More Undecidability 14-0 Computation Histories Sometimes - - PDF document

✬ ✫ ✩ ✪ Griffith University 3130CIT Theory of Computation (Based on slides by Harald Søndergaard of The University of Melbourne)

More Undecidability

14-0

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Computation Histories

Sometimes it is useful to consider the language of computation histories for a machine. Some important reduction proofs rely on this. An accepting computation history for M on w is a sequence C1, . . . , Ck of configurations, with C1 the start configuration and Ck an accepting configuration. A rejecting computation history is defined similarly. Note that these are finite sequences. A deterministic machine has at most one computation history (none if it fails to halt).

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Linear-Bounded Automata

A linear-bounded automaton (LBA) is a Turing machine which can use only a finite segment of its tape, say, the part occupied by input. The acceptance problem for LBAs is decidable: ALBA = {M, w | M is an LBA that accepts w} can be decided because there are only finitely many different configurations for a given LBA. Let q = |Q|, n = |w|, t = |Γ|, and k = qntn. The TM that decides ALBA simulates M on w for up to k steps, accepting or rejecting according to M. If M has not halted by then, the simulating machine rejects.

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LBA Emptiness is Undecidable

Theorem: ELBA = {M | M is an LBA and L(M) = ∅} is undecidable. We reduce ATM to ELBA. For a given M and w, we can construct an LBA B which accepts exactly the accepting computation histories for M on w. #

C1

#

C2

# · · · #

Ck

# B has to check that

• C1 is the start configuration for M on w,
• each Ci+1 follows legally from Ci,
• Ck is an accepting configuration.

All this can be done in bounded space.

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LBA Emptiness is Undecidable (cont.)

Now we are ready for the reduction. Assume that R decides ELBA. We can decide ATM as follows:

• 1. Given M and w, construct the LBA B.
• 2. Run R on B.
• 3. If R rejects, accept; if it accepts, reject.

But we already know that ATM is undecidable. Hence ELBA is undecidable.

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CFG Exhaustiveness is Undecidable

Theorem: AllCFG = {G | G is a CFG and L(G) = Σ∗} is undecidable. For a proof, see Sipser page 181. Is this result surprising? Recall that {G | G is a CFG and L(G) = ∅} is decidable.

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Post’s Correspondence Problem

Or just PCP. Given a finite set of “dominos” such as b ca

• ,

a ab

• ,

ca a

• ,

abc c

• Formally, a set of pairs (ti, bi) with ti, bi ∈ ΣΣ∗.

(There is an infinite supply of each domino.) Is there a sequence of dominos that produce a match, that is, the top halves spell the same string as the bottom halves? In this case, yes: a ab b ca ca a a ab abc c

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PCP (cont.)

How about this case? a cb

• ,

bc ba

• ,

c aa

• ,

abc c

• And this?

ab aba

• ,

bba aa

• ,

aba bab

• And this?

baa abaaa

• ,

aaa aa

• Yes,

aaa aa baa abaaa aaa aa

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PCP Is Undecidable

Theorem: PCP is undecidable. The proof has tedious details, but the idea is simple. We reduce ATM to PCP via computation histories. That is, for given M and w we construct an instance P of PCP such that P has a solution iff M accepts w. A solution to P will effectively simulate the running of M on w. Let M = (Q, Σ, Γ, δ, q0, qa, qr).

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PCP Is Undecidable (cont.)

Dominos for the problem: To start the simulation:

• #

#q0w1w2 · · · wn#

• To simulate the tape head moving right:

qa br

• whenever δ(q, a) = (r, b, R).

To simulate the tape head moving left: cqa rcb

• whenever δ(q, a) = (r, b, L).

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PCP Is Undecidable (cont.)

To simulate the tape not affected by a transition: a a

• for each a ∈ Γ.

To make trailing blanks explicit: #

• #
• To let the top string “catch up” once we get to qa:

aqa qa

• ,

qaa qa

• ,

qa## #

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PCP Is Undecidable (cont.)

Let us see how this works. Let w be 010 and assume M just moves through the input, reversing the bits. So δ(q0, 0) = (1, q0, R) δ(q0, 1) = (0, q0, R) δ(q0, ) = (qa, , L) The match then begins

• #

#q0010#

• The only way we can continue the match is
• #

#q0010# q00 1q0 1 1 # #

• We are then forced to add

1 1 q01 0q0 # # 1 1 q00 1q0 # #

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PCP Is Undecidable (cont.)

At this point the upper sequence is #q0010#1q010#10q00# and the lower is #q0010#1q010#10q00#101q0# Using simple dominos and adding #

• #
• we extend the sequences to

#q0010#1q010#10q00#101q0# and #q0010#1q010#10q00#101q0#101q0 #

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PCP Is Undecidable (cont.)

Now we can use the last transition domino to get: #q0010#1q010#10q00#101q0#101q0 # and #q0010#1q010#10q00#101q0#101q0 #10qa1 # The match can then be completed using the three “catch up” dominoes.

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PCP Is Undecidable (cont.)

We have cheated slightly, in that we required the match to begin with a designated domino. This is not a serious flaw. We reduced ATM to this modified PCP, and it is quite easy to reduce the latter to PCP proper.

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Domino Snakes

Consider a finite set of types of tiles ⊠. There are infinitely many tiles of each type. Can points X and Y in the plane be connected?

X Y

If there are no constraints on where the snake can go, the problem is decidable.

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Domino Snakes (cont.)

Can X and Y be connected?

X Y

In finite segment of plane: also decidable.

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Domino Snakes (cont.)

Can points X and Y be connected?

X Y

In half-plane: Undecidable! Sometimes our intuition is no guide at all.

14-17