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\deflang1031\pard\ri4\plain\f3\fs20\cf0\b _________________________________________________________________________________
\par
\par \plain\f3\fs20\cf0 Inhalt....: Einf\'fchrung in das RSA-Krypto-Verfahren
\par Kategorie.: Unterrichtsmaterial
\par Mathematik: Kryptographie, Zahlentheorie
\par MuPAD.....: 3.0.0
\par Datum.....: 2002-06-10
\par Autoren...: Annamaria Pelles \plain\lang2057\f3\fs20\cf0 <\plain\f3\fs20\cf0 pellesa\plain\lang2057\f3\fs20\cf0 @freemail.hu>
\par Autoren...: Kai Gehrs
\par Funktionen: random, isprime, float, igcd, igcdex, mod, powermod, factor,
\par Funktionen: nextprime, time\plain\f3\fs20\cf0
\par \plain\f3\fs20\cf0\b _________________________________________________________________________________
\par \plain\f4\fs20\cf0\b
\par \plain\f4\fs36\cf0\b Einf\'fchrung in das RSA-Krypto-Verfahren
\par \plain\f4\fs28\cf1\b
\par \plain\f4\fs24\cf1 RSA ist derzeit quasi der weltweite Standard f\'fcr public-key-Verfahren.
\par Wir werden sehen wie RSA f\'fcr die zwei wesentlichen Aufgaben der Kryptographie benutzt werden
\par kann. Hierbei handelt es sich um die
\par
\par \tab \tab \tab \tab \tab 1. Verschl\'fcsselung von Nachrichten
\par \tab \tab \tab \tab \tab 2. Entschl\'fcsselung von Nachrichten
\par
\par Dieses Notebook pr\'e4sentiert stellenweise mathematische Sachverhalte, die \'fcber den Rahmen
\par dessen, was man in der Schule behandelt hinausgehen. Diese komplizierteren Sachverhalte
\par kann man jedoch getrost \'fcberlesen, wenn man sich nur daf\'fcr interessiert, wie mit Hilfe des
\par RSA-Verfahrens Botschaften ver- oder entschl\'fcsselt werden. Dar\'fcber hinaus kann dieses
\par Notebook eventuell als Anreiz f\'fcr eine eventuelle Facharbeit verwendet werden.
\par \plain\f5\fs28\cf0
\par \plain\f4\fs28\cf2\b
\par Wozu dient das RSA-Krypto-Verfahren?
\par \plain\f4\fs28\cf0
\par Wir erkl\'e4ren das Verfahren an einem Beispiel: B\plain\f4\fs28\cf0\i o\plain\f4\fs28\cf0 b m\'f6chte an Alice eine
\par Nachricht M senden, ohne dass eine dritte Person, wir nennen sie Eve, diese
\par Nachricht lesen kann, obwohl Eve \tab unbeschr\'e4nkten Zugang zum Kommunikations-
\par kanal zwischen Alice und Bob hat.
\par
\par Die L\'f6sung besteht darin die Nachricht M zu verschl\'fcsseln. Das Verfahren kann
\par man sich grafisch wie folgt vorstellen:
\par \tab
\par
\par \plain\f3\fs24\cf3\b Kanal
\par Bob ----\plain\lang2057\f3\fs24\cf3\b ->\plain\f3\fs24\cf3\b Verschl\'fcsselung -------------\plain\lang2057\f3\fs24\cf3\b -> \plain\f3\fs24\cf3\b Entschl\'fcsselung ----\plain\lang2057\f3\fs24\cf3\b ->\plain\f3\fs24\cf3\b Alice
\par \plain\lang1038\f6\fs24\cf3\b \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab Eve
\par \plain\f4\fs28\cf2\b
\par
\par Wie funktioniert dieses Verfahren im Detail?
\par \plain\f4\fs28\cf0\i
\par \plain\f4\fs28\cf0 Bob m\'f6chte also an Alice eine Nachricht senden, so dass nur Alice diese
\par Nachricht lesen kann, und niemand anderes.
\par
\par Aus diesem Grund erzeugt Alice einen sogenannten \plain\f4\fs28\cf0\b private key\plain\f4\fs28\cf0 \plain\f4\fs28\cf0\b S\plain\f4\fs28\cf0 (privater
\par Schl\'fcssel) und einen \plain\f4\fs28\cf0\b public key K \plain\f4\fs28\cf0 (\'f6ffentlicher Schl\'fcssel). Jeder kann \plain\f4\fs28\cf0\b K\plain\f4\fs28\cf0\i \plain\f4\fs28\cf0 lesen,
\par aber \plain\f4\fs28\cf0\b S\plain\f4\fs28\cf0\i \plain\f4\fs28\cf0 ist ein Geheimnis.
\par
\par Auch Bob kennt nur \plain\f4\fs28\cf0\b K\plain\f4\fs28\cf0 . Wenn nun Bob eine Nachricht an Alice schickt, mu\'df er zur
\par Verschl\'fcsselung \plain\f4\fs28\cf0\b K \plain\f4\fs28\cf0 verwenden: Bei Empfang der Nachricht kann Alice mit ihrem
\par \plain\f4\fs28\cf0\b privaten Schl\'fcssel \plain\f4\fs28\cf0 die Nachricht entschl\'fcsseln.
