Complex numbers polynomial multiplication

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Complex numbers polynomial multiplication

  1. 1. Multiplying Polynomials Fast Why do we need Complex Numbers?
  2. 2. The Problem• (an xn + an-1 xn-1+…..+ a0 x0) * (bn xn + bn-1 xn-1+…..+ b0 x0)• O(n^2) time – (Σi=0..k ai * bk-i) is coefficient of xk• Can one do better?
  3. 3. Applications• Where all does a pattern string P appear in a text string T? – P 0’s, 1’s and don’t cares. – T 0’s and 1’s• Easy in O(|P|*|T|) time – Can one do better?
  4. 4. Conversion to Polynomial Multiplication• Treat P and T as polynomials – T=0101 1 x0 + 0 x1 + 1 x2 + 0 x3 tk-i tk-2 tk-1 tk – P=01D1 0 x0 + 1 x1 + 0 x2 + 1 x3 1 0 1 0 1 1 0 0 1• Multiply Prev and T 1 0 0 0 1 0 0 p-i p-1 p-0 – (Σi=0..k previ * tk-i) is coefficient of xk – (Σi=0..k p|P|-i * tk-i) is coefficient of xk• >=1 if and only if a 1 in P aligns with a 0 in T when P is placed with end at tk – 2 polynomial multiplications suffice to find all matches of P in T
  5. 5. Other Applications• Image Processing At this location Slide this mask all over the Multiply each bit bigger one in the mask with the corresponding bit in the image, sum these up
  6. 6. Polynomial Multiplication An Equivalent Form• Evaluate each polynomial at 2n+1 distinct x’s – A(x) = (an xn + an-1 xn-1+…..+ a0 x0) -> A(v0)…..A(v2n) – B(x) = (bn xn + bn-1 xn-1+…..+ b0 x0) -> B(v0)…..B(v2n)• A(x) * B(x) -> A(v0)*B(v0)………..A(v2n)*B(v2n) – Convolution in one domain=Simple multiplication in another – O(n) time!!!
  7. 7. Multi-Point Polynomial Evaluation• Evaluate A(x) at v0 ……vn – O(n) time per vi using Horner’s rule• Problems – O(n2) time – Large numbers with n log vi bits
  8. 8. Multi-Point Polynomial Evaluation Speed Up• A(x) mod (x-v) – A(v) – O(n) time using high school polynomial division• A’(x) = A(x) mod (x-v0) (x-v1) [how fast?] – A’(x) mod (x-v0) = A(v0) [O(1)] – A’(x) mod (x-v1) = A(v1) [O(1)] – 2 expensive polynomial divisions could potentially be replaced by 1
  9. 9. Fast Multi-Point Polynomial Evaluation T0(x)=A(x) mod (x-v0) (x-v1).. (x-vn)T1(x)=T0 (x) mod (x-v0)..(x-v(n+1)/2-1) T2 (x)=T0 (x) mod (x-v(n+1)/2) (x-vn) • If T(x) mod (x-vi).. (x-vj) can be done in O(deg(T)) time, then what is the total time taken? – O(n log n)!!
  10. 10. Computing T(x) mod (x-v1).. (x-vk)• High school algorithm – Time taken: O( deg(T) * k ) – How do we make this faster?• Stroke of genius • Can we choose vi’s so (x-v1).. (x-vk) = xk-v? • Then we get O(deg(T) ) time
  11. 11. Choosing vi’s• When is (x-v1)(x-v2) = x2+v1v2 ? – v2+v1=0• When is (x-a)(x+a)(x-b)(x+b) = x4+ a2b2? – b2-a2=0 => b2=-a2 • => b= sqrt(-1) a • the 4 numbers are 1 –i -1 i • alternatively (–i)0 (–i)1 (–i)2 (–i)3
  12. 12. Choosing vi’s• Choose vi to be roots of xn -1 = Cos(2Πk/n) + i Sin(2Πk/n) = e i 2Πk/n – Powering just goes around the unit circle • Computing with log n bits suffices
  13. 13. Exercise• Choose vi to be roots of xn -1 = Cos(2Πk/n) + i Sin(2Πk/n) = e i 2Πk/n – Organize these roots so we get polynomials with just 2 terms in every node of the tree on slide 9 • Assume n+1 is a power of 2
  14. 14. Conclusion• Also called : Fast Fourier Transform• Complex Numbers are interesting! • Reality can be explained elegantly only with complex numbers!!

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