Question

Help me answer this question plz!

4. Let L = { (A) M is a Turing machine that accepts more than one string } a) Define the notions of Turing-recognisable language and undecidable language. b) Is L Turing-recognisable? Justify your answer with an informal argument. c) Justify with a formal proof your answer to b) d) Prove that L is undecidable. (Hint: use Rice's theorem.) e) Modify your answer to b) when instead of L you have the language Ln = {(M): M is a Turing machine that accepts more than n strings). where n is a fixed positive integer.

Answer 1

given data:

L = ((M):M is a Turing machine that accepts more than one string}

a) condition for decidability

considering the language of a Turing Machine M, denoted as L(M), is the set of all strings w on which M accepts.

property-1

A language L is decidable if there is a Turing machine M such that L(M) = L and M halts on every input. (which means it needs a prompt to accept the next input;in this case a delimiter like space or any other special symbol)

property-2

and also , A language L is recognizable if there is a Turing Machine M such that L(M) = L. whch means if M accepts the language L then it must satisfy the codition the all the input symbols supported by the machine gives the output symbols exactly similar to input symbols

Thus, if L is decidable then L is recursively enumerable.

condition for undecidability

Simply, A language L is undecidable if L is not decidable.

Thus, in this case there is no Turing machine M that halts on every input and L(M) = L. (This statement might seem funny but this is a standard definition seriously )

b) Condition for turing recognisability

L is turing recognisable as it satisfies the property-1(decidability) and property-2(recognizability) so we can say M recognises the given Language L ;

c) formal proff of recognisability

If A is Turing-decidable,

then it is also Turing-recognizable.

Let A ,A' be the symbols in the given language {$A,A'\subset L$ }For all x$\varepsilon A$ and x'$\varepsilon A$' holds that M,M'(turing machines dealing with separate strings derived from the same language L) halts on input x. Combining machines A and A' to run parallel gives a total Turing machine for recognizing A.

So, we can say L isa recognisable over M

proof for L being Undecidable

This can be proven by reducing the halting problem to L: with an assumption that there are non zero halts for an input 's'

1. For the halting problem instance (L,s) where s is a symbol belonging to L, create a new machine M for the L problem.
2. On input n, M simulates (L,s) for length(n) steps.
3. If the simulation halted within that number of steps, then M halts. Otherwise, M will go into infinite loop.

This reduction is valid because:

• If (L,s) does halt eventually in k steps, then M will halt for all inputs of length k or greater, thus M is in L.
• Otherwise (L,s) does not halt, then M will not halt for any input string no matter how long it is, thus M is not in L.

Finally, the halting problem is undecidable, therefore L is undecidable.

e)Considering this version of turing machine M where s

L,-{(A1): M is a Turing machine that accepts more than n strings}

since n is a fixed value  the simulations just iterate through all possible input strings of length N. the simulations will iterate through more strings, the machine doesnot acquire more knowledge, because only the first N symbols will be considered rest of the inputs will iterate through pre transited symbols. So the simulation can't go any further than N places(cells) as the Turing machines only move one cell per step.

resulting L being a turing recognisable.