Give a Turing Machine which accepts strings that have the sa

Give a Turing Machine which accepts strings that have the same number of a\'s as b\'s. You can assume the TM SHIFT module has been run to put a # at the beginning of the string.

Solution

Your intuition is correct: your LL is unrecognizable. Here\'s a hint on how to proceed. If we define ATM={(M,w)M accepts w}ATM={(M,w)M accepts w}, then it\'s known that its complement, ATM¯¯¯¯¯¯¯¯¯¯ATM¯, is unreconizable so if we can construct a mapping reduction ATM¯¯¯¯¯¯¯¯¯¯MLATM¯ML, then we\'ll have shown that LL is unrecognizable. Define the mapping f(M,w)Nf(M,w)N where the TM NN depends on MM and ww and is defined by N(x) = run M on w for |x| moves if M hasn\'t accepted yet, accept The key idea here is that if MM accepts ww, then it will do so in kk moves, for some number kk, so if MM accepts ww, then NN will accept all and only those strings of length less than kk, and so the language of NN will be L(N)=k1L(N)=k1. On the other hand, if MM doesn\'t accept ww, then NN will accept all strings.