Lecture Notes 24: Reductions and Ennumeration

Outline

This class we'll discuss:

• Recap: More Reductions

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ATM is UNdecidable

We proved this by Contradiction.
We assumed ATM was decidable and built a helper machine to arrive at a contradiction:


Consider the Machine \(M_{OPPOSITE} ( < M> )\)

\[ \begin{align*} &M_{OPPOSITE}:\\ & \text{ On input $ < M > $ }:\\ & \text{ Simulate $D_{ATM} \; on \; < M , < M > > $}\\ & \text{ If $D_{ATM}$ accepts, REJECT. }\\ & \text{ If $D_{ATM}$ rejects, ACCEPT. }\\ \end{align*} \]

Now, what happens when we run \(M_{OPPOSITE} ( < M_{OPPOSITE}> )\) ?

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Reducing ATM to HALT



Assume HALT is decidable, and so there exists some \(D_{HALT}\) that decides, for any input \(< M,w>\), whether \(M\) halts on \(w\).

We'll design the Machine \(D_{ATM} \) as follows:

\[ \begin{align*} &D_{ATM}:\\ & \text{ On input $ < M, w > $ }:\\ & \text{ Simulate $D_{HALT} \; on \; < M, w > $}\\ & \text{ If $D_{HALT}$ rejects, $M$ doesn't halt, so it did not accept: REJECT. }\\ & \text{ If $D_{HALT}$ accepts, we know $M$ won't loop forever, so }\\ & \quad \text{ Simulate $M$ on $w$ ADWID }\\ \end{align*} \]







Reducing ATM to ATM-01














In terms of how to write this reduction properly:



Assume ATM-01 is decidable, and so there exists some \(D_{ATM-01}\) that decides, for any input \(< M>\), whether \(M\) Accepts \(01\).

We'll design the Machine \(D_{ATM} \) as follows:

\[ \begin{align*} &D_{ATM}:\\ & \text{ On input $ < M, w > $ }:\\ & \text{ Create (but don't run) $HELPER_{M,w}$ such that}\\ & \quad \text{ On input $ < X > $ }:\\ & \quad \quad \text{ Ignore $ < X > $ }\\ & \quad \quad \text{ Run $M$ on $w$ ADWID}\\ & \text{ Now Run $D_{ATM-01}$ on $HELPER_{M,w}$}\\ & \text{ If $D_{ATM-01}$ accepts, it is only because M accepts w so our machine ACCEPTS}\\ & \text{ If $D_{ATM-01}$ rejects, , it is only because M rejects w so our machine REJECTS}\\ \end{align*} \]

And the MOST IMPORTANT PART:

Because this is a VALID construction of \(D_{ATM}\) with the sole assuption that \(D_{ATM-01}\) Exists... this means that because we KNOW that \(D_{ATM}\) doesn't exist, \(D_{ATM-01}\) MUST ALSO NOT EXIST.


















This can be written like this:

Assume EMPTY-TM is decidable, and so there exists some \(D_{EMPTY-TM}\) that decides, for any input \(< M >\), whether \(L (M) \) is empty.

We'll design the Machine \(D_{ATM} \) as follows:

\[ \begin{align*} &D_{ATM}:\\ & \text{ On input $ < M, w > $ }:\\ & \text{ Create (but don't run) $HELPER_{M,w}$ such that}\\ & \quad \text{ On input $ < X > $ }:\\ & \quad \quad \text{ Ignore $ < X > $ }\\ & \quad \quad \text{ Run $M$ on $w$ ADWID}\\ & \text{ Now Run $D_{EMPTY-TM}$ on $HELPER_{M,w}$}\\ & \text{ If $D_{EMPTY-TM}$ rejects, invert the result and our machine ACCEPTS}\\ & \text{ If $D_{EMPTY-TM}$ accepts, invert the result and our machine REJECTS}\\ \end{align*} \]

As we saw above, the ONLY way \(D_{EMPTY-TM}\) Rejects is if \(HELPER_{M,W}\) Accepts, which happens ONLY when \(M\) accepts \(w\).

This means we CAN make \(D_{ATM}\) as long as \(D_{EMPTY-TM}\) exists.

However, \(D_{ATM}\) doesn't exist...which means \(D_{EMPTY-TM}\) CANNOT EXIST EITHER.




Is there a simple way we could prove \(\overline{EMPTY-TM}\) is NOT Decidable?







Is there a simple way we could prove \(\overline{EMPTY-TM}\) is Recognizable?

This will lead us to a new way of looking at Recognizers: Enumeration.

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Enumeration

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Homework

[Due for everyone]

Get together to study Reductions!!
You need to be able to master reducing the examples we've seen so far.

[Optional]

TODO