Applied Symbolic Computation (CS 300) Karatsuba

1
Applied Symbolic Computation (CS
300)Karatsubas Algorithm for Integer
Multiplication

• Jeremy R. Johnson

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Introduction

• Objective To derive a family of asymptotically
  fast integer multiplication algorithms using
  polynomial interpolation
• Karatsubas Algorithm
• Polynomial algebra
• Interpolation
• Vandermonde Matrices
• Toom-Cook algorithm
• Polynomial multiplication using interpolation
• Faster algorithms for integer multiplication
• References Lipson, Cormen et al.

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Karatsubas Algorithm

• Using the classical pen and paper algorithm two n
  digit integers can be multiplied in O(n2)
  operations. Karatsuba came up with a faster
  algorithm.
• Let A and B be two integers with
• A A110k A0, A0 lt 10k
• B B110k B0, B0 lt 10k
• C AB (A110k A0)(B110k B0)
• A1B1102k (A1B0 A0 B1)10k A0B0
• Instead this can be computed with 3
  multiplications
• T0 A0B0
• T1 (A1 A0)(B1 B0)
• T2 A1B1
• C T2102k (T1 - T0 - T2)10k T0

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Complexity of Karatsubas Algorithm

• Let T(n) be the time to compute the product of
  two n-digit numbers using Karatsubas algorithm.
  Assume n 2k. T(n) ?(nlg(3)), lg(3) ? 1.58
• T(n) ? 3T(n/2) cn
• ? 3(3T(n/4) c(n/2)) cn
  32T(n/22) cn(3/2 1)
• ? 32(3T(n/23) c(n/4)) cn(3/2
  1)
• 33T(n/23) cn(32/22 3/2 1)
• ? 3iT(n/2i) cn(3i-1/2i-1
  3/2 1)
• ...
• ? c3k cn((3/2)k - 1)/(3/2 -1)
  --- Assuming T(1) ? c
• ? c3k 2c(3k - 2k) ? 3c3lg(n)
  3cnlg(3)

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Divide Conquer Recurrence

• Assume T(n) aT(n/b) ?(n)
• T(n) ?(n) a lt b
• T(n) ?(nlog(n)) a b
• T(n) ?(nlogb(a)) a gt b

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Polynomial Algebra

• Let Fx denote the set of polynomials in the
  variable x whose coefficients are in the field F.
• Fx becomes an algebra where , are defined by
  polynomial addition and multiplication.

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Interpolation

• A polynomial of degree n is uniquely determined
  by its value at (n1) distinct points.
• Theorem Let A(x) and B(x) be polynomials of
  degree m. If A(?i) B(?i) for i 0,,m, then
  A(x) B(x).
• Proof.
• Recall that a polynomial of degree m has m roots.
• A(x) Q(x)(x- ?) A(?), if A(?) 0, A(x)
  Q(x)(x- ?), and deg(Q) m-1
• Consider the polynomial C(x) A(x) - B(x).
  Since C(?i) A(?i) - B(?i) 0, for m1 points,
  C(x) 0, and A(x) must equal B(x).

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Lagrange Interpolation Formula

• Find a polynomial of degree m given its value at
  (m1) distinct points. Assume A(?i) yi
• Observe that

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Matrix Version of Polynomial Evaluation

• Let A(x) a3x3 a2x2 a1x a0
• Evaluation at the points ?, ?, ?, ? is obtained
  from the following matrix-vector product

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Matrix Interpretation of Interpolation

• Let A(x) anxn a1x a0 be a polynomial of
  degree n. The problem of determining the (n1)
  coefficients an,,a1,a0 from the (n1) values
  A(?0),,A(?n) is equivalent to solving the linear
  system

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Vandermonde Matrix
V(?0,, ?n) is non-singular when ?0,, ?n are
distinct.
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Polynomial Multiplication using Interpolation

• Compute C(x) A(x)B(x), where degree(A(x)) m,
  and degree(B(x)) n. Degree(C(x)) mn, and
  C(x) is uniquely determined by its value at mn1
  distinct points.
• Evaluation Compute A(?i) and B(?i) for distinct
  ?i, i0,,mn.
• Pointwise Product Compute C(?i) A(?i)B(?i)
  for i0,,mn.
• Interpolation Compute the coefficients of C(x)
  cnxmn c1x c0 from the points C(?i)
  A(?i)B(?i) for i0,,mn.

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Interpolation and Karatsubas Algorithm

• Let A(x) A1x A0, B(x) B1x B0, C(x)
  A(x)B(x) C2x2 C1x C0
• Then A(10k) A, B(10k) B, and C C(10k)
  A(10k)B(10k) AB
• Use interpolation based algorithm
• Evaluate A(?), A(?), A(?) and B(?), B(?), B(?)
  for ? 0, ? 1, and ? ?.
• Compute C(?) A(?)B(?), C(?) A(?) B(?), C(?)
  A(?)B(?)
• Interpolate the coefficients C2, C1, and C0
• Compute C C2102k C110k C0

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Matrix Equation for Karatsubas Algorithm

• Modified Vandermonde Matrix
• Interpolation

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Integer Multiplication Splitting the Inputs into
3 Parts

• Instead of breaking up the inputs into 2 equal
  parts as is done for Karatsubas algorithm, we
  can split the inputs into three equal parts.
• This algorithm is based on an interpolation based
  polynomial product of two quadratic polynomials.
• Let A(x) A2x2 A1x A0, B(x) B2x2 B1x
  B, C(x) A(x)B(x) C4x4 C3x3 C2x2 C1x
  C0
• Thus there are 5 products. The divide and
  conquer part still takes time O(n). Therefore
  the total computing time T(n) 5T(n/3) O(n)
  ?(nlog3(5)), log3(5) ? 1.46

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Asymptotically Fast Integer Multiplication

• We can obtain a sequence of asymptotically faster
  multiplication algorithms by splitting the inputs
  into more and more pieces.
• If we split A and B into k equal parts, then the
  corresponding multiplication algorithm is
  obtained from an interpolation based polynomial
  multiplication algorithm of two degree (k-1)
  polynomials.
• Since the product polynomial is of degree 2(k-1),
  we need to evaluate at 2k-1 points. Thus there
  are (2k-1) products. The divide and conquer part
  still takes time O(n). Therefore the total
  computing time T(n) (2k-1)T(n/k) O(n)
  ?(nlogk(2k-1)).

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Asymptotically Fast Integer Multiplication

• Using the previous construction we can find an
  algorithm to multiply two n digit integers in
  time ?(n1 ?) for any positive ?.
• logk(2k-1) logk(k(2-1/k)) 1 logk(2-1/k)
• logk(2-1/k) ? logk(2) ln(2)/ln(k) ? 0.
• Can we do better?
• The answer is yes. There is a faster algorithm,
  with computing time ?(nlog(n)loglog(n)), based
  on the fast Fourier transform (FFT). This
  algorithm is also based on interpolation and the
  polynomial version of the CRT.