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I am absolutely stumped on this one. I am unsure of how to start with this one. I have thought to reducing the problem to Atm. Another thought I have had is to convert M1 to a PDA and use the complement of EQPDA to solve this but I am honestly really unsure. Any tips that could point me in the right direction or solutions to this or similar problems?

Here is one attempt to solve this problem:
DifferentPDADFA = {<M1, M2> | M1 is a PDA and M2 is a PDA, L(M1) != L(M2)}
Write a turing machine, C, to decide the language of M1
Write a turing machine, D, to decide the language of M2
ALLCFG = {<G, w> | G is a CFG, L(G)=Σ*}
ALLCFG' = {<G, w> | G is a CFG, L(G) != Σ*}
Since ALLCFG is known to be undecidable, its complement ALLCFG' is also undeciable.
Use turing machine A, which simulates ALLCFG'.
AllCFG' <ₘ L
F -> Input <D, P, w>:
1. Construct Turing Machine Y to simulate PDA P
Y<P, w>:
On Input X:
1. Ignore Input X
2. Convert our PDA P to a CFG G (page 147)
3. Simulate A with <G, w> and return what A returns.
2. Construct Turing Machine Z to simulate DFA D
Z<D, w>:
On Input X:
1. Ignore Input X, accept.

Output <Z, Y>

Since Z ignores all inputs the language accepted by L(Z) is always Σ*.
If A accepts <G, w> then L(Y) != Σ*, therefore L(Y) != L(Z).
If A rejects <G, w> then L(Y) must be Σ*, therefore L(Y) = L(Z).
Therefore A accepting <G, w> <=> L(Y) != L(Z)
Since we know that ALLCFG' is undeciable, L is also undeciable.

akysu
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1 Answers1

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Let $Different_{DFA,PDA} = \{<M_1,M_2> \vert \ M_1 \text{ is a DFA and } M_2 \text{ is a PDA where } L(M_1) \neq L(M_2)\}$, we want to show that this language is undecidable

Let $ALL_\text{CFG} = \{<G> | \text{ G is a CFG and } L(G) = Σ^*\}$, as we know this language is undecidable, hence its complement, ${\overline{ALL}}_\text{CFG} = \{<G> | \text{ G is a CFG and } L(G) \neq Σ^*\}$, is also undecidable

We will reduce ${\overline{ALL}}_\text{CFG}$ to $Different_{DFA,PDA}$, hence showing that the latter is undecidable

The following machine $F$ computes the reduction $f$

$F$ = “On input :

  1. Convert $G$ into an equivalent $PDA$ $P$
  2. Let $D$ be the $DFA$ recognising the language $\Sigma^*$
  3. output $<D,P>$”

If $L(G) \neq \Sigma^*$, then $L(P) \neq L(D)$, and so $<D,P> \in Different_{DFA,PDA}$

If $L(G) = \Sigma^*$, then $L(P) = L(D)$, and so $<D,P> \notin Different_{DFA,PDA}$

Anwar
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