1. §5 Structural Complexity: NPc
Comparing Problems: polynomial-time reduction
Reductions: CLIQUE p IS ≼p SAT p 3SAT ≼p IS
NP-completeness, Master Reduction
coNP, Ladner's Theorem

2. Comparing Problems, again
CLIQUE = { G,k | G contains a k-clique }
IS={ G,k : G has k pairwise non-connected vertices}
p N N f L L' LC
L'C
For L,L'N write L≼ L' if exists a computable f: NN such that x: xL  f(x)L'.
polynomial-time p
a) L'P  LP
a) L' decidable  so L.
b) L ≼ L' ≼ L''  L≼ L'' p p p

3. Reduction IS ≼p SAT Let G consist of vertices V={1,..,n} and edges E.
Goal: Upon input of (the encoding of) a graph G and kN,
produce in polynomial time a CNF formula  such that:
 satisfiable iff G contains ≥k independent vertices
Let G consist of vertices V={1,..,n} and edges E.
Consider Boolean variables xv,i, v  V, i=1...k
and clauses Ki := vV xv,i, i=1...k
and xv,i  xv,j, v  V, 1≤i<j≤k
and xu,i  xv,j, {u,v}  E, 1≤i<j≤k
Length of : O(k·n+n·k2+n2k2)=O(n2k2)
Computational cost of (G,k) → : polyn. in n+log k
Vertex v is #i among the k independent.
There is an i-th vertex
Vertex v cannot be both #i and #j.
No adjacent vertices are
independent.
since k≤n.

4. Example Reduction: 4SAT vs. 3SAT
4-SAT: Is formula Φ(Y) in 4-CNF satisfiable?
3-SAT: Is formula Φ(Y) in 3-CNF satisfiable?
Given  = (a  b  c  d)  (p  q  r  s)  …
with literals a,b,c,d, p,q,r,s,….
Introduce new variables u,v,… and consider
' := ( a  b  u )  (u  c  d )
 ( p  q  v)  (v   r  s )  …
variables, possibly negated
f:   '
For L,L'N write L≼ L' if exists a computable f: NN such that x: xL  f(x)L'.

5. Reduction 3SAT ≼p IS (k,1) (1,1) (k,2) (1,2) (k,3) (1,3)
Produce, given a 3-CNF term , within polynomial time a graph G and integer k such that it holds:
iff G contains k pairwise non-adjacent vertices.
 is satisfiable
e.g. ( u .... )  ( .. u .. )  ( .. .. u )  ( u  .. .. )
 = C1  C2 …  Ck, Ci = xi1  xi2  xi3, xis literals
V:= { (i,1),…(i,3): i≤k }, E:= { {(i,s),(j,t)} : i=j or xis= xjt }
(k,1)
(k,2)
(k,3)
(1,1)
(1,2)
(1,3)
Richard Karp

6. Problems of similar complexity
unknown yet
Showed: CLIQUE p IS ≼p SAT p 3SAT ≼p IS.
These 4 problem have about same complexity:
Either all are belong to P, or none of them.
We will show: Also TSP, HC, VC and many further problems in NP belong to this class called NPc.
And will show: These are ‘hardest‘ problems in NP.
Cook–Levin Theorem: L 2 NP : L ≼p SAT.
That is, either all or none of the problems in NPc can be decided in polynomial time.
A deterministic WHILE+ program could simulate any non-deterministic one with polynomial slowdown!
In the first case:

7. EXP Complexity Class Picture PSPACE CH #P PH PNP co-NP NP P
Def: ANP is NP-complete if
L ≼p A holds for every LNP.
PSPACE complete
PSPACE
Theorem (Cook'72/Levin'71):
SAT is NP-complete! CH #P PH
Lemma: For A
NP-complete
and A ≼p B NP,
B is also NPc.
PNP
coNP-complete
NP-
complete
co-NP NP
Now know ≈500
natural problems
NP-complete… P

8. unSAT := {  Boolean term, x: (x)=0 }  coNPc
Scenarios for P≠NP
coNP := { LN : N\L  NP }
unSAT := {  Boolean term, x: (x)=0 }  coNPc
Theorem [Ladner'75]: If P≠NP, there exists
LNP \ (PNPc)
NP = coNP
P = coP
coNPc
NPc NP
coNP
NP coNP P NP
coNP
P = coP

9. Master Reductions L = {xN: y, ℓ(y)≤poly(ℓ(x)), x,yV }, VP
SAT = {  :  Boolean term, y1,…ym: (y1,…ym)=1}
Cook/Levin Theorem: SAT is NP-complete!
Proof (Sketch): Fix LNP and VP.
Fix WHILE+ program B deciding V in time poly(n).
Express "bin(z0,…zk-1)V" as Boolean term "k(z0,…zk-1)=1" of length poly(k).
Then bin(x0,…xn-1)L  y0,…ym-1{0,1}: n+k(x, y)=1.
N x →  := n+k(bin(x), · )
Thm: The following problem UNP is NP-complete:
{ A,x,2N : nondetermin. WHILE+ program A accepts input x within at most N steps }
L = {xN: y, ℓ(y)≤poly(ℓ(x)), x,yV }, VP

10. SubsetSum is NP-complete
{ a1,…aN,b | a1,…aN,bN, 1,…N{0,1} : b=i ai·i }
SubsetSumNP √ Show: 3SAT ≼p SubsetSum
In polyn.time: 3CNF  → AN and bN s.t.
satisf. assignm. of   BA: b=aB a
Eg. Φ = (x1  x3  x5)  (x1  x5  x4)  (x2  x2  x5)
v1 :=
v2 :=
v3 :=
v4 :=
v5 :=
v1' :=
v2' :=
v3' :=
v4' :=
v5' :=
b :=
c1:=
d1:=
c2:=
d2:=
c3:=
d3:=
m clauses in n var.s → 2n+2m+1 values à n+m dec.digits
A NP-complete, BNP and A ≼p B  B also NP-complete.