\par
\par Die Nachricht kann aus Texten, Bildern, Dateien oder Programfiles etc. bestehen;
\par die Nachricht soll aber immer in der oben genannten Form verschl\'fcsselt werden.
\par
\par Im folgenden werden wir Schritt f\'fcr Schritt das Verfahren der \plain\f4\fs28\cf0\b Verschl\'fcsselung\plain\f4\fs28\cf0
\par (damit niemand anderer die Nachricht lesen kann) und der \plain\f4\fs28\cf0\b Entschl\'fcsselung\plain\f4\fs28\cf0
\par an einem Beispiel verstehen lernen.
\par \plain\f4\fs28\cf0\ul
\par \plain\f4\fs28\cf6 Schritt (1): \plain\f4\fs28\cf0 Alice w\'e4hlt zuf\'e4llig zwei Primzahlen p und q.
\par \plain\f5\fs28\cf4
\par \plain\f5\fs28\cf0 Zur Erinnerung: Eine Primzahl ist eine positive ganze Zahl, die nur durch 1 und sich
\par selbst (ganzzahlig) teilbar ist.
\par
\par Zum Beispiel ist 25 keine Primzahl, denn 5 * 5 ist 25 und damit 5 ein Teiler von 25.
\par Dagegen ist 7 eine Primzahl, denn 2,3,4,5 und 6 sind keine Teiler von 7.
\par
\par Mit MuPAD k\'f6nnen wir Zufallszahlen mit Hilfe der Funktion \plain\f5\fs28\cf4 random() \plain\f5\fs28\cf0 erzeugen.
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}Zufall:= random(1..500)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw4577\pich901\picwgoal2594\pichgoal510
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070000005346204D6174682045787400E8EC1200B8A4F177C1A4F1772030F3773A0A664F040000
002D010400040000002D0103000500000009020000FF00070000002105010070015E0165000700
00002105010072015E01F10007000000210501006F015E014E01070000002105010063015E01DA
01070000002105010020015E015602070000002105010072015E019C0207000000210501006101
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0607000000210501002E015E01130707000000210501002E015E01590707000000210501002E01
5E019F07070000002105010020015E01E507070000002105010065015E012B0807000000210501
006E015E01A708070000002105010064015E01330908000000FA02000000000000000000000400
00002D01050004000000F001000007000000FC020000FFFFFF000000040000002D010000040000
00F00101001C000000FB021000070000000000BC02000000000102022253797374656D0000DC0A
0A78E8EC1200B8A4F177C1A4F1772030F3773A0A664F040000002D010100040000002701FFFF04
000000F001020004000000F001030004000000F0010400040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par Eine Zufallszahl wird nun immer aus der Menge \{1,2,3,...,499,500\} gew\'e4hlt:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}p:= Zufall()
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
0100090000031E01000005001C0000000000050000000B0200000000050000000C0285034F0303
0000001E00030000001E00050000000C02FF01E001050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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020005000000020101000000050000000102FFFFFF00050000002E011800000005000000090200
00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
73204E657720526F6D616E00B8A4F177C1A4F1772030F377A60A6628040000002D0103000B0000
0026060F000C004D61746854797065000080000500000009020000FF0007000000210501003801
5E016400070000002105010032015E01F00008000000FA0200000000000000000000040000002D
01040004000000F001000007000000FC020000FFFFFF000000040000002D01000004000000F001
01001C000000FB021000070000000000BC02000000000102022253797374656D0000BD090A16E8
EC1200B8A4F177C1A4F1772030F377A60A6628040000002D010100040000002701FFFF04000000
F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par Wie k\'f6nnen wir nun schnell herausfinden, ob die erzeugte Zufallszahl eine Primzahl
\par ist?
\par
\par Die MuPAD Funktion \plain\f5\fs28\cf4 isprime\plain\f5\fs28\cf0 testet, ob p eine Primzahl ist oder nicht, d.h.
\par \plain\f5\fs28\cf4 isprime(p)\plain\f5\fs28\cf0 liefert \plain\f5\fs28\cf5 true\plain\f5\fs28\cf0 , wenn p eine Primzahl ist, und \plain\f5\fs28\cf5 false\plain\f5\fs28\cf0 , wenn p eine
\par zusammengesetzte Zahl ist.
\par
\par Wenn p keine Primzahl ist, dann m\'fcssen wir das obige Verfahren nochmals
\par wiederholen, bis wir mit p eine Primzahl gefunden haben.
