cs3102: Theory of Computation

Class 25: NP-Complete Appetizers

Spring 2010University of VirginiaDavid Evans

PS6 is due Tuesday, April 27 (but don’t wait until then to start PS7!)

Recap

To prove B is in NP-Hard, show that some known NP-Hard problem can be reduced to B.

3SAT NP-Hard

X

NP

B

A problem/language is in NP, if there is a certificate that allows you to verify the answer in polynomial time.

A problem/language is NP Hard‑ if every problem in NP can be reduced to it.

If a problem is in NP and NP Hard‑ , it is in NP Complete‑ .

Today’s Menu

NP-Hardness Reduction Proofs

SATSubset SumKnapsack(Pre-)History of P

= NP Question

SATisfiability

Is SAT in NP? Yes, the variable assignments are apolynomial-time verifiable certificate.

SAT is NP-Complete

To prove SAT is in NP-Hard, show that some known NP-Hard problem can be reduced to SAT.

3SAT

NP

SAT

3SAT NP-Hard Assume this is already

known

Proof: SAT is NP-Complete

MSAT

Polynomial-time decider for SAT3S

AT in

put

Polynomial-time Input

Transformer

Poly-time Output

Transformer

3SAT

out

put

If we can use MSAT, a polynomial-time decider for SAT to build a polynomial-time decider for 3SAT, this proves that SAT is at least as “hard” as 3SAT.

Show that if SAT is in P, so is 3SAT.

Proof: SAT is NP-Complete

MSAT

Polynomial-time decider for SAT3S

AT in

put

3SAT

out

put

Show that if SAT is in P, so is 3SAT.

Identify transform:3SAT input is a valid SAT input

Identify transform:SAT output is correct 3SAT output

Subset Sum Language

How many test of SUBSET-SUM do we to solve the subset sum search problem?

Can you order the same appetizer more than once?

SUBSET-SUM is in NP

SUBSET-SUM is in NP-Hard

MSS

Polynomial-time decider for SUBSET-

SUM

3SAT

inpu

t

Polynomial-time Input

Transformer

Poly-time Output

Transformer

3SAT

out

put

To finish the proof, we need to describe the transformers!

Transform Goal

Transform Ideax1 x2 x3

...

y11 0 0 ...

z11 0 0 ...

y20 1 0 ...

z20 1 0 ...

y30 0 1 ...

z30 0 1 ...

... ... ... ... ...

x1 x2 x3 ... xl

y1 1 0 0 ...

z1 1 0 0 ...

y2 0 1 0 ...

z2 0 1 0 ...

... ... ... ... ...

yl 0 0 0 ... 1

zl 0 0 0 ... 1

t

Variables

S

x1 x2 x3 ... xl c1 c2 c3 ... ck

y1 1 0 0 ...

z1 1 0 0 ...

y2 0 1 0 ...

z2 0 1 0 ...

... ... ... ... ...

yl 0 0 0 ... 1

zl 0 0 0 ... 1

t 1 1 1 ... 1

Variables

S

Clauses

x1 x2 x3 ... xl c1 c2 c3 ... ck

y1 1 0 0 ... 1 0

z1 1 0 0 ... 0 1

y2 0 1 0 ... 0 1

z2 0 1 0 ... 1 0

... ... ... ... ... ... ...

yl 0 0 0 ... 1 ... ...

zl 0 0 0 ... 1 ... ...

t 1 1 1 ... 1

Variables

S

Clauses

We need to make at least one term in each clause true!

x1 x2 x3 ... xl c1 c2 c3 ... ck

y1 1 0 0 ... 1 0

z1 1 0 0 ... 0 1

y2 0 1 0 ... 0 1

z2 0 1 0 ... 1 0

... ... ... ... ... ... ...

yl 0 0 0 ... 1 ... ...

zl 0 0 0 ... 1 ... ...

t 1 1 1 ... 1 1 1 1 1 ...

Variables

S

Clauses

We need to make at least one term in each clause true!

