NP-Completeness Part One

Transcription

1 NP-Completeness Part One

2 Recap from Last Time

3 The Limits of Efficient Computation P NP R

4 P and NP Rresher The class P consists of all problems solvable in terministic polynomial time. The class NP consists of all problems solvable in nonterministic polynomial time. Equivalently, NP consists of all problems for which there is a terministic, polynomial-time verifier for the problem.

5 Reducibility

6 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges.

7 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges.

8 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges. A matching, matching, but but not not a a maximum maximum matching. matching.

9 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges. A maximum maximum matching. matching.

10 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges.

11 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges.

12 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges.

13 Maximum Matching Given an undirected graph G, a matching in G is a set of edges such that no two edges share an endpoint. A maximum matching is a matching with the largest number of edges. Maximum Maximum matchings matchings are are not not necessarily necessarily unique. unique.

14 Maximum Matching Jack Edmonds' paper Paths, Trees, and Flowers gives a polynomial-time algorithm for finding maximum matchings. (This is the same Edmonds as in Cobham- Edmonds Thesis. Using this fact, what other problems can we solve?

15 Domino Tiling

16 Domino Tiling

17 Domino Tiling

18 Domino Tiling

19 Domino Tiling

20 Domino Tiling

21 Solving Domino Tiling

22 Solving Domino Tiling

23 Solving Domino Tiling

24 Solving Domino Tiling

25 Solving Domino Tiling

26 Solving Domino Tiling

27 Solving Domino Tiling

28 Solving Domino Tiling

29 The Setup To termine whether you can place at least k dominoes on a crossword grid, do the following: Convert the grid into a graph: each empty cell is a no, and any two adjacent empty cells have an edge between them. Ask whether that graph has a matching of size k or greater. Return whatever answer you get. Claim: This runs in polynomial time.

30 In Pseudoco boolean canplacekdominos(grid G, int k { } return hasmatching(gridtograph(g, k;

31 Another Example

32 Reachability Consir the following problem: Given an directed graph G and nos s and t in G, is there a path from s to t? As a formal language: REACHABILITY = { G, s, t G is a directed graph, s and t are nos in G, and there's a path from s to t } Theorem: REACHABILITY P. Given that we can solve the reachability problem in polynomial time, what other problems can we solve in polynomial time?

33 Converter Conundrums Suppose that you want to plug your laptop into a projector. Your laptop only has a VGA output, but the projector needs HDMI input. You have a box of connectors that convert various types of input into various types of output (for example, VGA to DVI, DVI to DisplayPort, etc. Question: Can you plug your laptop into the projector?

34 Converter Conundrums Connectors RGB RGB to to USB USB VGA VGA to to DisplayPort DB13W3 DB13W3 to to CATV CATV DisplayPort to to RGB RGB DB13W3 DB13W3 to to HDMI HDMI DVI DVI to to DB13W3 DB13W3 S-Vio S-Vio to to DVI DVI FireWire FireWire to to SDI SDI VGA VGA to to RGB RGB DVI DVI to to DisplayPort USB USB to to S-Vio S-Vio SDI SDI to to HDMI HDMI

35 Converter Conundrums Connectors RGB RGB to to USB USB VGA VGA to to DisplayPort DB13W3 DB13W3 to to CATV CATV DisplayPort to to RGB RGB DB13W3 DB13W3 to to HDMI HDMI DVI DVI to to DB13W3 DB13W3 S-Vio S-Vio to to DVI DVI FireWire FireWire to to SDI SDI VGA VGA to to RGB RGB DVI DVI to to DisplayPort USB USB to to S-Vio S-Vio SDI SDI to to HDMI HDMI VGA RGB USB DisplayPort DB13W3 CATV HDMI X FireWire DVI S-Vio SDI

36 Converter Conundrums VGA RGB USB DisplayPort DB13W3 CATV HDMI X DVI S-Vio FireWire SDI

37 Converter Conundrums VGA RGB USB DisplayPort DB13W3 CATV HDMI X DVI S-Vio FireWire SDI

38 Converter Conundrums VGA RGB USB DisplayPort DB13W3 CATV HDMI X DVI S-Vio FireWire SDI

39 Converter Conundrums Given a a list of plug converters, here's a algorithm for termining whether you can plug your computer into the projector: Create a graph with one no per connector type and an edge from one type of connector to another if there's a converter from the first type to the second. Use the reachability algorithm to see whether you can get from VGA to HDMI. Return whatever the result of that algorithm is. Claim: This runs in polynomial time.

