Chapter Introduction to Cryptography1 Caesar Cipher2 Affine Cipher3 Vigenere Cipher4 Autokey Cipher5 Hill Cipher6 Verman’s Telegraph Cipher7 Exponential Cipher8 RSA Cipher9 Knapsack Cipher10 ElGamal Cipher

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..

Hello

.

(Plaintext)

.

§♯♯♣♯

.

(Ciphertext)

.

Hello

.

(Plaintext)

.

Enciphering

.

Deciphering

.Key

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Caesar Cipher (Shift cipher)

Plaintext: A B C D . . .X Y ZCiphertext: D E F G . . .A B C

A B C D E F G H I J K L M00 01 02 03 04 05 06 07 08 09 10 11 12N O P Q R S T U V W X Y Z13 14 15 16 17 18 19 20 21 22 23 24 25

C = P + 3 (mod 26)P = C − 3 (mod 26)

.Example..

......

Plaintext: GOOD = 06 14 14 03C=P+3 (mod 26)Ciphertext: 09 17 17 06 = JRRH

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Problem of Caesar Cipher

.Example..

......

Decipher

”FKHQBLBDQJ”by Caesar Cipher.

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Affine Cipher

C = aP + b (mod 26)P = a−1(C − b) (mod 26), gcd(a, 26) = 1

.Remark:..

......

♯{(a, b) ∈ Z26 × Z26 | gcd(a, 26) = 1} = 12× 26 = 312

aφ(26) = a12 = 1(mod 26)a−1 = a11(mod 26)P = a11(C − 3)(mod 26)

.Example..

......

Plaintext: GOOD = 06 14 14 03a=5, b=7 C=5P+7 (mod 26)Ciphertext: 11 25 25 22 = LZZW

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Problem of Affine Cipher

.Example..

......

Decipher

”RTUEDWJ”by Affine Cipher using C ≡ 9P + 9(mod 26).

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Vigenere Cipher

Keyword : READY= 17 04 00 03 24 (no repeated).Example..

......

Plaintext: ATTACK AT ONCE00 19 19 00 02 10 00 19 14 13 02 04

Key +) 17 04 00 03 24 17 04 00 03 24 17 0417 23 19 03 00 01 04 19 17 11 19 08

Ciphertext: RXTDAB ET RLTI

.Vigenere Cipher..

......

Keyword: b1, b2, . . . , bn, where 0 ≤ bi ≤ 25

Ci ≡ Pi + bi(mod 26), 1 ≤ i ≤ nPi ≡ Ci − bi(mod 26), 1 ≤ i ≤ n

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Problem of Vigenere Cipher

.Example..

......

Decipher

”XIGOGIPLZ”by Vigenere Cipher using the Key ”MATH”.

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Autokey Cipher (Modifying Vigenere Cipher)

Keyword = seed +Plaintext.Example..

......

Plaintext: ONE IF BY DAWNseed = KKeyword K ONE IF BY DAWN

14 13 04 08 05 01 24 03 00 22 13Key +) 10 14 13 04 08 05 01 24 03 00 22

24 01 17 12 13 06 25 01 03 22 09Ciphertext: YBR MN GZ BDWJ

Enciphering:={

C1 ≡ P1 + S(mod 26)Ck ≡ Pk + Pk−1(mod 26) 2 ≤ k ≤ n

Deciphering:={

P1 ≡ C1 − S(mod 26)Pk ≡ Ck − Pk−1(mod 26) 2 ≤ k ≤ n

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Hill Cipher

Divid the plaintext into blocks of n letters.For n = 2, P1,P2 ⇒ C1,C2 by{

C1 ≡ aP1 + bP2(mod 26)C2 ≡ cP1 + dP2(mod 26) gcd(ad − bc, 26) = 1

.Example..

......

