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NP-Completeness and Decision Problems

This reference provides an overview of NP-completeness and decision problems in computer science, including the classes P and NP, polynomial transformations, and NP-complete problems. It also explores the question of whether P is equal to NP and the importance of SATISFIABILITY as the first NP-complete problem.

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NP-Completeness and Decision Problems

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  1. NP-CompletenessReference: Computers and Intractability: A Guide to the Theory of NP-Completenessby Garey and Johnson, W.H. Freeman and Company, 1979. NP-Completeness

  2. f(n) \ n 10 30 50 n 0.00001 sec 0.00003 sec 0.00005 sec n5 0.1 sec 24.3 sec 5.2 mins 2n 0.001 sec 17.9 mins 35.7 yrs General Problems, Input Size and Time Complexity • Time complexity of algorithms : polynomial time algorithm ("efficient algorithm") v.s. exponential time algorithm ("inefficient algorithm") NP-Completeness

  3. “Hard” and “easy’ Problems • Sometimes the dividing line between “easy” and “hard” problems is a fine one. For example • Find the shortest path in a graph from X to Y. (easy) • Find the longest path in a graph from X to Y. (with no cycles) (hard) • View another way – as “yes/no” problems • Is there a simple path from X to Y with weight <= M? (easy) • Is there a simple path from X to Y with weight >= M? (hard) • First problem can be solved in polynomial time. • All known algorithms for the second problem (could) take exponential time . NP-Completeness

  4. Decision problem: The solution to the problem is "yes" or "no". Most optimization problems can be phrased as decision problems (still have the same time complexity). Example : Assume we have a decision algorithm X for 0/1 Knapsack problem with capacity M, i.e. Algorithm X returns “Yes” or “No” to the question “is there a solution with profit  P subject to knapsack capacity  M?” NP-Completeness

  5. We can repeatedly run algorithm X for various profits(P values) to find an optimal solution. Example : Use binary search to get the optimal profit, maximum of lg pi runs. (where M is the capacity of the knapsack optimization problem) Min Bound Optimal Profit Max Bound 0 pi |___________________|_________________| Search for the optimal solution NP-Completeness

  6. The Classes of P and NP • The class P and Deterministic Turing Machine • Given a decision problem X, if there is a polynomial time Deterministic Turing Machine program that solves X, then X is belong to P • Informally, there is a polynomial time algorithm to solve the problem NP-Completeness

  7. The class NP and Non-deterministic Turing Machine • Given a decision problem X, if there is a polynomial time Non-deterministic Turing machine program that solves X, then X belongs to NP • Given a decision problem X. For every instance I of X, (a) guess solution S for I, and (b) check “is S a solution to I?”. If (a) and (b) can be done in polynomial time, then X belongs to NP. NP-Completeness

  8. NP • Obvious : P  NP, i.e. A problem in P does not need “guess solution”. The correct solution can be computed in polynomial time. P • Some problems which are in NP, but may not in P : • 0/1 Knapsack Problem • PARTITION Problem : Given a finite set of positive integers Z. Question : Is there a subset Z' of Z such that Sum of all numbers in Z' = Sum of all numbers in Z-Z' ? i.e.  Z' =  (Z-Z') NP-Completeness

  9. One of the most important open problem in theoretical compute science :Is P=NP ? Most likely “No”. Currently, there are many known problems in NP, and there is no solution to show anyone of them in P. NP-Completeness

  10. NP-Complete Problems • Stephen Cook introduced the notion of NP-Complete Problems. This makes the problem “P = NP ?” much more interesting to study. • The following are several important things presented by Cook : NP-Completeness

  11. Polynomial Transformation ("  ") • L1  L2 : There is a polynomial time transformation that transforms arbitrary instance of L1 to some instance of L2. • If L1  L2 then L2 is in P implies L1 is in P (or L1 is not in P implies L2 is not in P) • If L1  L2 and L2  L3 then L1  L3 NP-Completeness

  12. Focus on the class of NP – decision problems only. Many intractable problems, when phrased as decision problems, belong to this class. • L is NP-Complete if L NP and for all other L'  NP, L'  L • If a problem in NP-complete can be solved in polynomial time then all problems in NP can be solved in polynomial time. • If a problem in NP cannot be solved in polynomial time then all problems in NP-complete cannot be solved in polynomial time. Note that an NP-complete problem is one of those hardest problems in NP. L NP NP-Completeness

