Cook’s theorem and
NP-reductions
Pasi Fränti
25.10.2016
Cook’s theorem
Theorem:
Satisfiability problem (SAT) is NP-complete
Proof:
TM ≤p SAT
Satisfiability problem
f1
f2
f3
F=(X1 X2  X3)  (X1 X2)  (X1 X3)
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
f1
f2
f3
F
Non-deterministic algorithm
Satisfiability(F)
FOR i1 TO n DO
xi  Choose{0,1};
IF F(x)=1 THEN SUCCESS;
ELSE FAIL;
T(n)=O(n)  SAT  NP
Notations needed
Implication
x y  x y
x  y  x  y    y  x 
n
Equivalence
x
i
 x1    xn
All are true
x
 x1    xn
Some is true
i 1
n
i 1
i
 n

 
xi    xi     x i  x j

i 1
 i 1   i , j 1
 i j
n
n
At least one
is true

At most one
is true






Exactly one is true
Proof of Cook’s theorem
Accept ( x)  Confic  Initial  Transition  Final
  p  n  n s ,t   p  n  k s ,t  
Config       ai      q j  
t 1   s 1 i  0
  s 1 j 0

p n 
Only valid Turing machine configurations can have value TRUE
ai
= tape position s has symbol ai
qj
= machine is at state qj and R/W head is at position s
Cook’s theorem
Accept ( x)  Confic  Initial  Transition  Final
Initial  q01,1  ai11,1  ai22,1  ...  ainn,1  a0n1,1  ...  a0p n ,1
Machine is at state q0
Tape content is {ai1, ai2, …, ain}
Cook’s theorem
Accept ( x)  Confic  Initial  Transition  Final
Transit 

t 1

  k s ,t 
s ,t
s ,t 1




q

a

a
i
i


 j 

s 1 i  0   j  0

p n  p n  m
p n  p n  m
    a
k
t 1 s 1 i  0 j  0
s ,t
i

q
   a
l
s ,t
j
r 0
s ,t 1
ir





 q sjr d r ,t 1

If not at position s, then content does not change
If at position s and content is ai, then change state and content
Cook’s theorem
Accept ( x)  Confic  Initial  Transition  Final
Final 
p n 

s 1
qks , p n 
Can be in any position s but machine must be in final state qk
NP hard problems
1.
2.
3.
4.
5.
Satisfiability problem (SAT)
Coloring problem (Color)
Exact cover problem (EC)
Knapsack problem (KP)
Traveling salesman problem (TSP)
Satisfiability to Coloring
Complete k-clique
False color
c1
c2
c3
…
ck
…
xk
c0
Connect to all but ci
Literals
x1
x2
x1
x2
x3
x3
…
xk
Color for fi
xj
fi
fi
Connect to all but
those literals in fi
xj
SP to Coloring example
(X1 X2)  (X1 X3)
Additional example
f1
f2
f3
(X2  X3)  (X1 X3 )  (X1  X2)
x1
0
0
0
0
1
1
1
1
x2
0
0
1
1
0
0
1
1
x3
0
1
0
1
0
1
0
1
f1
1
0
1
1
1
0
1
1
f2
1
1
1
1
0
1
0
1
f3
1
1
0
0
1
1
1
1
F
1
0
0
0
0
0
0
1
Knapsack to TSP
Knapsack problem
• Input: knapsack instance {2,3,5,7,11}
• Size of the knapsack S=15.
Step 1: Create one node for every item
• Input: knapsack instance {2,3,5,7,11}
• Create a node for every knapsack element.
2
7
5
3
11
Step 2: Add start and end points
• Input: knapsack instance {2,3,5,7,11}
• Add node 0 as the home.
• Add node N+1 as the turning point.
2
7
5
0
3
n+1
11
N+2 nodes needed to represent the knapsack instance
Step 3: Create forward links
• Input: knapsack instance {2,3,5,7,11}
• Draw links from smaller nodes to bigger ones.
• Set weights according to the bigger node: w(i,j)=j.
2
7
2
7
7
7
7
5
5
0
5
n+1
5
3
3
3
11 11
11
11
11
Step 4: Create forward links for node N+1
• Input: knapsack instance {2,3,5,7,11}
• Draw links to N+1 with weights w(i,N+1)=0.
2
7
2
7
7
7
7
5
5
0
5
5
3
3
3
11 11
11
11
11
0
0
0
0
0
n+1
Step 5: Create backward links
• Input: knapsack instance {2,3,5,7,11}
• Draw backward links from bigger to smaller nodes.
• Set weight of the link as w(j,i)=0.
0
2
7
0
0
0
0
0
5
0
0
0
0
0
0
0
0
0
3
0
0
0
0
11
n+1
Solution for KP  Solution for TSP
• KP = {3,5,7}

• TSP = 0-3-5-7-(N+1)-11-2-0 (S=15)
Visit the nodes in an increasing order !
2
7
0
7
0
0
5
0
n+1
5
0
3
3
11
Solution for TSP  Solution for KP
• TSP = 0-3-5-7-(N+1)-11-2-0

• KP = {3,5,7} (all nodes which arrival cost > 0)
Select nodes with entrance w>0 !
2
0
7
77
0
0
0
5
55
n+1
0
3
33
11
Empty space for notes