40 In Pseudoco boolean canplugin(list<plug> plugs { } return isreachable(plugstograph(plugs, VGA, HDMI;

41 A Commonality Both of the solutions to our previous problems had the following form: boolean solveproblema(input { return solveproblemb(transform(input; } Important observation: Assuming that we already have a solver for problem B, the only work done here is transforming the input to problem A into an input to problem B. All the hard work is done by the solver for B; we just turn one input into another.

42 Mathematically Moling this Ia

43 Polynomial-Time Reductions Let A and B be languages. A polynomial-time reduction from A to B is a function f : Σ* Σ* such that For any w Σ*, w A iff f(w B. The function f can be computed in polynomial time. What does this mean?

44 Polynomial-Time Reductions If f is a polynomial-time reduction from A to B, then w Σ*. (w A f(w B If you want to know whether w A, you can instead ask whether f(w B. Every w A maps to some f(w B. Every w A maps to some f(w B.

45 Polynomial-Time Reductions If f is a polynomial-time reduction from A to B, then w Σ*. (w A f(w B Σ* Σ*

46 Polynomial-Time Reductions If f is a polynomial-time reduction from A to B, then w Σ*. (w A f(w B YES NO Σ* Σ*

47 Polynomial-Time Reductions If f is a polynomial-time reduction from A to B, then w Σ*. (w A f(w B YES f(w YES NO f(w Σ* Σ* NO

48 Reductions, Programmatically Suppose we have a solver for problem B that's dined in terms of problem A in this specific way: boolean solveproblema(input { return solveproblemb(transform(input; } The reduction from A to B is the function transform in the above setup: it maps yes answers to A to yes answers to B and no answers to A to no answers to B.

49 Reducibility among Problems Suppose that A and B are languages where there's a polynomial-time reduction from A to B. We'll note this by writing A p B You can read this aloud as A polynomialtime reduces to B or A poly-time reduces to B.

50 Reductions and P Theorem: If A p B and B P, then A P. Proof sketch: Show that this pseudoco runs in polynomial time: boolean solveproblema(input { return solveproblemb(transform(input; } Calling transform only takes polynomial time. Since it runs in polynomial time, the transform function can only produce an output that's polynomially larger than its input. Feeding that into a polynomial-time function solveproblemb then only takes polynomial time. Overall, this is a polynomial-time algorithm for A, so A P.

51 Time-Out for Announcements!

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53 Your Questions!

54 What's the best way to become a stronger cor? The summer seems like a great time to take advantage of bore the next school year. Seems like classes like CS 107 reward experience which is hard to compete with when you haven't cod for years! Find Find something something fun fun that that you you want want to to work work on, on, then then go go work work on on it. it. You'll You'll have have a a lot lot more more fun fun if if you you actually actually enjoy enjoy what what you're you're doing. doing. And And don't don't worry worry about about CS107. CS107. People People freak freak out out way way too too much much about about that that course. course. You You don't don't need need to to prepare prepare for for months months in in advance advance bore bore taking taking it. it. There There aren't aren't that that many many stunts stunts in in the the course course that that have have a a huge huge headstart headstart over over everyone. everyone. Just Just be be prepared prepared to to put put in in a a cent cent amount amount of of time time and and be be strategic. strategic.

55 I'm sticking around campus for summer any suggestions on what to put on my summer bucket list? Rent Rent a a car car (you (you get get a a discount discount for for being being a a Stanford Stanford stunt stunt and and drive drive up up Skyline Skyline and and walk walk around around the the open open space space preserves. preserves. It's It's beautiful beautiful there. there. Then Then go go to to Pescaro Pescaro and and visit visit a a world-class world-class goat goat dairy dairy farm farm while while eating eating artichoke-garlic artichoke-garlic bread. bread. Go Go to to the the Mission Mission in in SF. SF. Do Do something something totally totally hipsterish hipsterish (Urban (Urban Putt, Putt, Mission Mission Cheese, Cheese, Danlion Danlion Chocolate Chocolate and and something something more more local local (get (get pupusas. pupusas.

56 After your teaser during Wednesday's lecture and 2 days of waiting, we're even more excited for you to speak on something that fascinates you. We'd also like to thank you for answering these questions with such enthusiasm throughout the quarter. Go!

57

58 Back to CS103!

59 Reductions and NP

60 boolean solveproblema(input { return solveproblemb(transform(input; } What What happens if if transform runs runs in in terministic polynomial time, time, but but solveproblemb runs runs in in nonterministic polynomial time? time?