\par \plain\f3\fs28\cf5
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}isprime(p)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1863\pich901\picwgoal1056\pichgoal510
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0000001E00030000001E00050000000C02FF012004050000000B020000000014000000FF060000
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0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C000000A60A0A29E8EC1200B8A4F177C1A4F1772030F377100966B1040000002D01
020005000000020101000000050000000102FFFFFF00050000002E011800000005000000090200
00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
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5E016400070000002105010041015E01000107000000210501004C015E01CA0107000000210501
0053015E017502070000002105010045015E01110308000000FA02000000000000000000000400
00002D01040004000000F001000007000000FC020000FFFFFF000000040000002D010000040000
00F00101001C000000FB021000070000000000BC02000000000102022253797374656D0000BD09
0A17E8EC1200B8A4F177C1A4F1772030F377100966B1040000002D010100040000002701FFFF04
000000F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par Eine M\'f6glichkeit besteht nun darin, die obige Befehlszeile \plain\f3\fs28\cf4 p:= Zufall()
\par \plain\f5\fs28\cf0 so lange auszuf\'fchren, bis p eine Primzahl eine Primzahl ist.
\par
\par Wir verfahren wie folgt:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}p:= Zufall():
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4 while isprime(p) = FALSE do
\par p:= Zufall();
\par end_while:
\par p;
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1094\pich901\picwgoal620\pichgoal510
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0000001E00030000001E00050000000C02FF016C02050000000B020000000014000000FF060000
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0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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00
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0 Dieses Verfahren scheint nicht sehr elegant zu sein, aber es f\'fchrt nach kurzer
\par Zeit zum Ziel, denn der sogenannte \plain\f5\fs28\cf0\b Primzahlsatz\plain\f5\fs28\cf0 der Zahlentheorie besagt,
\par dass in einem vorgegebenen Intervall stets hinreichend "viele" Primzahlen
\par vorhanden sind.
\par
\par
\par \tab \plain\f5\fs28\cf0\b Primzahlsatz.\plain\f5\fs28\cf0 \tab Ist x eine positive Zahl und bezeichnet \plain\f7\fs28\cf0 p\plain\f5\fs28\cf0 (x) die Anzahl der
\par \tab Primzahlen, die kleiner oder gleich x sind, so gilt:
\par \tab
\par \tab \tab {\pict\wmetafile8\picw2830\pich1155\picscalex99\picscaley99\picwgoal1617\pichgoal658
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}\plain\f5\fs28\cf0
\par
\par
\par Was bedeutet dies nun in unserem Beispiel? In unserer Situation gilt x = 1000.
\par Also gilt:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}hold(PI(x)) = float(1000/ln(1000))
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw4406\pich901\picwgoal2497\pichgoal510
0100090000033C02000008001C0000000000050000000B0200000000050000000C028503361103
0000001E00030000001E00050000000C02FF01C209050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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F00101001C000000FB021000070000000000BC02000000000102022253797374656D00003A0A0A
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par Es gibt also n\'e4herungsweise 144 Primzahlen in der Menge \{1,2,3,...,999,1000\},
\par d.h. wiederum, dass die Wahrscheinlichkeit, dass eine gew\'e4hlte Zufallszahl eine
\par Primzahl ist, betr\'e4gt n\'e4herungsweise
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}float(144/1000)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1464\pich901\picwgoal829\pichgoal510
0100090000033301000005001C0000000000050000000B0200000000050000000C028503B80503
0000001E00030000001E00050000000C02FF013E03050000000B020000000014000000FF060000
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0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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0034015E01C201070000002105010034015E014E0208000000FA02000000000000000000000400
00002D01040004000000F001000007000000FC020000FFFFFF000000040000002D010000040000
00F00101001C000000FB021000070000000000BC02000000000102022253797374656D0000DC0A
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par d.h. 14,4%. Dies bedeutet, dass wir innerhalb von 10 Versuchen in der Regel
\par mindestens eine Primzahl finden werden.
\par
\par Nun zur\'fcck zu Bob und Alice: Jetzt muss Alice nach der gleichen Methode die
\par zweite Primzahl q w\'e4hlen.\plain\f8\fs28\cf0
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}q:= Zufall():
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4 while isprime(q) = FALSE do
\par q:= Zufall();
\par end_while:
\par q;
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1094\pich901\picwgoal620\pichgoal510
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00
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0 Schon haben wir p und q gefunden, n\'e4mlich
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}p,q
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw2131\pich901\picwgoal1208\pichgoal510
0100090000034101000005001C0000000000050000000B0200000000050000000C028503530803
0000001E00030000001E00050000000C02FF01B804050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf6
\par Schritt (2):\plain\f5\fs28\cf0 Alice muss N und Phi(N) berechnen, wobei N = p*q und Phi(N) wie
\par folgt erkl\'e4rt ist:
\par
\par \tab F\'fcr eine nat\'fcrliche Zahl k ist \plain\f5\fs28\cf0\b Phi(k)\plain\f5\fs28\cf0 definiert als die Anzahl der zu k
\par \tab teilerfremden Zahlen (dazu geh\'f6rt auch die 1), die kleiner sind als k.
\par
\par Zum Beispiel ist Phi(6) = 2, denn nur 1 und 5 sind kleiner als 6 und teilerfremd
\par zu 6. Andererseits ist Phi(7) = 6, denn alle Zahlen 1,2,3,4,5 und 6 sind teilerfremd
\par zu 7.
\par
\par Allgemein gelten die folgenden Rechenregeln f\'fcr die Eulersche Phi-Funktion.
