CSE115/ENGR160 Discrete Mathematics 03/17/11

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CSE115/ENGR160 Discrete Mathematics03/17/11

• Ming-Hsuan Yang
• UC Merced

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3.6 Integers and algorithms

• Base b expansion of n
• For instance, (245)8282485165
• Hexadecimal expansion of (2AE0B)16
• (2AE0B)162164101631416201611175627
• Constructing base b expansion

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Base conversion

• Constructing the base b expansion
• nbq0a0, 0 a0ltb
• The remainder a0, is the rightmost digit in the
  base b expansion of n
• Next, divide q0 by b to obtain
• q0bq1a1, 0a1ltb
• We see a1 is the second digit from the right in
  the base b expansion of n
• Continue this process, successively dividing the
  quotients by b, until the quotient is zero

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Example

• Find the octal base of (12345)10
• First, 12345815431
• Successively dividing quotients by 8 gives
• 154381927
• 1928240
• 24830
• 3803
• (12345)10(30071)8

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Modular expansion

• Need to find bn mod m efficiently
• Impractical to compute bn and then mod m
• Instead, find binary expansion of n first, e.g.,
  n(ak-1 a1 a0)
• To compute bn , first find the values of b, b2,
  , (b4)2b8,
• Next multiple the where aj1

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Example

• To compute 311
• 11(1011)2 ,So 31138 32 31 . First compute 329,
  and then 349281, and 38(34)2(81)26561, So
  311656193177147
• The algorithm successively finds b mod m, b2 mod
  m, b4 mod m, , mod m, and multiply
  together those terms

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Algorithm

• procedure modular exponentiation (binteger,
  n(ak-1ak-2a1a0, , an)2, mpositive integer)
• x 1
• powerb mod m
• for i0 to k-1
• if ai 1 then x(x power) mod m
• power(power power) mod m
• end
• x equals bn mod m
• It uses O((log m)2 long n) bit operations

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Example

• Compute 3644 mod 645
• First note that 644(1010000100)2
• At the beginning, x1, power3 mod 645 3
• i0, a00, x1, power32 mod 6459
• i1, a10, x1, power92 mod 64581
• i2, a21, x181 mod 64581, power812 mod
  6456561 mod 645111
• i3, a30, x81, power1112 mod 64512321 mod
  64566
• i4, a40, x81, power662 mod 6454356 mod
  645486
• i5, a50, x81, power4862 mod 645236196 mod
  645126
• i6, a60, x81, power1262 mod 64515876 mod
  645396
• i7, a71, x(81396) mod 645471, power3962 mod
  645156816 mod 64581
• i8, a80, x471, power812 mod 6456561mod
  645111
• i9, a91, x(471111) mod 64536
• 3644 mod 64536

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Euclidean algorithm

• Need more efficient prime factorization algorithm
• Let abqr, where a, b, q, and r are integers.
  Then gcd(a,b)gcd(b,r)
• Proof Suppose d divides both a and b. Recall if
  da and db, then da-bk for some integer k. It
  follows that d also divides a-bqr. Hence, any
  common division of a and b is also a common
  division of b and r
• Suppose that d divides both b and r, then d also
  divides bqra. Hence, any common divisor of b
  and r is also common divisor of a and b
• Consequently, gcd(a, b)gcd(b,r)

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Euclidean algorithm

• Suppose a and b are positive integers, ab. Let
  r0a and r1b, we successively apply the division
  algorithm
• Hence, the gcd is the last nonzero remainder in
  the sequence of divisions

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Example

• Find the GCD of 414 and 662
• 662414 1248
• 414248 1166
• 248166 182
• 16682 2 2
• 822 41
• gcd(414,662)2

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The Euclidean algorithm

• procedure gcd(a, b positive integers)
• x a
• yb
• while (y?0)
• begin
• rx mod y
• xy
• yr
• end gcd(a,b)x
• The time complexity is O(log b) (where a b)

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3.7 RSA cryptosystem

• Each individual has an encryption key consisting
  of a modulus npq, where p and q are large
  primes, say with 200 digits each, and an exponent
  e that is relatively prime to (p-1)(q-1) (i.e.,
  gcd(e, (p-1)(q-1))1)
• To transform M Encryption CMe mod n,
  Decryption CdM (mod pq)
• The product of these primes npq, with
  approximately 400 digits, cannot be factored in a
  reasonable length of time (the most efficient
  factorization methods known as of 2005 require
  billions of years to factor 400-digit integers)