P
vs N P
“A rough summary” (2-11-2012, Georg Still)
Consider two decision problems:
P1: (Partition problem)
Given n ∈ N and a set of numbers {a1 , . . . , an }, ai ∈ N.
Question: Does there exist a subset S ⊂ N := {1, . . . , n}
such that
X
X
ai =
aj
i∈S
P2:
j∈N\S
Answer: yes or no
As in P1 but with the additional constraints
(?)
ai ≤ aj ∀i ∈ S, j ∈ N \ S
Def. 1 The set P is the set of decision problems which are
solvable in polynomial time.
(In time t(s) ≤ O(poly(s)) depending on the input size s of
the problem)
Exercise: Problem P2 is in P. It is solvable in polynomial
time: t(n) ≤ O(n2 ) (even t(n) ≤ O(n · log n))
CO, Chapter0,1
p1
What about P1 ? At least we are sure that P1 belongs to the
large set of N P-problems defined as follows:
Def. 2 A decision problem is in N P (nondeterministically
polynomial solvable) if it has a “polynomially verifiable proof
system” (for the yes instances (see [G,p61]).
Roughly: for any yes instance there is a witness which can be
checked in polynomial time.
P1 is in N P: Take a solution S ⊂ N of the corresponding
search problem as a witness.
CO, Chapter0,1
p2
Def. 3 A decision problem P is N P-complete if “a
polynomial solution algorithm for P would provide a
polynomial solution algorithm for any problem in N P .
(Any problem in N P can be polynomially reduced to P).
An astonishing result
(Cook 71, Karp 72)
There exist (many interesting) N P-complete problems.
(The Partition Problem P1 is one of them)
Observation A polynomial algorithm for P1 (or for any
N P-complete problem) would prove: P = N P.
Conjecture: P =
6 N P.
Meaning: “ It is more difficult to find a proof of a “statement”
than to check whether a given “proof” is correct.”
CO, Chapter0,1
p3
Def. 4 A decision problem is in coN P if it has a
polynomially verifiable “proof system for the no
instances”.
Example: The following problem P1’ is in coN P.
P1’: Same input as inP
P1 but with
Pa question: Does for all
subsets S ⊂ N hold:
i∈S ai 6=
j∈N\S aj .
Conjecture: N P 6= coN P.
Fact.
P ⊂ N P ∩ coN P.
(A polynomial solution algorithm
yields a witness for yes and no.)
CO, Chapter0,1
p4
Consider the problem
P3: Given n, m ∈ N, 1 ≤ m ≤ n. Question: Does there exist
a divisor d of n with 1 < d < m?
Remarks.
P3 is in N P ∩ coN P.
(Yes and no is witnessed by a factorization of n.)
Conjectures: P3 ∈
/ P, P3 is not N P-complete, P3 is not
coN P-complete.
P3 is N P-complete (or coN P-complete) would imply
N P = coN P.
Some warnings: We did not consider details such as “length
of input”, “what are basic computation steps in an algorithm”
(Turing-machine or equivalent concepts), etc .
CO, Chapter0,1
p5
Some references:
[FKS]: “Faigle,Kern,Still, Algorithmic principles of mathematical
programming.”
[GJ]: “Garey,Johnson, Computers and Intractability.”
[Go]: “Goldreich, P,NP, and NP-completeness.”
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