  13. So, if an NP-complete problem is in P then P=NP • if P != NP then all NP-complete problems are in NP-P Question : how can we obtain the first NP-complete problem L? • Cook Theorem : SATISFIABILITY is NP-Complete. (The first NP-Complete problem) Instance : Given a set of variables, U, and a collection of clauses, C, over U. Question : Is there a truth assignment for U that satisfies all clauses in C? NP-Completeness

  14. Example : U = {x1, x2} C1 = {(x1, ¬ x2), (¬ x1, x2)} = (x1 OR ¬ x2) AND (¬ x1 OR x2) if x1 = x2 = True  C1 = True C2 = (x1, x2) (x1, ¬ x2) (¬ x1)  not satisfiable “¬ xi ”= “not xi ” “OR”= “logical or” “AND”= “logical and” This problem is also called “CNF-Satisfiability” since the expression is in CNF – Conjunctive Normal Form (the product of sums). NP-Completeness

  15. With the Cook Theorem, we have the following property : Lemma : If L1 and L2 belong to NP, L1 is NP-complete, and L1  L2 then L2 is NP-complete. i.e. L1, L2 NP and for all other L'  NP, L'  L1 and L1  L2  L'  L2 NP-Completeness

  16. So now, to prove a problem L to be NP-complete problem , we need to • show L is in NP • select a known NP-complete problem L' • construct a polynomial time transformation f from L' to L • prove the correctness of f (i.e. L’ has a solution if and only if L has a solution) and that f is a polynomial transformation NP-Completeness

  17. A group of problems, including all of the ones we have discussed (Satisfiability, 0/1 Knapsack, Longest Path, Partition) have an additional important property: If any of them can be solved in polynomial time, then they all can! • P: Problems solvable by deterministic algorithms in polynomial time • NP: Problems solved by non-deterministic algorithms in polynomial time NP-Complete NP P • These problems are called NP-complete problems. NP-Completeness

  18. Some NP-complete problems : • SATISFIABILITY • 0/1 Knapsack • PARTITION • Two-Processor Non-Preemptive Schedule Length • CLIQUE : An undirected graph G=(V, E) and a positive integer J  |V| Question : Does G contain a clique (complete subgraph) of size J or more? NP-Completeness

  19. Proving NP-Completeness Results • Example 1 : Show that the PARTITION problem is NP-complete. Given a known NPC problem - Sum of Subset Problem (SS), show that PARTITION problem is NPC. SS Problem Instance : Let A = {a1, a2, …, an} be a set of n positive numbers. Question : Given M, is there a subset A'  A such that A' = M NP-Completeness

  20. PARTITION Problem Instance : Given a finite set of m positive integers Z. Question : Is there a subset Z'  Z such that  Z' =  (Z-Z') • PARTITION is in NP guess a subset Z' O(m) verify  Z' =  (Z-Z')? O(m) Total O(m) NP-Completeness

  21. SS  PARTITION Given an arbitrary instance of SS, i.e. A = {a1, a2, …, an} and M, Construct an instance of PARTITION as follows : B={b1, b2, …, bn, bn+1, bn+2} of m = n+2 positive numbers where  bi = ai for 1  i  n bn+1 = M + 1 bn+2 =A + (1 – M) Note :  bi = 2 A + 2. Also, the transformation can be done in polynomial time (based on input size of A & M) NP-Completeness

  22. To show the transformation is correct : The SS problem has a solution if and only if the PARTITION problem has a solution. • If SS problem has a solution, then the PARTITION problem has a solution assume A' is the solution for SS problem then Z' = A'  {bn+2} and Z-Z' =A-A'  {bn+1}  Z' = M + A + (1 – M) = A + 1 =  (Z-Z') NP-Completeness

  23. If the PARTITION problem has a solution then the SS problem has a solution if Z' is the solution then  Z' = A + 1  exactly one of bn+2 orbn+1 Z'  if bn+2  Z' then Z' – { bn+2 } = A' and  A' = M if bn+1 Z', then use Z - Z' to obtain A' NP-Completeness

  24. Example 2 : Show that the Traveling Salesman (TS) Problem is NP-complete. Given a known NPC problem - Hamiltonian Circuit (HC), show that TS problem is NPC. Hamiltonian Circuit (HC) problem Instance : Give an undirected graph G=(V, E) Question : Does G contain a Hamiltonian circuit, i.e. a sequence < v1, v2, …, vn > of all vertices in V which is a simple cycle. NP-Completeness

