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NP-Complete Problems

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NP-Complete Problems

Polynomial time vs exponential time

Polynomial O(nk),

where n is the input size

e.g., number of nodes in a graph, the length of strings , etc

k is a constant

e.g., k=2 in LCS, k=1 in KMP, etc.

Exponential time: 2n or nn

n= 2 10 20, 30

2n4 1024 1 million 1000 million

If a computer solves a problem of size n in one hour, now you have a computer 1,000,000 faster, what size of the problem can you solve in one hour?

n+20 (2n+201,000,0002n)

The improvement is small.

Hardware improves little on problems of exponential running time

Exponential running time is considered as “not efficient”.

Story

All algorithms we have studied so far are polynomial time algorithms (unless specified).

Facts: people have not yet found any polynomial time algorithms for some famous problems, (e.g., Hamilton Circuit, longest simple path, Steiner trees).

Question: Do there exist polynomial time algorithms for those famous problems？

Answer: No body knows.

Story

Research topic: Prove that polynomial time algorithms do not exist for those famous problems, e.g., Hamilton circuit problem.

Turing award

Vinay Deolalikar attempted 2010, but failed.

An article on The New Yorker http://www.newyorker.com/online/blogs/elements/2013/05/a-most-profound-math-problem.html

To answer the question, people define two classes

P class

NP class.

To answer if PNP, a rich area, NP-completeness theory is developed.

Class P and Class NP

Class P contains problems which are solvable in polynomial time.

The problems have algorithms in O(nk) time, where n is the input size and k is a constant.

Class NP consists of those problem that are verifiable in polynomial time.

we can verify that the solution is correct in time polynomial in the input size to the problem.

algorithms produce an answer by a series of “correct guesses”

Example: Hamilton Circuit: given an order of the n distinct vertices(v1, v2, …, vn), we can test if (vi, v i+1) is an edge in G for i=1, 2, …, n-1and (vn, v1) is an edge in G in time O(n) (polynomial in the input size).

Class P and Class NP

PNP

If we can design a polynomial time algorithm for problem A, then A is in P.

However, if we have not been able to design a polynomial time algorithm for A, then two possibilities:

No polynomial time algorithm for A exists or

We are not smart.

Open problem: PNP?

Polynomial-Time Reductions

Suppose a black box (an algorithm) can solve instances of problem X. If we give an instance of X as input the black box will return the correct answer in a single step.

Question: Can we “transform” problem Y into X. Can an arbitrary instance of problem Y be solved by the black box?

We can use the black box polynomial number of times.

We can use polynomial number of standard computational steps

If yes, then Y is polynomial-timereducible to X.

NP-Complete

A problem X is NP-complete if

it is in NP, and

any problem Y in NP has a polynomial time reduction to X.

it is the hardest problem in NP.

If ONE NP-complete problem can be solved in polynomial time, then any problem in class NP can be solved in polynomial time.

The first NPC problem is Satisfiabilityproblem

Proved by Cook in 1971 and obtains the Turing Award for this work

Boolean formula

A boolean formula f(x1, x2, …xn), where xi are boolean variables (either 0 or 1), contains boolean variables and boolean operations AND, OR and NOT .

Clause: variables and their negations are connected with OR operation, e.g., (x1ORNOTx2OR x5)

Conjunctive normal form of boolean formula:

contains m clauses connected with AND operation.

Example:

(x1OR NOT x2) AND (x1OR NOT x3OR x6) AND (x2OR x6) AND (NOT x3OR x5).

Here we have four clauses.

Satisfiability problem

Input: conjunctive normal form with n variables, x1, x2, …, xn.

Problem: find an assignment of x1, x2, …, xn(setting each xi to be 0 or 1) such that the formula is true (satisfied).

Example: conjunctive normal form is

(x1OR NOTx2) AND (NOT x1OR x3).

The formula is true for assignment

x1=1, x2=0, x3=1.

Note: for n Boolean variables, there are 2n assignments.

Testing if formula=1 can be done in polynomial time for any given assignment.

Given an assignment that satisfies formula=1 is hard.

The First NP-complete Problem

Theorem: Satisfiability problem is NP-complete.

It is the first NP-complete problem.

Stephen. A. Cook in 1971 http://en.wikipedia.org/wiki/Stephen_Cook

Won Turing prize for his work.

Significance:

If Satisfiability problem is in P, then ALL problems in class NP are in P.

To solve PNP, you should work on NPC problems such as satisfiability problem.

We can use the first NPC problem, Satisfiability problem, to show other NP-complete problems.

How to show that a problem is NPC?

To show that problem A is NP-complete, we can

First, find a NP-complete problem B.

Then, show that

IF problem A is P, then B is in P.

that is, to give a polynomial time reduction from B to A.

Remarks:Since a NPC problem, problem B, is the hardest in class NP, problem A is also the hardest

- Hamilton circuit : a circuit uses every vertex of the graph exactly once except for the last vertex, which duplicates the first vertex.
- It was shown to be NP-complete.

