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# Cryptography Kinder Garden Number theory and Classical Cryptosystems - PowerPoint PPT Presentation

INCS 741: Cryptography. Cryptography Kinder Garden Number theory and Classical Cryptosystems. Dr. Monther Aldwairi New York Institute of Technology- Amman Campus 10/10/2010. Basic Number Theory. Divisibly. Definition Let a and b be integers with a≠0

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INCS 741: Cryptography

### Cryptography Kinder GardenNumber theory and Classical Cryptosystems

Dr. Monther Aldwairi

New York Institute of Technology- Amman Campus

10/10/2010

Dr. Monther Aldwairi

### Basic Number Theory

Dr. Monther Aldwairi

Definition

Let a and b be integers with a≠0

a divides b if there is an integer q such that b=aq

a divides b is denoted bya|b,

Proposition

• For every a≠0, a|0, a|aand 1|b

• If a|b and b|c, then a|c

• If a|b and a|c, then a|(sb+tc) for all integers s, t.

Dr. Monther Aldwairi

Definition

A number a >1is a prime number if it is only divisible by 1 and itself.

Examples: 2, 3, 5, 7, 11, 13

• Numbers that are not prime are composites = (ab)

Theorem

Every positive integer can be expressed as a unique multiplication of prime numbers raised to different powers

Example 504=23327

Corollary

Assuming p is prime, Ifp|abc…z then p must divide one of the factors a, b, c, …, z

Dr. Monther Aldwairi

Definition

The greatest common divisor of a and b is the largest positive integer dividing both a and b and is denoted by gcd(a,b)

Examples: gcd(12,15)=3, gcd(13,7)=1

• a and b are relatively prime if the gcd(a,b) =1

Euclidean Algorithm

Suppose a>b

a =q1b + r1

b =q2r1 + r2

r1=q3r2 + r3 ,until

rk-1=qk+1rk ,then

gcd(a,b)=rk

Dr. Monther Aldwairi

Modulo operation

6 mod 4 = 2 or 6 ≡ 2 (mod 4)

read 6 is congruent to 2 mod 4

Definition

Let a, r, m ∈ Ζ (where Ζ is a set of all integers) and m≠0.

a ≡ r (mod m), if m|(a – r)

m is called the modulus.

r is called the remainder

• In other words, a and r differ by multiple of m

a = q · m + r 0 ≤ r < m

Dr. Monther Aldwairi

Propositions

• a ≡ a (mod m)

• a ≡ 0 (mod m) if and only if m|a

• a ≡ r (mod m) if and only if r ≡ a (mod m)

• If a ≡ r and r ≡ c (mod m), then a ≡ c (mod m)

Examples

12 ≡ 7 mod 5

9 ≡ 3 (mod 6) if and only if 3 ≡ 9 (mod 6)

14 ≡ 8 and 8 ≡ 2 (mod 6), then 14 ≡ 2 (mod m)

Dr. Monther Aldwairi

Work on the set of integers mod m, denoted as Zm ={0,1,2,…, m-1} and perform addition, subtraction and multiplication with congruencies

Proposition

Let a,b,c,d,m ∈ Ζ with m≠0. and suppose

a ≡ b (mod m), and c ≡ d (mod m), then

a+c ≡ b+d (mod m), a ‒ c ≡ b ‒ d (mod m), ac ≡ bd (mod m)

If the result of addition or multiplication is larger than m‒ 1 take the Modula

Example: m = 9  Ζ9 = {0, 1, 2, 3, 4, 5, 6, 7, 8}

6 + 8 = 14 ≡ 5 mod 9

6 × 8 = 48 ≡ 3 mod 9

Dr. Monther Aldwairi

1. The additive identity is 0: a + 0 = a

2. The additive inverse of a is -a = m – a

s.t. a + (-a) ≡ 0 mod m

3. Addition is closed i.e if a, b ∈ Ζmthen a + b ∈ Ζm

4. Addition is commutative a + b = b + a

5. Addition is associative (a + b) + c = a + (b + c)

6. Multiplicative identity is 1: a × 1 ≡ a mod m

7. The multiplicative inverse of a exists if gcd(a, m) = 1 and

denoted as a-1s.t. a-1 × a ≡ 1 mod m

8. Multiplication is closed i.e if a, b ∈ Ζmthen a × b ∈ Ζm

9. Multiplication is commutative a × b = b × a

10. Multiplication is associative (a × b) × c = a × (b × c)

Dr. Monther Aldwairi

If a and m are relatively prime we can divide both sides of the congruence by a.

