MELJUN CORTES AUTOMATA THEORY chapter19

8/10/2019 MELJUN CORTES AUTOMATA THEORY chapter19

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CSC 3130: Automata theory and formal languages

Undecidable problems for CFGsand descriptive complexity

MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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Decidable vs. undecidable

TMM accepts w

TMMaccepts some input

TMMand M acceptsame inputs

TMM accepts all inputs

undecidable

TMM halts on wPDAM accepts w

DFAM accepts wdecidable

PDA Paccepts all

inputsCFG Gis ambiguous

other kinds of problems?

?

DFAMaccepts allinputs


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0q0l0o0t0u0s0

0o0q6o0t0u0s00o0k6q3t0u0s

0o0k6r0q0u0s00o0k6r0a0q1s

0o0k6r0qaa0

Computation is local

lotus

ootus

oktus

okrus

okras

okra

Mq6

q0

q3

q0

q1

qacc

The changes between rows occur in a 2x3 window

computation

tableau


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Computation histories as strings

IfMhalts on w, We can represent thecomputation tableau by a string t over alphabetGQ{#, }0q0l0o0t0u0s0



0o0k6r0qaa0

q0lotus#oq6otus#...#okrqaa#

Maccepts w qaoccurs in string t

Mrejects w qadoes not occur in t


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Undecidable problems for PDAs

Theorem

Proof: We will show that

ALLPDA= {P: Pis a PDA that accepts allinputs}

The languageALLPDAis undecidable.

IfALLPDAcan be decided, so canATM.


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Undecidable problems for PDAs

M, w

reject

ifMaccepts w

accept

ifMrej/loops w

Areject if not

accept ifPaccepts all inputsP

AP

P accepts all inputs ifMrejects or loops on w

Pdoes not accept some input ifMaccepts w


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Undecidability via computation histories

Task: Design a PDA Psuch that

Pcandidate computationhistory tofMon w

reject acceptinghistories

0q0l0o0t0u0s0


0o0k6r0q0u0s0

0o0k6r0a0q1s0o0k6r0qaa0

Expect tof the form w1#w2#...#wk#

If w1q0w, accept t.

If tdoes not contain qa, accept t.

If two consecutive blocks wi#wi+1do not correspond to a propertransition ofM, accept t.


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Implementing P

If w1q0w, accept t.

If two consecutive blocks wi#wi+1do notrepresent a valid transition ofM, accept t.

On input t:

Nondeterministically make one of the following choices

If tdoes not contain qa, accept t.

Look for the beginning of the ith block of t

Look in the first block w1of t

Look for the appearance of qa

0o0k6q3t0u0s #

0o0k6r0q0u0s0

wi#wi+1represents a valid transition if all 3x2windows correspond to possible transitions ofM

valid transition


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Valid and invalid windows

6c3a0t0 0c6a0p00

6t3t0u0 0t6t0u00

valid window

invalid window

6t3t0u0 0t6t0q3 0

valid window

6t3q3u0 0t6a0q7 0

valid if d(q3, u) = (q7, a, R)

6q3t0u0 0k6t0q0 0

invalid window

6c3a0t0 0b6a0t00

valid window


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Implementing P

To check this it is better to write tin

boustrophedon

wi#wi+1represent a valid transition ofM

0q0l0o0t0u0s00o0q6o0t0u0s00o0k6q3t0u0s


0o0k6r0qaa0

q0lotus#oq6otus#...#okrqaa#

q0lotus#sutoq6o#...#okrqaa#

Alternate rows are written in reverse


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Implementing P

0o0k6q3t0u0s #

0o0k6r0q0u0s0

proper transition

#okq3tus#suq0rko#

wi wi+1

Nondeterministically look for beginning of 3x2 window

wi#wi+1represent a valid transition ofM

#

Remember first row of window in state

Use stack to detect beginning of second row

Remember second row of window in state

If window is not valid, accept, otherwise reject.


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The Post Correspondence Problem

Input: A set of tiles like this

Given an infinite supply of such tiles, can youmatch top and bottom?

bab

cc

c

ab

a

ab

baa

a

bab

cc

c

ab

a

ab

baa

a

c

ab

a

baba

a

baba

bab

e

bab

e


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Undecidability of PCP

Theorem


PCP= {D: Dis a collection of tiles thatcontains a top-bottom match}

The language PCPis undecidable.

If PCPcan be decided, so canATM.


