Satisfiability Problem - Automata and Complexity Theory - Lecture Slides, Slides of Theory of Automata

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1

The Satisfiability Problem

Cook’s Theorem: An NP-Complete

Problem

Restricted SAT: CSAT, 3SAT

2

Boolean Expressions

Boolean, or propositional-logic expressions are built from variables andconstants using the operators AND, OR,and NOT.



Constants are true and false, representedby 1 and 0, respectively.



We’ll use concatenation (juxtaposition) forAND, + for OR, - for NOT, unlike the text.

4

The Satisfiability Problem (

SAT )

Study of boolean functions generally is concerned with the set of

truth

assignments

(assignments of 0 or 1 to

each of the variables) that make thefunction true.

NP-completeness needs only a simpler question (SAT): does there exist a truthassignment making the function true?

5

Example: SAT

(x+y)(-x + -y) is satisfiable.

There are, in fact, two satisfying truthassignments:

x=0; y=1.

x=1; y=0.

x(-x) is not satisfiable.

7

Example: Encoding for SAT

(x+y)(-x + -y) would be encoded by the string

(x1+x10)(-x1+-x10)

8

SAT is in

NP

There is a multitape NTM that can decide if a Boolean formula of length n is satisfiable.

The NTM takes O(n

2

) time along any path.

Use nondeterminism to guess a truth assignment on a second tape.

Replace all variables by guessed truth values.

Evaluate the formula for this assignment.

Accept if true.

10

Assumptions About NTM for L

One tape only.

Head never moves left of the initialposition.

States and tape symbols are disjoint.

Key Points: States can be namedarbitrarily, and the constructionsmany-tapes-to-one and two-way-infinite-tape-to-one at most squarethe time.

11

More About the NTM M for L

Let p(n) be a polynomial time bound for M.

Let w be an input of length n to M.

If M accepts w, it does so through a sequence I

0

I

1

I

p(n)

of p(n)+1 ID’s.



Assume trivial move from a final state.

Each ID is of length at most p(n)+1, counting the state.

13

Picture of Computation as an Array

Initial ID

X

00

X

01

…

X

0p(n)

X

10

X

11

…

X

1p(n)

I

1

I

p(n)

X

p(n)

X

p(n)

…

X

p(n)p(n)

... ...

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Intuition

From M and w we construct a boolean formula that forces the X’s to representone of the possible ID sequences ofNTM M with input w, if it is to besatisfiable.

It

is

satisfiable iff some sequence leads

to acceptance.

16

Points to Remember

The boolean function has components thatdepend on n.



These must be of size polynomial in n.

Other pieces depend only on M.



No matter how many states/symbols m has,these are of constant size.

Any logical formula about a set ofvariables whose size is independent of ncan be written somehow.

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Designing the Function

We want the Boolean function thatdescribes the X

ij

’s to be satisfiable if

and only if the NTM M accepts w.

Four conditions:

Unique: only one symbol per position.

Starts right: initial ID is q

0

w.

Moves right: each ID follows from thenext by a move of M.

Finishes right: M accepts.

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Starts Right

The Boolean Function needs to assertthat the first ID is the correct one withw = a

1

…a

n

as input.

X

00

= q

0

.

X

0i

= a

i

for i = 1,…, n.

X

0i

= B (blank) for i = n+1,…, p(n).

Formula is the AND of y

0iZ

for all i,

where Z is the symbol in position i.

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Finishes Right

Somewhere, there must be an accepting state.

Form the OR of Boolean variables y

ijq

where i and j are arbitrary and q is anaccepting state.

Note: differs from text.