If gcd(a,m) =1 then their exist a multiplicative inverse for a mod m denoted as a-1.

Example:

What does the division 4/5 mod 7 mean?

4/5 mod 7 ≡ 4 × 5-1 mod 7

Does 5-1 mod 7 exist ?

It exists because gcd(5,7) = 1.

5-1 mod 7 = 3

therefore, 4/5 mod 7= 4 × 3 = 12 mod 7 ≡ 5 mod 7

Dr. Monther Aldwairi

Proposition

Let a,b,c,m ∈ Ζ with m≠0

If ab ≡ ac (mod m), Then b ≡ c (mod m) only if gcd(a,m)=1

Example: Solve 2x+11 ≡ 5 (mod 15)

2x ≡ ‒6(mod 15)

gcd(2,15)=1 then division is allowed

x ≡ ‒ 3 ≡ 12(mod 15)

Dr. Monther Aldwairi

Proposition

The modulo operation can be applied whenever we want

(a + b) mod m = [(a mod m) + (b mod m) ] mod m

(a × b) mod m = [(a mod m) × (b mod m) ] mod m

Exponentiation Example

Example: 38 mod 7 = ?

38 mod 7 = 6561 mod 7 = 2 since 6561 = 937 × 7 + 2.

Or

38 = 34 × 34 = 32 × 32 × 32 × 32

38 mod 7 = [(32 mod 7)×(32 mod 7)×(32 mod 7)×(32 mod 7)] mod 7

38 mod 7 = 2 × 2 × 2 × 2 mod 7 = 16 mod 7 = 2

Dr. Monther Aldwairi

### Substitution Ciphers Monoalphabetic Ciphers

Classical Crypto Systems

Dr. Monther Aldwairi

plaintext in lowercase and CIPHERTEXT in caps

Letters of the alphabet and assigned number as follows

Spaces and punctuation are omitted

Makes decryption easier

Gives information about the structure of the message

Spaces would dominate the leter fequency counts

Dr. Monther Aldwairi

• Shift each letter by k places, k is the key

Let P = C = K= Ζ26and x ∈ P, y ∈ C, k ∈ K

Encryption: y=Ek(x) = x + k mod 26.

Decryption: x=Dk(y) = x ‒ k mod 26.

• When k = 3 the shift cipher is called Caesar Cipher.

• Example

Let the key k = 3

Plaintext : x = A T T A C K = (0, 19, 19, 0, 2, 10).

Ciphertext : y = (0+3 mod 26, 19+3 mod 26, …)

y = (3, 22, 22, 3, 5, 13) = D W W D F N

Dr. Monther Aldwairi

• Known plaintext, chosen plain text or ciphertext easily break the Shift Cipher

• Ciphertext only; Eve has two choice

• Exhaustive search, only 25 possible keys

• Frequency count if the message is long enough

• In english e is by far the most common letter, then T,A,R,N,I,O, S.

Dr. Monther Aldwairi

• The key k = (α, β) and α, β ∈ Ζ26

• Assume gcd(α,26)=1

• α∈ {1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25}

• Encryption: Ek(x) = y = α · x + β (mod 26).

• Example: k = (α, β) = (3, 2) y = 3x +2

Plaintext : X = A T T A C K = (0, 19, 19, 0, 2, 10).

Ciphertext : Y = (3×0 + 2 mod 26, 3×19+2 mod 26, …)

Y = (2, 7, 7, 2, 8, 6) = C H H C I G

• Decryption: Dk(y) = x = α -1(y – β) (mod 26).

