Chapter 3: Proving NP-completeness Results

1. Chapter 3: Proving NP-completenessResults • Six Basic NP-Complete Problems • Some Techniques for Proving NP-Completeness • Some Suggested Exercises

2. Six Basic NP-Complete Problems • 3-SATISFIABILITY (3SAT) • 3-DIMENSIONAL MATCHING (3DM) • VERTEX COVER (VC) • CLIQUE • HAMILTONIAN CIRCUIT (HC) • PARTITION

3. 3-Satisfiability • A restricted version of satisfiability in which all instances have exactly 3 literals per clause. 3-SATISFIABILITY (3-SAT) INSTANCE: Collection C = {c1, c2,…,cm} of clauses on a finite set U of variables such that {cj} = 3, for 1 ≤ j ≤ m. QUESTION: Is there a truth assignment for U that satisfies all the clauses in C?

4. A I’ = U’,C’ 3-SAT instance I = U,C SAT instance 3-Satisfiability Theorem 3.1: 3-SAT is NP-Complete. • 3-SAT  NP • SAT  3-SAT I is satisfiable iff I’ is

5. 3-Satisfiability • Each clause cj in I => a set Cj’ of clauses with new variables Uj’ • Let cj = {z1, z2,…,zk} where zi’s are literals derived from variables in U • Case 1) k=1 cj = {z1} Uj’ = {yj1, yj2} Cj’ = { {z1, yj1, yj2}, {z1, yj1, yj2}, {z1, yj1, yj2}, {z1, yj1, yj2} } • Case 2) k=2 cj = {z1, z2} Uj’ = {yj1} Cj’ = { {z1, z2, yj1}, {z1, z2, yj1} }

6. 3-Satisfiability • Case 3) k=3 cj = {z1, z2, z3} Uj’ = {} Cj’ = { {z1, z2, z3} } • Case 4) k>=4 cj = {z1, z2,…,zk} Uj’ = {yji | 1 ≤ i ≤ k-3} Cj’ = { {z1, z2, yj1} } U {yji, zi+2, yji+1 } | 1 ≤ i ≤ k-4} U {yjk-3, zk-1, zk } } Example: cj = {z1, z2, z3, z4, z5, z6} Uj’ = { {z1, z2, a}, {a, z3, b}, {b, z4, c}, {c, z5, z6} }

7. 3-Satisfiability • Finally, let: U’ = U  { Uj’ | 1 ≤ j ≤ m } C’ = { Cj’ | 1 ≤ j ≤ m } • Claim: 1) I = U,C is satisfiable iff I’ = U’,C’ is satisfiable. 2) I’ can be computed in polynomial time. Let m = |C|, n = |U| case 1) creates 4 clauses case 2) creates 2 clauses case 3) creates 1 clause case 4) creates k-1 clauses, where k is the number of literals in the clause This gives a total of at most k-1 new clauses in I’ for each clause in I Therefore, there is a total of m*(k-1) clauses in I’ Since k<=2n, it follows that there are at most 2nm clauses in I’, which is O(mn)

8. Some More New Problems Not All Equal 3-SAT INSTANCE: Collection C = {c1, c2,…,cm} of clauses on a finite set U of variables such that {cj} = 3, for 1 ≤ j ≤ m. QUESTION: Is there a truth assignment for U such that each clause in C has at least one true literal and at least one false literal? • Examples: { (a, b, c), (a, b, c), (a, b, c) } a = T, b = T, c = F { (a, b, c), (a, b, c), (a, b, c), Not even satistfiable (a, b, c), (a, b, c), (a, b, c) (a, b, c), (a, b, c) } { (a, b, c), (a, b, c), (a, b, c), Satistfiable, but not with (a, b, c), (a, b, c), (a, b, c) at least one literal true and false (a, b, c) } per clause

9. More Problems, cont. Hypergraph 2-Colorability (H2C) INSTANCE: Hypergraph H = (V, E), where 2 ≤ |ei| ≤ 3, for all ei E. QUESTION: Is H 2-colorable? In other words, is there a function f : V  {0, 1} such that for all ei E there exist vertices u,v  ei such that f(u)  f(v)? Examples: No Yes No 0 1 1 0 0 0

