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Fall 2013 CMU CS 15- 855 Computational Complexity Lecture 5 Savich’s theorem, and IS theorems. se slides are mostly a resequencing of Chris Umans’ slides m CS 151: Complexity Theory (Spring 2013) at CalTech.

Fall 2013 CMU CS 15-855 Computational Complexity

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Fall 2013 CMU CS 15-855 Computational Complexity. Savich’s theorem, and IS theorems. Lecture 5. These slides are mostly a resequencing of Chris Umans ’ slides From CS 151: Complexity Theory (Spring 2013) at CalTech . World Picture. coNEXP. NEXP. EXP. PSPACE. coNP. NP. P. L. - PowerPoint PPT Presentation

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Page 1: Fall 2013 CMU CS 15-855 Computational Complexity

Fall 2013 CMU CS 15-855Computational Complexity

Lecture 5

Savich’s theorem, and IS theorems.

These slides are mostly a resequencing of Chris Umans’ slides From CS 151: Complexity Theory (Spring 2013) at CalTech.

Page 2: Fall 2013 CMU CS 15-855 Computational Complexity

World Picture

EXPPSPACE

PL

NEXP

NP coNP

coNEXP

9/24/2013

Page 3: Fall 2013 CMU CS 15-855 Computational Complexity

Ladner’s Theorem• Try to “merge”:

• on input x, either– answer SAT(x)– answer TRIV(x)

• if can decide which one in P, L NP

SAT TRIVMi

M0

L

N0

Ni

: :

: :

. . . ≠

9/24/2013

Page 4: Fall 2013 CMU CS 15-855 Computational Complexity

Ladner’s Theorem

• Key: n eventually large enough to notice completed previous stage

Lf 0 0 1 1 k k k

. . .

. . .

Mi

input xinput z < x

suppose k = 2i

• I’m sup-posed to ensure Mi is killed

• I finally have enough time to check input z

• I notice z did the job, increase f to k+1

k k

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Page 5: Fall 2013 CMU CS 15-855 Computational Complexity

April 4, 2013 5

At most 2O(f(n)) bits to describe a SPACE(fn)) configuration.

.• # configurations for a q-tape TM (work tape + read-only input) that runs in space f(n)

n x f(n)q x |Q| x |∑|f(n)q = O( cqf(n) ) = 2O(f(n))

input-tape head position work-tape head

positions

state work-tape contents

Page 6: Fall 2013 CMU CS 15-855 Computational Complexity

At most 2O(f(n)) bits to describe a SPACE(fn)) configuration.

.• Pick any consistent way of interpreting binary strings as configurations of a TM.

n x f(n)q x |Q| x |∑|f(n)q = O( cqf(n) ) = 2O(f(n))

Page 7: Fall 2013 CMU CS 15-855 Computational Complexity

April 4, 2013 7

Configuration graph for NTMs• configuration graph of M on input x: • nodes are all sequences of 2O(f(n)) bits • configurations, edge (C, C’) • iff the config of M(x) represented by C, and the

config of M(x) represented by C’ are consistent with the transition relation of M.

Page 8: Fall 2013 CMU CS 15-855 Computational Complexity

• NSPACE(f(n)) = languages decidable by a multi-tape NTM that touches at most f(n) squares of its work tapes along any computation path, where n is the input length, and f :N ! N

Let C configuration graph for a space f(n) NTM on input x.

C has cf(n) = 2kf(n) nodes (Exponential in f(n))f(n) = k’log(n) means POLY-SIZED graph.

9/24/2013

Page 9: Fall 2013 CMU CS 15-855 Computational Complexity

Do DFS or BFS on C

• most nk configurations of given NTM M running in NSPACE(log n).

qstartx1x2x3…xn

qaccept qreject

xL

qaccept qreject

xL• easy to

determine if C yields C’ in one step

• configuration graph for M on input x:

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Page 10: Fall 2013 CMU CS 15-855 Computational Complexity

Reachability

• Conclude: NL P – and NPSPACE EXP

• S-T-Connectivity (STCONN): given directed graph G = (V, E) and nodes s, t, is there a path from s to t ?

Theorem: STCONN is NL-complete under logspace reductions.

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Page 11: Fall 2013 CMU CS 15-855 Computational Complexity

Reachability

• Proof:– given L NL decided by NTM M construct

configuration graph for M on input x (can be done in logspace)

– s = starting configuration; t = qaccept

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Page 12: Fall 2013 CMU CS 15-855 Computational Complexity

STCONN is in NL

• NUMSTEPS = 0 (number of steps taken.)• C = s (current node)• FLAG=False

• Until NUMSTES = n do– GUESS Z from 1 to n– Increment NUMSTEPS– If (c,z) is an edge in G, set c=z – If c==t set FLAG= True.

