NP Completeness Lecture 3 September 3, 2013 Sariel (UIUC) CS573 - PowerPoint PPT Presentation

CS 573: Algorithms, Fall 2013 NP Completeness Lecture 3 September 3, 2013 Sariel (UIUC) CS573 1 Fall 2013 1 / 36

Part I . NP Completeness . Sariel (UIUC) CS573 2 Fall 2013 2 / 36

Certifiers . Definition . An algorithm C ( · , · ) is a certifier for problem X if for every s ∈ X there is some string t such that C ( s , t ) = ”yes”, and conversely, if for some s and t , C ( s , t ) = ”yes” then s ∈ X . The string t is called a certificate or proof for s . . . Definition (Efficient Certifier.) . A certifier C is an efficient certifier for problem X if there is a polynomial p ( · ) such that for every string s , we have that ⋆ s ∈ X if and only if ⋆ there is a string t : . . . | t | ≤ p ( | s | ) , 1 . . . C ( s , t ) = ”yes”, 2 . . . and C runs in polynomial time. 3 . Sariel (UIUC) CS573 3 Fall 2013 3 / 36

NP-Complete Problems . Definition . A problem X is said to be NP-Complete if . . X ∈ NP , and 1 . . (Hardness) For any Y ∈ NP , Y ≤ P X . 2 . Sariel (UIUC) CS573 4 Fall 2013 4 / 36

Solving NP-Complete Problems . Proposition . Suppose X is NP-Complete . Then X can be solved in polynomial time if and only if P = NP . . . Proof. . ⇒ Suppose X can be solved in polynomial time . . . Let Y ∈ NP . We know Y ≤ P X . 1 . . . We showed that if Y ≤ P X and X can be solved in polynomial 2 time, then Y can be solved in polynomial time. . . . Thus, every problem Y ∈ NP is such that Y ∈ P ; NP ⊆ P . 3 . . . Since P ⊆ NP , we have P = NP . 4 ⇐ Since P = NP , and X ∈ NP , we have a polynomial time algorithm for X . . Sariel (UIUC) CS573 5 Fall 2013 5 / 36

NP-Hard Problems . Definition . A problem X is said to be NP-Hard if . . (Hardness) For any Y ∈ NP , we have that Y ≤ P X . 1 . An NP-Hard problem need not be in NP ! Example: Halting problem is NP-Hard (why?) but not NP-Complete . Sariel (UIUC) CS573 6 Fall 2013 6 / 36

Consequences of proving NP-Complete ness If X is NP-Complete . . Since we believe P ̸ = NP , 1 . . and solving X implies P = NP . 2 X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X . (This is proof by mob opinion — take with a grain of salt.) Sariel (UIUC) CS573 7 Fall 2013 7 / 36

Consequences of proving NP-Complete ness If X is NP-Complete . . Since we believe P ̸ = NP , 1 . . and solving X implies P = NP . 2 X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X . (This is proof by mob opinion — take with a grain of salt.) Sariel (UIUC) CS573 7 Fall 2013 7 / 36

Consequences of proving NP-Complete ness If X is NP-Complete . . Since we believe P ̸ = NP , 1 . . and solving X implies P = NP . 2 X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X . (This is proof by mob opinion — take with a grain of salt.) Sariel (UIUC) CS573 7 Fall 2013 7 / 36

Consequences of proving NP-Complete ness If X is NP-Complete . . Since we believe P ̸ = NP , 1 . . and solving X implies P = NP . 2 X is unlikely to be efficiently solvable. At the very least, many smart people before you have failed to find an efficient algorithm for X . (This is proof by mob opinion — take with a grain of salt.) Sariel (UIUC) CS573 7 Fall 2013 7 / 36

NP-Complete Problems . Question . Are there any problems that are NP-Complete ? . . Answer . Yes! Many, many problems are NP-Complete . . Sariel (UIUC) CS573 8 Fall 2013 8 / 36

Circuits . Definition . A circuit is a directed acyclic graph with . Output: ∧ Input vertices (without 1 incoming edges) labelled with 0 , 1 or a distinct variable. ¬ ∧ . . Every other vertex is labelled 2 ∨ , ∧ or ¬ . ∧ ∨ ∨ . . Single node output vertex 3 with no outgoing edges. Inputs: 1 ? ? 0 ? . Sariel (UIUC) CS573 9 Fall 2013 9 / 36

Cook-Levin Theorem . Definition (Circuit Satisfaction ( CSAT ).) . Given a circuit as input, is there an assignment to the input variables that causes the output to get value 1 ? . . Theorem (Cook-Levin) . CSAT is NP-Complete . . Need to show . . CSAT is in NP . 1 . . every NP problem X reduces to CSAT . 2 Sariel (UIUC) CS573 10 Fall 2013 10 / 36

