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Cryptography : From Demaratus to RSA

Date post: 29-Jun-2015
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From Demaratus in ancient Sparta using wax covered tablets to the German Enigma to Diffie-Helman and RSA cryptography has always been at the cutting edge.
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Cryptography 101
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Page 1: Cryptography : From Demaratus to RSA

Cryptography 101

Page 2: Cryptography : From Demaratus to RSA

What is Cryptography?Encryption is the process of: 1.Transforming data (x) 2.Using an algorithm (e)3.To make it unreadable to anyone (y)4.except those possessing the key. (k)

K={k1,…,kn}

The Quick Brown Fox

Me (mod N) where M=The Quick Brown Fox

Uif Rvjdl Cspxo Gpy

Kerchhoffs’ principle: A cryptosystem should be secure even if the Attacker knows all the details about the system, with the exception of The key.

Page 3: Cryptography : From Demaratus to RSA

We shall extend the empire of Persia such that its boundaries will be God's own sky, so the sun will not look down upon any land beyond the boundaries of what is our own

-Xerxes (Ahasuerus) ~450 B.C.

Page 4: Cryptography : From Demaratus to RSA

(Spartan) Scytale

Rail Fence Cipher

Route Cipher

Transposition Ciphers

Page 5: Cryptography : From Demaratus to RSA

The Quick Brown FoxGSV JFRXP YILDM ULC

Substitution Ciphers

The Quick Brown FoxZIT JXOEA WKGVF YGB

Shift Cipher (Caesar)

The Quick Brown FoxSGD PTHBJ AQNVM ENW

At-Bash

Page 6: Cryptography : From Demaratus to RSA

Modular ciphers

a = r mod m42 = 9*4 + 6r = a – m*q

42 = 6 mod 942 = q*9 + 66 = 42 – q*9

q = 0, r = 42q = 1, r = 33q = 2, r = 24q = 3, r = 15q = 4, r = 6 (0<q<m-1)q = 5, r = -3q = 6, r = -12

12 + 7 = 19 => 1 mod 914 – 2 = 12 => 3 mod 911 * 8 = 88 => 7 mod 9

15/5 = 3 !=> 3 mod 9

If the multiplicative inverse exists for a number then we can divide by that number5*2=10 => 1 mod 9 2 is the multiplicative inverse of 5 (and vice versa)15*2 = 30 => 3 mod 9If x is coprime with modulus then it has an inverse.

Page 7: Cryptography : From Demaratus to RSA

Caesar CipherEncryption: ek (x) = x + k mod 26Decryption: ek (y) = y – k mod 26

The quick brown foxk=3

t=20, 20 + 3 = 23 mod 26h=8, 8 + 3 = 11 mod 26e=5, 5 + 3 = 8 mod 26

Affine Cipher k=(a,b)Encryption: ek (x) = a*x + b mod 26Decryption: ek (y) = a-1 * (y – b) mod 26

The quick brown foxk=(5, 3)

t=20, 5*20 + 3 = 103 = 25 mod 26h=8, 5*8 + 3 = 43 = 17 mod 26e=5, 5*5 + 3 = 28 = 2 mod 26

21 * 5 = 105 = 1 mod 26

21 * 25-3 = 462 = 20 mod 2621 * 17-3 = 294 = 8 mod 2621 * 2-3 = -21 = 5 mod 26

Page 8: Cryptography : From Demaratus to RSA

Brute-Force Attacks

Given: y = SGD PTHBJ AQNVM ENWKeyspace = {1,…,25}Decryption : ki(y) =? x

Page 9: Cryptography : From Demaratus to RSA

Abu Yūsuf Yaʻqūb ibn ʼIsḥāq aṣ-Ṣabbāḥ al-Kindī 850 C.E.

Cryptanalysis

Letter Frequency Short word and letter combinations

thebetoofandthathavefornot

http://www.richkni.co.uk/php/crypta/

Page 10: Cryptography : From Demaratus to RSA

Normal English letter frequency

Ciphertext letter frequency

Page 11: Cryptography : From Demaratus to RSA

Vigenère cipherA polyalphabetic cipher

Key = KINGThe sun and the man in the moonDpr yev ntn buk wia ox buk wwbt

4 possible ways to spell the word “the”K – DPR I - BUKN – GNO G - ZRM

http://www.simonsingh.net/The_Black_Chamber/vigenere_cracking_tool.html

http://sharkysoft.com/vigenere/

Page 12: Cryptography : From Demaratus to RSA

Enigma

http://cryptoclub.math.uic.edu/shiftcipher/shiftcipher.phphttp://enigma.louisedade.co.uk/enigma.html

