The Timpson Lyceum · Computation Theory

Lecture 17 (Computation Theory) Reducibility and Undecidable Problems

Overview

This lecture introduces reducibility, the central technique for proving that problems about Turing machines are undecidable. We begin by reviewing the proof that the acceptance problem $A_{TM}$ is undecidable, then prove that a language is decidable exactly when both it and its complement are recognizable — which lets us exhibit our first non-recognizable language, $\overline{A_{TM}}$ . The heart of the lecture is the reduction method: if we can convert the acceptance problem into another problem $B$ , then $B$ must be undecidable too, because a solver for $B$ would solve the impossible $A_{TM}$ . We apply this to the halting, emptiness, and regularity problems, and finally prove Rice's theorem, which shows that every non-trivial property of a Turing machine's language is undecidable.

Key Concepts

The acceptance problem. $A_{TM} = \{\langle M, w\rangle \mid M \text{ is a Turing machine that accepts } w\}$ . This language is undecidable: no Turing machine can, for every $\langle M, w\rangle$ , halt and correctly say whether $M$ accepts $w$ . The proof assumes a decider $H$ for $A_{TM}$ exists, builds a machine $D$ that does the opposite of $H$ on its own input, and derives the contradiction that $D$ accepts $\langle D\rangle$ if and only if it does not.

Decidable vs. recognizable.

A language is recognizable (Turing-recognizable) if some machine accepts every string in the language; on strings outside the language it may reject or loop forever.
A language is decidable if some machine accepts every string in the language and rejects every string outside it — it always halts.
Every decidable language is recognizable, but not conversely. The Diophantine-equation problem (does a polynomial have a non-trivial integer root?) is recognizable but not decidable: you can confirm a solution by searching, but you can never be sure no solution exists.

Decidability theorem. A language $A$ is decidable if and only if both $A$ and its complement $\overline{A}$ are recognizable. ( $\Rightarrow$ ) A decider for $A$ already recognizes $A$ , and flipping its accept/reject behaviour recognizes $\overline{A}$ . ( $\Leftarrow$ ) Given recognizers $M_1$ for $A$ and $M_2$ for $\overline{A}$ , run them in parallel (alternating one step each) on input $w$ ; since $w$ lies in exactly one of $A$ , $\overline{A}$ , one machine must eventually halt — accept if $M_1$ accepts, reject if $M_2$ accepts.

Co-recognizable. A language is co-recognizable if its complement is recognizable. $\overline{A_{TM}}$ is not even recognizable: if it were, then $A_{TM}$ (which is recognizable) and $\overline{A_{TM}}$ would both be recognizable, making $A_{TM}$ decidable — a contradiction. A language is decidable exactly where recognizable and co-recognizable overlap.

Reducibility. A reduction from problem $A$ to problem $B$ is a conversion such that any solution to $B$ yields a solution to $A$ , written $A \le B$ . Logically, " $A$ reduces to $B$ " means (solution to $B$ ) $\Rightarrow$ (solution to $A$ ); the contrapositive is ( $A$ unsolvable) $\Rightarrow$ ( $B$ unsolvable). So to prove $B$ undecidable, reduce a known-undecidable problem (usually $A_{TM}$ ) to $B$ . Example: finding a rectangle's area reduces to finding its length and width — knowing length and width gives the area (but not conversely).

Problems proved undecidable by reduction.

Halting problem $HALT_{TM} = \{\langle M, w\rangle \mid M \text{ halts on } w\}$ . Reduce $A_{TM}$ to it: a halting-decider $R$ tells us whether $M$ halts on $w$ ; if not, reject (a non-halting machine does not accept), otherwise simulate $M$ on $w$ — now guaranteed to halt — and copy its accept/reject. This decides $A_{TM}$ , which is impossible.
Emptiness problem $E_{TM} = \{\langle M\rangle \mid L(M) = \varnothing\}$ . Build $M_1$ that rejects every input except $w$ , and on input $w$ runs $M$ on $w$ . Then $L(M_1) = \{w\}$ if $M$ accepts $w$ , and $L(M_1) = \varnothing$ otherwise. An emptiness-decider applied to $M_1$ thus decides $A_{TM}$ .
Regularity problem $REGULAR_{TM} = \{\langle M\rangle \mid L(M) \text{ is regular}\}$ . Build $M_1$ that accepts every string of the form $0^n 1^n$ outright, and accepts all other strings only when $M$ accepts $w$ . Then $L(M_1) = \Sigma^*$ (regular) if $M$ accepts $w$ , and $L(M_1) = \{0^n 1^n\}$ (non-regular) otherwise. A regularity-decider on $M_1$ decides $A_{TM}$ .

