Computational Complexity

Are there problem classes that have no good algorithms? How can ew prove it?

These are questions for complexity researchers.

Good Algorithms

Bob: What makes an algorithm ``good''?

Graph input size against computation time; if it can be bounded above by a polynomial (any polynomial will do), the algorithm is good.

       ^
       |          _                        _
       |          _ a polynomial          _
       |         _                       _
       |         _                    _-~
time   |        _                   _- algorithm
req'd  |       _                   -   performance
       |      _                   -
       |    _~                __-~
       |__-~        _------~~~
       |__---------~
       +----------------------------------------------------->
          bits required for input -->

Complaints

Bob: Polynomials can be much worse than 2^n for reasonable values of n!

Alice: Why not define good algorithms to be bounded by degree-6 polynomials?

• Natural problem classes required high-degree polynomials are rare and usually short-lived.
• Composability is important. A good algorithm used a good number of times should take a good about of time. But n^6 * n^6 = n^12 is not degree-6.
• It often works.

Easy Problems

A problem class is easy if a good algorithm exists for it. The set of easy problem classes is called P.

There is a much larger set of problem classes called NP.

  +-----------------------------------------------------------------+
  |   ___________________________________________________________   |
  |  /   ________________                                        \  |
  |  |  /                \                                   NP  |  |
  |  |  |             P  |                                       |  |
  |  |  |                |        .                              |  |
  |  |  | .ADDITION      |    PRIMALITY                          |  |
  |  |  |                |                                       |  |
  |  |  |                |                                       |  |
  |  |  |           .    |                                .      |  |
  |  |  | ALL-PAIRS-PATHS|                             HAM-PATH  |  |
  |  |  |                |                                       |  |
  |  |  |                |                                       |  |
  |  |  |                |                                       |  |
  |  |  |                |                                       |  |
  |  |  \________________/                                       |  |
  |  \___________________________________________________________/  |
  |                                                                 |
  +-----------------------------------------------------------------+

                        Problem class topology

The $1,000,000 Question

Does P=NP?

Are there problem classes in NP which are not in P?

(We know P is a subset of NP.)

This is the biggest open question is computer science (and probably mathematics too).

The $1,000 Answer

We think not: P != NP. No proof yet, but we have evidence.

We do know this:

If HAM-PATH is in P, then P = NP.

  +-----------------------------------------------------------------+
  |   ________________________________________________________      |
  |  /   ________________                                     \     |
  |  |  /                \                                NP  |     |
  |  |  |             P  |                     _______________|___  |
  |  |  |                |        .           /               |   \ |
  |  |  | .ADDITION      |    PRIMALITY       |           NP-hard | |
  |  |  |                |                    |               |   | |
  |  |  |                |                    |               |   | |
  |  |  |           .    |                    |        .      |   | |
  |  |  | ALL-PAIRS-PATHS|                    |     HAM-PATH  |   | |
  |  |  |                |                    |               |   | |
  |  |  |                |                    |               |   | |
  |  |  |                |                    |               |   | |
  |  |  |                |                    \_______________|___/ |
  |  |  \________________/                                    |     |
  |  \________________________________________________________/     |
  |                                                                 |
  +-----------------------------------------------------------------+

                   Revised problem class topology

Many very important problem classes are NP-complete, so people have nonetheless worken on them very hard.

After 30 years, we haven't found that any are in P. Hence our confidence in the P != NP conjecture.

Proving NP-hardness

Alice: How can I show that my favorite problem class is NP-hard?

To show A is NP-hard, we reduce a known NP-hard problem class B to A. That is, we show taht a good algorithm for A implies a good algorithm for B. Since this would imply P = NP, it follows that A is NP-hard.

(This only works when we know of other NP-hard problems. Take my word they exist.)

A First Reduction

We begin with the known NP-hard problem class VERTEX-COVER.

Input: a graph (V, E)

Output: a minimal subset V' of V so that every vertex in V is adjacent to a vertex in V'.

example:

      *---o---o---*---o
       \_  \_ | _/| _/
         \   \|/  |/
          o---o---o
           \_____/
* represents the output set

We show that SET-COVER is NP-hard.

Input: a base set S, a collection C of subsets of S.

Output: a minimal subcollection of C whose union is S.

example:

  ______
 /      \
 |  ____|_______
 | / o  |     o \
 | \____|_______/ <---\
 |  ____|_______       > The optimal set cover
 | / o  |     o \ <---/
 | \____|_______/
 \______/

The NP-hardness proof:

1. Start with a VERTEX-COVER problem.

         a---b---c---d---e
          \_  \_ | _/| _/
            \   \|/  |/
             f---g---h
              \_____/
   

2. Show how we can use a SET-COVER algorithm to solve it.

       S = { a, b, c, d, e, f, g, h }
       C = { {a, b, f}, {b, a, c, g}, {c, b, d, g}, {d, c, e, g, h},
             {e, d, h}, {f, a, g, h}, {g, b, c, d, f, h}, {h, d, e, g} }
   

3. So VERTEX-COVER is NP-hard implies SET-COVER is NP-hard.

Another Reduction

Problem class FACILITY-LOCATION

Input: a graph (V, E) with distances, a radius r, and a subset H of V.

Output: the smallest subset of V so that each element of H is within r of something in the subset.

example:

      x--1--o--1--o--1--x--1--*--2--x
       \___  \_1_       |
         1 \     \      | 1
            o--1--*--1--o
                         \1
                          \
                           x
x represents the houses (elements of H)
* represents the optimal facility placement (the output)

The proof:

1. Start with a SET-COVER problem.

     ______
    /   C  \
    |  ____|_______
    | / b  | A   a \
    | \____|_______/ <---\
    |  ____|_______       > The optimal set cover
    | / c  | B   d \ <---/
    | \____|_______/
    \______/
   

2. Show how we can use a FACILITY-LOCATION algorithm to solve it.

          A   C   B
         / \ / \ / \
        a   b   c   d
   
   All edges have length 1.
   H = {a, b, c, d}
   r = 1
   

   Any solution can be changed to a SET-COVER solution (just covering the set-vertices) by removing any house in the optimal cover and selecting a non-house adjacent to it. A facility only covers that house, and that house remains covered.
3. So SET-COVER is NP-hard implies FACILITY-LOCATION is NP-hard.