- Longest Simple Path:
- Input: V={v1, v2, ..., vn} be a set of nodes in a graph G and d(vi, vj) the distance between vi and vj,,
- Output: a longest simple path from u to v .

- Theorem 2: The longest simple path problem is NP-complete.

Proof:

Hamilton Circuit Problem (HC): Given a graph G=(V, E), find a Hamilton Circuit.

We want to show that if the longest simple path problem is in P, then the Hamilton circuit problem is in P.

Design a polynomial time algorithm to solve HC by using an algorithm for LSP.

Step 0: Set the length of each edge in G to be 1

Step 1: for each edge (u, v)E do

find the longest simple path P from u to v in G.

Step 2: if the length of P is n-1 thenby adding edge (u, v) we

obtain a Hamilton circuit in G.

Step 3: if no Hamilton circuit is found for every (u, v) then

print “no Hamilton circuit exists”

Conclusion:

- if LSP is in P, then HC is also in P.
- Since HC was proved to be NP-complete, LSP is also NP-complete.

- 3-Satisfiability : Each clause contains at most three variavles or their negations.
- Vertex Cover: Given a graph G=(V, E), find a subset V’ of V such that for each edge (u, v) in E, at least one of u and v is in V’ and the size of V’ is minimized.
- Hamilton Circuit: (definition was given before)
- History: Satisfiability3-Satisfiabilityvertex coverHamilton circuit.
- Those proofs are very hard.

Approximation Algorithms

Concepts

Knapsack

Steiner Minimum Tree

TSP

Vertex Cover

Optimization Problem:

The solution of the problem is associated with a cost (value).

We want to maximize the cost or minimize the cost.

Minimum spanning tree and shortest path are optimization problems.

Euler circuit problem is NOT an optimization problem.

An algorithm A is an approximation algorithm

- if given any instance I, it finds a candidate solution s(I)
How good an approximation algorithm is?

We use performance ratio to measure the quality of an approximation algorithm.

For minimization problem, the performance ratio r of algorithm A is defined as for any instance I of the problem,

where OPT(I) is the value of the optimal solution for instance I and A(I) is the value of the solution returned by algorithm A on instance I.

For maximization problem, the performance ratio r of algorithm A is defined as for any instance I of the problem,

OPT(I)

A(I)

is at most r (r1), where OPT(I) is the value of the optimal solution for instance I and A(I) is the value of the solution returned by algorithm A on instance I.

Given a finite set U of items, a size s(u) Z+, a capacity B, find a subset U'U such that

and such that the above summation is as large as possible. (It is NP-hard.)

- Sort u's based on s(u)'s in increasing order.
- Select the smallest remaining u until no more u can be added.
- Compare the total value of selected items with the item of the largest size, and select the larger one.
Theorem: The algorithm has performance ratio 2.

- Case 1: the total of selected items 0.5B (got it!)
- Case 2: the total of selected items < 0.5B.
- No remaining item left: we get optimal.
- There are some remaining items: the size of the smallest remaining item >0.5B. (Otherwise, we can add it in.)
- Selecting the largest item gives ratio-2.

- The 0-1 knapsack problem:
- N items: the i-th item is worth vi dollars and weight wi pounds.
- vi and wi are integers.

- A thief can carry at most W (integer) pounds.
- How to take as valuable a load as possible.
- An item cannot be divided into pieces.

- The fractional knapsack problem:
- The same setting, but the thief can take fractions of items.

- Delete the items i with wi>W.
- Sort items in decreasing order based on vi/wi.
- Select the first k items item 1, item 2, …, item k such that
w1+w2+…, wkW andw1+w2+…, wk +w k+1>W.

4. Comparevk+1 with v1+v2+…+vk and select the larger one.

Theorem: The algorithm has performance ratio 2.

- C(opt): the cost of optimum solution
- C(fopt): the optimal cost of the fractional version.
- C(opt)C(fopt).
- v1+v2+…+vk +v k+1> C(fopt). (from the greedy alg.)
- So, either v1+v2+…+vk >0.5 C(fopt)0.5c(opt)
or v k+1 >0.5 C(fopt)0.5c(opt).

- Since the algorithm chooses the larger one from v1+v2+…+vk and v k+1
- We know that the cost of the solution obtained by the algorithm is at least 0.5 C(fopt)c(opt).

Steiner minimum tree in the plane

- Input: a set of points R (regular points) in the plane.
- Output: a tree with smallest weight which contains all the nodes in R.
- Weight: weight on an edge connecting two points (x1,y1) and (x2,y2) in the plane is defined as the Euclidean distance

Example: Dark points are regular points.

Key for our approximation algorithm.

For any three points in the plane, we have:

dist(a, c ) ≤ dist(a, b) + dist(b, c).

Examples:

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Compute a minimum spanning tree for R as the approximation solution for the Steiner minimum tree problem.