\par
\par \plain\f5\fs28\cf0\b Rechenregeln f\'fcr die Eulersche Phi Funktion: \tab
\par \plain\f5\fs28\cf0 \tab \tab \tab \tab \tab
\par \tab \tab Phi(1) = 1
\par \tab \tab Phi(p) = p-1, wenn p eine Primzahl ist.
\par \plain\lang1038\f4\fs28 \tab \tab Phi(p\plain\f4\fs28 *q) = (p - 1) * (q - 1), wenn p und q Primzahlen sind.\plain\lang1038\f4\fs28
\par \plain\lang1038\f9\fs28\i\up12 \plain\f8\fs28\cf0\ul
\par \plain\lang1038\f4\fs28 Wir werden im folgenden nur die dritte Regel benutzen. Wer die Eulersche
\par Phi-Funktion nicht kennt, kann sich daher f\'fcr alles folgende merken:
\par Phi(N) ist stets das Produkt (p - 1) * (q - 1). \plain\f8\fs28\cf0
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}N:=p*q; Phi_N:= (p-1)*(q-1)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par
\par \pard\ri4\plain\f5\fs28\cf6
\par Schritt(3):\plain\f5\fs28\cf0 \plain\f4\fs28\cf0 Alice berechnet ihren \'f6ffentlichen Exponenten e auf folgende
\par Weise: \plain\f5\fs28\cf0 Sie w\'e4hlt e zuf\'e4llig und teilerfremd zu Phi(N).
\par \plain\f8\fs28\cf0
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}Zufall:= random(2..Phi_N-2)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw4577\pich901\picwgoal2594\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}e:= Zufall()
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
0100090000033301000005001C0000000000050000000B0200000000050000000C028503340603
0000001E00030000001E00050000000C02FF018403050000000B020000000014000000FF060000
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par In MuPAD berechnet \plain\f5\fs28\cf4 igcd \plain\f5\fs28\cf0 (integer greatest common divisor) den gr\'f6\'dften
\par gemeinsamen Teiler zweier Zahlen.
\par
\par Um zu pr\'fcfen, ob e und Phi(N) tats\'e4chlich teilerfremd sind, berechnen wir
\par einfach ihren gr\'f6\'dften gemeinsamen Teiler. Ist dieser 1, so haben wir
\par ein passendes e gefunden. Ist dieser gr\'f6\'dfer oder gleich 2, so w\'e4hlen wir
\par ein neues e. Auch dieses Verfahren liefert beweisbar innerhalb kurzer Zeit
\par einen passenden Wert f\'fcr e.
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}igcd(e,Phi_N)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw600\pich901\picwgoal340\pichgoal510
0100090000031701000005001C0000000000050000000B0200000000050000000C028503580203
0000001E00030000001E00050000000C02FF015401050000000B020000000014000000FF060000
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf6
\par Schritt(4):\plain\f5\fs28\cf0 Alice berechnet ihren privaten Exponenten d als Inverses von e
\par modulo Phi(N).
\par
\par Das \plain\f5\fs28\cf0\b modulare Inverse\plain\f5\fs28\cf0 berechnet man mit Hilfe des Euklidischen Algorithmus.
\par
\par F\'fcr zwei Zahlen a und m liefert der Euklidische Algorithmus nicht nur den
\par gr\'f6\'dften gemeinsamen Teiler g, sondern auch Zahlen k und l, so dass gilt:
\par
\par g = k * a + l * m
\par
\par Reduziert man diese Gleichung modulo m (d.h. man rechnet auf beiden
\par Seiten modulo m - man nimmt also einfach den Rest, der bei ganzzahliger
\par Division der Zahlen durch m verbleibt), so ergibt sich wegen m modulo m = 0:
\par
\par \tab g modulo m = (k * a) modulo m
\par
\par Ist nun g = 1, also der gr\'f6\'dfte gemeinsame Teiler von a und m ist 1, so folgt:
\par
\par \tab 1 modulo m = (k * a) modulo m
\par
\par d.h. k ist das modulare Inverse zu a (denn k * a ist 1 modulo m).
\par
\par In unserer Situation setzen wir f\'fcr m die Zahl Phi(N) ein und f\'fcr a die Zahl e.
\par Die Zahl k wird dann Alices privater Exponent, d.h. k = d.
\par \plain\f8\fs28\cf0
\par \plain\f5\fs28\cf0 Mit der MuPAD Funktion \plain\f5\fs28\cf4 igcdex\plain\f5\fs28\cf0 kann man die Werte g, k und l des Euklidischen
\par Algorithmus berechnen. \plain\f8\fs28\cf0
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}igcdex(e,Phi_N)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw3859\pich901\picwgoal2187\pichgoal510
0100090000039B01000006001C0000000000050000000B0200000000050000000C028503130F03
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000000
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}dd:= op(igcdex(e,Phi_N),2)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw2032\pich901\picwgoal1152\pichgoal510
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000000FC020000FFFFFF000000040000002D01000004000000F00101001C000000FB0210000700
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Da Alices privater Schl\'fcssel d eine positive Zahl sein muss, m\'fcssen wir zu dem
\par Wert dd einmal Phi(N) addieren, falls dd negativ ist. Andernfalls k\'f6nnen wir
\par dd als Wert f\'fcr d w\'e4hlen. Dies h\'f6rt sich jetzt unglaublich kompliziert an, aber
\par wir m\'fcssen alle F\'e4lle ber\'fccksichtigen, denn wir haben e ja als zuf\'e4lligen Wert
\par gew\'e4hlt.
