Cook

1
Cooks Theorem

• Lecture 40
• Section 7.4
• Mon, Dec 3, 2007

2
Cooks Theorem

• Cooks Theorem SAT ? P if and only if P NP.
• Proof idea
• First show that SAT is in NP.
• Then show that every problem in NP is
  polynomial-time reducible to SAT.

3
Proof of Cooks Theorem

• Proof
• It is easy to see that SAT is in NP.
• Simply nondeterministically choose Boolean values
  for each variable.
• Then verify that the expression is true.

4
Proof of Cooks Theorem

• Let A be a problem in NP.
• Then there is a nondeterministic Turing machine M
  that decides A in polynomial time.
• Let the run time of M be nk.
• When M is run, it can access at most the first nk
  cells.

5
Proof of Cooks Theorem

• Represent the execution of M by an nk ? nk array.
• Each row contains a configuration of M.
• The first row is the initial configuration, with
  input w.
• The sequence of rows represents a computation of
  M that reaches a halting state.

6
Proof of Cooks Theorem

• We will place the symbol in the first and last
  cells of each row, to help identify the ends of
  the tape.
• In each row, each cell between the s will
  contain
• A tape symbol x, or
• A state q, or
• A blank.

7
Proof of Cooks Theorem

• That is, each cell contains one symbol from the
  set
• C Q ?? ? ? .
• We will create a Boolean expression from the
  table that is satisfiable if and only if the
  table represents a valid accepting computation of
  M.

8
Proof of Cooks Theorem

• Our Boolean expression ? will be of the form
• ?cell ? ?start ? ?accept ? ?move,
• where
• ?cell true if and only if every cell contains
  exactly one symbol.
• ?start true if and only if row 1 is a valid
  start configuration.
• ?accept true if and only if one of the rows is
  a valid accept configuration.

9
Proof of Cooks Theorem

• ?move true if and only if each row yields the
  configuration in the next row.
• These expressions will all be true if and only if
  the table represents a valid computation
  resulting in acceptance.

10
Proof of Cooks Theorem

• For each row number i from 1 to nk, and for each
  column number j from 1 to nk, and for each symbol
  s in C, let xi, j, s be a Boolean variable that
  is true if cell (i, j) contains symbol s.

11
Example

• Consider the Turing machine below processing the
  input 0110.

1 ? 0, R
1 ? 0, R
0 ? 1, R
r accept s reject
p
q
0 ? 1, R
_ ? _, R
_ ? _, R
s
r
12
Example

• The computation table.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
13
Example

• For example,
• x1,1, true
• x1,1,0 false
• x1,2,p true
• x1,3,0 true

14
The Expression ?cell

• The expression ?cell consists of two parts.
• The first part is true if and only if cell (i, j)
  contains at least one symbol.

15
The Expression ?cell

• The second part is true if and only if cell (i,
  j) contains at most one symbol.

16
The Expression ?cell

• Therefore,

17
The Expression ?cell

• Therefore,

Every cell
18
The Expression ?cell

• Therefore,

Every cell contains at least one symbol
19
The Expression ?cell

• Therefore,

Every cell contains at least one symbol and
20
The Expression ?cell

• Therefore,

Every cell contains at least one symbol
and contains at most one symbol.
21
The Expression ?cell

• Therefore,

Every cell contains exactly one symbol.
22
The Expression ?start

• The expression ?start is easy because we know
  what the start configuration looks like.
• q0w1w2wn _ _ _..._
• We simply verify that the cells contain those
  symbols.

23
The Expression ?start

• Therefore,

24
The Expression ?start

• Therefore,

Start state
25
The Expression ?start

• Therefore,

Initial tape contents w1w2wn
26
The Expression ?accept

• The expression ?accept is also very simple.
• All we need to do is to guarantee that qacc
  occurs somewhere in the table.

27
The Expression ?accept

• Therefore,

28
The Expression ?move

• The expression ?move will take a bit more effort.
• We will consider 2 ? 3 windows in the table.
• Theorem The table represents a valid computation
  if and only if ?cell and ?start are true and
  every 2 ? 3 window is valid.

29
Example

• The computation table.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
30
Example

• Various 2 x 3 windows.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
31
Example

• Various 2 x 3 windows.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
32
Example

• Various 2 x 3 windows.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
33
Example

• Various 2 x 3 windows.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
34
Example

• Various 2 x 3 windows.

p 0 1 1 0 _ _ _
1 q 1 1 0 _ _ _
1 0 q 1 0 _ _ _
1 0 0 q 0 _ _ _
1 0 0 1 p _ _ _
1 0 0 1 _ r _ _
35
The Expression ?move

• For any symbol x and any state q, the legal moves
  are
• Write x, move left, and change state to q.
• Write x, move right, and change state to q.
• Recall that if we try to move left from the first
  cell, we remain in the first cell.

36
The Expression ?move

• Valid 2 ? 3 windows for moving left (states p, q
  symbols a, b, c, x).

p a b
c x b
a p b
q a x
a b p
a q b
p a
q x
a b c
a b c
a b c
a b q
a p
q a
a b
a q
a b
a b
37
The Expression ?move

• There is a similar set for a move to the right.
• Let S be the set of all sextuples in C6
  representing valid 2 ? 3 windows.
• Then S includes (p, a, b, c, x, b), (a, p, b, q,
  a, x), etc., for all a, b, c, and x in ?? and for
  all p and q in Q.

38
The Expression ?move

• Then the expression
• is true if and only if (x1, x2, x3, x4, x5, x6)
  is a valid sextuple in window (i, j).

39
The Expression ?move

• So the expression
• is true if and only if window (i, j) is valid.

40
The Expression ?move

• Then the expression
• is true if and only if every 2 ? 3 window in the
  table is valid.

41
Proof of Cooks Theorem

• Then the Boolean expression
• ?cell ? ?start ? ?accept ? ?move
• is satisfiable if and only if w is accepted by
  M.
• Therefore, SAT is NP-complete.

42
Building from Cooks Theorem

• Now we may show that other problems are
  NP-complete by showing that SAT is
  polynomial-time reducible to them.

43
3SAT is NP-Complete

• Theorem 3SAT is NP-complete.
• Proof
• We need only show that SAT is polynomial-time
  reducible to 3SAT.
• Any boolean expression may be put into
  3-conjunctive normal form by the following
  technique.

44
3SAT is NP-Complete

• Rewrite the each clause
• x1 ? x2 ? x3 ? ? xn
• as
• (x1 ? x2 ? y1) ? (?y1 ? x3 ? y2) ? ? ? (?yn 3 ?
  xn 1 ? ?xn).

45
3SAT is NP-Complete

• The expressions??cell, ??start, and ??accept are
  already in conjunctive normal form.
• Apply the above technique to put them in
  3-conjunctive normal form.
• The expression ??move is not in conjunctive
  normal form.

46
3SAT is NP-Complete

• However, it may be rewritten in conjunctive
  normal form by repeatedly applying the
  distributive laws
• x ? (y ? z) (x ? y) ? (x ? z)
• and
• x ? (y ? z) (x ? y) ? (x ? z).

47
3SAT is NP-Complete

• For example,
• (x ? y) ? (z ? w)
• ((x ? y) ? z) ? ((x ? y) ? w)
• ((x ? z) ? (y ? z)) ? ((x ? w) ? (y ? w))
• (x ? z) ? (y ? z) ? (x ? w) ? (y ? w).

48
3SAT is NP-Complete

• Therefore, the expression
• ?cell ? ?start ? ?accept ? ?move
• can be rewritten in 3-conjunctive normal form.
• Therefore, 3SAT is NP-complete.