Dr. Monther Aldwairi

• 12 choice of α and 26 choices for β 312keys.

• Ciphertext only: exhaustive search or frequency analysis

• Known plaintext: two letters in the plaintext and corresponding ciphertext letters would suffice to find the key.

Example : plaintext: IF=(8, 5) and ciphertext PQ=(15, 16)

8 · α + β ≡ 15 mod 26

5 · α + β ≡ 16 mod 26 ⇒ α = 17 and β = 9

• Chosen plaintext ab y1 = α · 0 + β  y2= α+ β

• Chosen ciphertext AB yields the decryption function

Dr. Monther Aldwairi

• key concept - monoalphabetic substitution ciphers do not change relative letter frequencies

• calculate letter frequencies for ciphertext compare counts/plots against known values

Dr. Monther Aldwairi

Dr. Monther Aldwairi

### Ployalphabetic CiphersSubstitution Ciphers

Classical Crypto Systems

Dr. Monther Aldwairi

• Designed to flatten the frequency distribution by using multiple encryption functions

Example E1(x)= 3x (mod 26),

E2(x)=5x+13 (mod 26)

• Encrypt odd positioned letters using E1 and even positioned letters using E2

• Ultimately, have 26 different functions

• Use a keyword to know what function to use

Dr. Monther Aldwairi

• Multiple caesar ciphers

• Use a key to select which alphabet is used for each letter of the message

• key is multiple letters long K= k1 k2 ... Kd

• write the plaintext out

• write the keyword repeated above it with numeral values

• use each key letter as a caesar cipher key

• encrypt the corresponding plaintext letter

• keyword deceptive (3,4,2,4,15,19,8,21,4)

k: deceptivedeceptivedeceptive

p: wearediscoveredsaveyourself

c: ZICVTWQNGRZGVTWAVZHCQYGLMGJ

Dr. Monther Aldwairi

• Chosen plaintext will easily yield the key

• Try aaaaaaa….

• With enough known plaintext k=y-x

• Chosen ciphertext try AAAAA… results in key negative

• Known ciphertext only

• Find the key length

• Find the key

Dr. Monther Aldwairi

• Write the ciphertext twice (under each other) with one shifted by the potential key length.

• Mark repeated letters (matching the letter below) and count the number of coincidences

• The shift with the most repeated coincidences is the best guess for the key length

• Now examine the frequencies for 1st , 10th , 19th … and 2nd, 11th,20th …

Dr. Monther Aldwairi

Shift of 4

ZICVTWQNGRZGVTWAVZHCQYGLMGJ

ZICVTWQNGRZGVTWAVZHCQYGLMGJ

2

Shift 9

ZICVTWQNGRZGVTWAVZHCQYGLMGJ

ZICVTWQNGRZGVTWAVZHCQYGLMGJ

3

Dr. Monther Aldwairi

• Most common letters in English e, t, a, o, i, n, s, h, r

• Other letters are fairly rare: z, q, x, j, k, v

• Guess the most common such as E and use it to figure the key

• Verify the key is correct by making sure rare letters have low frequencies

• Tables of common pairs/triple letters?!

• Dr. Monther Aldwairi

• Entry 12 in row WR column means that combination appears 12 times

• Entry 14 in N row and W column means NW appears 14 times

Dr. Monther Aldwairi

• Most common was W, B, R, S, I  W=e

• The vowels a, i, o tend to avoid each other

• S, I, P

• 80% of letters that precede n are vowels (WSIP)

• R, A

• The letter h often appears before e and rarely after it

• N=h

• Most common combination is th

• B=t

• Continue the analysis

Dr. Monther Aldwairi

• An encryption in which the letters of the plaintext are rearranged.

• Columnar Transportation as an example

• The plaintext are arranged into n rows

• The resulting ciphertext is formed by traversing the columns x1 x4 x7 x2 x1…

• Use key to rearrange columns

Dr. Monther Aldwairi

• In Stream ciphers every letter plaintext corresponds to a letter in CIPHERTEXT such as shift and Affine

• Block ciphers encrypt a block of letters to overcome this problem.