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}if dd < 0 then
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4 d:= dd + Phi_N;
\par else
\par d:= dd;
\par end_if:
\par print(Unquoted, "d = ".d);
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect d = 60227
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par d.h. es gilt
\par 1 = (d * e) modulo Phi(N)
\par
\par Wir k\'f6nnen das Ergebnis \'fcberpr\'fcfen:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}(d * e) mod Phi_N
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw600\pich901\picwgoal340\pichgoal510
0100090000031701000005001C0000000000050000000B0200000000050000000C028503580203
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par \plain\f4\fs28\cf6 Schritt(5): \plain\f4\fs28\cf0 Alice ver\'f6ffentlicht das Paar (N,e) (\'f6ffentlicher Schl\'fcssel) und beh\'e4lt
\par (N,d) als Geheimnis (geheimer Schl\'fcssel). Schauen wir uns die bisher
\par berechneten Werte einmal an:
\par \plain\f8\fs28\cf0
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}N,e
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw3119\pich901\picwgoal1768\pichgoal510
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}N,d
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw3119\pich901\picwgoal1768\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf6
\par Schritt (6):\plain\f5\fs28\cf0 Alice l\'f6scht p, q und Phi(N) vom Computer - sie ben\'f6tigt diese Zahlen
\par nicht mehr. Sie sollte sie auch l\'f6schen, damit diese Eve nicht auf ihrem Computer
\par finden kann.
\par
\par Jetzt m\'f6chte Bob an Alice eine Nachricht senden; was mu\'df er jetzt tun?
\par
\par \plain\f5\fs28\cf6 Schritt(7):\plain\f5\fs28\cf0 Bob m\'f6chte Alice zum Beispiel eine Zahl x senden, x soll zwischen
\par 1 und N sein, also z.B. x = 13
\par \plain\f5\fs28\cf0\ul
\par \plain\f5\fs28\cf0 Bob soll x zu einem y verschl\'fcsseln. y ist x^e mod N. Bob berechnet also die
\par Zahl y und sendet sie an Alice. Auf diese Weise hat er x zu y verschl\'fcsselt.
\par Die folgenden Rechnungen ben\'f6tigen durchaus etwas mehr Zeit. Man
\par m\'f6ge sich gedulden. Wir werden uns sp\'e4ter damit besch\'e4ftigen, warum
\par diese Rechnung so zeitintensiv ist und wie wir sie effizienter durchf\'fchren
\par k\'f6nnen.
\par \plain\f8\fs28\cf0\ul
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}x:= 13
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}y:=x^e mod N
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par \plain\f4\fs28\cf6 Schritt(8): \plain\f4\fs28\cf0 Alice berechnet den Wert xx mit Hilfe ihres eigenen geheimen
\par Exponenten d. Dazu bestimmt sie y^d modulo N (dass dieses Verfahren
\par bei korrekter Anwendung immer funktioniert und wieder den urspr\'fcnglichen
\par Wert von x liefert, kann man mit Mitteln aus der Zahlentheorie beweisen,
\par die allerdings den Rahmen dieses Notebooks sprengen).
\par \plain\f8\fs28\cf0
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}xx:= y^d mod N
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
0100090000031E01000005001C0000000000050000000B0200000000050000000C0285034F0303
0000001E00030000001E00050000000C02FF01E001050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C000000DC0A0A8DE8EC1200B8A4F177C1A4F1772030F377BD09662F040000002D01
020005000000020101000000050000000102FFFFFF00050000002E011800000005000000090200
00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
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0026060F000C004D61746854797065000080000500000009020000FF0007000000210501003100
5E016400070000002105010033005E01F00008000000FA0200000000000000000000040000002D
01040004000000F001000007000000FC020000FFFFFF000000040000002D01000004000000F001
01001C000000FB021000070000000000BC02000000000102022253797374656D0000F40A0A38E8
EC1200B8A4F177C1A4F1772030F377BD09662F040000002D010100040000002701FFFF04000000
F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Probe: Ist x gleich xx?\plain\f10\fs28\cf0
\par \plain\f8\fs28\cf0
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}x = xx
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1909\pich901\picwgoal1082\pichgoal510
0100090000036B01000006001C0000000000050000000B0200000000050000000C028503750703
0000001E00030000001E00050000000C02FF013A04050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C000000BD090A30E8EC1200B8A4F177C1A4F1772030F377100966CB040000002D01
020005000000020101000000050000000102FFFFFF00050000002E011800000005000000090200
00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
73204E657720526F6D616E00B8A4F177C1A4F1772030F377100966CB040000002D0103000B0000
0026060F000C004D61746854797065000080001C000000FB02E8FE000000000000900100000002
0700000053796D626F6C0000F40A0A39E8EC1200B8A4F177C1A4F1772030F377100966CB040000
002D010400040000002D010300040000002D0104000500000009020000FF00040000002D010300
070000002105010031005E016400070000002105010033005E01F000040000002D010400070000
00210501003D005E01D001040000002D010300070000002105010031005E01BE02070000002105
010033005E014A0308000000FA0200000000000000000000040000002D01050004000000F00100
0007000000FC020000FFFFFF000000040000002D01000004000000F00101001C000000FB021000
070000000000BC02000000000102022253797374656D0000A50A0A21E8EC1200B8A4F177C1A4F1
772030F377100966CB040000002D010100040000002701FFFF04000000F001020004000000F001
030004000000F0010400040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f8\fs28\cf0
\par \plain\f5\fs28\cf0 Das Verfahren hat also funktioniert: Bob konnte seine Nachricht an Alice
\par schicken, Alice konnte mit Hilfe ihres eigenen Geheimschl\'fcssels die
\par Nachricht entschl\'fcsseln.