Doesn’t work: more than one term in a clause can be true!

x1 x2 x3 ... xl c1 c2 c3 ... ck

y1 1 0 0 ... 1 0

z1 1 0 0 ... 0 1

y2 0 1 0 ... 0 1

z2 0 1 0 ... 1 0

... ... ... ... ... ... ...

yl 0 0 0 ... 1 ... ...

zl 0 0 0 ... 1 ... ...

ff1 1 0 0 ... 0

fff1 1 0 0 ... 0

ff2 0 1 0 ... 0

fff2 0 1 0 ... 0

... 0 0 ... ... ...

t 1 1 1 ... 1

Variables

S

Clauses

Sele

ct tr

ue/f

alse

for e

ach

Fudg

e fa

ctor

s

x1 x2 x3 ... xl c1 c2 c3 ... ck

y1 1 0 0 ... 1 0

z1 1 0 0 ... 0 1

y2 0 1 0 ... 0 1

z2 0 1 0 ... 1 0

... ... ... ... ... ... ...

yl 0 0 0 ... 1 ... ...

zl 0 0 0 ... 1 ... ...

ff1 1 0 0 ... 0

fff1 1 0 0 ... 0

ff2 0 1 0 ... 0

fff2 0 1 0 ... 0

... 0 0 ... ... ...

t 1 1 1 ... 1 3 3 3 3 3

Variables

S

Clauses

Sele

ct tr

ue/f

alse

for e

ach

Fudg

e fa

ctor

s

x1 x2 x3... xl c1 c2 c3

... cl

y11 0 0 ...

z11 0 0 ...

y20 1 0 ...

z20 1 0 ...

y30 0 1 ...

z30 0 1 ...

... ... ... ... ...

yl 0 0 0 ... 1

zl 0 0 0 ... 1

Variables

3SAT reduces to SUBSET-SUM

MSS

Polynomial-time decider for SUBSET-

SUM

3SAT

inpu

t

Polynomial-time Input

TransformerIdentity

3SAT

out

put

RejectSolutions involve multiple orders of the same appetizer.

Knapsack

KNAPSACK is in NP

Example

Is KNAPSACK in NP-Hard?

MSS

Polynomial-time decider for

KNAPSACK

3SAT

inpu

t

Polynomial-time Input

Transformer

Output transformer

3SAT

out

put

Could prove by reducing 3SAT to KNAPSACK.

Any easier way?

Is KNAPSACK in NP-Hard?

MSS

Polynomial-time decider for

KNAPSACK

SUBS

ET-S

UM

inpu

t

Polynomial-time Input

Transformer

Output transformer

SUBS

ET-S

UM

out

put

Prove by reducing SUBSET-SUM to KNAPSACK

Input Transformation

Input Transformation

S = { 102.15, 1002.75, 10003.35, 100003.55,

1000004.20, 1000005.80, 100.00, 1000.00, 10000.00, 100000.00, 1000000.00, 1000000.00 }t = 11111115.05

Opps! Iit was pointed out in class that this doesn’t work since the KNAPSACK solver might include 10x or 100x of one of the numbers and still get the right sum. This should be fixable…if you can fix it, a good answer replaces Problem 5 on PS6.

Chain of Reductions

KNAPSACK

SUBSET-SUM

3SAT

reduces to

reduces to

Every problem in NPreduces to

Hints for PS6, Problem 5

• Prove Genome Assembly is NP-Complete• Solve this by thinking, not googling• You can (of course!) do the reduction from any

NP-Complete problem to Genome Assembly– But, some will be easier than others– Some of the problems described in Sipser are

better choices than ones we have covered in class

History of P = NP

Stephen Cook (US, 1971): The complexity of theorem proving proceduresRichard Karp (US 1972): List of 21 NP-Complete problems

Leonid Levin (USSR 1969-73): Universal search problems

Cook-Levin Theorem: SAT is NP-Complete

Stephen Cook Leonid Levin

(Pre)-History of P=NP

Princeton, 20 March 1956 Dear Mr. von Neumann:

With the greatest sorrow I have learned of your illness. … I hope and wish for you that your condition will soon improve even more and that the newest medical discoveries, if possible, will lead to a complete recovery.