61 Reductions and NP We can reduce problems in NP to one another using polynomial-time reductions. The reduction itself must be computable in terministic polynomial time. The output of that reduction is then fed in as input to a nonterministic, polynomialtime algorithm. Remember the goal of the reduction is to transform the problem, not solve it!

62 A Sample Reduction

63 Satisfiability A propositional logic formula φ is called satisfiable if there is some assignment to its variables that makes it evaluate to true. p q is satisfiable. p p is unsatisfiable. p (q q is satisfiable. An assignment of true and false to the variables of φ that makes it evaluate to true is called a satisfying assignment.

64 SAT The boolean satisfiability problem (SAT is the following: Formally: Given a propositional logic formula φ, is φ satisfiable? SAT = { φ φ is a satisfiable PL formula } Claim: SAT NP. (Do you see why?

65 What other problems can we solve with a solver for SAT?

66 Perfect Matchings A perfect matching in a graph G is a matching where every no is adjacent to some edge in the matching. Claim: We can reduce the problem of termining whether a graph has a perfect matching to the boolean satisfiability problem.

67 Perfect Matchings A perfect matching in a graph G is a matching where every no is adjacent to some edge in the matching. Claim: We can reduce the problem of termining whether a graph has a perfect matching to the boolean satisfiability problem.

68 a b c d e f

69 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching

70 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching p ab ab p ad ad p bc bc p p ce ce p p p

71 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching p ab ab p ad ad p bc bc p p ce ce p p p

72 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

73 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

74 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

75 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

76 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

77 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

78 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

79 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

80 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

81 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

82 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

83 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

84 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p (p (p ab ab p ad ad (p (p ab ab p bc bc (p (p ab ab p (p (p bc bc p (p (p bc bc p ce ce (p (p bc bc p (p (p ce ce p (p (p ad ad p (p (p ad ad p (p (p p (p (p ce ce p (p (p ce ce p (p (p p (p (p p p ab ab p ad ad p bc bc p p ce ce p p p

85 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p p ab ab p ad ad p bc bc p p ce ce p p p

86 a b Notice Notice that that this this formula formula is is the the many-way many-way AND AND of of lots lots of of smaller smaller formulas formulas of of the the form form (l₁ (l₁ l₂ l₂ lₙ lₙ where d e f where each each lₖ lₖ is is either either a variable variable or or its its negation. negation. Each Each lₖ lₖ is is For called each called edge a literal, literal, {u, v}, and and we'll the the introduce multi-way multi-way a propositional OR OR of of literals literals variable is is called called p a uv meaning edge {u, v} is includ clause. clause. in the perfect matching. We To To need satisfy satisfy to enforce this this formula, formula, the following: we we need need to to choose choose truth truth values values so so that that Every no every every is adjacent clause clause has to has at at at least least least one one one edge. true true ( perfect literal. literal. Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p p ab ab (p (p ad ad p p (p (p ce ce p p (p (p p p ad ad ( p ( p ab ab p p ad ad p bc bc ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p p ce ce ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p p ( p ( p p p p c

87 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p p ab ab p ad ad p bc bc p p ce ce p p p

88 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p p ab ab p ad ad p bc bc p p ce ce p p p

89 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p p ab ab p ad ad p bc bc p p ce ce p p p

90 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p T p ab ab F p ad ad F p bc bc F p F p ce ce T p T p F p

91 a b c d e f For each edge {u, v}, we'll introduce a propositional variable p uv meaning edge {u, v} is includ in the perfect matching. We need to enforce the following: Every no is adjacent to at least one edge. ( perfect Every no is adjacent to at most one edge. ( matching (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p T p ab ab F p ad ad F p bc bc F p F p ce ce T p T p F p

92 (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p p ab ab p ad ad p bc bc p p ce ce p p p

93 (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p F p ab ab T p ad ad T p bc bc F p F p ce ce F p F p T p

94 (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p F p ab ab T p ad ad T p bc bc F p F p ce ce F p F p T p

95 a b c d e f (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p F p ab ab T p ad ad T p bc bc F p F p ce ce F p F p T p

96 a b c d e f (p (p ab ab p ad ad (p (p ab ab p bc bc p (p (p bc bc p ce ce p (p (p ad ad p p (p (p ce ce p p (p (p p ( p ( p ab ab p p ad ad ( p ( p ab ab p p bc bc ( p ( p ab ab p p ( p ( p bc bc p p ( p ( p bc bc p p ce ce ( p ( p bc bc p p ( p ( p ce ce p p ( p ( p ad ad p p ( p ( p ad ad p p ( p ( p p p ( p ( p ce ce p p ( p ( p ce ce p p ( p ( p p p ( p ( p p p F p ab ab T p ad ad T p bc bc F p F p ce ce F p F p T p

97 The Reduction Suppose we have a graph with n nos. For each of the n nos, we introduce a clause of at most n literals to force each no to be adjacent to at least one chosen edge. Total size: O(n 2 For each of the n nos, we introduce O(n 2 clauses with two literals each to force each no to be adjacent to at most one chosen edge. Total size: O(n 3 Total size of the generated formula: at most O(n 3. Unsubstantiated but true claim: This can be computed in polynomial time.

