Polynomials and FFT

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Polynomials and FFT
Lecture 11 Prof. Dr. Aydin Öztürk
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Polynomials

• A polynomial in the variable x is defined as
• where are
  the
• coefficiens of the polynomial.

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Polynomials(cont.)

• A polynomial A(x) is said have degree k if its
  highest non-zero coefficient is ak.
• Any integer strictly greather than the degree of
  a polynomial is a degree bound of that
  polynomial.
• Therefore, the degree of a polynomial of degree
  bound n may be any integer between 0 and n-1.

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Polynomial addition

• If A(x) and B(x) are polynomials of degree bound
  n, their sum is a polynomial C(x) s.t.
• C(x) A(x) B(x)
• For all x underlying their field F that is if

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Polynomial addition(cont.)

• Example

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Polynomial multiplication

• If A(x) and B(x) are polynomials of degree bound
  n, their product C(x) is a ppolynomial of
  degree-bound 2n-1 s.t.
• C(x) A(x) B(x)
• for all x underlying their field F.

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Polynomial multiplication(cont.)

• Example

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Representation of polynomials
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Coefficient representation

• A coefficient representation of a polynomial
• of degree bound n is a vector of coefficients

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Coefficient representation(cont.)

• A coefficient representation is convenient for
  certain operations on polynomials.
• Example Evaluating A(x) at x0. Evaluation takes
  time T(n) when Horners rule is used

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Point-value representation

• A point-value representation of a polynomial A(x)
  of degree-bound n is a set of poin-value pairs.
• s.t. all of the xk are distinct and
• With Horners method, an n-point evaluation takes
  T(n2).

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Point-value representation(cont.)

• Addition based on point-value representation
• If we have a point value representation for A(x)
• and for B(x)
• Then a point-value representation for C(x) A(x)
  B(x) is

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Point-value representation(cont.)

• The time to add two polynomials of degree bound
  point-value representation is T(n).

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Point-value representation(cont.)

• Multiplication based on point-value
  representation
• If we have a point value representation for A(x)
• and for B(x)
• then a point-value representation for C(x)
  A(x)B(x) is

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Point-value representation(cont.)

• The time to multiply two polynomials of degree
  bound point-value representation is T(n).

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Fast multiplication of polinomials

• Can we use the linear-time multiplication method
  for polynomials in point-value form to expedite
  polynomial multiplication in coefficient form?
• The answer hinges on our ability to convert a
  polynomial quickly from coefficient form to
  point-value form and vice-versa.

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Fast multiplication of polinomials

• We can choose the evaluation points carefully,
  we can convert between representations in T(nlg
  n) time.
• If we choose complex roots of unity as the
  evaluation points, we can produce a point-value
  representation by taking the Discrete Fourier
  Transform(DFT) of a coefficient vector
• The inverse operation can be performed by taking
  the inverse DFT of point-value pairs in T(nlg n)
  time.

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Fast multiplication of polinomials

Ordinary multiplication
Coefficient representation
T(n2 )
Interpolation time T(nlg n)
T(nlg n)
Pointwise multiplication
Point-value representation
T(n)
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The DFT and FFT

• In this section we define
• Complex roots of unity,
• Define the DFT
• Show how the FFT computes the DFT

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Complex roots of unity

• A complex nth root of unity is a complex number ?
  s.t.
• The n roots are
• To interpret this formula we use the defination
  of exponential of complex number

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Complex roots of unity

• The value is called the
  principal root of unity.
• All of the other complex nth roots of unity are
  powers of .
• The n complex nth roots of unity are

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Complex roots of unity

• Lemma-1 For any integers n 0 and d gt 0,
• Proof

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Complex roots of unity

• Corollary For any even integer n gt 0,
• Lemma-2 If n gt 0 is even then the squares of the
  nth roots of unity are the (n/2)th roots of
  unity.
• Proof We have (by
  lemma-1)

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Complex roots of unity

• Lemma-3 For any integer n 1 and nonnegative
  integer k not divisible by n,
• Proof

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The DFT

• We wish to evaluate a polynomial
• of degree bound n at
• Without loss of generality, we assume that n is
  a power of 2.
• (We canalways add new high order zero coefficient
  as necessary)

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The DFT

• Assume that A(x) is given in coefficient form
• For k0,1, ..., n-1 we define
• The vector
  is called the Discrete Fourier Transform(DFT) of
  the coefficient vector
  We also write
  yDFTn(a).

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The FFT(cont.)

• Let
• It follows that

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The FFT(cont.)

• The problem of evaluating A(x) at
• reduces to
• Evaluating the degree-bound n/2 polynomials
• and at the points
• as
• 2. Combining the results.

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Recursive FFT
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Running time of RECURSIVE-FFT

• We note that exclusive of the recursive calls,
  each invocation takes time T(n).
• The recurrence for the running time is therefore

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Interpolation

• We can write the DFT as the matrix product y Va
  that is

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Interpolation(cont.)

• Theorem For j,k0,1, ..., n-1, the ( j,k) entry
  of the inverse of matrix is
• Given the inverse matrix V -1, we have that
  
• is given by

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Interpolation(cont.)

• By using the FFT and the inverse FFT, we can
  transform a polynomial of degree-bound n back and
  forth between its coefficient representation and
  a point-value representation in time T(n lg n).

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Interpolation(cont.)

• Theorem For any two vectors a and b of length n
  is a power of 2,
• Where the vectors a and b are padded with zeros
  to length 2n and
• . denotes the componentwise product of two 2n
  element vectors.

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Example

• Multiply the following polynomials.
• Run time

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Example

• Multiply the polynomials in
• The Discrete Fourier Transform(DFT) of the
  coefficient vectors
• DFT(a)
• 4.000, (-0.309 - 2.126i), (0.809 1.314i),
  (0.809 - 1.314i), ( -0.309 2.126i)
• DFT(b)
• 6.000, (-0.809 - 3.665i), (0.309 1.677i),
  (0.309 -1.677i), (-0.809 3.665i)
• Run time

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Example

• DFT(a)DFT(b)
• 24.00, (-7.545 2.853i), (-1.954 1.763i),
  ( -1.954 - 1.763i), (-7.545 - 2.853i)