• DES operates on 64 bit and AES on 128 bit blocks

• Electronic Codebook (ECB) mode converts a block of plaintext to a block of CIPHERTEXT independently and at onetime.

• Cipher Block Chaining (CBC) and Cipher Feedback (CFB) modes use feedback from cipher block to encrypt subsequent blocks.

Dr. Monther Aldwairi

• The key is a word with repeated letters removed

• Schoolbus Scholbu

• A 5 × 5 matrix based on the key is constructed with the remaining alphabets with( i and j) treated as one letter.

Dr. Monther Aldwairi

Playfair Cipher/Ek(x)

• Remove spaces and divide plaintext into groups of two letters.

• If double letters insert x and regroup

• Add extra x to complete the last group?!

• dinner is ready  dinx ne risr ea dy

• If both letters fall in the same row, replace each with letter to right

• “ea" becomes“BD“

• If both letters fall in the same column, replace each with the letter below it

• “dy" encrypts as “KO“

• Otherwise each letter is replaced by the one in its row in the column of the other letter of the pair

• “di" encrypts as “AK“

Dr. Monther Aldwairi

• Susceptible to frequency count attacks

• Count common digrams! 26 × 26 = 676 digrams table

• Corresponds to the English most common digrams such as: th, he, an , in, re, es

• Each letter has only 5 possible corresponding ciphertext letters (4 in row and one below)

• Common pairs XY and YX giveup corners of rectangles with er and re (most common digrams)

Dr. Monther Aldwairi

Hill Cipher/Ek(x)

Example: Let n=3 and the key is an n × n matrix whose entries are integers in Ζ26.

and the plaintext be ABC = (0, 1, 2) then the encryption operation is a vector-matrix multiplication

 AXW

Dr. Monther Aldwairi

Hill Cipher/Dk(y)

In order to decrypt we need the inverse of key matrix M, which is

Multiple the encrypted text by the inverse N

 ABC

Dr. Monther Aldwairi

Matrix Inverse in mod m

For a matrix M to have in Inverse mod m.

gdc(Determinant(M), m) = 1

?

Dr. Monther Aldwairi

• CIPHERTEXT only is difficult!

• Changing one letter in plaintext changes n letters in CIPHERTEXT making frequency count less effective.

• Known plaintext attack can easily find M given n

• Chosen plaintext of baa.., abaa...,…, aa…b

• i.e. chosentext = I

• CIPHERTEXT=M

• Chosen CIPHERTEXT of BAA.., ABAA…,…, AA…B

• i.e. CIPHERTEXT= I

• plaintext=N

Dr. Monther Aldwairi

• Diffusion

• Changing one character in plaintext results in several ciphertext changing and visa versa.

• Hill Cipher has that property

• Confusion

• The key doesn’t relate in a simple way to ciphertext. Each character of the ciphertext should depend on several parts of the key.

• In Hill Cipher one character in ciphertext depends on a key matrix column

Dr. Monther Aldwairi

• Unbreakable cryptosystem!

• Represent the message as binary (ASCII)

• Ek(x)=K+x mod 2

• Ek(x)=K x

• Decryption uses the same key Dk(y)=K y

• Key is used once and discarded

• Unbreakable for a ciphertext only attack.

• Known plain and cipher texts reveal the one time key only. Which is never used again

Dr. Monther Aldwairi

• Keys generation and exchange

• Long and expensive to transmit

• Pseudo random number generators are not secure

• One way functions such as DES and SHA

• xj=f(s+j) for j =1,2,3…

• bjis the least significant bit of xj.

Dr. Monther Aldwairi

• Keys generation and exchange

• Long and expensive to transmit

• Pseudo random number generators are not secure

• One way functions such as DES and SHA

• xj=f(s+j) for j =1,2,3…

• bjis the least significant bit of xj.

Dr. Monther Aldwairi