{C1 ≡ 2P1 + 3P2(mod 26)C2 ≡ 5P1 + 8P2(mod 26)

Plaintext: BUY NOW ≡ BU YN OW ≡ 01 20 24 13 14 22Ciphertext: 10 09 09 16 16 12 ≡ KJJQQMFor (C1,C2)=(10,09),{

P1 ≡ 8C1 − 3C2(mod 26) ≡ 8(10)− 3(09) ≡ 01P2 ≡ −5C1 + 2C2(mod 26) ≡ −5(10) + 2(09) ≡ 20

Decipher to BU.

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Problem of Hill Cipher

.Example..

......

The ciphertext

”ZZ”has been enciphered by Hill cipher{

C1 ≡ 5P1 + 2P2(mod 26)C2 ≡ 3P1 + 7P2(mod 26)

Given the plaintext.

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Verman’s Telegraph Cipher (Nonalphabetic Cpher)

Baudot code:

A = 11000 B = 10011 C = 01110 D = 10010 E = 10000F = 10110 G = 01011 H = 00101 I = 01100 J = 11010K = 11110 L = 01001 M = 00111 N = 00110 O = 00011P = 01101 Q = 11101 R = 01010 S = 10100 T = 00001U = 11100 V = 01111 W = 11001 X = 10111 Y = 10101Z = 10001

.Example..

......

Plaintext=YES 10101 10000 10100Keyword(any) +) 10101 01010 10101Ciphertext 00000 11010 00001

Enciphering:= C ≡ P + K (mod 2)Deciphering:= P ≡ C + K = P + K + K ≡ P (mod 2)

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Problem of Verman Cipher

.Example..

......

Decipher

”10010 01001 10011 11010 00000”by Verman’s Telegraph Cipher using Keyword

”1010101010101010101010101”

.

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Exponential Cipher

1 p is a prime, k exponent with gcd (k, p-1)=1.

2 Encode: A B C . . . X Y Z00 01 02 . . . 23 24 25

3 Group the plaintext in blocks of m letters, where2525 . . . 25︸ ︷︷ ︸

m times< p < 2525 . . . 25︸ ︷︷ ︸

m+1 timesp = 4283, m = 2 by 2525 < 4283 < 252525.p = 670417, m = 3 by 252525 < 670417 < 25252525.

4 Encode a block B using

C ≡ Bk (mod p), 0 ≤ C < p

Remark: it cannot covert to letters.5 Let d be inverse of k modulo p-1. (i.e.dk ≡ 1 (mod p − 1)).

dk = 1 + r(p − 1)

Cd = Bdk = B1+r(p−1) = B(Bp−1)r ≡ B1r = B (mod p)

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Encipher of Exponential Cipher

.Example..

......

p=2621, p − 1 = 2620 = 22 ∗ 5 ∗ 131. take k=11m=2, since 2525 < 2621 < 252525

Plaintext= DE EP YO GU RT0304 0415 2414 0620 1719

(0304)11 ≡ 0065 (mod 2621)

(0415)11 ≡ 0415 (mod 2621)

(2414)11 ≡ 1323 (mod 2621)

(0620)11 ≡ 1567 (mod 2621)

(1719)11 ≡ 0150 (mod 2621)

ciphertext= 0065 0415 1323 1567 0150

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Decipher of Exponential Cipher

.Example..

......

p=2621, k=11, where gcd(2620,11)=1

By extended Euclidean algorithm,

a q y2620 − 119111 238 52 5 11 2 0

1=(-5)(2620)+1191*111191 ∗ 11 ≡ 1 (mod 2620)

decode C = 0065

(0065)1191 ≡ 0304 (mod 2621)

0304 = DE

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Problem of Exponential Cipher

.Example..

......

The ciphertext from an exponential cipher with key (p, k) = (41, 27) is

”13 25 37”Find the plaintext.

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RSA Cipher(generlization of exponential cipher)

1 n=p*q, p and q are large prime number, n is a key2 Find an exponent k with gcd(k, ϕ(n))=1.