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Page 13: Fall 2013 CMU CS 15-855 Computational Complexity

I started at S – I got drunk – and now I am at T

I wandered, therefore, my path from S to T exists. 9/24/2013

Page 14: Fall 2013 CMU CS 15-855 Computational Complexity

Two startling theorems

• Strongly believe P ≠ NP• nondeterminism seems to add enormous

power• for space: Savitch ‘70:

NPSPACE = PSPACEand

NL SPACE(log2n)

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Page 15: Fall 2013 CMU CS 15-855 Computational Complexity

Two startling theorems

• Strongly believe NP ≠ coNP• seems impossible to convert existential

into universal

• for space: Immerman/Szelepscényi ’87/’88:

NL = coNL

9/24/2013

Page 16: Fall 2013 CMU CS 15-855 Computational Complexity

Savitch’s Theorem

Theorem: STCONN SPACE(log2 n)

• Corollary: NL SPACE(log2n)

• Corollary: NPSPACE = PSPACE

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Page 17: Fall 2013 CMU CS 15-855 Computational Complexity

Proof of Theorem– input: G = (V, E), two nodes s and t– recursive algorithm:

/* return true iff path from x to y of length at most 2i */PATH(x, y, i) if i = 0 return ( x = y or (x, y) E ) /* base case */ for z in V if PATH(x, z, i-1) and PATH(z, y, i-1) return(true); return(false);end

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Page 18: Fall 2013 CMU CS 15-855 Computational Complexity

Proof of Theorem– answer to STCONN: PATH(s, t, log n)– space used:

• (depth of recursion) x (size of “stack record”)– depth = log n– claim stack record: “(x, y, i)” sufficient

• size O(log n)– when return from PATH(a, b, i) can figure out

what to do next from record (a, b, i) and previous record

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Page 19: Fall 2013 CMU CS 15-855 Computational Complexity

Nondeterministic space

• Robust nondeterministic space classes:

NL = NSPACE(log n)

NPSPACE = k NSPACE(nk)

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Page 20: Fall 2013 CMU CS 15-855 Computational Complexity

Second startling theorem

• Strongly believe NP ≠ coNP• seems impossible to convert existential

into universal

• for space: Immerman/Szelepscényi ’87/’88:

NL = coNL

9/24/2013

Page 21: Fall 2013 CMU CS 15-855 Computational Complexity

I started at S – I got drunk NL lucky, tiny brain – and now I am NOT at T

How might I know that NO PATH from S to T exists ???? !

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Page 22: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem

Theorem: ST-NON-CONN NL• Proof: slightly tricky setup:

– input: G = (V, E), two nodes s, t

t

s“yes”

t

s“no”

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Page 23: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– want nondeterministic procedure using only

O(log n) space with behavior:

“yes” input

“no” input

qaccept qreject qaccept qreject

t

s

t

s

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Page 24: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– observation: given count of # nodes

reachable from s, can solve problem• for each v V, guess if it is reachable• if yes, guess path from s to v

– if guess doesn’t lead to v, reject. – if v = t, reject. – else counter++

• if counter = count accept

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Page 25: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– every computation path has sequence of

guesses…– only way computation path can lead to

accept:• correctly guessed reachable/unreachable

for each node v• correctly guessed path from s to v for each

reachable node v• saw all reachable nodes• t not among reachable nodes

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Page 26: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– R(i) = # nodes reachable from s in at most i

steps– R(0) = 1: node s

– we will compute R(i+1) from R(i) using O(log n) space and nondeterminism

– computation paths with “bad guesses” all lead to reject

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Page 27: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– Outline: in n phases, compute

R(1), R(2), R(3), … R(n) – only O(log n) bits of storage between phases– in end, lots of computation paths that lead to

reject– only computation paths that survive have

computed correct value of R(n)– apply observation.

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Page 28: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– computing R(i+1) from R(i):

– Initialize R(i+1) = 0– For each v V, guess if v reachable from s in

at most i+1 steps

R(i) = R(2) = 6

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Page 29: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem– if “yes”, guess path from s to v of at most i+1

steps. Increment R(i+1)– if “no”, visit R(i) nodes reachable in at most i

steps, check that none is v or adjacent to v• for u V guess if reachable in ≤ i steps;

guess path to u; counter++• KEY: if counter ≠ R(i), reject• at this point: can be sure v not reachable

9/24/2013

Page 30: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem

• correctness of procedure:• two types of errors we can make • (1) might guess v is reachable in at most

i+1 steps when it is not– won’t be able to guess path from s to v of

correct length, so we will reject.

“easy” type of error

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Page 31: Fall 2013 CMU CS 15-855 Computational Complexity

I-S Theorem• (2) might guess v is not reachable in at

most i+1 steps when it is – then must not see v or neighbor of v while

visiting nodes reachable in i steps.– but forced to visit R(i) distinct nodes– therefore must try to visit node v that is not

reachable in ≤ i steps– won’t be able to guess path from s to v of

correct length, so we will reject.“easy” type of error9/24/2013

Page 32: Fall 2013 CMU CS 15-855 Computational Complexity

Summary

• nondeterministic space classesNL and NPSPACE

• ST-CONN NL-complete

9/24/2013