Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 12H (Automata and Formal Languages)

Fix an acceptable effective enumeration $f_{e,n}$ of the $n$-ary partial computable functions. The S-m-n theorem states that for every $m,n\ge 1$ there is a total computable function $s_n^m$ such thatwhere $\simeq$ means that both sides have the same value when defined and are otherwise both undefined.

The recursion theorem states that every total computable transformation $h$ of program indices has a semantic fixed point: there is an $e$ such that

Rice theorem says that every nontrivial extensional property of programs computing partial functions is undecidable. Equivalently, a nontrivial index set cannot be a computable set.

Let $A$ be such an index set. Replacing $A$ by its complement if necessary, assume that the nowhere-defined function has no index in $A$. Since the property is nontrivial, choose an index $a\in A$. If membership in $A$ were decidable, we could decide the diagonal halting problem as follows. For each $x$, defineBy the S-m-n theorem, a total computable function $q$ produces an index $q(x)$ for ${\psi }_x$. If $f_{x,1}(x)$ diverges, then ${\psi }_x$ is nowhere defined and $q(x)\notin A$. If it halts, then ${\psi }_x\simeq f_{a,1}$ and extensionality gives $q(x)\in A$. A decider for $A$ would therefore decide whether $f_{x,1}(x)$ halts, a contradiction. Hence $A$ is undecidable.

Apply the S-m-n theorem to $g$. It gives a total computable function $h$ such thatThe recursion theorem supplies an index $e$ satisfying $f_{e,1}\simeq f_{h(e),1}$. Consequently

For each $x$, let $h(x)$ be an effectively obtained index of the partial functionThe existence of the total computable index transformation $h$ follows from the S-m-n theorem. By the recursion theorem, some $m$ satisfies $f_{m,1}\simeq f_{h(m),1}$. Therefore its domain isand hence

No. The propertyis an index set: it depends only on the domain $W_n$ of the partial function computed, not on the syntax of its program. It is nontrivial. For example, an index whose domain is a singleton belongs to $\mathcal{S}$, whereas an index whose domain has two elements does not. The Rice theorem therefore shows that $\mathcal{S}$ is not a computable set, so no algorithm can decide the requested property for arbitrary $n$.

Invoking the Church–Turing thesis, this also rules out any effective procedure in the informal sense, because such a procedure would be represented by a computable function.