Page 13: Cryptography : From Demaratus to RSA

AXP AVC .. IOV NKZ .. HSA PYT .. PPZ LEXFZD YQO .. IZL NQL .. NNQ CMA .. GUH BISFGT YHD .. KDY GNV .. NBJ COQ .. GOI BKKMIW MRI .. VWG EZG .. SYX SJB .. TVB KFMDJG UDG .. OJN QDE .. SNH SMS .. TLI KPKLNK TMF .. ZAO RXJ .. SXV SVZ .. TYO KJJXKN JAE .. CTL OUL .. ERS XWU .. WHJ WBQBHG DBG .. CMM OTY .. EAA XXT .. JQR ISHRZU ZQN .. UKM HAY .. YCE FGR .. JEY ICVRTC ZUW .. QFF VLP .. PII LRK .. JCE IGP

Loops(1,4) (LTKGBDUHP) (XJINCOQVE) (FY) (RZ) (A) (M) (S) (W)

Loops(2,5) (XVFLPECGHBOKA) (ZQSYJDNMTUIRW)

Loops(3,6) (PCWIKF) (DOJQAT) (NERHSU) (VZXBMY) (L) (G)

Loops(1,4) 8, 9, 9, 2, 2, 1, 1, 1, 1 Loops(2,5) 2, 13, 13 Loops(3,6) 6, 6, 6, 6, 6, 1, 1

By the end of WWII enigma had a key space of 159 sextillion (159*1021)

Page 14: Cryptography : From Demaratus to RSA

Confusion and Diffusion

Claude Shannon

ConfusionThe relationship between the key and the ciphertext as complex and as involved as possible.e.g. Enigma & complex substitution (S-boxes)

011011

Diffusion Statistics of the plaintext is "dissipated" in the statistics of the ciphertext. If we change a character of the plaintext, then several characters of the ciphertext should change.

http://en.wikipedia.org/wiki/Permutation_box

P-Box

Page 15: Cryptography : From Demaratus to RSA

Left Right

ABCDEF GHIJKL

ABCDEF F() = HJLGIK

Xor = JIHGKL

JIHGKL ABCDEF

DESData Encryption Standard (1973)56 bit (Lucifer cipher)

Key Length Security Estimation

56-64 bits A few hours or days

112-128 bits Several decades (w/o QC)

256 bits Several decades (w QC)

Page 16: Cryptography : From Demaratus to RSA

AESAdvance Encryption Standard (2001)Currently accepted industry standardSupports 128, 192 and 256 bit keys.

In 1997 National Institute of Standards and Technology (NIST) Called for proposals for AES• Rijandel• Mars• RC6• Serpent• Twofish

In 2001 Rijandel was adopted and renamed AES.

Page 17: Cryptography : From Demaratus to RSA

Diffie-Hellman Key Exchange (DHKE)Discrete Logarithm Problem

Used in:SSHTLSIPSec

Page 18: Cryptography : From Demaratus to RSA

Diffie-Hellman Key Exchange (DHKE)Discrete Logarithm Problem

1. Choose a prime modulus P. 172. Choose an integer A that will be known as the generator. 33. Alice and Bob both choose a private number

Ax mod PAlice a – 15 Bob b – 13315 mod 17 = 6 313 mod 17 = 12

6<- 12

12 15 mod 17 = 10 6 13 mod 17 = 10

Hacker knows: Alice - Ax mod P = 6Bob - Ax mod P = 12

A is specially chosen to induce the discrete logarithm problem and ensurea one way function.Exponentiation is commutative: k = (Ax)y = (Ay)x

Page 19: Cryptography : From Demaratus to RSA

RSARivest, Shamir, Adleman

Discrete logarithm and integer factorization

Set up1. Choose two large primes: p=3 and q=112. n = p*q = 33 3. Θ(n) = (p-1)(q-1)=(3-1)(11-1)=204. Find a number e where gcd(e, Θ(n)) = 1 e=35. Find the number d where e*d = 1 mod Θ(n) d=7

Public key (n, e) = (33, 3) Private key(d)c = me mod nm = cd mod n

Alice Bobm=443 mod 33 = 31

31->317 mod 20 = 4


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