  25. Traveling Salesman (TS) Problem Instance : Give an undirected complete graph G=(V, E) with distance d(i,j)  0 for each edge (i,j) for i  j and a positive integer B. Question : Is there a tour of all cities (a simple cycle with all vertices) having total distance no more than B. • TS is in NP guess a tour, i.e. sequence of all vertices O(|V|) verify that it is a cycle covering all vertices and total distance  B O(|V|) NP-Completeness

  26. HC  TS Given arbitrary instance of HC, i.e. G=(V, E). Construct an instance of TS as follows : G’ = (V , E’ ), where (u,v)  E’ for all u, v  V and u  v d(u,v) = 0 if (u,v)  E d(u,v) = 1 if (u,v)  E and B = 0 Note : The transformation can be done in polynomial time (based on input size of V and E) NP-Completeness

  27. To show the transformation is correct : The HC problem has a solution if and only if the TS problem has a solution. • If HC problem has a solution, then TS problem has a solution Assume a < v1, v2, …, vn > is the solution for HC  It is a simple cycle which contains all vertices  Each edge (u,v) in this cycle has d(u,v) = 0  Total distance is 0  Solution for TS • If TS problem has a solution then HC problem has a solution Obvious to see. NP-Completeness

  28. Example 3 : Show that the Vertex Cover (VC) Problem is NP-complete. Given 3SAT problem is NPC, show that VC problem is NPC. 3SAT Problem Instance : Given a set of variables U = {u1, u2, …, un} and a collection of clauses C = {c1, c2, …, cm} over U such that | ci | = 3 for 1  i  m. Question : Is there a truth assignment for U that satisfies all clauses in C? Note : 3SAT problem is a restricted problem of SATISFIABILITY problem. NP-Completeness

  29. Vertex Cover (VC) Problem Instance : Given an undirected graph G=(V, E) and a positive integer K  |V| Question : Is there a vertex cover of size K or less for G, i.e. a subset V’  V such that |V’|  K and, for each (u,v)  E, at least one of u or v  V’. • VC is in NP guess a set of vertices V’  V O(|V|) verify that |V’|  K and, for each (u,v)  E, u  V’ or v  V’ O(|V|+|E|) NP-Completeness

  30. 3SAT  VC Given arbitrary instance of 3SAT, i.e. U = {u1, u2, …, un} and C = {c1, c2, …, cm}. Construct an instance of VC as follows : • G = (V , E ) and K = n+2m • V = Vu Vc Vu = {u1t, u1f, u2t, u2f, …, unt, unf} and Vc = {a11, a12, a13}  {a21, a22, a23}  …  {am1, am2, am3} NP-Completeness

  31. E = Eu Ec Euc Eu = {(u1t, u1f) , (u2t, u2f) , …, (unt, unf)} Ec = {(a11, a12), (a12, a13), (a13, a11)}  …  {(am1, am2), (am2, am3), (am3, am1)} Assume ci = (xi, yi, zi) for 1  i  m, find the corresponding vertices, xi, yi, zi, in Vu Euc = {(x1, a11), (y1, a12), (z1, a13)}  …  {(xm, am1), (ym, am2), (zm, am3)} NP-Completeness

  32. |V| = 2n+3m and |E| = n+3m+3m • The transformation can be done in polynomial time (based on input size n and m) Example : U = {u1, u2, u3 , u4} and C = {{u1, ¬ u3, ¬ u4}, {u1, u2, ¬ u4}} Eu u1f u2t u2f u3t u3f u4t u4f u1t Euc E  a22 a12 Ec a11 a13 a21 a23 NP-Completeness

  33. Major Property : if there is a vertex cover set V’  K = n+2m, then • |V’| = n+ 2m, and • V’ must include exactly 1 vertex in {uit, uif} for 1  i  n from Vu and at exactly 2 vertices in {ai1, ai2, ai3} for 1  i  m from Vc, i.e. n vertices from Vu and 2m vertices from Vc • Look at edges in Eu and Ec, a vertex cover set V’ must include at least 1 vertex {uit, uif} for 1  i  n, and at least 2 vertices from {ai1, ai2, ai3} for 1  i  m. Since |V’|  K  |V’| = K NP-Completeness

