Computational Complexity Lecture 3 in which we come across Diagonalization and Time-hierarchies (But first some more of NP-completeness) 1 NP-Complete Languages 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) Trivial example: L1 = TMSAT 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) Trivial example: L1 = TMSAT First interesting example: L1 = CKT-SAT 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) Trivial example: L1 = TMSAT First interesting example: L1 = CKT-SAT Convert x into a circuit C s.t. C is satisfiable iff x in L 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) Trivial example: L1 = TMSAT First interesting example: L1 = CKT-SAT Convert x into a circuit C s.t. C is satisfiable iff x in L More examples, bootstrapping from CKT-SAT 2 NP-Complete Languages A language L1 is NP-complete if L1 is in NP and any NP language L can be reduced to L1 (Karp reduction: polynomial time many-one reduction) Trivial example: L1 = TMSAT First interesting example: L1 = CKT-SAT Convert x into a circuit C s.t. C is satisfiable iff x in L More examples, bootstrapping from CKT-SAT If L ≤p L1 and L1 ≤p L2, then L ≤p L2 2 CKT-SAT ≤p SAT 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) Converting a circuit to a collection of clauses: 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) Converting a circuit to a collection of clauses: For each wire (connected component), add a variable 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) Converting a circuit to a collection of clauses: For each wire (connected component), add a variable Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate: 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) Converting a circuit to a collection of clauses: For each wire (connected component), add a variable Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate: x AND ⇒ ⇒ ⇒ ∨ e.g. y z : (z x), (z y), (¬z ¬x ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y). 3 CKT-SAT ≤p SAT SAT: Are all given “clauses” simultaneously satisfiable? (Conjunctive Normal Form) Converting a circuit to a collection of clauses: For each wire (connected component), add a variable Add output variable as a clause. And for each gate, add a clause involving variables for wires connected to the gate: x AND ⇒ ⇒ ⇒ ∨ e.g. y z : (z x), (z y), (¬z ¬x ¬y). i.e., (¬z ∨ x), (¬z ∨ y), (z ∨ ¬x ∨ ¬y). x and y OR z : (z⇒x ∨ y), (¬z⇒¬x), (¬z⇒¬y). 3 SAT ≤p 3SAT 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. More directly: 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT 2SAT is in fact in P! [Exercise] 4 SAT ≤p 3SAT Previous reduction was to 3SAT, so 3SAT is NP-complete. And SAT is in NP. So SAT ≤p 3SAT. More directly: (a ∨ b ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ d ∨ e) → (a ∨ b ∨ x), (¬x ∨ c ∨ y), (¬y ∨ d ∨ e) Reduction needs 3SAT 2SAT is in fact in P! [Exercise] Reduction not parsimonious (can you make it? [Exercise]) 4 3SAT ≤p CLIQUE 5 3SAT ≤p CLIQUE CLIQUE: Does graph G have a clique of size m? 5 3SAT ≤p CLIQUE CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) have a clique of size m? (w ∨ y) (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) have a clique of size m? Clauses → Graph (w ∨ y) (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) have a clique of size m? Clauses → Graph (w ∨ y) 1*0* 1*1* 0*1* (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) have a clique of size m? Clauses → Graph (w ∨ y) 1*0* 1*1* 0*1* (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) 1*0* 1*1* 0*1* (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) 1*0* 1*1* 0*1* (w ∨ x ∨ ¬z) 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) 1*0* 1*1* 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* 00*0 (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* edges between 00*0 consistent assignments (w ∨ x ∨ ¬z) 11*1 10*0 01*1 01*0 10*1 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* edges between 00*0 consistent assignments (w ∨ x ∨ ¬z) 11*1 10*0 01*1 m-clique iff all m 01*0 10*1 clauses satisfiable 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* edges between 00*0 consistent assignments (w ∨ x ∨ ¬z) 11*1 10*0 01*1 m-clique iff all m 01*0 10*1 clauses satisfiable 11*0 5 3SAT ≤p CLIQUE *000 CLIQUE: Does graph G (x ∨ ¬y ∨ ¬z) *111 *100 have a clique of size m? *101 *010 *110 *001 Clauses → Graph (w ∨ y) vertices: each clause’s satisfying assignments 1*0* 1*1* (for its variables) 0*1* edges between 00*0 consistent assignments (w ∨ x ∨ ¬z) 11*1 10*0 3-Clique 1*1* 01*1 m-clique iff all m *110 01*0 10*1 clauses satisfiable 11*0 ---- 11*0 sat assignment 1110 5 INDEP-SET and VERTEX-COVER 6 INDEP-SET and VERTEX-COVER CLIQUE ≤p INDEP-SET 6 INDEP-SET and VERTEX-COVER CLIQUE ≤p INDEP-SET G has an m-clique iff Gc has an m-independent-set 6 INDEP-SET and VERTEX-COVER CLIQUE ≤p INDEP-SET G has an m-clique iff Gc has an m-independent-set INDEP-SET ≤p VERTEX-COVER 6 INDEP-SET and VERTEX-COVER CLIQUE ≤p INDEP-SET G has an m-clique iff Gc has an m-independent-set INDEP-SET ≤p VERTEX-COVER G has an m-indep-set iff G has an (n-m)-vertex-cover 6 NP, P, co-NP and NPC NP coNP NPC coNPC P 7 NP, P, co-NP and NPC We say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X NP coNP NPC coNPC P 7 NP, P, co-NP and NPC We say class X is “closed under polynomial reductions” if (L1 ≤p L2 and L2 in class X) implies L1 in X e.g.