Intractable Problems - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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1

Intractable Problems

Time-Bounded Turing Machines

Classes

P

and

NP

Polynomial-Time Reductions

2

Time-Bounded TM’s

A Turing machine that, given an input of length n, always halts within T(n)moves is said to be

T(n)-time bounded.



The TM can be multitape.



Sometimes, it can be nondeterministic.

The deterministic, multitape case corresponds roughly to “an O(T(n))running-time algorithm.”

4

Polynomial Equivalence of

Computers and TM’s

A multitape TM can simulate a computer that runs for time O(T(n)) inat most O(T

2

(n)) of its own steps.

If T(n) is a polynomial, so is T

2

(n).

5

Examples of Problems in

P

Is w in L(G), for a given CFG G?



Input = w.



Use CYK algorithm, which is O(n

3

).

Is there a path from node x to node y in graph G?



Input = x, y, and G.



Use Dijkstra’s algorithm, which is O(n log n)on a graph of n nodes and arcs.

7

A Tricky Case: Knapsack

The

Knapsack Problem

is: given positive

integers i

1

, i

2

,…, i

n

, can we divide them

into two sets with equal sums?

Perhaps we can solve this problem in polytime by a dynamic-programmingalgorithm:



Maintain a table of all the differences we can

achieve by partitioning the first j integers.

8

Knapsack – (2)

Basis: j = 0.

Initially, the table has

“true” for 0 and “false” for all otherdifferences.

Induction: To consider i

j

, start with a

new table, initially all false.

Then, set k to true if, in the old table, there is a value m that was true, and kis either m+i

j

or m-i

j

10

Knapsack – (4)

Since n < m, we can build the final table in O(m

2

) time.

From that table, we can see if 0 is achievable and solve the problem.

11

Subtlety: Measuring Input Size 

“Input size” has a specific meaning: the length of the representation of theproblem instance as it is input to a TM.

For the Knapsack Problem, you cannot always write the input in a number ofcharacters that is polynomial in eitherthe number-of or sum-of the integers.

13

Bad Case – (2)

Thus, the proposed “polynomial” algorithm actually takes time O(n

2

n

) on

an input of length O(n

2

Or, since we like to use n as the input size, it takes time O(n

sqrt(n)

) on an

input of length n.

In fact, it appears no algorithm solves Knapsack in polynomial time.

14

Redefining Knapsack

We are free to describe another problem, call it

Pseudo-Knapsack,

where integers are represented inunary.

Pseudo-Knapsack is in

P

16

Example:

NP

The Knapsack Problem is definitely in NP

, even using the conventional binary

representation of integers.

Use nondeterminism to guess one of the subsets.

Sum the two subsets and compare.

17

P

Versus

NP

Originally a curiosity of Computer Science, mathematicians now recognizeas one of the most important openproblems the question

P

NP

There are thousands of problems that are in

NP

but appear not to be in

P

But no proof that they aren’t really in

P

19

Complete Problems – Intuition

A complete problem for a class embodies every problem in the class,even if it does not appear so.

Compare: PCP embodies every TM computation, even though it does notappear to do so.

Strange but true: Knapsack embodies every polytime NTM computation.

20

Polytime Reductions

Goal: find a way to show problem

L

to

be NP-complete by reducing everylanguage/problem in

NP

to

L

in such a

way that if we had a deterministicpolytime algorithm for

L, then we could

construct a deterministic polytimealgorithm for any problem in

NP