\par
\par Folgende Fragen sind nun offensichtlich:
\par
\par \tab (1) Warum sollte Eve das Geheimnis nicht entsch\'fcsseln k\'f6nnen,
\par \tab wo sie doch N kennt?
\par
\par \tab (2) Warum ben\'f6tigen die letzten Rechnungen so viel Zeit?
\par
\par \tab (3) Kann man das Suchen von Primzahlen mit MuPAD nicht
\par \tab noch geschickter machen?
\par
\par \plain\f5\fs28\cf0\b Zu Frage (1):\plain\f5\fs28\cf0 Eve k\'f6nnte bei Kenntnis von N versuchen, die Zahl in ihre Faktoren
\par p und q zerlegen. Mit Hilfe von p und q k\'f6nnte Eve Phi(N) = (p-1) * (q-1) berechnen.
\par Da sie auch e kennt (e wurde von Alice ver\'f6ffentlicht) k\'f6nnte sie genau wie Alice
\par den Wert von d bestimmen (\'fcber den Euklidischen Algorithmus, wie wir es oben
\par f\'fcr Alice getan haben). Wenn Eve d kennt, so kann Eve die Nachricht von Bob
\par auch genauso leicht entschl\'fcsseln wie Alice.
\par
\par Wo k\'f6nnte es Probleme f\'fcr Eve geben? Eves gro\'dfes Problem ist es, N zu
\par faktorieren, d.h. die Zahlen p und q zu finden. In der Realit\'e4t wird man N als
\par Produkt zweier riesig gro\'dfer Primzahlen w\'e4hlen, so dass die Faktorisierung
\par von N viel zu lange dauert.
\par
\par Man kann zeigen, dass alle anderen Operationen auch bei sehr gro\'dfem N
\par immer noch schnell genug durchgef\'fchrt werden k\'f6nnen (z.B. der Euklidische
\par Algorithmus zur Berechnung des modularen Inversen funktioniert auch bei
\par gro\'dfen Zahlen sehr schnell).
\par \plain\f8\fs28\cf0
\par \plain\f3\fs28\cf3\b
\par \plain\f3\fs24\cf3\b leicht \tab Kanal leicht
\par Bob ------\plain\lang2057\f3\fs24\cf3\b ->\plain\f3\fs24\cf3\b Verschl\'fcsselung ------\plain\lang2057\f3\fs24\cf3\b -> \plain\f3\fs24\cf3\b Entschl\'fcsselung ----\plain\lang2057\f3\fs24\cf3\b --->\plain\f3\fs24\cf3\b Alice
\par \plain\lang1038\f6\fs24\cf3\b \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab \tab | extrem
\par \tab \tab \tab \tab \tab | schwer
\par \tab \tab \tab \tab \tab \tab |
\par \tab \tab \tab \tab \tab Eve
\par \plain\f8\fs28\cf0
\par
\par \plain\f5\fs28\cf0 Ein Beispiel:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}NN:= 112589990654654760673 * 112589990654654760791;
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw10478\pich901\picwgoal5940\pichgoal510
0100090000032F02000005001C0000000000050000000B0200000000050000000C028503EE2803
0000001E00030000001E00050000000C02FF013417050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C00000010090ACCE8EC1200B8A4F177C1A4F1772030F377F40A663B040000002D01
020005000000020101000000050000000102FFFFFF00050000002E011800000005000000090200
00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
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0026060F000C004D61746854797065000080000500000009020000FF0007000000210501003100
5E016400070000002105010032005E01F000070000002105010036005E017C0107000000210501
0037005E010802070000002105010036005E019402070000002105010035005E01200307000000
2105010030005E01AC03070000002105010035005E013804070000002105010039005E01C40407
0000002105010039005E015005070000002105010035005E01DC05070000002105010036005E01
6806070000002105010031005E01F406070000002105010035005E018007070000002105010032
005E010C08070000002105010034005E019808070000002105010036005E012409070000002105
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002105010031005E01540B070000002105010030005E01E00B070000002105010039005E016C0C
070000002105010034005E01F80C070000002105010030005E01840D070000002105010031005E
01100E070000002105010031005E019C0E070000002105010030005E01280F0700000021050100
39005E01B40F070000002105010030005E014010070000002105010037005E01CC100700000021
05010032005E015811070000002105010031005E01E411070000002105010036005E0170120700
00002105010039005E01FC12070000002105010031005E018813070000002105010037005E0114
14070000002105010032005E01A014070000002105010033005E012C1507000000210501003400
5E01B815070000002105010033005E01441608000000FA0200000000000000000000040000002D
01040004000000F001000007000000FC020000FFFFFF000000040000002D01000004000000F001
01001C000000FB021000070000000000BC02000000000102022253797374656D0000BD090A31E8
EC1200B8A4F177C1A4F1772030F377F40A663B040000002D010100040000002701FFFF04000000
F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0 Die Zahl N kann man in MuPAD mit dem Befehl \plain\f5\fs28\cf4 factor \plain\f5\fs28\cf0 in ihre Faktoren zerlegen:
\par Wir kennen das Ergebnis bereits, denn oben stehen die beiden Primfaktoren von
\par N. Wie lange ben\'f6tigt MuPAD nun ungef\'e4hr f\'fcr die Faktorisierung? Wir testen dies...
