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CSC =========================================================================== 363 Lecture Summary for Week 10 Fall 2008 =========================================================================== Examples of NP-completeness: - Definition: a propositional formula is in Conjunctive Normal Form (CNF) if it is written as a conjunction ("and") of one or more "clauses": C_1 /\ C_2 /\ ... /\ C_r where each clause is a disjunction ("or") of one or more "literals": C_j = (a_{j,1} \/ a_{j,2} \/ ... \/ a_{j,s_j}) where each literal is either a propositional variable or the negation of a propositional variable. For example, (p \/ ~q) /\ q is in CNF, but ~(p /\ q) /\ (r \/ ~p) is NOT in CNF (it is equivalent to some CNF formula, but the property of being in CNF is all about the way the formula is written, *not* what it is equivalent to). The proof of the Cook-Levin Theorem actually shows that CNF-SAT is NPc (where CNF-SAT = { <F> : F is a formula in CNF that is satisfiable }), because it's possible to construct the formula F_x to be in CNF. - Definition: a propositional formula is in 3-CNF if it is in CNF where each clause contains exactly 3 literals, e.g., (p \/ q \/ ~r) /\ (p \/ p \/ ~q) 3SAT = { <F> : F is a formula in 3-CNF that is satisfiable } is NPc: 3SAT (- NP because it's a special case of SAT (same verifier works). 3SAT is NP-hard because CNF-SAT <=p 3SAT: reduction written up and posted along with summary of lecture notes for week 9. Note: Careful with directions! Trivially, 3SAT <=p CNF-SAT (3SAT is special case of CNF-SAT). But we need other direction to conclude 3SAT is NP-hard, transforming instances of general problem into instances of restricted problem (trickier). - VERTEX-COVER (VC for short) = { <G,k> : G is a graph that contains a vertex cover of size k, i.e., a set C of k vertices such that each edge of G has at least one endpoint in C } . VC (- NP: certificate = vertex cover of size k. . VC is NP-hard: 3SAT <=p VC. Given F = (a1 \/ b1 \/ c1) /\ ... /\ (ar \/ br \/ cr), where ai,bi,ci in {x1,~x1,x2,~x2,...,xs,~xs}, construct G=(V,E) and k such that F satisfiable iff G contains vertex cover of size k, as follows: k = s + 2r V = { a1,b1,c1, ..., ar,br,cr, x1,~x1, ..., xs,~xs } E = { (xi,~xi) : 1 <= i <= s } U { (ai,bi),(bi,ci),(ci,ai) : 1 <= i <= r } U { (l,x) : l = ai or bi or ci, and x = xj or ~xj corresponding to l } For example, if F = (x1 \/ ~x2 \/ ~x4) /\ (x2 \/ ~x3 \/ x1) /\ (~x3 \/ x4 \/ ~x2), then a1=x1, b1=~x2, c1=~x4, a2=x2, b2=~x3, c2=x1, a3=~x3, b3=x4, c3=~x2 so k = 4 + 2*3 = 10 V = {a1,b1,c1, a2,b2,c2, a3,b3,c3, x1,~x1, x2,~x2, x3,~x3, x4,~x4} E = { (x1,~x1), (x2,~x2), (x3,~x3), (x4,~x4), (a1,b1), (b1,c1), (c1,a1), (a1,x1), (b1,~x2), (c1,~x4), (a2,b2), (b2,c2), (c2,a2), (a2,x2), (b2,~x3), (c2,x1), (a3,b3), (b3,c3), (c3,a3), (a3,~x3), (b3,x4), (c3,~x2) } Clearly, construction can be done in polytime (with one scan of F). Also, if F is satisfiable, then there is an assignment of truth values that make at least one literal in each clause true. Pick a cover C as follows: for each variable, C contains xi or ~xi, whichever is true under the truth assignment; for each clause, C contains every literal except one that's true (pick arbitrarily if more than one true literal). C contains exactly s+2r vertices and is a cover: all edges (xi,~xi) are covered; all edges in clause triangles are covered (because we picked two vertices from each triangle); all edges between "clauses" and "variables" are covered (two from inside triangle, one from true literal for that clause). Finally if G contains a cover C of size k=s+2r, C must contain at least one of xi or ~xi for each i (because of edges (xi,~xi)) and at least two of ai,bi,ci for each i (because of triangle), so only way for C to have size s+2r is to contain exactly one of xi or ~xi and exactly two of ai,bi,ci, for each i. Since C covers all edges with only two vertices per triangle, the third vertex in each must triangle have its "outside" edge covered because of xi or ~xi. If we set literals according to choices of xi or ~xi in C, this will make formula F true: at least one literal will be true in each clause (because at least one edge from "variables" to "clauses" is covered by the variable in C). - INDEPENDENT-SET = { <G,k> | G is an undirected graph that contains an independent set of size k -- a subset of vertices with NO edge between any two of them } INDEPENDENT-SET (IS) is NPc: . in NP: certificate = independent set . NP-hard: VC <=p IS: exercise... - CLIQUE is NPc: Already known in NP. For NP-hardness, use reduction from INDEPENDENT-SET: On input <G,k>, construct <G',k'> as follows: G' = complement of G (same vertices, edge in G' iff edge not in G) k' = k This construction can obviously be carried out in polytime. Moreover, if G contains an I.S. of size k, then the same set in G' will be a clique of size k; conversely, if G' contains a clique of size k, the same set in G will be an I.S. of size k. Note: textbook uses different reduction from 3SAT. - SUBSET-SUM is NPc: Already known in NP. For NP-hardness, show 3SAT <=p SS: Given formula F = (a1 \/ b1 \/ c1) /\ ... /\ (ar \/ br \/ cr) where ai,bi,ci in {x1,~x1,...,xs,~xs}, construct numbers as follows: . For j = 1,...,s, number xj = 1 followed by s-j 0s followed by r digits where k-th next digit equals 1 if xj appears in clause C_k, 0 otherwise; number ~xj = 1 followed by s-j 0s followed by r digits where k-th next digit equals 1 if ~xj appears in clause C_k, 0 otherwise. . For j = 1,...,r, number Cj = 1 followed by r-j 0s and number C'j = 2 followed by r-j 0s. . Target t = s 1s followed by r 4s. Clearly, this can be constructed in polytime. Example of reduction for F = (x1 \/ ~x2 \/ ~x4) /\ (x2 \/ ~x3 \/ x1) /\ (~x3 \/ x4 \/ ~x2): S = { x1 = 1000110, ~x1 = 1000000, x2 = 0100010, ~x2 = 0100101, x3 = 0010000, ~x3 = 0010011, x4 = 0001001, ~x4 = 0001100, C1 = 0000100, C'1 = 0000200, C2 = 0000010, C'2 = 0000020, C3 = 0000001, C'3 = 0000002 } t = 1111444 Note: slightly different from book to ensure S contains all distinct numbers (book's reduction constructs S with repeated numbers, so S not really a set). If F is satisfiable, then there is a setting of variables such that each clause of F contains at least one true literal. Consider the subset S' = {numbers that correspond to true literals}. By construction, \sum_{x (- S'} x = s 1s followed by r digits, each one of which is either 1, 2, or 3 (because each clause contains at least one true literal). This means it is possible to add suitable numbers from {C1,C'1,...,Cr,C'r} so that the last r digits of the sum are equal to 4, i.e., there is a subset S' such that \sum_{x (- S'} x = t. If there is a subset S' of S such that \sum_{x (- S'} x = t, then S' must contain exactly one of {xj,~xj} for j = 1,...,n, because that is the only way for the numbers in S' to add to the target (with a 1 in the first s digits). Then, F is satisfied by setting each variable according to the numbers in S': for each clause j, the corresponding digit in the target is equal to 4 but the numbers Cj and C'j together only add up to 3 in that digit; this means that the selection of numbers in S' must include some literal with a 1 in that digit, i.e., clause j contains at least one true literal. ----------------- Self-reducibility ----------------- Note: This material is not in the textbook. This summary will therefore be slightly more detailed than usual. Problem of deciding language A sometimes called "decision problem": given input x, solution = yes/no answer. But many problems are more naturally "search problems": given input x, find solution y. Examples: - Given prop. formula F, find satisfying assignment, if one exists. - Given graph G, integer k, find a clique of size k in G, if one exists. - Given graph G, find a Ham. path in G, if one exists. - Given set of numbers S, target t, find subset of S whose sum equals t, if one exists. - etc.

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