Have a decider decides {|M is TM and |L(M)|=n}, build a decider decides n-1

Question

Is this possible? Suppose that we have a decider decides {|M is a TM and |L(M)|=n} Want to build a decider decides {|M is a TM and |L(M)|=n-1} If possible, how?

I'm voting to close this question as off-topic because it is about pure theoretical CS, which is a better fit at cs.stackexchange.com. — templatetypedef, Nov 23 '16 at 19:52

score 0 · Accepted Answer · answered Oct 19 '17 at 13:08

To see that we can decide |L(M)| = n - 1, imagine constructing a TM N for a language L(N) = L(M) U {a}, where a is a symbol not in the alphabet of the language L(M). N will accept exactly the strings that M accepts, plus this one other string that M could not possible accept. Therefore, N accepts n strings if and only if M accepts n - 1 strings. To decide whether |L(M)| = n - 1, then, it suffices to run our decider on N and see whether |L(N)| = n.

Have a decider decides {|M is TM and |L(M)|=n}, build a decider decides n-1

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