CSCI 3130: Formal languages and automata theory

1. Fall 2011 The Chinese University of Hong Kong CSCI 3130: Formal languages and automata theory The Cook-Levin Theorem Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

2. NP-completeness IS reduces to VC CLIQUE Cook-Levin theorem: 3SAT SAT NP-complete Everythingin NP reduces to SAT P NP (easy to solve) (easy to verify solution)

3. Polynomial-time reductions • Language Lpolynomial-time reduces to L’ if there exists a polynomial-time computable map R that takes an instance x of L into instance y of L’ s.t. x ∈ L if and only if y ∈ L’ L (CLIQUE) L’ (IS) R (G, k) x y (G’, k’) x ∈ L y ∈ L’ (G has clique of size k) (G’ has IS of size k’)

4. 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)

5. The Cook-Levin Theorem • To show this, we need a reduction R such that: Every L∈NP reduces to SAT R z Boolean formula f f is satisfiable z ∈ L

6. NP-completeness of SAT • All we know: L has a polynomial-time verifierV • Look at the computation tableau of V on input 〈z, s〉 S z | s V … q0 z1 z2 S-th configuration symbol at time T T … qacc a b

7. NP-completeness of SAT S … q0 z1 n = length of z z2 height of tableau isO(nc) for some constantc T u width is at most nc k possible tableau symbols … qacc a b true, if the (T, S) cell of tableau contains u 1 ≤ S ≤ nc xT, S, u = 1 ≤ T ≤ nc false, if not 1 ≤ u ≤ k

8. NP-completeness of SAT • We will design a formula f such that: R z Boolean formula f f is satisfiable z ∈ L variables of f : xT, S, u assignment to xT, S, u way to fill up the tableau accepting computation tableau satisfying assignment f is satisfiable V accepts 〈z, s〉for some s

9. NP-completeness of SAT • We want to construct (in time O(n2c)) a formula f : true, if the (T, S) cell of tableau contains u 1 ≤ S ≤ nc xT, S, u = 1 ≤ T ≤ nc false, if not 1 ≤ u ≤ k true, if the xs represent a legal and accepting tableau f(x1, 1, 1, ..., xnc, nc, k) = false, if not

10. NP-completeness of SAT f = fcell ∧ f0 ∧ fmove ∧ facc … q0 z1 z2 fcell: “Every cell contains exactly one symbol” f0: “The first row is q0z|s☐...☐” for some s … qacc a b fmove: “The moves between rows follow the transitions of V” facc: “qacc appears somewhere in the last row”

11. NP-completeness of SAT fcell: “Every cell contains exactly one symbol” “Exactly one of xS, T, 1 ∨ ... ∨ xS, T, k is true” fcell = fcell1, 1 ∧ ... ∧ fcellnc,nc where fcellT, S = (xT, S, 1 ∨ ... ∨ xT, S, k) ∧ (xT, S, 1 ∧ xT, S, 2) ∧ (xT, S, 1 ∧ xT, S, 3) ∧ ... ∧ (xT, S, k-1 ∧ xT, S, k) at least one symbol no cell containstwo or more symbols

12. NP-completeness of SAT f0: “The first row is q0z1z2...zn|s1s2...sm for some (possible solution) s1s2...sm” f0 = x1, 1, q0 ∧ x1, 2, z1 ∧ ... ∧ x1, n+1,zn ∧ x1, n+2, |

13. NP-completeness of SAT facc: “qacc appears somewhere in the last row” facc = xnc, 1, qacc ∨ xnc, 2, qacc ∨ ... ∨ xnc, nc, qacc

14. Legal and illegal windows legal windows illegal windows … 6a3b0x0 … … 0a6b0x0…0 … 6q2a0b0 … … 0a6b0q2 …0 … 6a3q2a0 … … 0q5a6x0…0 … 6q2q2a0 … … 0q2q2x3…0 q2 a/xL … 6#3b0a0 … … 0#6b0q5 …0 … 6a3q2a0 … … 0q5a6b0…0 q5 … 6a3a0☐0 … … 0x6a0☐0…0 … 6#3q2a0 … … 0q5#6x0…0

15. NP-completeness of SAT … q0 z1 z2 fmove: “The moves between rows follow transitions of V” a2 a3 a1 a4 a5 a6 LW = list of all legal windows fmove2, 2 = {(a1, a2, a3, a4, a5, a6): is legal} … qacc a b fmove = fmove1, 1 ∧ ... ∧ fmovenc-3, nc-3 ∨ fmoveT, S = (xT, S, a1 ∧ xT, S+1, a2 ∧ xT, S+2, a3 ∧ xT+1, S, a4 ∧ xT+1, S+1, a5 ∧ xT+2, S+1, a6) a1 a2 a3 (a1, a2, a3, a4, a5, a6) ∈ LW a4 a5 a6

16. NP-completeness of SAT R z Boolean formula f ✔ f is satisfiable z ∈ L Let V be a poly-time verifier TM for L. R := On input z, Construct the formulas fcell,f0,fmove, andfacc Output the formula f = fcell ∧ f0 ∧ fmove ∧ facc. Then R is computable in time O(n2c) and V accepts z if and only if f is satisfiable.

17. 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

18. 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.

19. How to argue NP-completeness • Step 2: Some NP-complete Lreduces to SUBGRAPH Direction of reduction is important! 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?

20. 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

21. 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

22. 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

23. 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)

24. Cover for triangles • Step 2: Some NP-complete Lreduces to TRICOVER TRICOVER = {〈G, k〉:G has a k-coverfor triangles} SUBGRAPH 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

25. 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’

26. 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

27. 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

28. 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

29. 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

30. 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

31. 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.

32. Coloring • Step 2: Some NP-complete Lreduces to 3COL 3COL = {〈G〉:G has a valid 3-coloring} SUBGRAPH 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

33. 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

34. 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 )

35. 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 )

36. 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

37. 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

38. 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

39. 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 )

40. 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