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The Theory of NP-Completeness 1

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What is NP-completeness? Consider the circuit satisfiability problem Difficult to answer the decision problem in polynomial time with the classical deterministic algorithms 2

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Nondeterministic algorithms A nondeterminstic algorithm consists of phase 1: guessing phase 2: checking If the checking stage of a nondeterministic algorithm is of polynomial time-complexity, then this algorithm is called an NP (nondeterministic polynomial) algorithm. 3

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4 N ondeterministic searching algorithm Search for x in an array A Choice(S) : arbitrarily chooses one of the elements in set S Failure : an unsuccessful completion Success : a successful completion Nonderministic searching algorithm (which will be performed with unbounded parallelism): j ← choice(1 : n) /* guessing */ if A(j) = x then success /* checking */ else failure

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5 A nondeterministic algorithm terminates unsuccessfully iff there exist not a set of choices leading to a success signal. A deterministic interpretation of a non- deterministic algorithm can be made by allowing unbounded parallelism in computation. The runtime required for choice(1 : n) is O(1). The runtime for nondeterministic searching algorithm is also O(1)

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6 Nondeterministic sorting B ← 0 /* guessing */ for i = 1 to n do j ← choice(1 : n) if B[j] ≠ 0 then failure B[j] = A[i] /* checking */ for i = 1 to n-1 do if B[i] > B[i+1] then failure success Perform the above with unbounded parallelism

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Exercise 1 How to handle the circuit satisfiablity problem? 7

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NP : the class of decision problem which can be solved by a non-deterministic polynomial algorithm. P: the class of problems which can be solved by a deterministic polynomial algorithm. NP-hard: the class of problems to which every NP problem reduces. NP-complete (NPC): the class of problems which are NP-hard and belong to NP. 8

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Some concepts of NP Complete Definition of reduction: Problem A reduces to problem B (A B) iff A can be solved by a deterministic polynomial time algorithm using a deterministic algorithm that solves B in polynomial time. B is harder. Up to now, none of the NPC problems can be solved by a deterministic polynomial time algorithm in the worst case. It does not seem to have any polynomial time algorithm to solve the NPC problems. 9

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If A, B NPC, then A B and B A Theory of NP-completeness If any NPC problem can be solved in polynomial time, then all NP problems can be solved in polynomial time. (NP = P) 10

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The circuit satisfiability problem The logical formula : x 1 v x 2 v x 3 & - x 1 & - x 2 the assignment : x 1 ← F, x 2 ← F, x 3 ← T will make the above formula true. (-x 1, -x 2, x 3 ) represents x 1 ← F, x 2 ← F, x 3 ← T 11

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If there is at least one assignment which satisfies a formula, then we say that this formula is satisfiable; otherwise, it is unsatisfiable. An unsatisfiable formula : x 1 v x 2 & x 1 v -x 2 & -x 1 v x 2 & -x 1 v -x 2 12

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Definition of the satisfiability problem: Given a Boolean formula, determine whether this formula is satisfiable or not. A literal : x i or -x i A clause : x 1 v x 2 v -x 3 C i A formula : conjunctive normal form C 1 & C 2 & … & C m 13

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Cook ’ s theorem Circuit satisfiablity problem (circuit SAT) is NP-complete. It is the first NP-complete problem. Every NP problem reduces to circuit SAT. To prove the other problems to be NP- complete, just need to show that they are as hard as circuit SAT problem. 14

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All the NP problems reduce to circuit SAT The proof is complicated Any problem in NP can be computed with a Boolean combination circuit (i.e., a computer) This circuit has a polynomial number of elements and can be constructed in polynomial time The circuit runs in polynomial time so we can check the result in polynomial time 15

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Decision problems The solution is simply “ Yes ” or “ No ”. Optimization problems are more difficult. e.g. the traveling salesperson problem Optimization version: Find the shortest tour Decision version: Is there a tour whose total length is less than or equal to a constant c ? 16

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Solving an optimization problem by a decision algorithm Solving minimization problem by decision algorithm Give c 1 and test (decision algorithm) Give c 2 and test (decision algorithm) Give c n and test (decision algorithm) We can find the smallest c i 17

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Toward NP-Completeness Once we have found an NP-complete problem, proving that other problems are also NP-complete becomes easier. Given a new problem Y, it is sufficient to prove that Cook’s problem, or any other NP-complete problems, is polynomially reducible to Y. Known problem -> unknown problem 18

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NP-Completeness Proof: CLIQUE Given that SAT problem is NP-complete, to prove that CLIQUE problem is NP-complete Problem: Does G=(V,E) contain a clique of size k? Theorem: Clique is NP-Complete. (reduction from SAT) Idea: Make “column” for each of k clauses. No edge within a column. All other edges present except between x and x’ Proof: (Reduction from SAT) CLIQUE is in NP. This is trivial since we can check it easily in polynomial time Goal: Transform arbitrary SAT instance into CLIQUE instance such that SAT answer is “yes” iff CLIQUE answer is “yes 19

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NP-Completeness Proof: CLIQUE Example: G = G has m-clique (m is the number of clauses in E), iff E is satisfiable. (Assign value 1 to all variables in clique) 20