CSAT : Circuit Satisfaction . Claim . CSAT is in NP . . . . Certificate: Assignment to input variables. 1 . . Certifier: Evaluate the value of each gate in a topological sort of 2 DAG and check the output gate value. Sariel (UIUC) CS573 11 Fall 2013 11 / 36

CSAT : Circuit Satisfaction . Claim . CSAT is in NP . . . . Certificate: Assignment to input variables. 1 . . Certifier: Evaluate the value of each gate in a topological sort of 2 DAG and check the output gate value. Sariel (UIUC) CS573 11 Fall 2013 11 / 36

CSAT is NP -hard: Idea Need to show that every NP problem X reduces to CSAT . What does it mean that X ∈ NP ? X ∈ NP implies that there are polynomials p () and q () and certifier/verifier program C such that for every string s the following is true: . . If s is a YES instance ( s ∈ X ) then there is a proof t of length 1 p ( | s | ) such that C ( s , t ) says YES. . . If s is a NO instance ( s ̸∈ X ) then for every string t of length 2 at p ( | s | ) , C ( s , t ) says NO. . . C ( s , t ) runs in time q ( | s | + | t | ) time (hence polynomial time). 3 Sariel (UIUC) CS573 12 Fall 2013 12 / 36

Reducing X to CSAT X is in NP means we have access to p () , q () , C ( · , · ) . What is C ( · , · ) ? It is a program or equivalently a Turing Machine! How are p () and q () given? As numbers. Example: if 3 is given then p ( n ) = n 3 . Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or a TM . Sariel (UIUC) CS573 13 Fall 2013 13 / 36

Reducing X to CSAT Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or TM . Problem X: Given string s , is s ∈ X ? Same as the following: is there a proof t of length p ( | s | ) such that C ( s , t ) says YES. How do we reduce X to CSAT ? Need an algorithm A that . . takes s (and ⟨ p , q , C ⟩ ) and creates a circuit G in polynomial 1 time in | s | (note that ⟨ p , q , C ⟩ are fixed). . . G is satisfiable if and only if there is a proof t such that 2 C ( s , t ) says YES. Sariel (UIUC) CS573 14 Fall 2013 14 / 36

Reducing X to CSAT Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or TM . Problem X: Given string s , is s ∈ X ? Same as the following: is there a proof t of length p ( | s | ) such that C ( s , t ) says YES. How do we reduce X to CSAT ? Need an algorithm A that . . takes s (and ⟨ p , q , C ⟩ ) and creates a circuit G in polynomial 1 time in | s | (note that ⟨ p , q , C ⟩ are fixed). . . G is satisfiable if and only if there is a proof t such that 2 C ( s , t ) says YES. Sariel (UIUC) CS573 14 Fall 2013 14 / 36

Reducing X to CSAT Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or TM . Problem X: Given string s , is s ∈ X ? Same as the following: is there a proof t of length p ( | s | ) such that C ( s , t ) says YES. How do we reduce X to CSAT ? Need an algorithm A that . . takes s (and ⟨ p , q , C ⟩ ) and creates a circuit G in polynomial 1 time in | s | (note that ⟨ p , q , C ⟩ are fixed). . . G is satisfiable if and only if there is a proof t such that 2 C ( s , t ) says YES. Sariel (UIUC) CS573 14 Fall 2013 14 / 36

Reducing X to CSAT Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or TM . Problem X: Given string s , is s ∈ X ? Same as the following: is there a proof t of length p ( | s | ) such that C ( s , t ) says YES. How do we reduce X to CSAT ? Need an algorithm A that . . takes s (and ⟨ p , q , C ⟩ ) and creates a circuit G in polynomial 1 time in | s | (note that ⟨ p , q , C ⟩ are fixed). . . G is satisfiable if and only if there is a proof t such that 2 C ( s , t ) says YES. Sariel (UIUC) CS573 14 Fall 2013 14 / 36

Reducing X to CSAT Thus an NP problem is essentially a three tuple ⟨ p , q , C ⟩ where C is either a program or TM . Problem X: Given string s , is s ∈ X ? Same as the following: is there a proof t of length p ( | s | ) such that C ( s , t ) says YES. How do we reduce X to CSAT ? Need an algorithm A that . . takes s (and ⟨ p , q , C ⟩ ) and creates a circuit G in polynomial 1 time in | s | (note that ⟨ p , q , C ⟩ are fixed). . . G is satisfiable if and only if there is a proof t such that 2 C ( s , t ) says YES. Sariel (UIUC) CS573 14 Fall 2013 14 / 36