Parallel Recursion: Powerlists. Greg Plaxton Theory in Programming Practice, Fall 2005 Department of Computer Science University of Texas at Austin

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1 Parallel Recursion: Powerlists Greg Plaxton Theory in Programming Practice, Fall 2005 Department of Computer Science University of Texas at Austin

2 Overview Definition of a powerlist Basic operations on powerlists Powerlist formulation of Batcher s bitonic merge Powerlist formulation of FFT

3 Powerlist A powerlist is a list of length 2 d for some nonnegative integer d We will refer to d as the dimension of the powerlist As in Haskell, all of the elements of a powerlist are required to be of the same type Two powerlists p and q are similar if they have equal dimension and the elements of p are of the same type as the elements of q Here are a couple of examples: We write the dimension 0 powerlist containing the sole element x as x We write the dimension 2 powerlist corresponding to the sequence of integers 1, 7, 5, 3 as

4 Basic Operations on Powerlists Given two similar powerlists p and q, we can form a larger powerlist (dimension one higher) in one of two basic ways The powerlist p q is the powerlist obtained by concatenating p and q The powerlist p q is the powerlist obtained by alternately taking elements from p and q, starting with p Here are a couple of examples: = =

5 Deconstruction of Powerlists For any non-singleton powerlist p, there is a unique set of similar powerlists a, b, c, d such that p = a b and p = c d

6 Batcher s Bitonic Sort Revisited Recall that Batcher s bitonic sort is a comparator network, so by the zero-one principle it is sufficient to study its behavior on zero-one inputs Throughout our discussion of bitonic sort, we restrict our attention to zero-one powerlists Recall that the key component of Batcher s bitonic sort is the bitonic merge We will give a powerlist-based presentation and analysis of bitonic merge

7 Bitonic Merge The bitonic merge takes as input a bitonic sequence and permutes it into sorted order If the input sequence is of the form 0 a 1 b 0 c, the output is 0 a+c 1 b If the input sequence is of the form 1 a 0 b 1 c, the output is 0 b 1 a+c Before presenting a powerlist-based version of bitonic merge, we define some useful auxiliary functions

8 Bitonic Merge: Auxiliary Functions It will be useful to define the min function over an arbitrary pair of similar powerlists We can do so as follows: x min y = min(x, y) (p q) min(r s) = (p min r) (q min s) We define max similarly The following operator will be particularly useful in our presentation of bitonic merge p q = (p min q) (p max q) Finally, we define z(p) as the number of zeros in the powerlist p

9 Bitonic Merge Let f denote the powerlist function defined as follows f( x ) = x f(p q) = f(p) f(q) In the slides that follow, we prove that for any bitonic powerlist p, f(p) is sorted (ascending) and z(f(p)) = z(p) The proof is by induction on the dimension of the argument of f The base case, where the dimension is zero, is easy We focus on the induction step in what follows First we establish a few useful lemmas

10 Lemma 1: If p q is bitonic, then so are p and q To see this, note that any subsequence of a bitonic sequence is bitonic

11 Lemma 2: If p q is bitonic, then z(p) z(q) 1 If p q is of the form 0 a 1 b 0 c, then z(p) = a/2 + c/2 and z(q) = a/2 + c/2 Thus z(p) z(q) 1 A symmetric argument may be used for the case where p q is of the form 1 a 0 b 1 c

12 Lemma 3: If p is sorted, q is sorted, and z(p) z(q) 1, then p q is sorted By symmetry (note that is a symmetric operator, since min and max are symmetric operators), it is sufficient to consider the following two cases Case 1: z(p) = z(q) Thus p and q are both of the form 0 a 1 b, so p min q = p max q = 0 a 1 b Thus p q = 0 a 1 b 0 a 1 b = 0 2a 1 2b, which is sorted Case 2: z(p) = z(q) + 1 Thus p and q are of the form 0 a+1 1 b 1 and 0 a 1 b, respectively, so p min q = 0 a+1 1 b 1 and p max q = 0 a 1 b Thus p q = 0 a+1 1 b 1 0 a 1 b = 0 2a+1 1 2b 1, which is sorted

13 Lemma 4: z(p q) = z(p) + z(q) We have z(p q) = z((p min q) (p max q)) = z(p min q) + z(p max q) It is easy to prove by induction on the dimension of p (and q) that z(p min q) + z(p max q) = z(p) + z(q)