3 Encode: A B C . . . X Y Z00 01 02 . . . 23 24 25

4 Group the plaintext in blocks of m letters, where2525 . . . 25︸ ︷︷ ︸

m times< n < 2525 . . . 25︸ ︷︷ ︸

m+1 times5 Encode a block B using

C ≡ Bk (mod n)

6 Decode: Let d be inverse of k modulo ϕ(n). (i.e.dk ≡ 1 (mod ϕ(n))).dk = 1 + rϕ(n)

Cd = Bdk = B1+rϕ(n) = B(Bϕ(n))r ≡ B1r = B (mod n)

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Encipher of RSA Cipher.Example..

......

n=p*q=37*73=2701, ϕ(n) = 36 ∗ 72 = 2592

k=47, gcd(47,2592)=1m=2, since 2525 < 2701 < 252525

Plaintext= NO WAY TODAYB= NO WA YT OD AY

1314 2200 2419 1403 0024(1314)47 ≡ 1241 (mod 2701)

(2200)47 ≡ 1993 (mod 2701)

(2419)47 ≡ 1044 (mod 2701)

(1403)47 ≡ 2081 (mod 2701)

(0024)47 ≡ 0873 (mod 2701)

ciphertext = 1241 1993 1044 2081 0873

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Decipher of RSA Cipher.Example..

......

n=2701, k=47, where gcd(47,ϕ(n))=1 and ϕ(n) = 2592

By extended Euclidean algorithm,

a q y2592 − 110347 55 207 6 35 1 22 2 11 2 0

1=(-20)*2592+1103*471103 ∗ 47 ≡ 1 (mod 2592)

decode C = 1241

(1241)1103 ≡ 1314 (mod 2701)

1314 = NO

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Problem of RSA Cipher

.Example..

......

The ciphertext from a RSA cipher with key (n, k) = (35, 5) is

”09 20 16 09 12”Find the plaintext. (Note: 2573 = 31× 83)

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Knapsack cryposystem

.Knapsack Problem..

......

Given a set of positive integers a1, a2, . . . , an and an integer S, the problemasks that which of those integers add together to S? i.e. solve the equation

S = a1x1 + a2x2 + · · ·+ anxn, xi = 0 or 1

.Example..

......

For (a1, a2, a3, a4, a5) = (2, 7, 8, 11, 12) and S = 21,

21 = 2 + 7 + 12 = 2 + 8 + 11

For system 21 = 2x1 + 7x2 + 8x3 + 11x4 + 12x5 with solutionsx1 = x2 = x5 = 1, x3 = x4 = 0 or x1 = x3 = x4 = 1, x2 = x5 = 0

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Knapsack cryposystem

.Example..

......

For ai = 2i−1 and S < 2n, the system

S = a1x1 + a2x2 + · · ·+ anxn

with solution: Binary expansion of S

.Definition..

......A sequence a1, a2, . . . , an is superincreasing if ai > a1 + a2 + · · ·+ ai−1 fori = 2, 3, . . . , n

Rremark: Above example is superincresing by

2i > 1 + 2 + 22 + . . . 2i−1 = 2i − 1

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Knapsack cryposystem

.Example..

......

Solve the superincreasing Knapsack problem

28 = 3x1 + 5x2 + 11x3 + 20x4 + 41x5

Solution:1 41 > 28 ⇒ x5 = 0

2 20 < 28 and 3 + 5 + 11 < 20 < 28 ⇒ x4 = 1

8 = 3x1 + 5x2 + 11x3

3 11 > 8 ⇒ x3 = 08 = 3x1 + 5x2

4 x1 = 1 and x2 = 1

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In grneral, for superincreasing Knapsack problem

