NP-Complete Problems

1
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) of our problem and k is a constant
  (e.g., k1, 2, 3, etc).
• Exponential time 2n or nn.
• n 2, 10, 20,
  30
• 2n 4 1024 1 million
  1000 million
• Suppose our computer can solve a problem of size
  k (i.e., compute 2k operations) in a
  hour/week/month. If the new computer is 1024
  times faster than ours, then the new computer can
  solve the problem of size k10 in the same time.
  The improvement is very little.
• Hardware improvement has little use for solving
  problems
• that require exponential running time.
• Exponential running time is considered as
  not efficient.

2
Story

• All algorithms we have studied so far are
  polynomial time algorithms.
• 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.

3
Story

• Research topic Prove that polynomial time
  algorithms do not exist for those famous
  problems, e.g., Hamilton circuit problem.
• You can get Turing award if you can give the
  proof.
• In order to answer the above question, people
  define two classes of problems, P class and NP
  class.
• To answer if P?NP, a rich area, NP-completeness
  theory is developed.

4
Class P and Class NP

• Class P contains those problems that are solvable
  in polynomial time.
• They are problems that can be solved 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.
• What we mean here is that if we were somehow
  given a solution, then we can verify that the
  solution is correct in time polynomial in the
  input size to the problem.
• Example Hamilton Circuit given an order of the
  n distinct vertices (v1, v2, , vn), we can test
  if (vi, v i1) is an edge in G for i1, 2, ,
  n-1 and (vn, v1) is an edge in G in time O(n)
  (polynomial in the input size).

5
Class P and Class NP

• Based on definitions, P?NP.
• If we can design a polynomial time algorithm for
  problem A, then problem A is in P.
• However, if we have not been able to design a
  polynomial time algorithm for problem A, then
  there are two possibilities
• polynomial time algorithm does not exist for
  problem A or
• we are not smart.
• Open problem P?NP?
• Clay 1 million prize.

6
Polynomial-Time Reductions

• Suppose we have a black box (an algorithm) that
  could solve instances of a problem X If we give
  the input of an instance of X, then in a single
  step, the black box will return the correct
  answer.
• Question
• Can arbitrary instances of problem Y be solved
  using polynomial number of standard computational
  steps, plus a polynomial number of calls to a
  black box that solves problem X?
• If yes, then Y is polynomial-time reducible to X.
  

7
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 an 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 Satisfiability probelm
• Proved by Cook in 1971 and obtains the Turing
  Award for this work

8
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., (x1 OR NOTx2
  OR x5)
• Conjunctive normal form of boolean formula
• contains m clauses connected with AND
  operation.
• Example
• (x1 OR NOT x2) AND (x1 OR NOT x3 OR x6) AND
  (x2 OR x6) AND (NOT x3 OR x5).
• Here we have four clauses.

9
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
• (x1 OR NOTx2) AND (NOT x1 OR x3).
• The formula is true for assignment
• x11, x20, x31.
• Note for n Boolean variables, there are 2n
  assignments.
• Testing if formula1 can be done in polynomial
  time for any given assignment.
• Given an assignment that satisfies formula1 is
  hard.

10
The First NP-complete Problem

• Theorem Satisfiability problem is NP-complete.
• It is the first NP-complete problem.
• S. A. Cook in 1971 http//en.wikipedia.org/wiki/St
  ephen_Cook
• Won Turing prize for his work.
• Significance
• If Satisfiability problem can be solved in
  polynomial time, then ALL problems in class NP
  can be solved in polynomial time.
• If you want to solve P?NP, then you should work
  on NPC problems such as satisfiability problem.
• We can use the first NPC problem, Satisfiability
  problem, to show that other problems are also
  NP-complete.

11
How to show that a problem is NPC?

• To show that problem A is NP-complete, we can
• First find a problem B that has been proved to be
  NP-complete.
• Show that if Problem A can be solved in
  polynomial time, then problem B can also be
  solved in polynomial time.
• 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

12
Hamilton circuit and Longest Simple Path

• 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 Vv1, v2, ..., vn be a set of nodes
  in a graph and d(vi, vj) the distance between
  vi and vj,, find a longest simple path from u to
  v .
• Theorem 2 The longest simple path problem is
  NP-complete.

13
Theorem 2 The longest simple path (LSP) 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 we can solve the longest
  simple path problem in polynomial time, then we
  can also solve the Hamilton circuit problem in
  polynomial time.
• 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 then by
  adding edge (u, v) we
• obtain an 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 can be solved in polynomial time, then HC
  can also be solved in polynomial.
• Since HC was proved to be NP-complete, LSP is
  also NP-complete.

14
Some basic NP-complete problems

• 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 Satisfiability?3-Satisfiability?vertex
  cover?Hamilton circuit.
• Those proofs are very hard.

15
Approximation Algorithms

• Concepts
• Knapsack
• Steiner Minimum Tree
• TSP
• Vertex Cover

16
Concepts of Approximation Algorithms

• 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. (it is a decision problem.)

17
Approximation Algorithm

• 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.

18
Performance ratio

• For minimization problem, the performance ratio
  of algorithm A is defined as a number r such that
  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.

19
Performance ratio

• For maximization problem, the performance ratio
  of algorithm A is defined as a number r such that
  for any instance I of the problem,
• OPT(I)
• A(I)
• is at most r (r?1), 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.

