This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6.045J/18.400J: Automata, Computability and Complexity Nancy Lynch Homework 11.1 (FAKE) Due: Never Elena Grigorescu This fake homework is intended as a study guide covering the material on class 22 (NPcomplete problems). Readings: Sipser, Section 7.5. Also (optionally) see Garey and Johnson’s book, “Computers and Intractability: a Guide to NPCompleteness”. Problem 1 : In class, we covered constructions reducing 3SAT directly to four other problems: • CLIQUE = {( G,k ) G is an undirected graph with a kclique } , • HAMPATH = {( G,s,t ) G is a directed graph with a Hamiltonian path from s to t } , • SUBSETSUM = {( S,t ) S = { x 1 ,...,x k } and for some { y 1 ,...,y ℓ } ⊆ { x 1 ,...,x k } , we have Σ y i = t } , and • 3DIMENSIONALMATCHING = {( A,B,C,M ) A,B,C are disjoint sets of size n , M ⊆ A × B × C , a set of acceptable triples, such that ∃ M ′ ⊆ M,  M ′  = n, and each element of A,B,C appears exactly once in M ′ } . In this problem, we propose variations on the constructions that were presented and ask you whether they work or not, and why. 1. We modify the construction reducing 3SAT to CLIQUE by adding an edge between each pair of nodes in the same triple, unless the pair is contradictory (e.g., x and x ). No! This reduction maps a Boolean formula φ to a graph G and a value k . (See page 252 of Sipser.) In the original reduction, we set k to be the number of clauses in φ . Adding these additional edges to G might render a kclique for an unsatisfiable formula, because it would then be possible to “skip” a clause when selecting nodes for the kclique (i.e., take two true literals from one clause, but skip a clause with three false literals). Obviously, this is not acceptable. 2. In the construction reducing 3SAT to HAMPATH , we constructed a diamond for each variable. The horizontal row contains 3 k + 1 nodes in addition to the two nodes on the ends belonging to the diamond (here k is the number of clauses in φ ). Now, we try to make the reduction more efficient by cutting out the “separator” nodes in the diamond, reducing the size of the horizontal row by 1 3 . No! Actually, a similar error appeared in the first printing of Sipser’s book. If your textbook was printed before 2001, you might want to check the HAMPATH proof in a friend’s copy. So, why would removing the separator nodes be a problem? The separator nodes help force the zigzag (or zagzig) traversal of each diamond, which corresponds to a true or false setting of that variable. If they were not present, a situation, like that illustrated in Figure 7.19 on page 267, might occur. Exercise for the reader: can you think of an example where this would happen? In such a case, it might be possible for a Hamiltonian graph to exist without φ being satisfiable....
View
Full
Document
This note was uploaded on 08/12/2009 for the course CS 430 taught by Professor Nancylynch during the Spring '07 term at New Mexico Junior College.
 Spring '07
 NancyLynch

Click to edit the document details