10. A H = (V, E) H2C instance I = U,C 3-SAT instance H2C is NP-Complete Fact: Not All Equal 3-SAT is NP-complete Theorem: H2C is NP-complete • H2C  NP • Not All Equal 3-SAT  H2C

11. H2C is NP-Complete, cont. • Suppose I = U,C where: U = {u0, u1,…,un-1} C = {c0, c1,…,cm-1} • Construct H = (V, E) where: V = { vi1 | ui U}  {vi2 | ui U} vi1 corresponds to ui vi2 corresponds to the complement of ui E = E1 E2 where E1 = { (v1, v2, v3) | ci C, ci = (z1, z2, z3) and v1, v2, v3 are the vertices corresponding to the literals z1, z2 and z3, respectively} E2 = { (vi1, vi2) | 0 ≤ i ≤ n-1}

12. H2C is NP-Complete, cont. • Example: U = {u0, u1, u2, u3} C = { (u0, u1, u2), (u0,u2, u3), (u1, u2 u3) }

13. 3-Dimensional Matching 3-DIMENSIONAL MATCHING (3-DM) INSTANCE: A set M  W  X  Y, where W, X and Y are disjoint sets having the same number q of elements. QUESTION: Does M contain a matching, that is, a subset M’  M sucht at |M| = q and no two elements of M’ agree in any coordinate? Example #1: q = 3 (not part of input technically) W = {a1, a2, a3} X = {b1, b2, b3} Y = {c1, c2, c3} M = { (a1, b2, c1), (a2, b1, c3), (a3, b3, c2), (a1, b1, c3), (a2, b2, c1) } Yes! The first 3 form a matching. Note that every element in each set will appear in exactly one triple in M’.

14. 3-Dimensional Matching Example #2: q = 3 (not part of input technically) W = {a1, a2, a3} X = {b1, b2, b3} Y = {c1, c2, c3} M = { (a1, b1, c1), (a1, b1, c2), (a2, b1, c3), (a3, b1, c1) } No! In fact b1 and b3 do not appear in any triple.

15. A I’ = W,X,Y,M 3DM instance I = U,C 3-SAT instance 3-Dimensional Matching Theorem: 3DM is NP-complete. Proof: • 3DM  NP • 3-SAT  3DM

16. 3-Dimensional Matching • Suppose I = U,C where: U = {u1, u2,…,un} C = {c1, c2,…,cm} • Construct M as follows. • M will consists of three types of “components:” • Truth Setting and fan-out • Satisfaction testing • Garbage collection

17. 3-Dimensional Matching • Suppose I = U,C where: U = {u1, u2,…,un} C = {c1, c2,…,cm} • Construct M as follows. • M will consists of three types of “components:” • Truth Setting and fan-out • Satisfaction testing • Garbage collection

18. Truth Setting and Fan Out Components • For each variable ui U, create the following elements: Tit = { (ui[j], ai[j], bi[j]) | 1 ≤ j ≤ m } U Tif = { (ui[j], ai[j+1], bi[j]) | 1 ≤ j ≤ m-1 } U { (ui[m], ai[1], bi[m]) } a big “blob” for each Boolean variable (see page 51) • Observation #1: Creates 2mn elements in W, mn in X and mn in Y. • ui[j] represents the fact that variable ui could occur in clause cj. • Similarly for ui[j] • Observation #2: ai[j] and bi[j] both occur in exactly two triples and nowhere else. • Observation #3: This tells us that in any matching M’  M we must have all white or all shaded triples, in other words, all triples from Tit or all triples from Tif • This corresponds to setting the variable false or true, respectively.

19. Satisfaction Testing Components • For each clause cj C, create three triples: Cj = { (ui[j], s1[j], s2[j]) | ui cj} U { (ui[j], s1[j], s2[j]) | ui cj } • Observation #4: This adds m elements to X, and m elements to Y. • Observation #5: For each j, 1 ≤ j ≤ m, s1[j] and s2[j] both appear in exactly three triples and nowhere else. • Observation #6: Any matching must choose exactly one triple from Cj • This corresponds to making that literal true.

20. Garbage Collection Components • For each clause cj C, create three triples: G = { (ui[j], g1[k], g2[k]), (ui[j], g1[k], g2[k]) | 1 ≤ k ≤ m(n-1), 1 ≤ i ≤ n, 1 ≤ j ≤ m} • Observation #7: This adds m(n-1) elements to X, and to Y. • Observation #8: For each k, 1 ≤ k ≤ m(n-1), g1[k] and g2[k] both appear in exactly 2mn triples and nowhere else. • Observation #9: Exactly m(n-1) triples from G must occur in any matching for M (why?). • Observation #10: |W| = |X| = |Y| = 2mn = q.

