Costas Busch - RPI1 Decidability. Costas Busch - RPI2 Another famous undecidable problem: The halting problem.

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Presentation transcript:

Costas Busch - RPI1 Decidability

Costas Busch - RPI2 Another famous undecidable problem: The halting problem

Costas Busch - RPI3 The Halting Problem Input:Turing Machine String Question: Does halt on input ?

Costas Busch - RPI4 Theorem: The halting problem is undecidable Proof:Assume for contradiction that the halting problem is decidable (there are and for which we cannot decide whether halts on input )

Costas Busch - RPI5 Thus, there exists Turing Machine that solves the halting problem YEShalts on doesn’t halt on NO

Costas Busch - RPI6 Input: initial tape contents Encoding of String YES NO Construction of

Costas Busch - RPI7 Construct machine : If returns YES then loop forever If returns NO then halt

Costas Busch - RPI8 NO Loop forever YES

Costas Busch - RPI9 Construct machine : Input: If halts on input Then loop forever Else halt (machine )

Costas Busch - RPI10 copy

Costas Busch - RPI11 Run machine with input itself: Input: If halts on input Then loop forever Else halt (machine )

Costas Busch - RPI12 on input If halts then loops forever If doesn’t halt then it halts : NONSENSE !!!!!

Costas Busch - RPI13 Therefore, we have contradiction The halting problem is undecidable END OF PROOF

Costas Busch - RPI14 Another proof of the same theorem: If the halting problem was decidable then every recursively enumerable language would be recursive

Costas Busch - RPI15 Theorem: The halting problem is undecidable Proof:Assume for contradiction that the halting problem is decidable

Costas Busch - RPI16 Let be a recursively enumerable language Let be the Turing Machine that accepts We will prove that is also recursive: we will describe a Turing machine that accepts and halts on any input

Costas Busch - RPI17 halts on ? YES NO Run with input reject accept reject Turing Machine that accepts and halts on any input Halts on final state Halts on non-final state

Costas Busch - RPI18 Thereforeis recursive But there are recursively enumerable languages which are not recursive Contradiction!!!! Since is chosen arbitrarily, every recursively enumerable language is also recursive

Costas Busch - RPI19 Therefore, the halting problem is undecidable END OF PROOF