Transcript pps - University of Virginia

Theory of Computation
CS3102 – Spring 2014
A tale of computers, math, problem solving, life,
love and tragic death
Nathan Brunelle
Department of
Computer Science
University of Virginia
www.cs.virginia.edu/~njb2b/theory
Complexity Classes
Def: DTIME(t(n))={L | L is decidable within
time O(t(n)) by some deterministic TM}
Def: NTIME(t(n))={L | L is decidable within
time O(t(n)) by some non-deterministic TM}
Def: DSPACE(s(n))={L | L decidable within
space O(s(n)) by some deterministic TM}
Def: NSPACE(s(n))={L | L decidable within
space O(s(n)) by some non-deterministic TM}
Non-Robustness of TM Complexity
• Computability: all variations on TMs have the same
computing power
– If there is a multi-tape TM that can decide L, there is a
regular TM that can decide L.
– If there is a nondeterministic TM that can decide L, there is a
deterministic TM that can decide L.
• Complexity: variations on TM can solve problems in
different times
– Is a multi-tape TM faster than a regular TM?
– Is a nondeterministic TM faster than a regular TM?
Complexity Class P
P =U
k
k
DTIME(n )
P is the class of languages that can be decided
in Polynomial Time on a deterministic, singletape Turing machine.
Classes in P
TIME(n2)
Yes! We can simulate each step of a 2tape TM by making 2 passes over the
whole tape ~ 2(n+t(n))
(See Theorem 7.8)
a)
b) TIME(O(n7))
c) TIME(O(2n))
d) Class of languages that can be decided in
Polynomial Time by a 2-tape TM
e) Class of languages that can be decided in
Polynomial Time by a nondeterministic TM
Unknown! This is the P = NP question. Focus of this class…
P Examples
• What are some examples of problems in P?
– Arithmetic
– Matrix Multiplication
– Matrix Inversion
– Sorting
Non-deterministic polynomial time
• Deterministic Polynomial Time: TM takes at
most O(nc) steps to accept a string of length n
• Non-deterministic Polynomial Time: The TM
takes at most O(nc) steps on some path to
accept a string of length n
Complexity Class NP
NP = U
k
k
NTIME(n )
P is the class of languages that can be decided
in Polynomial Time on a non-deterministic
Turing Machine.
Alternative Definition
• A language is in NP if some execution path answers in
polynomial time
• What if we had a “list of directions”?
• Some information that told us which path accepted
quickly
• L is in NP if there is a “verifier” which given a string x in L
and a polynomial length “witness” it can verify x is a
member of L in polynomial time
• NP = “easy to verify” problems
Alternate
Definition of NP
Using Descriptive
complexity
Neil Immerman
Argue why these are in NP
• Sets of integers which sum to 0
– Add them together, check if sum is 0
• Sets of integers which have a subset with sum 0
– Given a “witness” w (a particular subset which sums to 0) we can:
• Verify w is indeed a subset of the input
• Verify that w sums to 0
• Sets of integers which have a subset of size 3 which sums to 0
– Try all subsets of size 3, check if any sum to 0
– Called “3-Sum”
• Primality testing: Given an integer, is it prime?
– Given a “witness” (a pair of numbers) we can:
• Verify the product of these numbers gives the input (just 1 multiplication)
• Surprisingly, this can be solved in Polynomial time (AKS, 2002)
The Class P and the Class NP
• P = { L | L is accepted by a deterministic Turing
Machine in polynomial time }
• NP = { L | L is accepted by a non-deterministic
Turing Machine in polynomial time }
• They are sets of languages
Most important CS Problem
• Does P = NP?
• Are all “easily verifiable problems” also “easily
computable”?
• How much more efficient are nondeterministic Turing Machines?
