Transcript PNP - College of Computer and Information Science

What's All This About P ≠ NP?
Ken Clarkson
Ron Fagin
Ryan Williams
P vs. NP
• A mathematical issue, not a legal one
• P and NP:
• Each is a set of computational problems
• Each is described differently
• Are they actually the same set?
• A million dollar problem
• A Clay Millennium Prize
• Most everyone thinks P ≠ NP
– the problem is to prove it
• On August 6, Vinay Deolalikar proposed a proof
Taking this proposed proof seriously
People claim proofs all the time
– Every couple months on ArXiV, P=NP
But:
• D. is a Principal Research Scientist at HP.
• Steve Cook:
“This appears to be a relatively serious claim...”
• Dick Lipton:
“...this is a serious effort...”
• Moshe Vardi:
“This looks like a serious paper...”
However:
It doesn't look like the proof goes through.
Finding flaws can take time
• Four-color Theorem
– Proven 1879
– Bug found 1890
– Proven 1976 (using a computer)
• Hilbert's 21st problem
– Solved 1908
– Counterexample 1990
• Hilbert's 16th problem, special case
– Solved 1923
– Gaps 1980
– Solved 1991
Finding flaws in internet time
• August 6: Manuscript is sent to 22 people,
including Ron Fagin, and put on webpage
• 7: Blog post [Greg Baker]
• 8: Slashdot, Lipton’s blog
• 9: Wikipedia article about D. (deleted later)
• 10: Wiki for technical discussion established
–
–
–
Based on comment thread on Lipton’s blog
About 340 edits since
Fields Medalists are involved
• 15: Commemorative blogpost:
The P≠NP “Proof” Is One Week Old
Updates in internet time
•
First draft, Aug 6
–
Overwritten several times
•
Second draft Aug 9 to Aug 10
•
Draft 2 + ε, Aug 9 to Aug 11
•
Third draft, Aug 11 to Aug 17
•
All drafts removed after Aug 17
–
•
D. says: the paper has been sent out for
refereeing
Three-page synopsis, Aug 13
–
Only current public version
Elements of the proposed proof
• Finite Model Theory
– Part of mathematical logic
– Impact on database theory, combinatorics, and complexity
theory
• Ron Fagin is the founder of FMT
– Ron will introduce P vs. NP, and explain the role of FMT
• Random k-SAT
– Analogs in statistical physics
– Ryan Williams was a key player in the on-line discussions
• Post-doc in K53, IBM Raviv Fellow
– Ryan gave a beautifully simple counter-argument to this part
Discovery vs. Verification
• Two important tasks for a scientist are discovery
of solutions, and verification of other people’s
solutions.
• It is easier to check that a solution, say to a
puzzle, is correct, rather than to find the
solution.
• That is, verifying a solution is easier than
discovering it.
• Example: Sudoku
Sudoku
Sudoku
The P vs. NP question asks whether
verification is easier than discovery
What is P?
• Polynomial Time
• The class of problems where the solution can
discovered “quickly”
– In time polynomial in the size of the input
• Example 1: Given a number, is it even?
• Example 2: Given a graph, is it connected?
What is NP?
• Nondeterministic Polynomial time
• The class of problems where the solution can
verified “quickly”
– In time polynomial in the size of the input
• Example 1: Sudoku
• A filled-in puzzle gives a quick verification.
• Example 2: 3-colorability
3-colorability
3-colorability
Quick verification of 3-colorability
Quick verification of 3-colorability
Does P = NP?
• For our examples (Sudoku and 3-coloring), it is
not known if they are in P.
P vs. NP
• Problems in P: efficient discovery of a solution
• Problems in NP: efficient verification of a solution
The problem of whether P = NP asks:
• Assume it easy to verify a solution.
• Is it easy to discover a solution?
Can always discover a solution by brute-force search
• But there are an exponential number of solutions to check
• Can we do better?
Consider the needle in a haystack metaphor.
NP-complete problems
• NP-complete problems are the “hardest”
problems in NP
– Examples: Sudoku and 3-colorability
• If there is a fast (polynomial-time) algorithm for
one NP-complete problem, then there is a fast
algorithm for every problem in NP!
– For example, a fast algorithm for Sudoku implies
P=NP.
Why is a proof that P ≠ NP important?
• A number of important problems in industry (such as
flight scheduling, chip layout, and protein folding) are
NP-complete. A proof that P ≠ NP would tell us that we
cannot expect to get optimal answers in practice.
• Cryptography is based on the assumption that P ≠ NP.
