Chapter 3: Proving NP-completeness Results

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

Six Basic NP-Complete Problems • 3-SATISFIABILITY (3SAT) • 3-DIMENSIONAL MATCHING (3DM) • VERTEX COVER (VC) • CLIQUE • HAMILTONIAN CIRCUIT (HC) • PARTITION First, a tour…

3-Satisfiability 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?

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 such 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’.

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 b2 and b3 do not appear in any triple.

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’? K = 3 u v w x z

Vertex Cover, Independent Setand Clique CLIQUE INSTANCE: A Graph G = (V, E) and a positive integer J <= |V|. QUESTION: Does G contain a clique of size J or more? J = 3 u v w x z

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. How is a Hamiltonian circuit different from a tour, as in TSP? TRAVELING SALESMAN INSTANCE: Set C of m cities, distance d(ci, cj)  Z+ for each pair of cities ci, cj C positive integer B. QUESTION: Is there a tour of C having length B or less, I.e., a permutation <c(1) , c(2),…, c(m)> of C such that:

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? • Contrasted with the more general SAT: SATISFIABILITY (SAT) INSTANCE: Collection C of clauses on a finite set U of variables. QUESTION: Is there a truth assignment for U that satisfies all the clauses in C?

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

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} }

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} }

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.

Six Basic NP-Complete Problems • A review of basic logic…and language… A if and only if B • Equates to: A if B The “if” part A only if B The “only if” part • Similarly: A if B ≈ B only if A ≈ B => A A only if B ≈ B if A ≈ A => B

Six Basic NP-Complete Problems • Lastly, to prove: X => Y • We frequently show: ~Y => ~X

3-Satisfiability • So lets go back to the basic claim: 1) I = U,C is satisfiable iff I’ = U’,C’ is satisfiable. a) (if) I = U,C is satisfiable if I’ = U’,C’ is… b) (only if) I = U,C is satisfiable only if I’ = U’,C’ is…

3-Satisfiability 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)

3-Satisfiability • 3-SAT is an example of what is called a “special case” of SAT. • Some special cases, like 3-SAT, are NP-complete • Others are solvable in polynomial time (chapter 4) • How about 4-SAT? • 5-SAT? • N-SAT, for any fixed N>=3? • 1-SAT? • 2-SAT?

Some More New Problems Not All Equal (NAE) 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? Fact: NAE 3-SAT is NP-complete

Some More New Problems • 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

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

A I = U,C NAE3SAT instance H = (V, E) H2C instance H2C is NP-Complete Theorem: H2C is NP-complete H2C  NP Not All Equal 3-SAT  H2C

H2C is NP-Complete, cont. • Suppose I = U,C where U = {u0, u1,…,un-1} and 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}

H2C is NP-Complete, cont. • Example: U = {u0, u1, u2, u3} C = { (u0, u1, u2), (u0,u2, u3), (u1, u2,u3) } • The resulting hyper-graph is 2-colorable IFF there is a NAE satisfying truth assignment for the Boolean expression. • (if) Suppose there is a NAE assignment, then there is a 2-coloring. • (only if) Suppose there is a 2-coloring, then there is a NAE satisfying assignment. 1 0 v0 1 1 0 v0 v2 v1 v3 1 v2 0 0 v1 v3

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 such 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’.

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 b2 and b3 do not appear in any triple.

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

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

Truth Setting and Fan Out Components • For each variable ui U, create the following (TS&FO) 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 (suppose m=4):

Truth Setting and Fan Out Components • 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.

Satisfaction Testing Components • For each clause cj C, create (ST) 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, thereby satisfying the clause. • This gives a total of m triples selected from the ST triples. • Observation #7: Any matching will contain exactlyone triple from Cjif and only if there is a satisfying truth assignment for the 3-SAT instance.

Garbage Collection Components • For each clause cj C, create 2mn(n-1) (GC) 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 #8: This adds m(n-1) elements to X, and to Y. • Observation #9: For each k, 1 ≤ k ≤ m(n-1), g1[k] and g2[k] both appear in exactly 2mn triples and nowhere else. • Observation #10: Consequently, exactly m(n-1) triples from G must occur in any matching for M. • Observation #11: |W| = |X| = |Y| = 2mn = q.

Garbage Collection Components • The resulting instance I’ of 3DM has a matching IFF the 3-SAT instance I is satisfiable. • (if) Suppose you have a satisfying truth assignment for I. How do you construct the matching? • (only if) Suppose you have a matching. What is the satisfying truth assignment?

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’? K = 3 u v w x z

Vertex Cover, Independent Setand Clique 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? J = 2 u v w x z

Vertex Cover, Independent Setand Clique CLIQUE INSTANCE: A Graph G = (V, E) and a positive integer J <= |V|. QUESTION: Does G contain a clique of size J or more? J = 3 u v w x z

Vertex Cover, Independent Setand Clique 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} u u u v v v w w w x x x z z z

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.

Vertex Cover 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 one of ui and its complement. 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’.

Vertex Cover 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.

Vertex Cover Example: U = {u1, u2,u3, u4} C = { (u1, u3, u4), (u1, u2, u4) } K = n + 2m = 8

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…

Vertex Cover, Independent Setand Clique • Recall Lemma 3.1: 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} • What does the previous result say about the independent set and clique problems?

Homework Problems 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? Prove IS NP-complete by giving a transformation from 3DM. 4-DIMENSIONAL MATCHING (4-DM) INSTANCE: A set M  W  X Y Z , where W, X, Yand Zare 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? Prove 4-DM NP-complete by giving a transformation from 3DM.

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. How is a Hamiltonian circuit different from a tour, as in TSP?

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’?

Hamiltonian Circuit (HC) Theorem: HC is NP-complete Proof: • HC  NP • VC  HC Let G = (V,E) and K <= |V| be an instance of VC. We will construct a graph G’ = (V’,E’) such that G has a vertex cover of size K or less if and only if G’ has a Hamiltonian circuit.

Hamiltonian Circuit (HC) 1) Add “selector” vertices a1, a2,…,akto V’ 2) For each edge e = {u,v} in E, construct the following “cover testing” component: this side this side corresponds to u corresponds to v More specifically, add the following 12 vertices: Ve’ = {(u,e,i), (v,e,i) : 1<=i<=6}} And 14 edges: Ee’ = {{(u,e,i), (u,e,i+1)}, {(v,e,i), (v,e,i+1)} : 1<=i<=5 } ∪ {{(u,e,3), (v,e,1)}, {(v,e,3), (u,e,1)}} ∪ {{(u,e,6), (v,e,4)}, {(v,e,6), (u,e,4)}}

Chapter 3: Proving NP-completeness Results

Chapter 3: Proving NP-completeness Results

Presentation Transcript

NP -Completeness

NP -Completeness

Chapter 3: Proving NP-completeness Results

NP-Completeness

NP-completeness

NP-Completeness

NP-Completeness

NP-completeness

NP-Completeness

NP-Completeness

NP-completeness

NP-Completeness

NP-Completeness

NP-Completeness

NP-Completeness

NP-Completeness

NP-Completeness

NP-completeness

NP-completeness

Chapter 34: NP-Completeness