  34. To show the transformation is correct : The 3SAT problem has a solution if and only if the VC problem has a solution. • If VC problem has a solution then 3SAT problem has a solution From the above property, V’ contains n vertices from Vu and 2m vertices from Vc From Vu , the truth assignment for {u1, u2, …, un} in 3SAT is ui = T if uit V’ ui = F if uif  V’ for 1  i  n NP-Completeness

  35. To see that this is a solution for 3SAT : we must show that for each ci = (xi, yi, zi), there is at least one variable i {xi, yi, zi} which set ci to TRUE, 1  i  n From the above property, exactly 2 vertices from {ai1, ai2, ai3} in V’, for 1  i  m only cover 2 edges from {(xi, ai1), (yi, ai2), (zi, ai3)} from Euc assume the edge (xi, ai1) is not covered by 2 vertices from {ai1, ai2, ci3} then xi  V’ since V’ is a vertex cover set xi set the clause ci to True for 1  i  n NP-Completeness

  36. If the 3SAT problem has a solution, then the VC problem has a solution The vertex cover set V’ with exactly n+2m vertices can be obtained as follows : From the truth assignment for {u1, u2, …, un} in 3SAT, we get n vertices from Vu, i.e. uit V’ if ui = T; otherwise uif V’ for 1  i  n This covers all edges in Eu and at least one edge in {(xi, ai1), (yi, ai2), (zi, ai3)} for 1  i  m From Vc, include 2 vertices from each {ai1, ai2, ai3} into V’, for 1  i  m. These 2 vertices cover all edges {(ai1, ai2), (ai2, ai3), (ai3, ai1)} and also cover edges in Euc that are not covered previously. NP-Completeness

  37. Example 4 : Show that the Square Packing (SP) Problem is NP-complete. Motivation : truck loading, processor allocation problem in a partitionable mesh connected system, the design of VLSI chips and etc. Square Packing Problem Given a packing square S and a set of packed squares L = {s1, s2, ..., sn}. Question : is there and orthogonal packing of L into S? Note : orthogonal packing  the sides of squares are parallel to the vertical and horizontal axes NP-Completeness

  38. 3-Partition Problem Given a list A = {a1, a2, ..., a3z} of 3z positive integers such that sum of all numbers is zB and B/4 < ai < B/2 for each 1  i  3z. Question : Can A be partitioned into z groups such that the sum of allnumbers in each group is B. Note : each group must have exactly 3 numbers Proof : Refer to the following research paper Leung, Tam, Wong, Young and Chin, “Packing Squares into a Square”, Journal of Parallel and Distributed Computing, 1990. NP-Completeness

  39. Coping with NP-Completeness Problem • NP-hard Problem Note : Refer to Chapter 5 of Garey and Johnson If L' L and L' is an NP-complete problem then L is called NP-hard problem. All NPC problems are NP-hard. NP-Completeness

  40. There are some NP-hard decision problems that are not in NP. Example : Kth Largest Subset Problem is not in NP Instance : Given a set of positive integers A = {a1, a2, …, an}, and two non-negative numbers, B A and K  2|A|. Question : Are there at least K distinct subsets A’  A such that each subset has total sum >= B. Note : PARTITION problem  Kth Largest Subset Problem NP-Completeness

  41. Pseudo-polynomial time algorithm Note : Refer to Chapter 4 of Garey and Johnson Some NP-complete problem may be solved in "polynomial" time (based on input size and magnitude). Example : PARTITION problem NP-Completeness

  42. Dynamic Programming Algorithm : • assume B = (sum of n integers)/2 • construct a table of size (approx.) n x B • fill in the table row by row • for each row, add a new element mark sum of all possible subsets if there is a subset with sum = B, stop. Time Complexity : O(nB) NP-Completeness

  43. NP-completeness in the strong sense Note : Refer to Chapter 4 of Garey and Johnson If L is NP-complete problem in the strong sense, then L cannot be solved by a pseudo-polynomial time algorithm unless P=NP L is NP-complete in the strong sense if L'  L , L' is NP-complete in the strong sense and L is in NP Example : 3-partition problem is NPC in the strong sense. NP-Completeness

  44. Solving more restricted problems If we restricted the problem L to problem L' i.e. L' is a special/restricted case of L, then we may solve L' in polynomial time. For example : PARTITION problem if we assume that each input integer ai n, where n is the number of input integers, then the pseudo polynomial time algorithm becomes polynomial time algorithm, i.e. O(n3) NP-Completeness

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