\par
\par (man kann die Faktorisierung per Mausklick auf das Abbruchsymbol - roter Kreis
\par mit wei\'dfem Kreuz in der obigen Befehlsleiste - vorzeitig abbrechen)
\par \plain\f3\fs32\cf4
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}time( factor(NN) )
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
0100090000033301000005001C0000000000050000000B0200000000050000000C028503340603
0000001E00030000001E00050000000C02FF018403050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C000000F40A0A3CE8EC1200B8A4F177C1A4F1772030F377DC0A6691040000002D01
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00000004000000080100001C000000FB02E8FE0000000000009001000000010700000054696D65
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5E016400070000002105010032005E01F000070000002105010039005E017C0107000000210501
0033005E010802070000002105010037005E01940208000000FA02000000000000000000000400
00002D01040004000000F001000007000000FC020000FFFFFF000000040000002D010000040000
00F00101001C000000FB021000070000000000BC02000000000102022253797374656D0000BD09
0A32E8EC1200B8A4F177C1A4F1772030F377DC0A6691040000002D010100040000002701FFFF04
000000F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}time(igcdex(NN, 112589990654654760673))
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw600\pich901\picwgoal340\pichgoal510
0100090000031701000005001C0000000000050000000B0200000000050000000C028503580203
0000001E00030000001E00050000000C02FF015401050000000B020000000014000000FF060000
0600F602000024000100020000800080FF7FFF7F02000080FF7F0080FF7F0200040000002D0100
0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
00417269616C000000DC0A0A92E8EC1200B8A4F177C1A4F1772030F377100966CF040000002D01
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5E01640008000000FA0200000000000000000000040000002D01040004000000F0010000070000
00FC020000FFFFFF000000040000002D01000004000000F00101001C000000FB02100007000000
0000BC02000000000102022253797374656D0000BD090A33E8EC1200B8A4F177C1A4F1772030F3
77100966CF040000002D010100040000002701FFFF04000000F001020004000000F00103000400
00002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f5\fs28\cf0
\par Wie man sieht: Die Berechnung des Euklidischen Algorithmus geht schnell,
\par die Faktorisierung dagegen nicht.
\par
\par \plain\f5\fs28\cf0\b Fazit:\plain\f5\fs28\cf0 Wenn Eve p und q nicht kennt, so hat sie f\'fcr gro\'dfe N keine Chance, die
\par Werte von p und q in angemessener Zeit zu berechnen. Dies zeigt die Sicherheit
\par des Verfahrens f\'fcr hinreichend gro\'dfe Werte von p, q und N.
\par \plain\f8\fs28\cf0
\par
\par \plain\f5\fs28\cf0\b Zu Frage (2): \plain\f5\fs28\cf0 Das Potenzieren x^d ist nicht kosteng\'fcnstig und ungeschickt,
\par weil erst potenziert und danach erst modulo N gerechnet wird.
\par
\par Besser ist es, nach jeder berechneten Potenz zun\'e4chst wieder modulo N
\par zu reduzieren, da dann die Zahlen viel kleiner sind. Wie wir unten sehen,
\par dauern die Berechnungen von oben, die mehrere Sekunden in Anspruch
\par genommen hatten, nur noch Sekundenbruchteile:
\par \plain\f3\fs22\cf5
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}xx:= powermod(y, d, N)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
0100090000031E01000005001C0000000000050000000B0200000000050000000C0285034F0303
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0004000000F001000008000000FA0200000000000000000000040000002D01000007000000FC02
0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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0026060F000C004D61746854797065000080000500000009020000FF0007000000210501003100
5E016400070000002105010033005E01F00008000000FA0200000000000000000000040000002D
01040004000000F001000007000000FC020000FFFFFF000000040000002D01000004000000F001
01001C000000FB021000070000000000BC02000000000102022253797374656D0000BD090A34E8
EC1200B8A4F177C1A4F1772030F377F40A663F040000002D010100040000002701FFFF04000000
F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Dabei berechnet \plain\f4\fs28\cf4 powermod(y, d, N)\plain\f4\fs28\cf0 nichts anderes als y^d modulo N.
