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Avalanche Effect in DES key desirable property of encryption algo where a change of one input or key bit results in changing approx half output bits making attempts to “home-in” by guessing keys impossible DES exhibits strong avalanche

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Avalanche Effect in DES DES exhibits a strong avalanche effect. Table shows some results. In Table (a), two plaintexts that differ by one bit were used: 00000000 00000000 00000000 00000000 10000000 00000000 00000000 00000000 00000000 00000000 00000000 00000000 with the key 0000001 1001011 0100100 1100010 0011100 0011000 0011100 0110010

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The Table (a) shows that after just three rounds, 21 bits differ between the two blocks. On completion, the two ciphertexts differ in 34 bit positions. Table (b) shows a similar test in which a single plaintext is input: 01101000 10000101 00101111 01111010 00010011 01110110 11101011 10100100 with two keys that differ in only one bit position: 1110010 1111011 1101111 0011000 0011101 0000100 0110001 11011100 0110010 1111011 1101111 0011000 0011101 0000100 0110001 11011100 Again, the results show that about half of the bits in the ciphertext differ and that the avalanche effect is pronounced after just a few rounds.

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Strength of DES – Key Size 56-bit keys have 2 56 = 7.2 x 10 16 values brute force search looks hard recent advances have shown is possible – in 1997 on Internet in a few months – in 1998 on dedicated h/w (EFF) in a few days – in 1999 above combined in 22hrs! still must be able to recognize plaintext must now consider alternatives to DES

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Key Size (bits)Number of Alternative Keys Time required at 1 decryption/µs Time required at 10 6 decryptions/µs 32 2 32 = 4.3 10 9 2 31 µs= 35.8 minutes 2.15 milliseconds 56 2 56 = 7.2 10 16 2 55 µs= 1142 years 10.01 hours 128 2 128 = 3.4 10 38 2 127 µs= 5.4 10 24 years 5.4 10 18 years 168 2 168 = 3.7 10 50 2 167 µs= 5.9 10 36 years 5.9 10 30 years 26 characters (permutation) 26! = 4 10 26 2 10 26 µs= 6.4 10 12 years 6.4 10 6 years Brute Force Search

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Chapter 4

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Modular Arithmetic define modulo operator “ a mod n” to be remainder when a is divided by n use the term congruence for: a = b mod n – when divided by n, a & b have same remainder – eg. 100 = 34 mod 11 b is called a residue of a mod n – since with integers can always write: a = qn + b – usually chose smallest positive remainder as residue ie. 0 <= b <= n-1 – process is known as modulo reduction eg. -12 mod 7 = -5 mod 7 = 2 mod 7 = 9 mod 7

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Divisors say a non-zero number b divides a if for some m have a=mb ( a,b,m all integers) that is b divides into a with no remainder denote this b|a and say that b is a divisor of a eg. all of 1,2,3,4,6,8,12,24 divide 24

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Modular Arithmetic Operations is 'clock arithmetic' uses a finite number of values, and loops back from either end modular arithmetic is when do addition & multiplication and modulo reduce answer can do reduction at any point, ie – a+b mod n = [a mod n + b mod n] mod n

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Modular Arithmetic can do modular arithmetic with any group of integers: Z n = {0, 1, …, n-1} form a commutative ring for addition with a multiplicative identity note some peculiarities – if (a+b)=(a+c) mod n then b=c mod n – but if (a.b)=(a.c) mod n then b=c mod n only if a is relatively prime to n

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Modular arithmetic Properties Modular arithmetic exhibits the following properties: [(a mod n) + (b mod n)] mod n = (a + b) mod n [(a mod n) (b mod n)] mod n = (a b) mod n [(a mod n) x (b mod n)] mod n = (a x b) mod

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Example 11 mod 8 = 3; 15 mod 8 = 7 [(11 mod 8) + (15 mod 8)] mod 8 = 10 mod 8 = 2 (11 + 15) mod 8 = 26 mod 8 = 2 [(11 mod 8) (15 mod 8)] mod 8 = 4 mod 8 = 4 (11 15) mod 8 = 4 mod 8 = 4 [(11 mod 8) x (15 mod 8)] mod 8 = 21 mod 8 = 5 (11 x 15) mod 8 = 165 mod 8 = 5

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Exponentiation is performed by repeated multiplication, as in ordinary arithmetic. (We have more to say about exponentiation in Chapter 8.) To find 11 7 mod 13, we can proceed as follows: 11 2 = 121 Ξ 4 (mod 13) 11 4 = (11 2 ) 2 Ξ 4 2 Ξ 3 (mod 13) 11 7 Ξ 11 x 4 x 3 Ξ 132 Ξ 2 (mod 13)

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Modulo 8 Addition Example +01234567 001234567 112345670 223456701 334567012 445670123 556701234 667012345 770123456

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Copyright © 2012, Elsevier Inc. All rights Reserved. 1 Chapter 7 Modeling Structure with Blocks.

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