Prove that the following problem is undecidable using a reduction: Given a Turing machine $S$, does $S$ accept a word $w$ iff it accepts its reverse $w^R$?
There is a solution here, which I don't understand. We reduce from the halting problem. Given a Turing machine $M$ and a word $w$, we construct a Turing machine $M_1$:
M1(x) = simulate M on w if M accepts w if x = 01 accept
The machine $M_1$ accepts only 01 and all other strings will be rejected. Then
MA(<M>, w) = Construct M1 as above if ML accepts <M1> reject else if ML rejects <M1> accept
so the Turing machine $S$ will accept all strings except $w=01$.
Why is the string 01 the only string which is accepted by $M_1$? Why it cannot be 10 instead of 01, for example? It doesn't make sense to me.
If $w$ were 01 and $L$ would also contain 10 and 01, then it should also be accepted, no?
Would anyone please explain me the whole process and especially the Turing machine $M_1$? Why does this work?