Since you now, as I hear, are feeling stronger, I would like to allow myself to write you about a mathematical problem, of which your opinion would very much interest me: One can obviously easily construct a Turing machine, which for every formula F in first order predicate logic and every natural number n, allows one to decide if there is a proof of F of length n (length = number of symbols). Let ψ(F,n) be the number of steps the machine requires for this and let ϕ(n) = maxF ψ(F, n). The question is how fast ϕ(n) grows for an optimal machine. One can show that ϕ(n) ≥ k ⋅ n. If there really were a machine with ϕ(n) ∼ k ⋅ n (or even ∼ k ⋅ n2), this would have consequences of the greatest importance. Namely, it would obviously mean that inspite of the undecidability of the Entscheidungsproblem, the mental work of a mathematician concerning Yes-or-No questions could be completely replaced by a machine. After all, one would simply have to choose the natural number n so large that when the machine does not deliver a result, it makes no sense to think more about the problem.

Gödel’s 1956 Letter (Translated)

Princeton, 20 March 1956 Dear Mr. von Neumann:

With the greatest sorrow I have learned of your illness. … I hope and wish for you that your condition will soon improve even more and that the newest medical discoveries, if possible, will lead to a complete recovery.

Since you now, as I hear, are feeling stronger, I would like to allow myself to write you about a mathematical problem, of which your opinion would very much interest me: One can obviously easily construct a Turing machine, which for every formula F in first order predicate logic and every natural number n, allows one to decide if there is a proof of F of length n (length = number of symbols). Let ψ(F,n) be the number of steps the machine requires for this and let ϕ(n) = maxF ψ(F, n). The question is how fast ϕ(n) grows for an optimal machine. One can show that ϕ(n) ≥ k ⋅ n. If there really were a machine with ϕ(n) ∼ k ⋅ n (or even ∼ k ⋅ n2), this would have consequences of the greatest importance. Namely, it would obviously mean that inspite of the undecidability of the Entscheidungsproblem, the mental work of a mathematician concerning Yes-or-No questions could be completely replaced by a machine. After all, one would simply have to choose the natural number n so large that when the machine does not deliver a result, it makes no sense to think more about the problem.

Gödel’s 1956 Letter (Translated)

Has-Short-Proof

“One can obviously easily construct a Turing machine, which for every formula F in first order predicate logic and every natural number n, allows one to decide if there is a proof of F of length n (length = number of symbols).”

Has Short Proof is Decidable

• Try all possible proofs of length n – How do we know this is finite?

• For each possible proof, check if it is a valid proof of F.– A proof is valid if:• It starts from the accepted axioms.• Each step follows a deduction rule (finite set of rules).• It reaches the statement F.

Princeton, 20 March 1956 Dear Mr. von Neumann: …One can obviously easily construct a Turing machine, which for every formula F in first order predicate logic and every natural number n, allows one to decide if there is a proof of F of length n (length = number of symbols). Let ψ(F,n) be the number of steps the machine requires for this and let ϕ(n) = maxF ψ(F, n). The question is how fast ϕ(n) grows for an optimal machine. One can show that ϕ(n) ≥ k ⋅ n. If there really were a machine with ϕ(n) ∼ k ⋅ n (or even ∼ k ⋅ n2), this would have consequences of the greatest importance. …

Princeton, 20 March 1956 Dear Mr. von Neumann: …One can obviously easily construct a Turing machine, which for every formula F in first order predicate logic and every natural number n, allows one to decide if there is a proof of F of length n (length = number of symbols). Let ψ(F,n) be the number of steps the machine requires for this and let ϕ(n) = maxF ψ(F, n). The question is how fast ϕ(n) grows for an optimal machine. One can show that ϕ(n) ≥ k ⋅ n. If there really were a machine with ϕ(n) ∼ k ⋅ n (or even ∼ k ⋅ n2), this would have consequences of the greatest importance. …

Let’s give Gödel the benefit of the doubt, and read this as, “(or even ∼kn2, ∼kn3, ∼kn4, ...)”

We’ll continue with this Tuesday…

Charge

NP-Complete problems are everywhere! Route planning, scheduling courses, matching medical students, register allocation, pancakes, sudoko, protein folding, soap bubbles, March madness, ...

Next week – When we want problems to be hard– Computing with photons and DNA

PS6 is due Tuesday, April 27 (but don’t wait until then to start PS7!)