S = a1x1 + a2x2 + · · ·+ anxn

xn =

{1 if S ≥ an0 if S < an

xj =

{1 if S −

∑ni=j+1 xiai ≥ aj

0 if S −∑n

i=j+1 xiai < aj j = n − 1, n − 2, . . . , 1

1 When S ≥ an, let xn = 0

n∑i=1

aixi =n−1∑i=1

aixi ≤n−1∑i=1

ai < an ≤ S ” ⇒⇐ ”

2 When S −∑n

i=j+1 xiai ≥ aj, let xj = 0

n∑i=1

aixi ≤j−1∑i=1

ai +n∑

i=j+1

aixi < aj +n∑

i=j+1

ai ≤ S ” ⇒⇐ ”

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Knapsack Cipher

1 take a superincreasing sequence a1, a2, . . . , an

2 take m s.t. m > 2an

3 take integer a s.t. gcd(a,m)=1 with inverse a.4 aai ≡ bi (mod m), 1 ≤ i ≤ n ⇒ 0 < bi < m

If bi = 0 i.e. aai ≡ 0 (mod m) ⇒ m|aai⇒ m|ai since gcd(a,m)=1 ” ⇒⇐ ” since m > 2an > ai

5 Enciphering: Covert the plaintext into binary sequence M using binaryequivalent of letter

A B C . . . X Y Z(0) (1) (2) . . . (23) (24) (25)

00000 00001 00010 . . . 10111 11000 11001Split M into blocks of n digits

6 For block x1x2 . . . xn, we obtain a cipertext by

S = b1x1 + b2x2 + · · ·+ bnxn

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Deciphering:S′ ≡ aS (mod m), 0 ≤ S′ < mS′ ≡ ab1x1 + ab2x2 + · · ·+ abnxn (mod m)

≡ aaa1x1 + aaa2x2 + · · ·+ aaanxn (mod m)≡ a1x1 + a2x2 + · · ·+ anxn (mod m)

Since 0 ≤ S′ < m and0 ≤ a1x1 + a2x2 + · · ·+ anxn ≤ a1 + a2 + · · ·+ an < 2an < mwe have

′S = a1x1 + a2x2 + · · ·+ anxn

Since a1, a2, . . . , an is superincreasing , it is solvable.

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Knapsack Cipher.Example..

......

secret key: superincresing seq 3, 5, 11, 20, 41m=85, a=44 gcd(a,m)=1public key bi ≡ aai (mod m)47, 50, 59, 30, 19Plaintext HELP USM= 00111 00100 01011 01111 10100 10010108=47*0+50*0+59*1+30*1+19*159=47*0+50*0+59*1+30*0+19*099=47*0+50*1+59*0+30*1+19*1158=47*0+50*1+59*1+30*1+19*1106=47*1+50*0+59*1+30*0+19*077=47*1+50*0+59*0+30*1+19*0Ciphertext: 108, 59, 99, 158, 106, 77

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Knapsack Cipher

Deciphering: 44 = 29 i.e. solution of 44x ≡ 1 (mod 85)

108 ∗ 29(mod85) = 72 = 3x1 + 5x2 + 11x3 + 20x4 + 41x5with solution x1 = x2 = 0 and x3 = x4 = x5 = 1First block is 00111 ⇔ H59 ∗ 29(mod85) = 11 = 3x1 + 5x2 + 11x3 + 20x4 + 41x5with solution x1 = x2 = x4 = x5 = 0 and x3 = 1Second block is 00100 ⇔ E99 ∗ 29(mod85) = 66 = 3x1 + 5x2 + 11x3 + 20x4 + 41x5with solution x1 = x3 = 0 and x2 = x4 = x5 = 1Second block is 01011 ⇔ L

...

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Knapsack Cipher.Example..

......

secret key: superincresing seq 3, 5, 11, 20, 41, 83, 179, 344, 690,1042m=2618, a=929 gcd(a,m)=1

By extended Euclidean algorithm,

a q y2618 − 31929 2 11760 1 9169 4 284 2 11 84 0

1 = (−11) ∗ 2618 + 31 ∗ 929 ≡ 31 ∗ 929 (mod 2618)

a = 929−1 = 31

public key: bi ≡ aai (mod m)169, 2027, 2365, 254, 1437, 1185, 1357, 180, 2218, 1976

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.Example..