21. A G = (V, E), J IS instance I = M (W,X,Y) 3DM instance Another Proof for Independent Set INDEPENDENT SET (IS) INSTANCE: Graph G = (V, E) and positive integer J <= |V|. QUESTION: Does G contain an independent set of size J or more? Theorem: IS is NP-complete • IS  NP • 3DM  IS

22. H2C is NP-Complete, cont. Proof: Let M be an instance of 3DM, and construct an instance of IS as follows. V = {vi | ti  M} E = {vi, vj | i  j and ti and tj agree on some coordinate} j = q Claims: • The transformation can be performed in polynomial-time • M contains a matching if and only if G=(V,E) contains a vertex cover of size j. (only if) Let M’ = {t0, t1,…,tq-1} be a matching, where M’  M. It can be easily verified that V’ = {v0, v1,…,vq-1} is an independent set of size j. (if) Let V’ = {v0, v1,…,vq-1} is an independent set of size j for G. Then it can be easily verified that M’ = {t0, t1,…,tq-1} is a matching.

23. Vertex Cover, Independent Setand Clique VERTEX COVER (VC) INSTANCE: A Graph G = (V, E) and a positive integer K <= |V|. QUESTION: Is there a vertex cover of size K or less for G, that is, a subset V’  V such that |V’| <= K and, for each edge {u,v}  E, at least one of u and v belongs to V’? INDEPENDENT SET (IS) INSTANCE: Graph G = (V, E) and positive integer J <= |V|. QUESTION: Does G contain an independent set of size J or more? CLIQUE INSTANCE: A Graph G = (V, E) and a positive integer J <= |V|. QUESTION: Does G contain a clique of size J or more? Lemma 3.1: For any graph G=(V,E) and subset V’  V, the following statements are equivalent: (a) V’ is a vertex cover for G (b) V-V’ is an independent set for G (c) V-V’ is a clique in the complement Gc of G, where Gc=(V,Ec) with Ec={{u,v} | u,v  V and {u,v}  E}

24. Vertex Cover Theorem: VERTEX COVER is NP-complete Proof: • VC  NP • 3SAT  VC Suppose I = U,C is an instance of 3SAT where: U = {u1, u2,…,un} C = {c1, c2,…,cm} We will construct a graph G=(V,E) and a positive integer K<=|V| such that G has a vertex cover of size K or less if and only if C is satistfiable. For each variable ui U, there is a truth-setting component Ti=(Vi,Ei) where: Vi = {ui, ui} and Ei = {{ui,ui}} *Note that any vertex cover will have to contain at least on of ui and its complement.

25. Vertex Cover For each clause cj C, create a satisfaction testing component Sj = (Vj’, Ej’): Vj’= { a1[j], a2[j], a3[j] } Ej’ = { {a1[j], a2[j]}, {a1[j], a3[j]}, {a2[j], a3[j]} } *Note that any vertex cover will have to contain at least two vertices from Vj’ in order to cover the edges in Ej’. For each clause cj C, let the three literals in cj be xj, yj, zj. Then add communication edges: Ej’’ = { {a1[j], xj}, {a2[j], yj}, {a3[j], zj} } Finally, let K = n + 2m. See page 55 for an example.

26. Vertex Cover Observations: • The transformation can be performed in polynomial-time. • G = (V, E) will have a vertex cover of size K or less IFF I = U,C is satisfiable. (if) Suppose I = U,C is satisfiable… (only if) Suppose G = (V, E) has a vertex cover of size K or less…

27. Hamiltonian Circuit (HS) HAMILTONIAN CIRCUIT (HC) INSTANCE: A Graph G = (V, E). QUESTION: Does G have a Hamiltonian circuit, that is an ordering < v1, v2,…,vn > of the vertices of G, where n = |V|, such that {vn, v1}  E and {vi, vi+1}  E for all i, 1 ≤ i  n. VERTEX COVER (VC) INSTANCE: A Graph G = (V, E) and a positive integer K <= |V|. QUESTION: Is there a vertex cover of size K or less for G, that is, a subset V’  V such that |V’| <= K and, for each edge {u,v}  E, at least one of u and v belongs to V’? Theorem: HC is NP-complete Proof: • HC  NP • VC  HC

