cs3102: Theory of Computation Class 25: NP-Complete Appetizers Spring 2010 University of Virginia David Evans PS6 is due Tuesday, April 27 (but don’t wait until then to start PS7!)
Cs3102: Theory of Computation Class 25: NP-Complete Appetizers Spring 2010 University of Virginia David Evans PS6 is due Tuesday, April 27 (but don’t wait.
Dec 15, 2015
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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
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
(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!)
Related Documents