......

Plaintext NOT NOWM= 0110101110, 1001101101, 01110101109584=169*0+2027*1+2365*1+254*0+1437*1

+1185*0+1357*1+180*1+2218*1+1976*05373=169*1+2027*0+2365*0+254*1+1437*1

+1185*0+1357*1+180*1+2218*0+1976*18229=169*0+2027*1+2365*1+254*1+1437*0

+1185*1+1357*0+180*1+2218*1+1976*0Ciphertext: 9584, 5373, 8229Deciphering: 929−1 = 31 i.e. solution of 929x ≡ 1 (mod 2618)9584 ∗ 31 = 1270 (mod 2618)1270 = 3x1 + 5x2 + 11x3 + 20x4 + 41x5 + 83x6

+179x7 + 344x8 + 690x9 + 1042x10with solution 0110101110 ⇔ NO

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Problem of Knapsack Cipher

.Example..

......

The ciphertext from a Knapsack cipher using the superincreasing sequence2, 3, 7, 13, 27, modulus m = 60, and multiplier a = 7 is

”49 63 61 49 23”Find the plaintext.

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ElGamal Cipher

1 Selecting a prime p and a primitive root r of p.2 A integer k, 2 ≤ k ≤ p − 2,

a ≡ rk(mod p); 0 ≤ a ≤ p − 1

3 (p, r, a) is public key4 k private key.

.Example..

......

Take p = 113 with primitive root r = 3. Choose k = 37

337 = 24 = a (mod 113)

(113, 3, 24) is public key37 private key.

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ElGamal Cipher

1 Encode: The message is first converted to its numerical M byA B C . . . X Y Z00 01 02 . . . 23 24 25

2 If M ≥ p, then M is split into blocks, each block containing the same(even) number of digits.

3 If B is first block, B =⇒ (C1,C2), where

C1 ≡ r j(mod p)

C2 ≡ Ba j(mod p)

2 ≤ j ≤ p − 2; For security, the choice of j can be changed from blockto block.

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ElGamal CipherDecode:

P = C2Cp−1−k1 (mod p)

by

P ≡ C2Cp−1−k1 ≡ (Ba j)(r j)p−1−k

≡ B(r k) j(r j(p−1)−jk)≡ B(r p−1) j

≡ B (mod p).Example..

......

Deliver the messageSELL NOW

to a person who has secret key k = 15 and public key

(p, r, a) = (43, 3, 22)

, where 22 ≡ 315 (mod 43)

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ElGamal Cipher

Encode:M = 18041111131422 with Blocks 18 04 11 11 13 14 22

Select an interger j, 2 ≤ j ≤ 41, take j = 23 by

C1 ≡ r j ≡ 3 23 ≡ 34(mod 43)

C2 ≡ Ba j ≡ 18 · (22) 23 = 17(mod 43)

ciphertext is of the form

(34, 17), (34, 42), (34, 08), (34, 08), (34, 29), (34, 18), (34, 16)

andM′ = 17420808291816

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ElGamal Cipher

Decode:(C1,C2) = (34, 17) =⇒ B = 18 by

P ≡ C2Cp−1−k1 ≡ 17 · (34)43−1−15

≡ 17 · (34)27≡ 18 (mod 43)

B= 18 04 11 11 13 14 22S E L L N O W

.Remark..

......

Encode In last block B = 22, take j = 31,

C1 ≡ r j ≡ 3 31 ≡ 33(mod 43)

C2 ≡ Ba j ≡ 18 · (22) 31 = 35(mod 43)

Decode P ≡ C2Cp−1−k1 ≡ 35 · (33)27 ≡ 22 (mod 43)

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Problem of ElGamal Cipher

.Example..