28. Partition PARTITION INSTANCE: A finite set A and a “size” s(a)  Z+for each a  A. QUESTION: Is there a subset A’  A such that 3-DIMENSIONAL MATCHING (3-DM) INSTANCE: A set M  W  X  Y, where W, X and Y are disjoint sets having the same number q of elements. QUESTION: Does M contain a matching, that is, a subset M’  M such at |M| = q and no two elements of M’ agree in any coordinate? Theorem: Partition is NP-complete Proof: • PARTITION  NP • 3DM  PARTITION

29. Techniques for ProvingNP-Completeness There are several “general” types of transformations: • Restriction • Local Replacement • Component Design

30. Restriction • Proving a problem   NPNP-complete by restriction consists of showing that  contains another known NP-complete problem ’ as a special case. • What is a “special case” of a problem? • Let I be all instances of a problem , and let I’ be all instances of a problem ’ • If I’  I then ’ is a special case of ’ • Furthermore, if  and ’ are both in NP, and if  is NP-complete, then so is ’ • The heart of a proof by restriction is the specification of additional restrictions to be placed on the instances of  so that the resulting problem will be identical to ’ • Note that a proof by restriction is still a transformation, i.e., it is not a different kind of NP-completeness proof

31. Hamiltonian Circuit (HS) DIRECTED HAMILTONIAN CIRCUIT (DHC) INSTANCE: A directed graph G = (V, E). QUESTION: Does G have a Hamiltonian circuit, that is an ordering < v1, v2,…,vn > of the vertices of G, where n = |V|, such that {vn, v1}  E and {vi, vi+1}  E for all i, 1 ≤ i  n. HAMILTONIAN CIRCUIT (HC) INSTANCE: A Graph G = (V, E). QUESTION: Does G have a Hamiltonian circuit, that is an ordering < v1, v2,…,vn > of the vertices of G, where n = |V|, such that {vn, v1}  E and {vi, vi+1}  E for all i, 1 ≤ i  n. Theorem: DHC is NP-complete Proof: • DHC  NP • HC  DHC, by restriction. Consider those instances of DHC where there is an edge from u to v if and only if there is an edge from v to u.

32. Invalid Restriction Subset Sum INSTANCE: Integers a1, a2,…,anand integer B. QUESTION: Is there a sequence of 0’s and 1’s, x1, x2,…,xnsuch that: Fact: Subset Sum is NP-complete Real Subset Sum INSTANCE: Integers a1, a2,…,anand integer B. QUESTION: Is there a sequence of real numbers x1, x2,…,xnsuch that:

33. Invalid Restriction, Cont. • Claim (from a Ph.D disseration): • Subset Sum is just a special case of Real Subset Sum • Therefore, Real Subset Sum is NP-complete • The proof is in error since it restricts the question, and not the instances. • The problem is actually trivially solvable in polynomial time simply taking x1 = B/a1 and xi = 0, for all i, where 2 ≤ i ≤ n.

34. Local Replacement • Transformations are sufficiently non-trivial to warrant spelling out details, in contrast to restriction. • An instance of a known NP-complete problem is broken up into some number of structurally similar “basic units” • During the transformation, each “basic unit” is replaced by a different structure, which is part of the target problems instance. • Each replacement constitutes only local modification.

35. Partition Exact Cover by 3-Sets (X3C) INSTANCE: Set X with |X| = 3q and a collection C of 3-element subsets of X. QUESTION: Does C contain a an exact cover for X, i.e., a subcollection C’  C such that every element of X occurs in exactly one member of C’? Fact: X3C is NP-complete (by restriction from 3DM) Partition into Triangles INSTANCE: A Graph G = (V, E), with |X| = 3q. QUESTION: Is there a partition of V into q disjoint sets V1, V2,…,Vqof three vertices each such that, for each Vi = {vi[1], vi[2], vi[3]} the three edges {vi[1], vi[2]} , {vi[1], vi[3]} and {vi[2], vi[3]} all belong to E’? Theorem: Partition into Triangles is NP-complete Proof: • Partition into Triangles  NP • X3C  Partition into Triangles