How good the algorithm is? (in terms of the quality of the solutions)

Theorem: The performance ratio of the approximation algorithm is 2.

We want to show that for any instance (input) I, A(I)/OPT(I) ≤r (r≥1)

- A(I) is the cost of the solution obtained from our spanning tree algorithm, and
- OPT(I) is the cost of an optimal solution.

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- Assume that T is the optimal solution for instance I. Consider a traversal of T.

- Each edge in T is visited at most twice. Thus, the total weight of the traversal is at most twice of the weight of T, i.e.,
w(traversal)≤2w(T)=2OPT(I). .........(1)

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- Based on the traversal, we can get a spanning tree ST as follows: (Directly connect two nodes in R based on the visited order of the traversal.)

From triangle inequality,

w(ST)≤w(traversal) ≤2OPT(I). ..........(2)

- Inequality(2) says that the cost of ST is no more than twice of that of an optimal solution.
- So, if we can compute ST, then we can get a solution with cost≤2OPT(I).(Great! But finding ST may also be very hard, since ST is obtained from the optimal solution T, which we do not know.)
- We can find a minimum spanning tree MST for R in polynomial time.
- By definition of MST, w(MST) ≤w(ST) ≤2OPT(I).
- Therefore, the performance ratio is 2.

- Input: a graph G=(V,E), a weight w(e) for each e∈E, and a subset R⊂V.
- Output: a tree with minimum weight which contains all the nodes in R.
- The nodes in R are calledregularpoints. Note that, the Steiner minimum tree could contain some nodes in V-R and the nodes in V-R are called Steiner points.

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Figure a

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Example: Let G be shown in Figure a. R={a,b,c}. A Steiner minimum tree T={(a,d),(b,d),(c,d)} is shown in Figure b.

Theorem: Graph Steiner minimum tree problem is NP-complete.

- For each pair of nodes u and v in R, computethe shortest path from u to v and assign the cost of the shortest path from u to v as the length of edge (u, v). (a complete graph is given)
- Compute a minimum spanning tree for the modified complete graph.
- Include the nodes in the shortest paths used.

Theorem: The performance ratio of this algorithm is 2.

Proof:

We only have to prove that Triangle Inequality holds. If

dist(a,c)>dist(a,b)+dist(b,c) ......(3)

then we modify the path from a to c like

a→b→c

Thus, (3) is impossible.

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Example II-1

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The given graph

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Example II-2

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Modified complete graph

Example II-3

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The minimum spanning tree

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Example II-4

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The approximate Steiner tree

- Given n points in a plane, find a tour to visit each city exactly once.
- Assumption: the triangle inequality holds.
- that is, d (a, c) ≤ d (a, b) + d (b, c).

- This condition is reasonable, for example, whenever the cities are points in the plane and the distance between two points is the Euclidean distance.
- Theorem: TSP with triangle inequality is also NP-hard.

Algorithm A:

- Compute a minimum spanning tree algorithm (Figure a)
- Visit all the cities by traversing twice around the tree. This visits some cities more than once. (Figure b)
- Shortcut the tour by going directly to the next unvisited city. (Figure c)

(a)

A spanning tree

(b)

Twice around the tree

(c)

A tour with shortcut

Example:

- The cost of a minimum spanning tree: cost(t), is not greater than opt(TSP), the cost of an optimal TSP.
- n-1 edges in a spanning tree. n edges in TSP. Delete one edge in TSP, we get a spanning tree. Minimum spanning tree has the smallest cost.)

- The cost of the TSP produced by our algorithm is less than 2×cost(T) and thus is less than 2×opt(TSP).

Center Selection Problem

Problem: Given a set of points V in the plane (or some other metric space), find k points c1, c2, .., ck such that for each v in V,

min { i=1, 2, …, k} d(v, ci) d

and d is minimized.

Farthest-point clustering algorithm

1: arbitrarily select a point in V as c1.

2: let i=2.

3: pick a point ci from V –{c1, c2, …, ci-1} to maximize min {|c1ci|, |c2ci|,…,|ci-1 ci|}.

4: i=i+1;

5: repeat Steps 3 and 4 until i=k.

Theorem: Farthest-point clustering algorithm has ratio-2.

Proof: Let c i be an point in V that maximize

i=min {|c1ci|, |c2ci|,…,|ci-1 ci|}.

We have i i-1 for any i.

Since two, say ci and cj (i>j), of the k+1 points must be in the same group (in an opt solution), t 2 opt.

Thus, k+1 t 2 opt.

For any v in V, by the definition of k+1 ,

min {|c1v|, |c2v|,…,|ck v|} k+1 .

So the algorithm has ratio-2.

- Given a graph G=(V, E), find V'⊆V with minimum number of vertices such that for each edge (u, v)∈E at least one of u and v is in V’.
- V' is called vertex cover.
- The problem is NP-hard.
- A ratio-2 algorithm exists for vertex cover problem.