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Some concept of Automata and Complexity Theory are Administrivia, Closure Properties, Context-Free Grammars, Decision Properties, Deterministic Finite Automata, Intractable Problems, More Undecidable Problems. Main points of this lecture are: Satisfiability Problem, Cook’S Theorem, Complete, Restricted, Expressions, Built From Variables, Operators, True and False, Represented, Concatenation
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Cook’s Theorem: An NP-Complete
Problem
Restricted SAT: CSAT, 3SAT
2
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
Study of boolean functions generally is concerned with the set of
(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
(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
(x+y)(-x + -y) would be encoded by the string
(x1+x10)(-x1+-x10)
8
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
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
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
1
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
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)
... ...
14
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
satisfiable iff some sequence leads
to acceptance.
16
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.
17
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.
19
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.
20
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.