20
Simplified Knapsack Problem

• Given a finite set U of items, a size s(u) ? Z,
  a capacity B?maxs(u)u ? U, find a subset U'?U
  such that and such that the above summation
  is as large as possible. (It is NP-hard.)

21
Ratio-2 Algorithm

• 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.

22
Proof

• Case 1 the total of selected items ? 0.5B (got
  it!)
• Case 2 the total of selected items lt 0.5B.
• No remaining item left we get optimal.
• There are some remaining items the size of the
  smallest remaining item gt0.5B. (Otherwise, we
  can add it in.)
• Selecting the largest item gives ratio-2.

23
The 0-1 Knapsack problem

• The 0-1 knapsack problem
• N items, where 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.

24
Ratio-2 Algorithm

• Delete the items i with wigtW.
• Sort items in decreasing order based on vi/wi.
• Select the first k items item 1, item 2, , item
  k such that
• w1w2, wk ?W and w1w2, wk w
  k1gtW.
• 4. Compare vk1 with v1v2vk and select the
  larger one.
• Theorem The algorithm has performance ratio 2.

25
Proof of ratio 2

• C(opt) the cost of optimum solution
• C(fopt) the optimal cost of the fractional
  version.
• C(opt)?C(fopt).
• v1v2vk v k1gt C(fopt).
• So, either v1v2vk gt0.5 C(fopt)?0.5c(opt)
• or v k1 gt0.5
  C(fopt)?0.5c(opt).
• Since the algorithm choose the larger one from
  v1v2vk and v k1
• We know that the cost of the solution obtained by
  the algorithm is at least 0.5 C(fopt)?c(opt).

26
Steiner Minimum Tree

• 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

27

• Example Dark points are regular points.

28
Triangle inequality

• Key for our approximation algorithm.
• For any three points in the plane, we have
• dist(a, c ) dist(a, b) dist(b, c).
• Examples

c
5
4
a
b
3
29
Approximation algorithm(Steiner minimum tree in
the plane)

• 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.

30
Proof

• We want to show that for any instance (input) I,
  A(I)/OPT(I) r (r1), where A(I) is the cost
  of the solution obtained from our spanning tree
  algorithm, and OPT(I) is the cost of an optimal
  solution.

31

• 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)

32

• 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)
33

• Inequality(2) says that the cost of the spanning
  tree ST is less than or equal to twice of the
  cost of an optimal solution.
• So, if we can compute ST, then we can get a
  solution with cost2OPT(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.

34
Story

• The method was known long time ago. The
  performance ratio was conjectured to be
• Du and Hwang (1990 ) proved that the conjecture
  is true.

35
Graph Steiner minimum tree

• 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 called regular points. Note
  that, the Steiner minimum tree could contain some
  nodes in V-R and the nodes in V-R are called
  Steiner points.

36

• Example Let G be shown in Figure a. Ra,b,c.
  The Steiner minimum tree T(a,d),(b,d),(c,d)
  which is shown in Figure b.
• Theorem Graph Steiner minimum tree problem is
  NP-complete.

37
Approximation algorithm(Graph Steiner minimum
tree)

1. For each pair of nodes u and v in R, compute the
   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)
2. Compute a minimum spanning tree for the modified
   complete graph.
3. Include the nodes in the shortest paths used.

38

• Theorem The performance ratio of this algorithm
  is 2.
• Proof
• We only have to prove that Triangle Inequality
  holds. If
• dist(a,c)gtdist(a,b)dist(b,c) ......(3)
• then we modify the path from a to c like
• a?b?c
• Thus, (3) is impossible.

39

• Example II-1

g
a
e
c
d
f
b
The given graph
40

• Example II-2

e-c-g /7
g /3
e /4
a
c
d
f/ 2
e /3
b
f-c-g/5
Modified complete graph
41

• Example II-3

g/3
a
c
d
f /2
e /3
b
The minimum spanning tree
42

• Example II-4

g
2
1
2
a
e
c
d
1
1
f
b
1
The approximate Steiner tree
43
Approximation Algorithm for TSP with triangle
inequality

• 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.

44
Ratio 2 Algorithm

• 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)

45

• Example

46
Proof of Ratio 2

1. The cost of a minimum spanning tree cost(t), is
   not greater than opt(TSP), the cost of an optimal
   TSP. (Why? 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.)
2. The cost of the TSP produced by our algorithm is
   less than 2cost(T) and thus is less than
   2opt(TSP).

47

• 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 i1, 2, , k d(v, ci) ? d
• and d is minimized.

48

• Farthest-point clustering algorithm
• Step 1 arbitrarily select a point in V as c1.
• Step 2 let i2.
• Step 3 pick a point ci from V c1, c2, ,
  ci-1 to maximize min c1ci, c2ci,,ci-1
  ci.
• Step 4 ii1
• Step 5 repeat Steps 3 and 4 until ik.

49

• Theorem Farthest-point clustering algorithm has
  ratio-2.
• Proof Let c i be an point in V that maximize
• ?imin c1ci, c2ci,,ci-1 ci.
• We have ?i ? ?i-1 for any i.
• Since two, say ci and cj (igtj), of the k1
  points must be in the same group (in an opt
  solution), ?i ?2 opt.
• Thus, ?k1 ? 2 opt.
• For any v in V, by the definition of ?k1 ,
• min c1v, c2v,,ck v ? ?k1 .
• So the algorithm has ratio-2.

50
Vertex Cover Problem

• 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.