 # The Theory of NP-Completeness

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The Theory of NP-Completeness
2010/11/30

Polynomial-time Reductions
We want to solve a problem R; we already have an algorithm for a problem S We have a transformation function T Correct answer for R on x is “yes”, iff the correct answer for S on T(x) is “yes” Problem R is polynomially reducible to (polynomially reduces to) problem S if such a transformation T can be computed in polynomial time (R  S) The point of reducibility: S is at least as hard to solve as R

Polynomial-time Reductions
We use reductions to prove that a problem is NP-complete x is an input for R; x’ is an input for S (R  S) x x’ Algorithm for S T Yes or No answer Algorithm for R

P NPC NP NP: Non-deterministic Polynomial P: Polynomial
? NP: Non-deterministic Polynomial P: Polynomial NPC: Non-deterministic Polynomial Complete

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. (It is “at least as hard as the hardest problems in NP.”) NP-complete: the class of problems which are NP-hard and belong to NP.

Nondeterministic algorithms
A nondeterministic algorithm is an algorithm consisting of two phases: guessing (or choice) and checking. Furthermore, it is assumed that a nondeterministic algorithm always makes a correct guessing.

Nondeterministic algorithms
Machines for running nondeterministic algorithms do not exist and they would never exist in reality. (They can only be made by allowing unbounded parallelism in computation.) (Quantum computation may be an approximation.) Nondeterministic algorithms are useful only because they will help us define a class of problems: NP problems

NP (Nondeterministic Polynomial) algorithm
If the checking stage of a nondeterministic algorithm is of polynomial time-complexity, then this algorithm is called an NP (nondeterministic polynomial) algorithm.

NP problem If a decision problem can be solved by a NP algorithm, this problem is called an NP (nondeterministic polynomial) problem. NP problems : must be decision problems

Decision problems The solution is simply “Yes” or “No”.
Optimization problem : harder Decision problem : easier E.g. the traveling salesperson problem Optimization version: Find the shortest tour (cycle) Decision version: Is there a tour whose total length is less than or equal to a constant C ?

Decision version of sorting
Given a1, a2,…, an and c, is there a permutation of ais ( a1, a2 , … ,an ) such that∣a2–a1∣+∣a3–a2∣+ … +∣an–an-1∣＜ C ?

Decision vs Original Version
We consider decision version problem D rather than the original problem O because we are addressing the lower bound of a problem and D ∝ O

To express Nondeterministic Algorithm by three special commands
Choice(S) : arbitrarily chooses one of the elements in set S at one step Failure : an unsuccessful completion Success : a successful completion

Example: Nondeterministic searching Algorithm
input: n elements and a target element x output: success if x is found among the n elements; failure, otherwise. j ← choice(1 : n) /* guess if A(j) = x then success /* check else failure A nondeterministic algorithm terminates unsuccessfully iff there exist no set of choices leading to a success signal. The time required for choice(1 : n) is O(1).

Relationship Between NP and P
It is known PNP. However, it is not known whether P=NP or whether P is a proper subset of NP It is believed NP is much larger than P We cannot find a polynomial-time algorithm for many NP problems. But, no NP problem is proved to have exponential lower bound. (No NP problem has been proved to be not in P.) So, “does P = NP?” is still an open question! Cook tried to answer the question by proposing NPC.

NP-hard A problem X is NP-hard if every NP problem (polynomially) reduces to X.

