hw03_sol

hw03_sol - CS170 Spring 2010 HW3 Solutions 1(DPV 2.7 For n...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CS170 - Spring 2010 HW3 - Solutions 1. (DPV 2.7) For n 6 = 0 and ω = e 2 π n : n- 1 X i =0 ω i = 1- ω n 1- ω = 0 n- 1 Y i =0 ω i = ω P n- 1 i =0 i = ω n ( n- 1) 2 The latter is 1 if n is odd and- 1 if n is even. 2. (DPV 2.9) a) Use N = 4 and ω = e 2 πi/ 4 = i . The FFT of x + 1 is FFT(1 , 1 , , 0) = (2 ,i + 1 , ,- i + 1). The FFT of x 2 + 1 is FFT(1 , , 1 , 0) = (2 , , 2 , 0). Hence, the FFT of their product is (4 , , , 0), corresponding to the polynomial 1 + x + x 2 + x 3 . b) Use ω = i . The FFT of 2 x 2 + x +1 is FFT(1 , 1 , 2 , 0) = (4 ,- 1+ i, 2 ,- 1- i ). The FFT of 3 x +2 is FFT(2 , 3 , , 0) = (5 , 2 + 3 i,- 1 , 2- 3 i ). The FFT of their product is then (20 ,- 5- i,- 2 ,- 5 + i ). This corresponds to the polynomial 6 x 3 + 7 x 2 + 5 x + 2. 3. (DPV 2.30) (a) Observe that taking ω = 3 produces the following powers : ( ω,ω 2 ,ω 3 ,ω 4 ,ω 5 ,ω 6 ) = (3 , 2 , 6 , 4 , 5 , 1). Verify that ω + ω 2 + ω 3 + ω 4 + ω 5 + ω 6 = 1 + 2 + 3 + 4 + 5 + 6 = 21 = 0 ( mod 7) (b) The matrix M 6 (3) is the following: 1 1 1 1 1 1 1 3 2 6 4 5 1 2 4 1 2 4 1 6 1 6 1 6 1 4 2 1 4 2 1 5 4 6 2 3 Multiplying with the sequence (0 , 1 , 1 , 1 , 5 , 2) we get the vector (3 , 6 , 4 , 2 , 3 , 3). (c) The inverse matrix of M 6 (3) is easily seen to be the matrix 6 · 1 1 1 1 1 1 1 5 4 6 2 3 1 4 2 1 4 2 1 6 1 6 1 6 1 2 4 1 2 4 1 3 2 6 4 5 1 Verify that multiplying these two matrices mod 7 equals the identity. Also multiply this matrix with vector (3 , 6 , 4 , 2 , 3 , 3) to get the original sequence. (d) We first express the polynomials as vectors of dimension 6 over the integers mod 7: (1 , 1 , 1 , , , 0), and (- 1 , 2 , , 1 , , 0) = (6 , 2 , , 1 , , 0) respectively. We then apply the matrix M 6 (3) to both to get the transform of the two sequences. That produces (3 , 6 , , 1 , , 3) and (2 , 4 , 4 , 3 , 1 , 1) respectively....
View Full Document

This note was uploaded on 04/30/2010 for the course CS 170 taught by Professor Henzinger during the Spring '02 term at Berkeley.

Page1 / 5

hw03_sol - CS170 Spring 2010 HW3 Solutions 1(DPV 2.7 For n...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online