Multiplying Polynomials Fast

Why do we need Complex Numbers?

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?

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?

Conversion to Polynomial Multiplication

• Treat P and T as polynomials

– T=0101 1 x0 + 0 x1 + 1 x2 + 0 x3

– P=01D1 0 x0 + 1 x1 + 0 x2 + 1 x3

• Multiply Prev and T

– (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

1 0 1 0 1 1 0 0 1

1 0 0 0 1 0 0

tktk-1tk-2tk-i

p-0p-1p-i

Other Applications

• Image Processing

Slide this mask all over the bigger one

At this location

Multiply each bit in the mask with

the corresponding bit

in the image, sum these up

Polynomial MultiplicationAn 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!!!

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

Multi-Point Polynomial EvaluationSpeed 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

Fast Multi-Point Polynomial Evaluation

T0(x)=A(x) mod (x-v0) (x-v1).. (x-vn)

T2 (x)=T0 (x) mod (x-v(n+1)/2) (x-vn) T1(x)=T0 (x) mod (x-v0)..(x-v(n+1)/2-1)

• 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)!!

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

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

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

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

Conclusion

• Also called : Fast Fourier Transform

• Complex Numbers are interesting!

• Reality can be explained elegantly only with complex numbers!!