\par Diesmal allerdings auf die etwas geschicktere, oben angedeutete Weise.
\par \plain\f3\fs28\cf0\b
\par \plain\f4\fs28\cf0 Auch das modulare Inverse l\'e4\'dft sich leichter berechnen. Wir wissen,
\par dass d das modulare Inverse zu e modulo Phi(N) ist. Mit MuPAD k\'f6nnen
\par wir dies auch unter Verzicht auf die Funktion igcd
\par direkt berechnen:
\par \plain\f3\fs28\cf0\b
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}d:= e^-1 mod Phi_N
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
0100090000033301000005001C0000000000050000000B0200000000050000000C028503340603
0000001E00030000001E00050000000C02FF018403050000000B020000000014000000FF060000
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0000000000000000040000002D0101001C000000FB0210FF000000000000900100000001070000
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000000F001020004000000F0010300040000002701FFFF030000000000
}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Die Probe zeigt auch dieses Mail wie oben, dass das Ergebnis korrekt ist:
\par \plain\f3\fs28\cf0\b
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}d * e mod Phi_N
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw600\pich901\picwgoal340\pichgoal510
0100090000031701000005001C0000000000050000000B0200000000050000000C028503580203
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f3\fs28\cf0\b
\par \plain\f4\fs28\cf0\b Zu Frage (3):\plain\f4\fs28\cf0 Das Finden von Primzahlen kann mit Hilfe der Funktion
\par \plain\f4\fs28\cf4 nextprime\plain\f4\fs28\cf0 verbessert werden. Diese Funktion liefert zu einer gegebenen
\par Zahl die n\'e4chst gr\'f6\'dfere Primzahl. Damit ist das Auffinden einer
\par zuf\'e4lligen Primzahl aus der Menge \{1,2,...,999,1000\} ganz einfach:
\par
\par Zuerst w\'e4hlen wir eine Zufallszahl a
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}a:= Zufall()
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Dann setzen wir p auf den Wert
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}pp:= nextprime(a)
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1588\pich901\picwgoal900\pichgoal510
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect
\par \pard\ri4\plain\f4\fs28\cf0
\par Dann ist pp sicherlich eine Primzahl und wir sind fertig.
\par
\par In der Praxis, wo man mit riesig gro\'dfen Zahlen rechnet, wird man aber
\par dennoch nicht \plain\f4\fs28\cf4 nextprime\plain\f4\fs28\cf0 benutzen, sondern eher, wie oben, Zufalls-
\par zahlen erzeugen und testen, ob diese Primzahlen sind. Dieses Verfahren
\par ist, so unglaublich es klingen mag, schneller.
\par \plain\f3\fs28\cf0\b
\par
\par \plain\f4\fs28\cf0\b Eine weitere Anwendung: \plain\f5\fs28\cf0 Nat\'fcrlich m\'f6chten Alice und Bob nicht nur
\par Zahlen aneinander verschicken. Man kann z.B. folgende Vereinbarung treffen:
\par
\par - Die Buchstaben des Alphabets werden mit 1 bis 26 durchnumeriert.
\par - W\'f6rter werden durch 27 getrennt oder separat verschickt.
\par
\par Eine Codierung des Worter "Hallo" w\'e4re also gegeben durch
\par
\par \tab \tab \tab \tab \tab 8 01 12 12 15
\par
\par Stellen wir uns vor, wir w\'e4ren Alice und Bob h\'e4tte uns die folgende
\par Nachricht geschickt:
\par
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\par
\par Unser privater Schl\'fcssel ist:
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}d:=155731;
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4 p:=271; q:=977;
\par N:=p*q;
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1834\pich901\picwgoal1039\pichgoal510
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}\plain\f3\fs28\cf5\protect
\par \pard\li50\ri2\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw1094\pich901\picwgoal620\pichgoal510
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00
}\plain\f3\fs28\cf5\protect
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00
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\par {\pict\wmetafile8\picw1834\pich901\picwgoal1039\pichgoal510
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\par
\par \pard\ri4\plain\f5\fs28\cf0 Wir entsch\'fcsseln die Nachricht sukzessiv wie oben: \plain\f3\fs28\cf0\b
\par
\par \pard\li300\ri5\fi-300{\*\pn\pnlvlblt\pnf1\pnindent300{\pntxtb\'b7}}\plain\f3\fs28\cf4 {\pntext\f1\'b7\tab}x1:= powermod(143723, d, N);
\par \pard\li600\ri1\fi-300\plain\f3\fs28\cf4 x2:= powermod(28057, d, N);
\par x3:= powermod(165879, d, N);
\par x4:= powermod(34978, d, N);
\par x5:= powermod(169363, d, N);
\par x6:= powermod(152576, d, N);
\par x7:= powermod(22350, d, N);
\par x8:= powermod(169363, d, N);
\par x9:= powermod(164850, d, N);
\par \plain\f3\fs22\cf4
\par \pard\li50\ri6\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
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\par \pard\li50\ri2\plain\f3\fs28\cf5\protect {\pict\wmetafile8\picw847\pich901\picwgoal480\pichgoal510
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}\plain\f3\fs28\cf5\protect
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