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# Administrative - PowerPoint PPT Presentation

Administrative. Oct. 2 Oct. 4 – QUIZ #2. (pages 45-79 of DPV). Polynomials. polynomial of degree d. p(x) = a 0 + a 1 x + ... + a d x d. Representing polynomial of degree d. the coefficient representation. (d+1 coefficients). evaluation. interpolation. the value representation.

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Oct. 2

Oct. 4 – QUIZ #2

(pages 45-79 of DPV)

polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

the coefficient representation

(d+1 coefficients)

evaluation

interpolation

the value representation

(evaluation on d+1 points)

polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

(Horner’s rule)

a0+x(a1+x(a2 + ... )))

R  0

for i from d to 0 do R  R*x + ai

a polynomial p of degree d such that

p(a0) = 1

p(a1) = 0

....

p(ad) = 0

(x-a1)(x-a2)...(x-ad)

(a0-a1)(a0-a2)...(a0-ad)

p1 = (1/2) x2 – (1/2) x

p2 = - x2 + 1

p3 = (1/2) x2 + (1/2) x

• compute p = (x-a0)...(x-ad)

• compute

• pi = p / (x-ai) for i=0,...,d

• ri = pi / pi(ai)

• 3) compute

• q = c0 r0 + ... + cd rd

Claim: q(ai) = ci for i=0,...,d

evaluation on 1 point O(d)

evaluation on d points O(d2)

interpolation O(d2)

Polynomial of degree d

p(x) = a0 + a1 x + ... + ad xd

Polynomial of degree d’

q(x) = b0 + b1 x + ... + bd’ xd’

p(x)q(x) = (a0b0) + (a0b1 + a1b0) x +

.... + (adbd’) xd+d’

>2d points

p,q in evaluation

representation

p,q

multiply

in

O(d) time

interpolate

pq in evaluation

representation

pq

p(x) = 7 + x + 5x2 + 3x3 + 6x4 + 2x5

p(z) = 7 + z + 5z2 + 3z3 + 6z4 + 2z5

p(-z) = 7 – z + 5z2 – 3z3 + 6z4 – 2z5

p(x) = (7+5x2 + 6x4) + x(1+3x2 + 2x4)

p( x) = pe(x2) + x po(x2)

p(-x) = pe(x2) – x po(x2)

p(x) = a0 + a1 x + a2 x2 + ... + ad xd

p(x) = pe(x2) + x po(x2)

p(-x) = pe(x2) – x po(x2)

To evaluate p(x) on

-x1,x1,-x2,x2,...,-xn,xn

we only evaluate pe(x) and po(x) on

x12,...,xn2

To evaluate p(x) on

-x1,x1,-x2,x2,...,-xn,xn

we only evaluate pe(x) and po(x) on

x12,...,xn2

To evaluate pe(x) on

x12,...,xn2

we only evaluate pe(x) on ?

2ik/n

= k

e

FACT 1:

n = 1

k . l = k+l

0 + 1 + ... + n-1 = 0

FACT 2:

FACT 3:

FACT 4:

k = -k+n/2

FFT (a0,a1,...,an-1,)

(s0,...,sn/2-1)= FFT(a0,a2,...,an-2,2)

(z0,...,zn/2-1) = FFT(a1,a3,...,an-1,2)

s0 + z0

s1 +  z1

s2 + 2 z2

....

s0 – z0

s1 -  z1

s2 - 2 z2

....

viewed as vector mutiplication

1

x

x2

.

.

xd

(a0,a1,a2,...,ad)

on multiple points

1

xn

xn2

.

.

xnd

1

x1

x12

.

.

x1d

1

x2

x22

.

.

x2d

(a0,a1,a2,...,ad)

. . .

Vandermonde matrix

1

d-1

2(d-1)

.

.

(d-1)

1

1

1

.

.

1

1

1

2

.

.

d-1

(a0,a1,a2,...,ad)

. . .

2

= e2 i / d

FT - matrix

1

d-1

2(d-1)

.

.

(d-1)

1

1

1

.

.

1

1

1-d

2(1-d)

.

.

-(d-1)

1

1

2

.

.

d-1

1

-1

-2

.

.

1-d

1

1

1

.

.

1

. . .

. . .

2

2

= e2 i / d

1

1

0

D

I

Fn

. . .

1

1

I

-D

Fn

0

. . .

D=diag(1,,...,d-1)

d=2n

string = x1,....,xn

pattern = p1,...,pk

k

 (xi – pi)2

TEST =

i=1

Occurs on position j ?

string = x1,....,xn

pattern = p1,...,pk

k

 (xi – pi)2

TEST =

i=1

Occurs on position j ?

k

 (xi+j-1 – pi)2

TESTj =

i=1

Occurs on position j ?

k

 (xi+j-1 – pi)2

TESTj =

i=1

pj = 0 DON’T CARE

k

 pj(xi+j-1 – pi)2

TESTj =

i=1