MA/CSSE 473 Day 19

1. MA/CSSE 473 Day 19 Subset Generation

2. MA/CSSE 473 Day 19 • HW 7 due today • Exam 1 Thursday • There will be significant time for your questions in tomorrow's class • HW 8 has grace days to allow you to finish it early in the break if you wish. • HW9 will be due on Tuesday after the break, HW 10 on Friday • With HW 11, we get back to Monday-Thursday schedule • Student Questions • Subset Generation

3. Preliminary: Reversing a string • Definition of reverse: • The reverse of the empty string is the empty string • If a is a character and w is a string, (aw)R = wRa. • Lemma: • For any two strings, x and y, (xy)R= yRxR • Proof by induction on length of y • Base case: y is empty • Otherwise, y = za for some string z and some character a. Assume by induction that the property is true for the shorter string z.

4. All Subsets of a Set • Sample Application: • Solving the knapsack problem • In the brute force approach, we try all subsets • If A is a set, the set of all subsets is called the power set of A, and often denoted 2A • If A is finite, then • So we know how many subsets we need to generate.

5. Generating Subsets of {a1, …, an} • Decrease by one: • Generate Sn-1, the collection of the 2n-1 subsets of {a1, …, an-1} • Then Sn = Sn-1 { s {an} : sSn-1} • Another approach: • Each subset of {a1, …, an} corresponds to an bit string of length n, where the ith bit it 1 iffai is in the subset

6. Another approach: • Each subset of {a1, …, an} corresponds to an bit string of length n, where the ith bit is 1 if and only if ai is in the subset defallSubsets(s): n = len(s) subsets=[] foriin range(2**n): subset = [] current = i for j in range (n): if current % 2 == 1: subset += [s[j]] current /= 2 subsets += [subset] return subsets Output:[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

7. Gray Codes • Named for Frank Gray • An ordering of the 2n n-bit binary codes such that any two consecutive codes differ in only one bit • Example: 000, 001, 011, 010, 110, 111, 101, 100 • Note also that only one bit changes between the last code and the first code. • A Gray code can be represented by its transition sequence: indicates which bit changes each timeIn above example: 0, 1, 0, 2, 0, 1, 0 • Traversal of the edges of a (hyper)cube. • In terms of subsets, the transition sequence tells which element to add or remove from one subset to get the next subset

8. Recursively Generating a Gray Code • Binary Reflected Gray Code • T1 = 0 • Tn+1 = Tn , n, Tnreversed • Show by induction that Tnreversed = Tn • Thus Tn+1 = Tn , n, Tn

9. Iteratively Generating a Gray Code • We add a parity bit, p. • Set all bits (including p) to 0. * Based on Knuth, Volume 4, Fascicle 2, page 6. Q4

10. Quote of the Day • There are 10^11 stars in the galaxy. That used to be a huge number. But it's only a hundred billion. It's less than the national deficit! We used to call them astronomical numbers. Now we should call them economical numbers.  - Richard Feynman