Proving that P ≠ NP is a stepping stone towards
provably secure cryptography.
• A proof that P ≠ NP would give us deep insight into the
nature of computation, which would have many ripple
effects
– For example, Wiles’ proof of Fermat’s Last Theorem led to other
fundamental advances in number theory.
Maybe P = NP?
Then the world is fundamentally different than is commonly
believed.
•
Bad news: P = NP would destroy the “standard model”
of complexity theory
–
Much previous research would become useless.
•
Good news: P = NP would probably imply that we can
solve problems efficiently that we can’t now.
•
Bad news: P = NP would probably imply that current
cryptographic systems can be broken.
–
Radically new approaches to security would be needed.
The P vs. NP problem has been called
“one of the deepest questions ever
asked by human beings”.
The blog author who said this “bet his
house” against Deolalikar’s proof.
SAT
• Given a logical formula that is an “and of ors”, is
there a solution (an assignment of 0’s and 1’s to
the variables that makes the formula true)?
– Example:
(x1 OR NOT(x2) OR x3)
AND (x2 OR NOT(x3))
AND (NOT(x ) OR NOT(x4))
1
• A solution: x1 = 1, x2 = 0, x3 = 0, x4 = 0.
• The set of such solutions is called the solution space.
Cook’s Theorem (1971): SAT is NP-complete.
• k-SAT: each clause has exactly k members.
This problem is also NP-complete for k ≥ 3.
Strategy of Deolalikar’s Proof
•
If k-SAT were in P, then the solution spaces for
all k-SAT formulas would have a “simple
structure”.
•
For some k-SAT formulas, the solution spaces
for these formulas do not have a simple
structure.
•
Therefore, k-SAT is not in P, and so P ≠ NP.
The proof Deolalikar gives for the first bullet uses
finite model theory
Existential second-order logic
3-colorability can be expressed quite informally as:
∃ a coloring (“the coloring is a 3-coloring of the graph”)
A little more formally as:
∃R∃G∃B (“Every point is in exactly one of the sets R, G, or
B, and no two points that are connected by an edge are
both in R, or both in G, or both in B”)
This formula can be expressed formally in existential
second-order logic (∃SO)
So 3-colorability can be expressed in ∃SO.
Capturing NP with logic
Fagin’s Theorem (1974): NP = ∃SO
Example: 3-colorability
Surprising, since characterizing a complexity class
in terms of logic, where there is no notion of
machine, computation, polynomial, or time.
How about P?
• Fagin’s Theorem captures NP in terms of logic.
• Can we also capture P in terms of logic?
– Answer: Yes (sort of).
Capturing P with logic
There is a logic called “least fixpoint logic” (LFP).
•
It is richer than first-order logic (it involves
“recursion”).
Immerman-Vardi Theorem (1982): P = LFP (over
ordered structures)
Back to Deolalikar’s proof strategy
Recall that the first part of Deolalikar’s proof strategy says that if k-SAT
were in P, then the solution spaces for all k-SAT formulas would
have a “simple structure”.
Deolalikar’s proof of this first part proceeds as follows:
1.
2.
3.
4.
Assume that k-SAT is in P.
So k-SAT can be expressed in LFP, by the Immerman-Vardi
Theorem.
LFP implies a simple structure for solution spaces.
So solution spaces for k-SAT formulas have a simple structure.
Unfortunately, Deolalikar’s proof of step 3 works only for a fragment of
LFP (the “monadic case”).
•
This was pointed out by Immerman in Lipton’s blog.
•
So k-SAT is not necessarily covered in step 3.
Strategy of Deolalikar’s Proof
• If k-SAT were in P, then the solution spaces for all kSAT formulas would have a “simple structure”.
• For some k-SAT formulas, the solution spaces for these
formulas do not have a simple structure.
• Therefore, k-SAT is not in P, and so P ≠ NP.
We just saw that there was an error in Deolalikar’s proof of
the first bullet.
But maybe the first bullet can be proven another way.
Ryan will now discuss the second bullet.
Strategy of Deolalikar’s Proof
• If k-SAT were in P, then the solution spaces for
all k-SAT formulas would have a “simple
structure”.
• For some k-SAT formulas, the solution spaces
for these formulas do not have a simple
structure.
• Therefore, k-SAT is not in P, and so P ≠ NP.