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Idea: Matches represent accepting histories

M, w T (collection of tiles)

IfMaccepts w, then Tcan be matched

IfMrej/loops on w, then Tcannot be matched

q0lotus#oq6otus#okq3t...#qa

q0

lotus#oq6

otus#okq3

r...#qa

e

q0lotus#q0l

oq6t

tu

us

s#

#

oq60

okq3o

o


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Some technicalities

We will assume that Before accepting, TMMerases its tape

One of the PCP tiles is marked as a starting tile

These assumptions can be made without loss ofgenerality (we will see why later)

bab

cc

c

ab

a

ab

baa

a

a

baba

s


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To decideATM, we construct these tiles for PCP

M, w T (collection of tiles)IfMaccepts w, then Tcan be matched

IfMrej/loops on w, then Tcannot be matched

e

q0w#

sa1qia3

b1b2b3for each valid

window of this form

a

afor all ainG{#, }

#qa

e

#

#

efinal tiles


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q0

lotus#oq6

otus#...#oq1

#qa

q0lotus#oq6otus#...#oq1#qa


q0lotus#oq6otus#...#oq1

#qa

eq0w#

s a1qia3b1b2b3

aa

#qa

e

#

#

e

accepting computation history


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IfMrejects on input w, then qr appears on bottomat some point, but it cannot be matched on top

IfMloops on w, then matching keeps going

forever

e

q0w#

sa1qia3

b1b2b3

a

a

#qa

e

#

#

e


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A technicality

We assumed that one tile marked as starting tile

We can remove assumption by changing tiles abit

bab

cc

c

ab

a

baba

s

b*a*b*

*c*c

c*

*a*b

*a*

*b*a*b*a

*

starting tilebegins with *

ending tilematches last *

middle tiles


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Ambiguity of CFGs

AMB= {G: Gis an ambiguousCFG}

Theorem


The languageAMBis undecidable.

IfAMBcan be decided, so can PCP.


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Ambiguity of CFGs

Proof:

Step 1: Number the tiles

T G

If Tcan be matched, then G isambiguous

If Tcannot be matched, then Gis unambiguous

(collection of tiles)

bab

cc

c

ab

a

ab

1 2 3

(CFG)


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Ambiguity of CFGs

Each sequence of tiles gives two derivations

If the tiles match, these two derive the samestring

bab

cc

c

ab

1 2c

ab

2

S T babT1 babcT21 babcc221

S B ccB1 ccabB21 ccabab221


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Ambiguity of CFGs

Argue by contradiction: If Gis ambiguous then ambiguity must look like this

T G

If Tcan be matched, then G isambiguous

If Tcannot be matched, then Gis unambiguous

(collection of tiles) (CFG)

S

T

Ta1 n1

ai ni

Ta2 n2

S

B

Bb1 m1

bj mj

Bb2 m2

Then n1...ni= m1mj

So there is a match

a1b1

a2b2

aibi

n1 n2 ni


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Descriptive complexity


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Roulette

In a game of roulette, you bet $1 on evenor odd

The outcome is a number between 1 and 36

If you guessed correctly, double your bet

Otherwise, you lose

17 6 5 16 5 2

11 8 31 18 7 4

5 2 29 8 1 12


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Randomness

If we write Efor even, Ofor odd, what we saw is

It seems the wheel is crooked. If it wasnt wewould expect something more like

But both sequences have same probability! Whydoes one appear less random than the other?

OEOEOEOEOEOEOEOEOEOE

OOOEEOEOOEOEOOOEEEOE


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Turing Machines with output

The goal of a Turing Machine with output is towrite something on the output tape and go intostate qhalt

M

output tape

0 1 0

work tape

0 1 0


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Descriptive complexity

The descriptive complexity K(x)of xis theshortestdescriptionof any Turing Machine thatoutputs x

We will assume xis longAndrey Kolmogorov(1903-1987)


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Example of descriptive complexity

Turing machine implementation:

x= OE...OE = (OE)n Repeat for nsteps:At odd step print OAt even step print E

Write nin binary on work tape

While work tape not equal to 0,Subtract 1from number on work tape

If number is odd, write OIf number is even, write E

(n= 1,000,000,000)

log2nstates

3states

15states

2states

log2n + 20K(x)


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Bounds on descriptive complexity

Theorem 1

Proof: Let x= x1...xnand consider the followingTM:

For every xof length n, K(x)is at most O(n)

Write x1to output tape and move rightWrite x2to output tape and move right

Write xnto output tape and halt.

...

n+ O(1)


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Descriptive complexity and randomness

Theorem 2For 99%of strings of length n, K(x) n10.

0 O(logn) n10

simple

strings111...1, OEOE...OE,

3.14159265, 1212321234321

random-looking

strings

n+O(1)

randomness-deficient

strings


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Evaluating randomness

How do we know if the casino is crooked?

Idea: Compute K(sequence).

If much less than n, indicates sequence is not

random

17 6 5 16 5 2

11 8 31 18 11 13

5 2 29 8 1 12

12

14

12

8

31

4


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Computing descriptive complexity

Proof:Suppose it is, fix nand consider this TM M:

Let x= output ofM, then

So (when nis large) we get K(x) > K(x),

i ibl !

It is not possible to compute K(x).

Output the first xof length n(in lexicographic order)such that K(x) n10

K(x) n10 K(x) |M| = log2n+ O(1)but

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MELJUN CORTES AUTOMATA THEORY chapter19