A Generalization of the Binary GCD Algorithm

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A Generalization of the Binary GCD Algorithm

• ???

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Outline

• The new algorithm
• Left shift GCD
• How are the a and b values found?
• Sequential Complexity
• Question

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The new algorithm(1)

• CXAyB/22m
• Let be d length(A)-length(B) and a,b
• the trailing d bits of A,B
• set c (ab-1) mod 2d
• then C (A-cB)/2d is d bits shorter than A

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The new algorithm(1)

• GCD(A,B)GCD(B , xAyB)
• GCD(A,B)GCD(G,A,B)
• GCD(GCD(G,A mod G),B mod G))
• Let x x1ßX0
• X-1 mod ß2 ((( 1-x x)x)x)
• modß2
• where x x0-1 mod ß

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The left-shift Binary

• 1.if ultv then swap(u,v)
• 2.while v?0 do
• 3. compute t2ev such that t?ult2t
• 4. uminu-t,2t-u
• 5. if ultv then swap(u,v)
• 6.return(u)
• ???
• 1.gcd(u,v)gcd(ubv,v),while b is any interger
• 2.gcd(u,0)u

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The left-shift k-ary GCD

• 1.if ultv then swap(u,v)
• 2.while v?0 do
• 3. compute tkev such that t?u?kt
• 4. find a,b?k such that t/ub/a
• uat-bu
• 5. if ultv then swap(u,v)

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The left-shift k-ary GCD

• Theorem
• Let k be a positive integer. For every real
  number x 0,1, there exist integers a and b
  satisfying 0lta?k and gcd(a,b)1 such that

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How are the a and b values found?

• For each fraction of the form i/k2,where is an
  integer between zero and k2 , position i will
  hold an approximation b/a such that i/k2 - b/a
  ? 1 / (a( k1 ))
• Ex. t/u 0.305
• i?k2 0.305 1/2?5
• so aA53, bB51

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How are the a and b values found?
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Sequential Complexity

• 1. n?m. Multiplication or division of n-bit
  integers by m-bit integers can be done in O(n)
  time using precomputed tables of size at most
  O(m22m) bits.

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Sequential Complexity

• 2.u and v be integer of length n in binary and
  assume that a list of primes up to B is
  available.Using trial division,all common prime
  divisors ?B of u and v can be found and removed
  using tables of at most O(B) entries in time
  O(n2/logB n(B/logB)BlogB).

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Sequential Complexity

• 3.u and v be the inputs to the left-shift k-ary
  GCD ,n be the sum of the lengths of u and v in
  binary.If kO(n/log1.5n),then takes at most
  O(n2/logk)bit operations.
• The total cost is O(n2/logkk2log2knk/logkklogk)
  
• O(n2/logk)

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Question

• One of the referees pointed out that a similar
  algorithm exists for removing the precomputation
  phase from the left-shift k-ary algorithm .
• The idea is to run Euclids algorithm on the
  input pair (i,k2) to generate a sequence of
  continuants pn/qn,where the qn increase and
  gcd(pn,qn)1

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Question

• Since i/k2-pn/qn 1/(qnqn1),one simply stops
  the algorithm when qn1gtk and uses pn/qn for b/a
• This method for computing b/a requires only
  O(log2k) bit operations, and so over the course
  of the algorithm this takes only O(nlogk) times