Deolalikar proposes to choose certain random
k-SAT formulas, and use known properties of
their solution spaces
Random k-SAT
Recall k-SAT: Satisfiability of Boolean formulas as AND of ORs
 n variables (0-1), m clauses, each clause has k literals
F = (x1 OR NOT(x2) OR x3)
AND (x2 OR NOT(x3) OR NOT(x4))
AND (NOT(x1) OR NOT(x2) OR NOT(x3))


Here we have n=4, m=3, k=3
Given a formula F, is F satisfiable?
Is there a setting of variables that makes F evaluate to 1?
Random k-SAT:
Fix n, m, k, and choose m clauses at random
Study the percentage of random formulas that are satisfiable
Random k-SAT
• Random k-SAT:
Fix n, m, k, and choose clauses at random
Looks like this
is where the
“hard”
formulas are!
Percent Satisfiable
Relative Run Time
• (Monasson et al., Nature 1999) As the clause-to-variable
ratio increases, we see a “phase transition” in SAT:
random formulas switch from being almost all satisfiable to
almost all unsatisfiable
100
Percent Satisfiable
Relative Run Time
80
60
40
20
0
0
5
10
Clause-to-variable ratio α  m
n
Random k-SAT
• What do the formulas undergoing this transition
from “almost all satisfiable” to “almost all
unsatisfiable” look like, on average?
• (Mezard et al. Science 2002) For random k-SAT, there
are actually three phases:
1. a “replica-symmetric” phase where the solutions are all in one
big “cluster” together, then
2. a “replica-symmetry-breaking satisfiable” (RSB) phase with
exponentially many clusters of solutions, each cluster being
“far” from all the others, and finally
3. a “replica-symmetry-breaking unsatisfiable” phase with no
solutions.
• Here the distance measure is Hamming distance:
e.g. (1,1,1,1) and (0,0,0,0) have distance 4,
(1,0,0,0) and (0,0,0,0) have distance 1
The RSB Satisfiable Phase of k-SAT
• Exponentially many clusters of solutions, each cluster
being “far” from all the others
• Deolalikar’s proof focuses on analyzing formulas
arising from this RSB satisfiable phase.
Certainly some complex-looking structure here…
Can this be the reason that k-SAT is hard?
The SAT0 Objection
• SAT0: Formulas that are satisfied when you set
every variable to zero.
• This problem is definitely in P. Very easy.
• However, we can show that for every k-SAT
formula, there is a SAT0 formula with an
isomorphic solution space. All distances
between solutions are preserved.
• So whatever complex structure you may have in
the solution space of a random k-SAT formula,
there are always SAT0 formulas with analogous
structure!
The SAT0 Objection
(0,0,…,0)
• Take any k-SAT formula F and one of its solutions
(A1,…,An) where Ai ∈ {0,1} for all i
• Create the formula F’ as follows:
for every Ai = 1,
change all xi in F to NOT(xi), and all NOT(xi) to xi
The SAT0 Objection
(0,0,…,0)
• What does this say?
The difficulty of k-SAT doesn’t arise from distinguishing
satisfiable formulas with “simple structure” from those with
“complex structure”, but rather from distinguishing satisfiable
formulas from unsatisfiable formulas.
• Still, this is just intuition...
The intuition is realized
Theorem (Proved by "vloodin" and Terence Tao)
Under the notion of “simple" given in the paper,
k-SAT does have simple solution spaces!
Proof Idea: First show that all SAT0 formulas have
"simple" solution spaces, then use the SAT0 objection
to translate this space over for an arbitrary k-SAT
instance.
So unfortunately the proof breaks in its current form.
Can we salvage something from it?
• Terence Tao's car analogy (paraphrased):
…the paper is like a lengthy blueprint for a revolutionary new car, that
somehow combines a high-tech engine with advanced fuel injection to
get 200 miles to the gallon.
The LFP objections are like a discovery of serious wiring faults in the
engine… but the inventor claims this can be fixed using a weak engine
The solution space objections are like a discovery that, according to
blueprints, the car would run just as well if gasoline was replaced with
ordinary tap water… D.’s response to this has been roughly “That
objection is invalid – everyone knows cars can’t run on water.”
The theorem (on the previous slide) is like a discovery that the fuel is in
fact being sent to a completely different component of the car than the
engine…"
• Can any parts of this car be salvaged?
Concluding remarks
• Deolalikar’s proof seems to be not only wrong,
but unfixable.
• Hardness and solution space complexity seem
to be orthogonal.
– New research question: can random k-SAT be
used to prove complexity results?
• There is a new world of community refereeing.
– Good: every part of the proof had corresponding
experts
– Bad: those experts spent a great deal of time
– The community is still learning how to work
effectively in this new world.