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Vertex Cover Given that CLIQUE problem is NP-complete, to prove that vertex cover (VC) problem is NP-complete. Definition: A vertex cover of G=(V, E) is V’ V such that every edge in E is incident to some v V’. Vertex Cover(VC): Given undirected G=(V, E) and integer k, does G have a vertex cover with k vertices? CLIQUE: Does G contain a clique of size k? 21

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NP-Completeness Proof: Vertex Cover(VC) Problem: Given undirected G=(V, E) and integer k, does G have a vertex cover with k vertices? Theorem: the VC problem is NP-complete. Proof: (Reduction from CLIQUE) VC is in NP. This is trivial since we can check it easily in polynomial time. Goal: Transform arbitrary CLIQUE instance into VC instance such that CLIQUE answer is “yes” iff VC answer is “yes”. 22

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NP-Completeness Proof: Vertex Cover(VC) Claim: CLIQUE(G, k) has same answer as VC (, n-k), where n = |V|. Observe: There is a clique of size k in G iff there is a VC of size n-k in. 23

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NP-Completeness Proof: Vertex Cover(VC) Observe: If D is a VC in, then has no edge between vertices in V-D. So, we have k-clique in G n-k VC in Can transform in polynomial time. 24

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More convenient to use 3SAT For a given Boolean formula in conjunctive normal form (CNF) where each clause contains three variables, find the assignment to make it true Example: Can we find an assignment to make E true? 25

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3SAT is NP Complete Just need to rewrite SAT Given a clause with k variables in circuit SAT When k = 1 Add two more literals to construct a clause with 3 literals Example: Original: c i = {x} Construction: c i_new = {(x, u1, u2)^(x, u1’, u2’)^(x, u1, u2’)^(x, u1’, u2)}, in which ’ means negation 26

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3SAT is NP Complete When k = 2 Add one literal so that the number of literals in each clause is 3 Example: Original: c i = {(x1, x2)} Add one literal c i_new = {(x1, x2, u)^(x1, x2, u’)} When k > 3 Arrange these literals as a cascade of three literal clauses Example: Original: c i = {(x1, x2, x3, …, xn)} Add one literal c i_new = {(x1, x2, u1)^(x3, u1’, u2)^ … ^(x k-2, u k-4 ’, u k-3 )^(x k-1, x k, u k-3 ’)} 27

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Subset sum problem Def: A set of positive integers A = { a 1, a 2, …, a n } a constant C Determine if A A s.t. a i = C e.g. A = { 7, 5, 19, 1, 12, 8, 14 } C = 21, A = { 7, 14 } C = 11, no solution 28

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Subset sum is NP complete Reduce from 3SAT problem E = (u 1 + u 3 ’ + u 4 ’)(u 1 ’ + u 2 + u 4 ’) There are 4 literals There are n = 2 clauses in the expression above Suppose the solution is u 1 = u 2 = u 3 = 1, u 4 = 0 29

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Table construction for subset sum Reduce from 3SAT select row T1, T2, T3, F4 according to solution Select S2 1 and S2 2 to make the sum of last two columns 4 Now we have found the solution for subset sum Table: 30

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Basic Construction Basic idea Create a table for the subset sum problem The first m columns of the table stand for each one of m literals Last n columns stand for each one of m clauses First 2m rows stand for TRUE and FALSE of each literal Last 2n rows stores additional number for each clause to make the sum of this column a constant 31

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Exercise 2 To prove the following partition problem to be NP complete Def: Given a set of positive integers A = { a 1,a 2, …,a n }, determine if a partition P, s.t. a i = a i i p i p 32

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Exercise 3 To prove the following bin packing problem to be NP complete Def: n items, each of size c i, c i > 0 bin capacity : C Determine if we can assign the items into k bins, s.t. c i C, 1 j k. i bin j 33

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Exercise 4 To prove the following knapsack problem to be NP complete Def: n objects, each with a weight w i > 0 a profit p i > 0 capacity of knapsack : M Maximize p i x i 1 i n Subject to w i x i M 1 i n x i = 0 or 1, 1 i n Decision version : Given K, p i x i K ? 1 i n Knapsack problem : 0 x i 1, 1 i n. 34

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Three dimensional matching problem is NP complete Reduce from 3SAT problem to show that three dimensional matching (3DM) problem is NP-complete. X, Y, and Z are finite disjoint sets T = X × Y × Z Find M ⊆ T such that for any two distinct triples (x 1, y 1, z 1 ) ∈ M and (x 2, y 2, z 2 ) ∈ M, we have x 1 ≠x 2, y 1 ≠y 2, and z 1 ≠z 2 M covers all elements in X, Y and Z 35

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Reduce from 3SAT by example Construct a gadget with 2k cores and 2k tips for each variable x Example: k = 2 This gadget can work as a Boolean variable: when x = 1, we choose cores and tips in light region; when x = 0, we choose the blue region 36

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Build Boolean expressions Construct a gadget for each literal in a clause Add two cores for each clause and enclose them with tips uncovered x1x1 x2x2 x3x3 37

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Proof Idea Basic idea We choose the wings based on whether we set a variable to true or false. We use the clean up gadgets to cover all the rest of the tips. 38

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Summary NP-hard and NP-complete NP-completeness proof Polynomial time reduction List of NP-complete problems Knapsack problem 39

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