CSCI 3130: Formal languages and automata theory

Fall 2010 The Chinese University of Hong Kong CSCI 3130: Formal languages and automata theory More NP-complete problems Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

NP-complete problems NP-complete DOM SET MIN COVER VERTEX COVER SUBSET SUM IND SET HAM CYCLE HAM PATH PARTITION CLIQUE 3SAT NP SAT P

SAT and 3SAT SAT = {〈f〉: fis a satisfiable Boolean formula} ((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧x3) 3SAT = {〈f〉: fis a satisfiable 3CNF formula} (x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

Reducing SAT to 3SAT • Example: f = (x2∨(x1∧x2 ))∧(x1∧(x1∨x2 )) x4x5 x7 x7 = x4 ∧ x5 x10 T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T AND (x4∨x5∨x7) x8 x9 (x4∨x5∨x7) OR NOT x6 x7 (x4∨x5∨x7) AND AND (x4∨x5∨x7) x4 x5 x3 NOT NOT OR (x4∨x5∨x7) ∧(x4∨x5∨x7) x2 x1 x2 x1 x1 x2 ∧(x4∨x5∨x7) ∧(x4∨x5∨x7) We add extra variables for every gate (“wire”)

Turning gates into 3CNFs z z z Gj: AND OR NOT x y x y x x y z = y ∧ x x y z = y ∨ x x z = x z z z T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T T T T TT T F FT F T TT F F FF T T TF T F FF F T FF F F T T T FT F T F T TF F F (x∨z)∧(x∨z) (x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z) (x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z) fj: (x∨x∨z)∧(x∨x∨z)

Reducing SAT to 3SAT R Boolean formula f 3CNF formula f’ R:= On input f, where f is a boolean formula Construct and output the following 3CNF formula f’: f’ has extra variables xn+1, ..., xn+t one for each gate Gj in f For each gate Gj, construct the formulafj Output f’ = fn+1 ∧ fn+2 ∧ ... ∧ ft ∧ (xn+t ∨xn+t ∨xn+t ) requires thatoutput of f is true

Example x4 AND f’ := (x2∨x2∨x3)∧(x2∨x2∨x3) NOT gate x3 (x1∨x3∨x4)∧(x1∨x3∨x4) ∧(x1∨x3∨x4)∧(x1∨x3∨x4) ∧ f = AND gate NOT (x4∨x4∨x4) ∧ output x1 x2 T AND T x1 = T, x2 = F, x3 = T, x4 = T NOT is a satisfying assignment of f’ T F

Reducing SAT to 3SAT • Every satisfying assignment of f extends uniquely to a satisfying assignment of f’ • In the other direction, in every satisfying assignment of f’ the part x1, ..., xnis a satisfying assignment of f R Boolean formula f 3CNF formula f’ ✔ f’ is satisfiable f is satisfiable

How to argue NP-completeness SUBGRAPH = {〈G, H〉: Gand H are graphs where H is a subgraph of G} 6 7 4 5 4 Every edge of H must be included in G 3 1 2 1 2 3 G H SUBGRAPH is NP-complete

How to argue NP-completeness ✔ • Step 1: SUBGRAPH is in NP What is a solution for SUBGRAPH? H G 6 7 4 A solution s is a map from vertices of H to vertices of G 5 4 3 (1➝4)(2➝2)(3➝5)(4➝7) 1 2 1 2 3 V := On input 〈G, H, s〉 Running time = O(m) m = number edges in H For every edge {u, v} of H: Find u’ and v’ such that u➝u’ and v➝v’ in s. If {u’, v’} is not an edge in G, reject. Otherwise, accept.

How to argue NP-completeness • Step 2: Some NP-complete Lreduces to SUBGRAPH Direction of reduction is important! DOM SET You want to show SUBGRAPH is harder than L VERTEX COVER So you assume you can solve SUBGRAPH, and show how to solve L IND SET CLIQUE SAT 3SAT This requires converting instances of L intoinstances of SUBGRAPH How to choose L?

Tips on choosing L • Both CLIQUE and SUBGRAPH are about graph containment, so their complexities may be related • It usually makes sense to try and reduce from the problem that looks the most related 4 G G H 6 7 4 CLIQUE 5 SUBGRAPH 4 3 1 2 1 2 3

How to argue NP-completeness ✔ • Step 3: Reduce CLIQUE to SUBGRAPH CLIQUE = {〈G, k〉: Ghas a k-clique} SUBGRAPH = {〈G, H〉:H is a subgraph of G} R := On input 〈G, k〉 Ghas a k-clique Construct the k-clique H Output 〈G, H〉 H is a subgraph of G

Cover for triangles • A graph has a k-cover for triangles if there are k vertices that touch all the triangles Does it have a 2-cover for triangles? Yes Does it have a 1-cover for triangles? No, because it has two vertex-disjoint triangles TRICOVER = {〈G, k〉:G has a k-coverfor triangles} TRICOVER is NP-complete

Cover for triangles 6 7 ✔ • Step 1: TRICOVER is in NP 5 4 What is a solution for TRICOVER? 1 2 3 A solution s is a subset of vertices like {2,5} V := On input 〈G, k, s〉, where s is a set of vertices of size k For every triple {u, v, w} of vertices: If {u, v}, {u, w}, {v, w}, are all edges in G: If none of u, v, orw are ins, reject. Otherwise, accept. Running time = O(n3)

