In Computational Complexity -- A Modern Approach, by Arora and Barak, they have the following claim (Example 3.6).

Let EXPCOM be the following language $$ \{ \langle M, x, 1^n\rangle \mid M \text{ outputs 1 on $x$ within $2^n$ steps} \} $$ Then $\mathbf{P}^{\mathrm{EXPCOM}} = \dots = \mathbf{EXP}$. [...]

Clearly, an oracle to EXPCOM allows one to perform an exponential-time computation at the cost of one call, and so $\mathbf{EXP} \subseteq \mathbf{P}^{\mathrm{EXPCOM}}$. [...]

I believe their reasoning is as follows:

  1. Suppose we have a language $L \in \mathbf{EXP}$.

  2. Thus there is a TM $M$ that decides it in time $2^{n^c}$.

  3. We want to create a poly-time oracle-TM $T^{\mathrm{EXPCOM}}$ that decides $L$.

  4. $T$ works as follows on input a string $x \in \{ 0, 1 \}^{*}$. It queries its EXPCOM oracle on input $\langle M, x, 1^{n^c} \rangle $, and outputs the same answer as the oracle.

  5. Clearly $T$ runs in polynomial time, and it decides the same language as $M$. QED.

Here is my confusion. For $T$ to call its EXPCOM oracle, it needs to know which $M$ that decides $L$ (in exponential time). However, 2. only promises the existence of a machine $M$; it doesn't actually tell you how to find it! (this problem also applies to the running time $2^{n^c}$ of $M$).

So clearly I'm misunderstanding something, but what? Is it my understanding of the language EXPCOM? Is it my description of $T$? Or have I misunderstood oracle-TMs altogether? (Or maybe all three?)


You are correct that $T$ needs to know which Turing machine accepts $L$. This Turing machine is $M$, and you can hardcode it into $T$. There is absolutely no problem with that.

Here is a similar example. Suppose that there is a proof that $P \neq NP$. Then there is a Turing machine that prints a proof of $P \neq NP$.

Proof: According to the assumption, there is a proof $\pi$ that $P \neq NP$. Construct a Turing machine $T$ that prints $\pi$. Then $T$ prints a proof of $P \neq NP$. $\quad\square$

What seems to worry you is that you think of $T$ as accepting $L$ as an input, from which it is supposed to come up with the Turing machine $T$. But this is not the case – all we have to do is to show that for each $L$ there exists an appropriate Turing machine $T$. Moreover, it is not clear how $T$ would accept $L$ as input – a language is, in general, an infinite object.

Sometimes we do need to be worried about the issue that you raised. Here is an example. Let $L_1,L_2,\dots$ be an infinite sequence of decidable languages. For each $L_i$, there is a Turing machine $T_i$ that decides $L_i$. But is there a Turing machine $T$ that accepts an index $i$ and a word $w$ and returns whether $w \in L_i$? Not necessarily (I'll let you come up with a counterexample). When such a machine $T$ does exist, we say that $L_1,L_2,\dots$ are uniformly decidable.

No such uniformity condition appears in your question. We could impose such a condition artificially by providing $L$ as an input via a Turing machine, not necessarily running in exponential time, that accepts $L$. In this case, your criticism would be valid – given a Turing machine that accepts a language in $\mathsf{EXP}$, it is not clear how to find a Turing machine accepting the same language and running in exponential time.

  • $\begingroup$ I didn't mean that $T$ takes $L$ as input no. Regarding your counter-example question, I guess the issue is that it gives you the ability to solve undecidable problems like the halting problem? (e.g., let $L$ be the unary language where $1^i \in L$ iff $M$ halts on input $x$ where $i$ encodes TM $M$ and input $x$. $L$ can be accepted by non-uniform TMs of course) $\endgroup$ – panto Sep 16 '18 at 18:02
  • $\begingroup$ My counterexample asks for a sequence of decidable languages $L_1,L_2,\ldots$ such that the language $\{(i,x) : x \in L_i\}$ is not decidable. $\endgroup$ – Yuval Filmus Sep 16 '18 at 23:02

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