|
I have a book that proves the halting problem with this simple statement: $$ A_\text{TM} \le_m \text{HALTING} \le_m \text{HALTING}^\varepsilon $$ It states that halting problem reduces to the language consisting of $\langle M, \omega \rangle$ for which a Turing machine $M$ accepts $\omega$ is undecidable. What does this mean? What does the notation $\le_m$ indicate? |
|||||||||||
|
|
The symbol $\leq_m$ means many-one reducible, contrast to reductions such as turing reducible (denoted by $\leq_T$) and one-one reducible (denoted $\leq_1$). A many-one reduction from $L$ to $L′$ (given an alphabet $\Sigma$) is a (computable) function $f:\Sigma^* \to \Sigma^*$ such that for every $w \in \Sigma^*$, we have that $w \in L$ if and only if $f(w) \in L′$. The "many-one" means that we allow $f(w) = f(w′)$ (it does not have to be injective). In your setting, it basically means that you can transform an input to $A_{TM}$ to $HALTING$ such that $A_{TM}$ accepts an input $w$ if and only if $HALTING$ accepts the transformed input $f(w)$. |
|||
|
|