14 Correctness of Bitonic Merge We need to prove that for any bitonic powerlist p q, f(p q) is sorted and z(f(p q)) = z(p q) By Lemma 1, p and q are each bitonic, so we can inductively assume that f(p) and f(q) are both sorted, z(f(p)) = z(p), and z(f(q) = z(q) By Lemma 2, z(p) z(q) 1 By Lemma 3, f(p) f(q) is sorted By Lemma 4, z(f(p) f(q)) = z(f(p)) + z(f(q)) = z(p) + z(q) This completes the proof since f(p q) is defined as f(p) f(q) and z(p q) = z(p) + z(q)

15 FFT Revisited: Overview Polynomial evaluation Some useful definitions Powerlist formulation of FFT Proof of correctness

16 Polynomial Evaluation We represent a polynomial with coefficients c 0,..., c n 1, where n is a power of 2, by the powerlist c 0 c n 1 For any pair of powerlists p and q (of possible unequal length), we define p q as the powerlist of the same length as q for which the ith component is equal to the value of the polynomial associated with p evaluated at the ith component of q Example: Suppose p = 1 2 and q = Then p corresponds to the polynomial 1 + 2x, and p q is equal to

17 Some Properties of Property 1: (p q) r = (p r 2 ) + (r (q r 2 )) Note that the addition and multiplication (including squaring) operations performed above are pointwise Example: The square of is Property 2: p (u v) = (p u) (p v)

18 Some Useful Definitions As in our earlier discussion of FFT, for any positive integer n, let ω n denote e 2πi/n Recall that ω n n = 1 and ω 2 2n = ω n For any length-n powerlist p, let f(p) denote the powerlist ω 0 n, ω 1 n,..., ω n 1 n Thus [f(p q)] 2 = f(p) f(q) For any length-n powerlist p, let g(p) denote the powerlist ω 0 2n, ω 1 2n,..., ω n 1 2n Thus f(p q) = g(p) g(p)

19 Fast Fourier Transform We wish to compute p f(p) efficiently We will show that the following powerlist function h (corresponding to the FFT) does the job h( x ) = x h(p q) = (a + c b) (a c b) where a and b are recursively defined as h(p) and h(q), respectively, and c = g(p) We prove the claim by induction on the dimension of the argument of h The base case, where the dimension is zero, is easy We focus on the induction step in what follows

20 FFT Correctness We now successively rewrite the expression (p q) f(p q) into the desired form By Property 1 of, we can rewrite the preceding expression as p [f(p q)] 2 + f(p q) (q [f(p q)] 2 ) Since [f(p q)] 2 = f(p) f(q), the preceding expression is equal to p (f(p) f(q)) + f(p q) (q (f(p) f(q))) By Property 2 of, the preceding expression is equal to ((p f(p)) (p f(q))) + f(p q) ((q f(p)) (q f(q)))

21 FFT Correctness (continued) Thus far we have argued that the Fourier transform of p q is equal to ((p f(p)) (p f(q))) + f(p q) ((q f(p)) (q f(q))) Note that f(p) = f(q) Furthermore, we may inductively assume that p f(p), the Fourier transform of p, is equal to h(p) Similarly, we may assume that q f(q) is equal to h(q) Recall that f(p q) = g(p) g(p) Thus the Fourier transform of p q is equal to (h(p) h(p)) + (g(p) g(p)) (h(q) h(q))

22 FFT Correctness (continued) Thus far we have shown that the Fourier transform of p q is equal to (a a) + (c c) (b b) where a = h(p), b = h(q), and c = g(p) Because is a pointwise operator, any expression of the form (p q) (r s) is equal to (p r) (q s) Thus the Fourier transform of p q is equal to (a a) + ((c b) ( c b))

23 FFT Correctness (continued) Thus far we have shown that the Fourier transform of p q is equal to (a a) + ((c b) ( c b)) where a = h(p), b = h(q), and c = g(p) Because + is a pointwise operator, any expression of the form (p q) + (r s) is equal to (p + r) (q + s) Thus we can rewrite the preceding expression as (a + c b) (a c b) The latter expression coincides with the definition of h(p q), completing the proof