Progress
• P = NP if every NP problem has a deterministic
polynomial time algorithm
• We could find an algorithm for every NP problem
• Seems… hard…
• Study an “archetypical” example
– An NP problem which is in P iff P=NP
– Called Complete
• We need: At least one such problem
NP Hardness & Completeness
Def: A problem L’ is NP-hard if:
(1) Every L in NP reduces to L’ in polynomial time.
Def: A problem L’ is NP-complete if:
(1) L is NP-hard; and
(2) L is in NP.
If one NPC problem is in P P=NP
NP-Hard
NP
P
NP Completeness Benefits
1.
2.
3.
4.
5.
Saves time & effort of trying to solve intractable problems
efficiently;
Saves money by not separately working to efficiently solve
different problems;
Helps systematically build on & leverage the work (or lack of
progress) of others;
Transformations can be used to solve new problems by reducing
them to known ones;
Illuminates the structure & complexity of seemingly unrelated
problems;
NP Completeness Benefits
6.
Informs as to when we should use approximate solutions vs.
exact ones;
7. Helps understand the ubiquitous concept of parallelism (via nondeterminism);
8. Enabled vast, deep, and general studies of other “completeness”
theories;
9. Helps explain why verifying proofs seems to be easier than
constructing them;
10. Illuminates the fundamental nature of algorithms and
computation;
NP Completeness Benefits
11. Gave rise to new and novel mathematical approaches, proofs,
and analyses;
12. Robustly decouples / abstracts complexity from underlying
computational models;
13. Gives disciplined techniques for identifying “hardest” problems /
languages;
14. Forged new unifications between computer science,
mathematics, and logic;
15. NP-Completeness is interesting and fun!
Our “Favorite” NP Complete Problem
Satisfiability (SAT): Given a Boolean expression in conjuctive
normal form, is there some assignment of T/F to its
variables such that the expression resolves to “True”?
Note: we may restrict each disjunction to have 3 variables (3-SAT)
Examples:
¬𝑎 ∨ 𝑏 ∨ ¬𝑐 ∧ 𝑎 ∨ 𝑑 ∨ 𝑐 ∧ (𝑑 ∨ 𝑏 ∨ ¬𝑒)
Satisfied if: 𝑎 = 𝐹, 𝑏 = 𝐹, 𝑐 = 𝑇, 𝑑 = 𝑇
𝑎 ∨ 𝑏 ∧ ¬𝑎 ∨ ¬𝑏 ∧ (¬𝑎 ∨ 𝑏) ∧ (𝑎 ∨ ¬𝑏)
Can’t be Satisfied!
The Cook/Levin Theorem
Theorem [Cook/Levin, 1971]: SAT is NP-complete.
Proof idea: given a non-deterministic polynomial
time TM M and input w, construct a CNF
formula that is satisfiable iff M accepts w.
Create boolean variables:
q[i,k]  at step i, M is in state k
h[i,k]  at step i, M’s RW head scans tape cell k
s[i,j,k]  at step i, M’s tape cell j contains symbol Sk
Stephen Cook
Leonid Levin
Qk
M halts in polynomial time p(n)
 total # of variables is polynomial in p(n)
The Cook/Levin Theorem
Add clauses to the formula to enforce necessary
restrictions on how M operates / runs:
• At each time i:
M is in exactly 1 state
r/w head scans exactly 1 cell
All cells contain exactly 1 symbol
• At time 0  M is in its initial state
• At time P(n)  M is in a final state
• Transitions from step i to i+1
all obey M's transition function
Resulting formula is satisfiable iff M accepts w!
Stephen Cook
Leonid Levin
Qk
“Guess and Verify” Approach
Note: SAT  NP.
Idea: Nondeterministically “guess” each Boolean
variable value, and then verify the guessed solution.
 polynomial-time nondeterministic algorithm  NP
This “guess & verify” approach is general.
Idea: “Guessing” is usually trivially fast ( NP)
 NP can be characterized by the “verify” property:
NP  set of problems for which proposed
solutions can be quickly verified
 set of languages for which string
membership can be quickly tested.