Cover for triangles • Step 2: Some NP-complete Lreduces to TRICOVER TRICOVER = {〈G, k〉:G has a k-coverfor triangles} SUBGRAPH DOM SET VERTEX COVER Some vertex in every triangle is covered CLIQUE IND SET VC = {〈G, k〉:G has a k-vertex cover} SAT 3SAT Some endpoint of every edge is covered

Cover for triangles • Step 3: Reduce VC to TRICOVER VC = {〈G, k〉:G has a k-vertex cover (for edges)} TRICOVER = {〈G’, k’〉:G’ has k’-coverfor triangles} Idea: Replace edges by triangles G G’ vertex-cover in G cover for triangles in G’

Cover for triangles • Step 3: Reduce VC to TRICOVER R := On input 〈G, k〉, where G has nvertices andmedges Construct the following graph G’: G’ has n + mvertices v1, ..., vn, anduij for every edge {vi, vj} of G For every edge {vi, vj} of G, include the edges {vi, vj}, {vi, uij}, {vj, uij} in G’ Output 〈G’, k〉. Running time = O(m) m = number of edges in G

Cover for triangles • Step 4: Argue that the reduction is correct 〈G, k〉 ∈ VC 〈G’, k〉 ∈ TRICOVER G ✔ G’ G has a k-vertex cover C C is a k-triangle cover of G’ Old triangles from G are covered New triangles in G’ also covered

Cover for triangles • Step 4: Argue that the reduction is correct ✔ 〈G, k〉 ∈ VC 〈G’, k〉 ∈ TRICOVER G G’ ✔ G has a k-vertex cover C’ G’ has a k-vertex cover C C’ is obtained after moving some vertices of C Some vertices in C may not come from G! Since C’ covers all triangles in G’, it must cover all edges in G But we can move them and still cover the same triangle

Coloring Suppose you need to schedule exams on same day in 3 time slots CSCI 3190 CSCI 3230 CSCI 2110 Y R B 1030-1230 130-330 430-630 CSCI 3130 CSCI 3160 Some people are enrolled in multiple classes, so there may be conflicts Task: Schedule the exams so there are no conflicts

Coloring Given a graph, we want to color the vertices so every edge has endpoints of different colors G Y R B This is called a valid 3-coloring 3COL = {〈G〉:G has a valid 3-coloring} 3COL is NP-complete

Coloring ✔ 5 • Step 1: 3COL is in NP 4 G 3 What is a solution for 3COL? 1 2 A solution s is a coloring of vertices like {1B,2R,3Y,4R,5B} V := On input 〈G, s〉 Running time = O(m) For every edge {u, v} of G: If uand v are assigned same color in s, reject. Otherwise, accept.

Coloring • Step 2: Some NP-complete Lreduces to 3COL 3COL = {〈G〉:G has a valid 3-coloring} SUBGRAPH DOM SET 3COL does not look like any of these problems VERTEX COVER CLIQUE IND SET In such a case it is a good idea to start with 3SAT SAT 3SAT

Coloring • Step 3: Reduce 3SAT to 3COL 3SAT = {〈f〉:f is a 3CNF that has a satisfying assignment} 3COL = {〈G〉:G has a valid 3-coloring} Idea: Assignment of f Coloring of G G X T F Part I: 3 special vertices T (true), F (false), and X

Coloring Either xi has color of T and xi has color of F X xi xi Or xi has color of F and xi has color of T Part 2: For each variable xi T F F T F T x1 x1 x2 x2 x3 x3 X T F Example: (x1∨x1∨x2)∧(x1∨x2∨x3 )

Coloring • To “encode” the clauses of f, we need a gadget x If xory is colored T, zcan be colored T z If xandy are colored F, zmust be colored F y x1 Can be colored if and only if at least one literal is colored T x2 T x3 Part 3: For each clause like (x1∨x2∨x3 )

Coloring reduction: An example x1 x1 x2 x2 x3 x3 X T F Example: (x1∨x1∨x2)∧(x1∨x2∨x3 ) T T T F F F

Coloring • Step 3: Reduce 3SAT to 3COL Running time = O(m + n) n = number of variables m = number of clauses R := On input 〈f〉, where f is a 3CNF: Construct the following graph G: Add 3 special vertices T, F, X connected by a triangle. For every variable xiof f, add vertices xi and xi and include the triangle xi, xi, X. u For every clause u∨v∨wof f, connect u, v, w,and T using the gadget below. v w Output 〈G〉. T

Coloring • Step 4: Argue that the reduction is correct 〈f〉 ∈ 3SAT 〈G〉 ∈ 3COL ✔ x1 x1 x2 x2 x3 x3 X f has a SAT assignment a G has a valid 3-coloring T F Each literal gets “color” as in a Each clause gadget contains a true literal, so it can be colored (x1∨x1∨x2)∧(x1∨x2∨x3 )

Coloring • Step 4: Argue that the reduction is correct 〈f〉 ∈ 3SAT 〈G〉 ∈ 3COL ✔ x1 x1 x2 x2 x3 x3 f has a SAT assignment G has a valid 3-coloring c X Each variable gets value as in G T F true, if node xi has same color as T xi = false, if node xi has same color as F (x1∨x1∨x2)∧(x1∨x2∨x3 ) Since each clause gadget is colored properly, each clause must contain a true literal

CSCI 3130: Formal languages and automata theory