{Z}^*_p$
+        \item $D((y',y''),b)\leftarrow y''*(x_0)^{-1}\ mod\ p$.
+    \end{itemize*}
 
     Behauptung: Wenn $Q=bP$, dann gilt $D(E^a (x,Q),b) =x$, für jedes $x\in X$ und jedes $a$, für das $(k,y) =aQ\in\mathbb{Z}^*_p \times\mathbb{Z}_N$. (Dies gilt nach den Anmerkungen in der Beschreibung von $E$ und $D$.)