Historical Note
The Cook/Levin theorem was independently proved by Stephen Cook
and Leonid Levin
•
•
•
Denied tenure at Berkeley (1970)
Invented NP completeness (1971)
Won Turing Award (1982)
•
•
Student of Andrei Kolmogorov
Seminal paper obscured by
Russian, style, and Cold War
An NP-Complete Encyclopedia
Classic book: Garey & Johnson, 1979
• Definitive guide to NP-completeness
• Lists hundreds of NP-complete problems
• Gives reduction types and refs
Michael Garey
David Johnson
NP-Completeness Proof Method
• To show that Q is NP-Complete:
• 1) Show that Q is in NP
– Usually by “guess and verify”
• 2) Pick an instance, R, of your favorite NPComplete problem (ex: 3-SAT)
• 3) Show a polynomial algorithm to transform R
into an instance of Q
Reducing Reduction Proofs
• Conjecture: A has some property Y.
• Proof by reduction from B to A:
– Assume A has Y. Then, we know there is an M that decides A.
– We already know B does not have property Y.
– Show how to build S that solves B using M.
• Since we know B does not have Y, but having S would
imply B has Y, S cannot exist. Therefore, M cannot exist,
and A does not have Y.
Undecidability Proofs
• Conjecture: A has some property Y.
• Proof by reduction from B to A:
– Assume A has Y. Then, we know an M exists.
– We already know B does not have property Y.
– Show how to build S that solves B using M.
• Since we know B does not have Y, but having S would imply B has Y,
S cannot exist. Therefore, M cannot exist, and A does not have Y.
Undecidability:
Y = “can be decided by a TM”
B = a known undecidable problem (e.g., ATM, HALTTM, EQTM, …)
M = “a TM that decides A”
NP-Hardness Proofs
• Conjecture: A has some property Y.
• Proof by reduction from B to A:
– Assume A has Y. Then, we know an M exists.
– We already know B does not have property Y.
– Show how to build S that solves B using M.
• Since we know B does not have Y, but having S would imply B has Y,
S cannot exist. Therefore, M cannot exist, and A does not have Y.
NP-Hardness:
Y = “is NP-Hard”
B = a known NP-Hard problem (e.g., 3-SAT, SUBSET-SUM, …)
M = “a TM that decides A in polynomial-time”
Example
• Suppose we know ATM is undecidable, but do
not yet know if EQTM is.
EQTM = { <M1, M2> | M1 and M2 are TMs where L(M1) = L(M2) }
ATM = { <M>, w | M is TM, w is string, w in L(M) }
Conjecture: EQTM is undecidable.
What do we need to do to prove conjecture?
Reduce from ATM to EQTM: show that a solver for EQTM could be used to
solve ATM.
Pitfall #1: Make sure you do reduction in right direction.
Showing how to solve B using MA, shows A is as hard as B.
Building Solvers
B=
ATM = { <M, w> | M is TM, w is string, w in L(M) }
A=
EQTM = { <M1, M2> | M1 and M2 are TMs where L(M1) = L(M2) }
Conjecture: EQTM is undecidable.
Reduce from EQTM to ATM: show that MEQ, a solver for EQTM can be used to
solve ATM.
MB(<M, w>): machine that decides ATM
Simulate MEQ on <M1, M2>:
M1 = a TM that simulates M running on w
M2 = a TM that always accepts
If it accepts, accept; if it rejects, reject.
Pitfall #2: Get the inputs to the solver to match correctly.
To solve B using MA, must transform inputs to B into inputs to A.
Legal Transformations
• Undecidability proofs: your transformation can do
anything a TM can do, but must be guaranteed to
terminate
– E.g., cannot include, “simulate M and if it halts, accept”
• NP-Hardness proofs: your transformation must finish in
polynomial time
– E.g., cannot include, “do an exponential search to find the
answer, and output that”