NP-complete (NPC) A problem X is NP-complete (NPC) if X∈NP and every NP problem (polynomially) reduces to X. So, X  NPC if XNP-hard and XNP

Cook’s Theorem (1971) Prof. Cook Toronto U.
Receiving Turing Award (1982) Discussing difficult problems whose worst case lower bound seems to be in the order of an exponential function The problems are called NP-complete (NPC) Problems

Cook’s Theorem (1971) NP = P iff SAT  P
That is, NP = P iff the satisfiability (SAT) problem is a P problem SAT is NP-complete It is the first NP-complete problem Every NP problem reduces to SAT

SAT is NP-complete Every NP problem can be solved by an NP algorithm
Every NP algorithm can be transformed in polynomial time to an SAT problem Such that the SAT problem is satisfiable iff the answer for the original NP problem is “yes” That is, every NP problem  SAT SAT is NP-complete

Proof of NP-Completeness
To show that X is NP-complete (I) Prove that X is an NP problem (II) Prove that  Y  NPC, Y  X  X  NPC Why ? Transitive property of polynomial-time reduction

Karp R. Karp showed several NPC problems, such as 3-STA, node (vertex) cover, and Hamiltonian cycle problems, etc. Karp received Turing Award in 1985

NP-Completeness Proof: Reduction
All NP problems SAT Clique 3-SAT Vertex Cover Chromatic Number Dominating Set

NP-Completeness “NP-complete problems”: the hardest problems in NP
Interesting property If any one NP-complete problem can be solved in polynomial time, then every problem in NP can also be solved in polynomial time (i.e., P=NP) Many believe P≠NP

Importance of NP-Completeness
NP-complete problems: considered “intractable” Important for algorithm designers & engineers Suppose you have a problem to solve Your colleagues have spent a lot of time to solve it exactly but in vain See whether you can prove that it is NP-complete If yes, then spend your time developing an approximation (heuristic) algorithm Many natural problems can be NP-complete

Some concepts 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. The lower bound of any NPC problem seems to be in the order of an exponential function. The theory of NP-completeness always considers the worst case.

Caution ! If a problem is NP-complete, its special cases may or may not be of exponential time-complexity. We consider worst case lower bound in NP-complete.

Even harder problems: Undecidable Problems
They cannot be solved by guessing and checking. They are even more difficult than NP problems. E.G.: Halting problem: Given an arbitrary program with an arbitrary input data (the number of input cases is infinite), will the program terminate or not? It is not NP It is NP-hard (SAT Halting problem)

Some known NPC problems
SAT problem 3SAT problem 0/1 knapsack problem Traveling salesperson problem Partition problem Art gallery problem Clique problem Vertex Cover problem Dominating Set problem Chromatic Coloring problem

The satisfiability (SAT) problem
Def : Given a Boolean formula, determine whether this formula is satisfiable or not. A literal (Boolean variable): xi or -xi A clause (disjunction of variables): ci  x1 v x2 v -x3 A formula : conjunctive normal form C1& c2 & … & cm

NPC Problems SAT: Give a Boolean expression (formula) in CNF (conjunctive normal form), determine if it is satisfiable. 3SAT: Give a Boolean expression in CNF such that each clause has exactly 3 variables (literals), determine if it is satisfiable.

The satisfiability (SAT) problem
The satisfiability problem The logical formula : x1 v x2 v x3 & - x1 & - x2 the assignment : x1 ← F , x2 ← F , x3 ← T will make the above formula true (-x1, -x2 , x3) represents

The satisfiability problem
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 : x1 v x2 & x1 v -x2 & -x1 v x2 & -x1 v -x2

Resolution principle Let formula F= c1 : -x1 v -x2 v x3 c2 : x1 v x4
 c3 : -x2 v x3 v x4 (resolvent) x1 cannot satisfy c1 and c2 at the same time, so it is deleted to deduce a resolvent of c1 and c2. This is called the resolution principle .

Satisfiable If no new clauses can be deduced  satisfiable
-x1 v -x2 v x (1) x (2) x (3) (1) & (2) x2 v x (4) (4) & (3) x (5) (1) & (3) x1 v x (6)

Unsatisfiable If an empty clause is deduced  unsatisfiable
- x1 v -x2 v x (1) x1 v -x (2) x (3) - x (4)    deduce   (1) & (2) x2 v x3 (5) (4) & (5) x (6) (6) & (3) □ (7)

Nondeterministic SAT Guessing for i = 1 to n do
xi ← choice( true, false ) if E(x1, x2, … ,xn) is true Checking then success else failure

Proof of Cook’s Theory Dr. Cook Prove the theory by showing that all instances of each NP algorithm can reduce to the SAT problem. Below, we show a simple example that the NP searching algorithm reduces to the SAT problem.