......

The ciphertext from ElGamal cipher with Public key (p, r, a) =(2017, 5, 303) and Private key k = 1999 is

C1 1318 1318 1318 1318 1318 1318 1318 1318C2 536 1381 1967 1699 904 854 1967 1063

Find the plaintext.

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Digital Signature of ElGamal Cipher

.Digital Signature(數位簽章)..

......

Digital Signature 的使用情境大概如下：假設 A 要傳訊息給 B，但是 B 要如何確認訊息真的是由 A 發送的呢?此時只要 A 在發送前, 利用自己的 private key 將訊息製作簽章碼，再傳給 B，B 再利用 A 的簽章碼進行驗証。如果訊息驗証通過，就可以確定訊息是由 A 所發出；即使訊息在傳送過程中被 C 所攔截，再使用 A 的 public key 還原成原本的訊息，還是沒辦法偽裝成 A 所發送的訊息 (因為這需要 A的 private key)。因此，Digital Signature 在實際應用上是很有意義的，因為這項技術代表了授權機制可以很容易建立起來。

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Digital Signature of ElGamal Cipher

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Digital Signature of ElGamal Cipher.Sender use public key (p, r, a) and private key k..

......

1 Selecting a integer j, 1 ≤ j ≤ p − 1 such that (j, p − 1) = 1

2 Take First block B of message M, computes

c ≡ r j (mod p); 0 ≤ c ≤ p − 1

a solution d of linear congruence

jd + kc ≡ B (mod p − 1); 0 ≤ d ≤ p − 2

3 The pair (c, d) is the digital signature appended to the message

.The recipient use the public key (p, r, a) to confirm the signature by..

......

1 V1 ≡ accd (mod p) and V2 ≡ rB (mod p)2 If V1 = V2, then the signature is accepted, otherwise it is

un-accepted. Why?

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Digital Signature of ElGamal Cipher

若 message M 與附加簽章碼 (c,d) 是 Sender 送出來的, 則 V1 = V2

V1 ≡ accd ≡ (rk)c(rj)d

≡ rkc+jd = rB+s(r−1)

≡ rBrs(r−1)

≡ rB ≡ V2 (mod p)若 message M 與附加簽章碼 (c,d) 被第三者攔截, 並送出假訊息 M’with first block B’ ,B = B′, 與附加簽章碼 (c,d), 則 V1 = V2

V1 ≡ accd ≡ (rk)c(rj)d

≡ rkc+jd = rB+s(r−1)

≡ rBrs(r−1) ≡ rB

≡ rB′ ≡ V2 (mod p)

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Problem of ElGamal Cipher

.Example..

......

1 Deliver the message SELL NOW; first block B = 18 to a person whohas secret key k = 15 and public key (p, r, a) = (43, 3, 22), where22 ≡ 315 (mod 43)

2 Take j = 25 such that (j, 42) = 1

3 c ≡ rj ≡ 325 ≡ 5 (mod 43)

4 jd + kc ≡ 25d + 15 · 5 ≡ B = 18 (mod 42) with solution d = 33

5 The digital signature (c, d) = (5, 33)

6 Recipient Check: V1 ≡ accd ≡ 225533 ≡ 35 (mod 43)

7 V2 ≡ rB ≡ 318 ≡ 35 (mod 43)

8 V1 = V2 Then the message come from Deliver.

June 8, 2017 43 / 44

Mod[2017, 54] PowerMod[2017, 34, 54]PowerMod[34,−1, 2017] GCD[2017, 1820]PrimitiveRoot[2017] PrimitiveRootList[2017]JacobiSymbol[45, 2017] (45/2017) Divisor[15]DivisorSigma[15] σ(15) Length[Divisors[15]] τ(15)MoebiusMu[15] µ(15) EulerPhi[15] ϕ(15)

��END��

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