Rice's theorem. Let $P$ be any non-trivial property of a Turing machine's language — formally, a set of machine descriptions such that (i) two machines with the same language are both in $P$ or both out (it is a property of the language, not the machine), (ii) at least one machine is in $P$ , and (iii) at least one is out. Then $P$ is undecidable. Informally: anything interesting you might want to know about the language a Turing machine recognizes is undecidable. Note properties of the machine itself (e.g. "has 28 states") are not language properties and are not covered.

Worked Example

Proof of Rice's theorem (reduction from $A_{TM}$ ). Let $P$ be non-trivial, with some machine $T \in P$ . Let $T_\varnothing$ be a machine with $L(T_\varnothing) = \varnothing$ ; without loss of generality assume $T_\varnothing \notin P$ (otherwise work with $\overline{P}$ , since $P$ is decidable iff $\overline{P}$ is). Assume for contradiction $R$ decides $P$ . Build $S$ deciding $A_{TM}$ : on input $\langle M, w\rangle$ , construct

\begin{aligned} M_1 \text{ on input } x:\quad & \text{run } M \text{ on } w;\ \text{reject if } M \text{ rejects},\\ & \text{otherwise run } x \text{ on } T \text{ and accept if } T \text{ accepts}. \end{aligned}

Tracing the language of $M_1$ against the property $P$ :

\begin{aligned} M \text{ accepts } w \;&\Longleftrightarrow\; L(M_1) = L(T) \;\Longleftrightarrow\; \langle M_1\rangle \in P,\\ M \text{ does not accept } w \;&\Longleftrightarrow\; L(M_1) = \varnothing = L(T_\varnothing) \;\Longleftrightarrow\; \langle M_1\rangle \notin P. \end{aligned}

So $S$ runs $R$ on $\langle M_1\rangle$ : accept if $R$ accepts (then $M$ accepts $w$ ), reject if $R$ rejects. This decides $A_{TM}$ — a contradiction. Hence no such $R$ exists and $P$ is undecidable. $\blacksquare$

A second reduction (equality problem, reducing to emptiness). The equality problem $EQ_{TM} = \{\langle M_1, M_2\rangle \mid L(M_1) = L(M_2)\}$ is undecidable. Assume $R$ decides it and build $S$ deciding emptiness: on input $\langle M\rangle$ , let $M_\varnothing$ be a machine with $L(M_\varnothing) = \varnothing$ , and observe

\begin{aligned} L(M) = L(M_\varnothing) \;&\Longleftrightarrow\; L(M) = \varnothing. \end{aligned}

Run $R$ on $\langle M, M_\varnothing\rangle$ and copy its answer; this decides emptiness, which is impossible. Hence $EQ_{TM}$ is undecidable.

Summary

$A_{TM}$ is undecidable; the self-reference machine $D$ gives the contradiction $D$ accepts $\langle D\rangle \iff D$ does not.
A language is decidable iff it and its complement are both recognizable; running two recognizers in parallel builds the decider.
$\overline{A_{TM}}$ is not recognizable — the first language we place strictly outside the recognizable class.
A reduction $A \le B$ means a solver for $B$ gives a solver for $A$ ; so reducing the undecidable $A_{TM}$ to $B$ proves $B$ undecidable.
The halting, emptiness, and regularity problems are all undecidable by reductions from $A_{TM}$ that build an intermediate machine $M_1$ whose language encodes whether $M$ accepts $w$ .
Rice's theorem: every non-trivial property of a Turing machine's language is undecidable — none of the earlier results are accidents; they are all special cases.