The Timpson Lyceum · Computation Theory

Lecture 18 (Computation Theory) Undecidable Problems and Mapping Reducibility

Overview

This lecture turns the informal "reduction" we have been using into a precise tool: mapping reducibility. We first warm up by re-proving that the equality problem for Turing machines is undecidable using a general reduction, then define a computable function and use it to give an exact definition of $A \le_m B$. From this single definition we get a family of transfer theorems: a mapping reduction lets us push decidability, undecidability, recognizability, and unrecognizability between problems. We re-prove the halting problem undecidable as a clean mapping-reduction example, and finish by showing that the equality problem $EQ_{TM}$ is neither recognizable nor co-recognizable — an entirely new complexity class beyond anything seen so far.

Key Concepts

Reduction (high-level idea). We say $A$ reduces to $B$ when any solution to $B$ yields a solution to $A$. The intuition: $B$ is at least as hard as $A$. So if $A$ is known to be unsolvable, $B$ must be unsolvable too — because a solution to $B$ would hand us one for $A$. This is exactly why the relation is written with a $\le$ sign.

The standard undecidable problems (reused as the "hard core" we reduce from):

• Acceptance problem: $A_{TM} = \{\langle M, w\rangle : M \text{ accepts } w\}$. Undecidable.
• Halting problem: $HALT_{TM} = \{\langle M, w\rangle : M \text{ halts on } w\}$. Undecidable.
• Emptiness problem: $E_{TM} = \{\langle M\rangle : L(M) = \varnothing\}$. Undecidable.
• Equality problem: $EQ_{TM} = \{\langle M_1, M_2\rangle : L(M_1) = L(M_2)\}$.

Computable function. A function $f : \Sigma^* \to \Sigma^*$ is computable if some Turing machine $M$, on every input $w$, halts with exactly $f(w)$ written on its tape. Example: squaring is computable — input $6$, the machine halts with $36$ on the tape.

Mapping reducibility (the exact definition). A language $A$ is mapping reducible to a language $B$, written $A \le_m B$, if there is a computable function $f : \Sigma^* \to \Sigma^*$ such that for every string $w$, $w \in A \iff f(w) \in B.$ Such an $f$ is called a reduction from $A$ to $B$. It relabels instances of $A$ as instances of $B$: things in $A$ map into $B$, and things not in $A$ map outside $B$.

The transfer theorems. Once $A \le_m B$ via some $f$:

• Decidability transfers downward: if $A \le_m B$ and $B$ is decidable, then $A$ is decidable.
• Undecidability transfers upward: if $A \le_m B$ and $A$ is undecidable, then $B$ is undecidable.
• Recognizability transfers downward: if $A \le_m B$ and $B$ is Turing-recognizable, then $A$ is recognizable.
• Unrecognizability transfers upward: if $A \le_m B$ and $A$ is unrecognizable, then $B$ is unrecognizable.

Why the directions work. To decide $A$ given a decider $M$ for $B$ and a reduction $f$: build $N$ that on input $w$ runs $M$ on $f(w)$ and copies its answer. Since $w \in A \iff f(w) \in B$, $N$ accepts exactly the strings of $A$. The recognizer version is identical (replace "decider" with "recognizer"). The upward (undecidable / unrecognizable) versions are the contrapositives.

Complement symmetry. $A \le_m B$ is the same statement as $\overline{A} \le_m \overline{B}$ — the same computable $f$ works, because $w \in A \iff f(w) \in B$ is logically identical to $w \notin A \iff f(w) \notin B$. This lets us flip a reduction to talk about complements.

Co-recognizable. A language is co-recognizable if its complement is Turing-recognizable. Showing a language is neither recognizable nor co-recognizable places it strictly above $A_{TM}$ (which is recognizable but not decidable) in the hierarchy.

Worked Example

Goal: prove the halting problem is undecidable via a mapping reduction $A_{TM} \le_m HALT_{TM}$.

We construct a computable $f$ that sends $\langle M, w\rangle$ to $\langle M', w\rangle$, where $M'$ is a new machine that halts on $w$ exactly when $M$ accepts $w$:

Now check the defining equivalence of a mapping reduction:

Since $f$ is computable and satisfies $\langle M,w\rangle \in A_{TM} \iff f(\langle M,w\rangle) \in HALT_{TM}$, we have $A_{TM} \le_m HALT_{TM}$. Because $A_{TM}$ is undecidable, upward transfer gives that $HALT_{TM}$ is undecidable. $\blacksquare$

Second example (reduction map for $EQ_{TM}$). To show $E_{TM} \le_m EQ_{TM}$, define $f(\langle M\rangle) = \langle M, M_\varnothing\rangle$, where $M_\varnothing$ is a fixed machine that rejects every input (its start state is also a reject state, so $L(M_\varnothing)=\varnothing$). Then $L(M)=\varnothing \iff L(M)=L(M_\varnothing) \iff \langle M, M_\varnothing\rangle \in EQ_{TM}$. So $E_{TM} \le_m EQ_{TM}$, and since $E_{TM}$ is undecidable, $EQ_{TM}$ is undecidable.

Summary

• A mapping reduction $A \le_m B$ is a computable function $f$ with $w \in A \iff f(w) \in B$; it makes the loose "$A$ reduces to $B$" idea exact.
• $A \le_m B$ means $B$ is at least as hard as $A$, justifying the $\le$ notation.
• Decidability and recognizability transfer downward ($B$ has it $\Rightarrow$ $A$ has it); undecidability and unrecognizability transfer upward ($A$ lacks it $\Rightarrow$ $B$ lacks it).
• Each transfer proof is the same pattern: run the reduction $f$, feed $f(w)$ to the machine for $B$, and echo its answer — using $w\in A \iff f(w)\in B$.
• Because $A \le_m B$ is equivalent to $\overline{A} \le_m \overline{B}$, the same map handles complements, letting us reason about co-recognizability.
• $EQ_{TM}$ is neither recognizable nor co-recognizable: $A_{TM} \le_m \overline{EQ_{TM}}$ shows $EQ_{TM}$ is unrecognizable, and $\overline{A_{TM}} \le_m \overline{EQ_{TM}}$ shows its complement is too — placing it strictly above the recognizable languages.