The NP searching algorithm reduces to the SAT problem
Does there exist a number in { x(1), x(2), …, x(n) }, which is equal to 7? For example, assume n = 2 The algorithm becomes:

The NP searching algorithm reduces to SAT
i=1 v i=2 & i=1 → i≠2 & i=2 → i≠1 & x(1)=7 & i=1 → SUCCESS & x(2)=7 & i=2 → SUCCESS & x(1)≠7 & i=1 → FAILURE & x(2)≠7 & i=2 → FAILURE & FAILURE → -SUCCESS & SUCCESS (Guarantees a successful termination when it is possible) & x(1)=7 (Input Data) & x(2)≠7

The NP searching algorithm reduces to SAT
CNF (conjunctive normal form) : i=1 v i= (1) i≠1 v i≠2 (2) x(1)≠7 v i≠1 v SUCCESS (3) x(2)≠7 v i≠2 v SUCCESS (4) x(1)=7 v i≠1 v FAILURE (5) x(2)=7 v i≠2 v FAILURE (6) -FAILURE v -SUCCESS (7) SUCCESS (8) x(1)= (9) x(2)≠ (10)

The NP searching algorithm reduces to SAT
Satisfiable at the following assignment : i=1 satisfying (1) i≠2 satisfying (2), (4) and (6) SUCCESS satisfying (3), (4) and (8) -FAILURE satisfying (7) x(1)=7 satisfying (5) and (9) x(2)≠7 satisfying (4) and (10)

The NP searching algorithm reduces to SAT
Searching for 7, but x(1)7, x(2)7 CNF :

The NP searching algorithm reduces to SAT
Apply resolution principle :

Searching in CNF with inputs

0/1 Knapsack problem Given M (weight limit) and P, is there is a solution with a profit larger than P? This is an NPC problem. P1 P2 P3 P4 P5 P6 P7 P8 Profit 10 5 1 9 3 4 11 17 Weight 7 22 15

Traveling salesperson problem
Given: A set of n planar points and a value L Find: Is there a closed tour which includes all points exactly once such that its total length is less than L? This is an NPC problem.

Partition problem Given: A set of positive integers S
Find: Is there a partition of S1 and S2 such that S1S2=, S1S2=S, iS1i=iS2 i (partition S into S1 and S2 such that element sum of S1 is equal to that of S2) e.g. S={1, 7, 10, 9, 5, 8, 3, 13} S1={1, 10, 9, 8} S2={7, 5, 3, 13} This problem is NP-complete.

Art gallery problem: *Given a constant C, is there a guard placement such that the number of guards is less than C and every wall is monitored? *This is an NPC problem.

NPC Problems CLIQUE(k): Does G=(V,E) contain a clique of size k?
Definition: A clique in a graph is a set of vertices such that any pair of vertices are joined by en edge.

NPC Problems Vertex Cover(k): Given a graph G=(V, E) and an integer k, does G have a vertex cover with k vertices? Definition: A vertex cover of G=(V, E) is V’V such that every edge in E is incident to some vV’.

NPC Problems Dominating Set(k): Given an graph G=(V, E) and an integer k, does G have a dominating set of size k ? Definition: A dominating set D of G=(V, E) is DV such that every vV is either in D or adjacent to at least one vertex of D.

NPC Problems Chromatic Coloring(k): Given a graph G=(V, E) and an integer k, does G have a coloring for k Definition A coloring of a graph G=(V, E) is a function f : V  { 1, 2, 3,…, k }  if (u, v)  E, then f(u)f(v).

Q&A