98 The Reduction boolean hasperfectmatching(graph G { return issatisfiable(graphtoformula(g; }

99 Reductions and NP Theorem: If A p B and B NP, then A NP. Proof sketch: Show that this pseudoco runs in nonterministic polynomial time: boolean solveproblema(input { return solveproblemb(transform(input; } Calling transform only takes polynomial time. The transform function can only produce an output that's polynomially larger than its input. Feeding that into a polynomial-time function solveproblemb then only takes nonterministic polynomial time. Overall, this is a nonterministic polynomial-time algorithm for A, so A NP.

100 Reducibility, Summarized A polynomial-time reduction is a way of transforming inputs to problem A into inputs to problem B that preserves the correct answer. If there is a polynomial-time reduction from A to B, we note this by writing A p B. Two major theorems: Theorem: If B P and A p B, then A P. Theorem: If B NP and A p B, then A NP.

101 NP-Hardness and NP-Completeness

102 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P

103 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P

104 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P

105 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P

106 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P NP

107 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P NP

108 Polynomial-Time Reductions If L 1 P L 2 and L 2 P, then L 1 P. If L 1 P L 2 and L 2 NP, then L 1 NP. P NP

109 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete iff L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP P

110 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete iff L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

111 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete iff L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

112 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. Intuitively: L has has to to be be at at least least as as hard as every problem in NP, since an an algorithm for for L can can be be used used to to ci all all problems in in NP. NP. A language in L is hard called as NP-complete every problem iff L is in NP-hard NP, since and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

113 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete iff L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

114 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete iff L What's What's is NP-hard in in here? here? and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

115 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete if L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP NP-Hard P

116 NP-Hardness A language L is called NP-hard if for every L' NP, we have L' P L. A language in L is called NP-complete if L is NP-hard and L NP. The class NPC is the set of NP-complete problems. NP NPC NP-Hard P

117 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP.

118 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP. NPC NP P

119 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP. NPC NP P

120 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP. NPC NP P

121 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP. P = NP

122 The Tantalizing Truth Theorem: If any NP-complete language is in P, then P = NP. Proof: Suppose that L is NP-complete and L P. Now consir any arbitrary NP problem X. Since L is NP-complete, we know that X p L. Since L P and X p L, we see that X P. Since our choice of X was arbitrary, this means that NP P, so P = NP. P = NP

123 The Tantalizing Truth Theorem: If any NP-complete language is not in P, then P NP. Proof: Suppose that L is an NP-complete language not in P. Since L is NP-complete, we know that L NP. Therore, we know that L NP and L P, so P NP. NP P NPC

124 A Feel for NP-Completeness If a problem is NP-complete, then unr the assumption that P NP, there cannot be an ficient algorithm for it. In a sense, NP-complete problems are the harst problems in NP. All known NP-complete problems are enormously hard to solve: All known algorithms for NP-complete problems run in worst-case exponential time. Most algorithms for NP-complete problems are infeasible for reasonably-sized inputs.

125 How do we even know NP-complete problems exist in the first place?

126 Satisfiability A propositional logic formula φ is called satisfiable if there is some assignment to its variables that makes it evaluate to true. p q is satisfiable. p p is unsatisfiable. p (q q is satisfiable. An assignment of true and false to the variables of φ that makes it evaluate to true is called a satisfying assignment.

127 SAT The boolean satisfiability problem (SAT is the following: Formally: Given a propositional logic formula φ, is φ satisfiable? SAT = { φ φ is a satisfiable PL formula }

128 Theorem (Cook-Levin: SAT is NP-complete. Proof: Take CS154!

129 Finding Additional NP-Complete Problems

130 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC.

131 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC

132 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC

133 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC

134 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC

135 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC

136 NP-Completeness Theorem: Let L₁ and L₂ be languages. If L₁ P L₂ and L₁ is NP-hard, then L₂ is NP-hard. Theorem: Let L₁ and L₂ be languages where L₁ NPC and L₂ NP. If L₁ P L₂, then L₂ NPC. NP P NPC