What is mapping reducible?

What is mapping reducible?

A mapping reduction A ≤m B (or A ≤P B) is an algorithm (respectively, polytime algorithm) that can transform any instance of decision problem A into an instance of decision problem B, in such a way that the answer correspondence property holds.

Does every language mapping reduces to its complement?

(j) Every language reduces to its complement. False.

Is ATM mapping reducible to ATM?

There is a mapping reduction f : ATM → ETM. Since f is a mapping reduction ATM → EQTM, it is also a mapping reduction ATM → EQTM. Hence, if EQTM were Turing-recognizable, the existence of f would prove that ATM would be Turing-recognizable, implying that ATM is decidable, which was proved to be wrong.

Is the complement of EQTM recognizable?

Note that for Turing reductions you can add complements arbitrarily. For mapping reduction you can only add complement to both sides, not just one at a time – otherwise it may no longer be true. Example 13. EQTM = {〈M1,M2〉 : M1,M2 are TMs and L(M1) = L(M2)} is neither recognizable nor co-recognizable.

Why is ATM not decidable?

D rejects (D), but then H accepted (D,(D)) and hence D accepted (D), contradiction! So D cannot exist, so H cannot exist either (D was built from H). This means that ATM is undecidable.

Is halt TM recognizable?

There is no way to decide whether a TM will accept or eventually terminate. and HALT are recognizable. We can always run a TM on a string w and accept if that TM accepts or halts.

What is Eqdfa?

EQDFA = {〈B, C〉 | B and C are DFAs and L(B) = L(C)} EQDFA is a decidable language. Proof Sketch: Construct a DFA D for (B ∩ C) ∪ (B ∩ C) and test if 〈D, w〉 is in EDFA. CS 4313/5353 Theory of Computation.

Is ATM mapping reducible to ETM?

But, ¬ETM is Turing recognizable and ¬ATM is not, which contradicts Theorem 5.22. This is a contradiction. Therefore, ATM is not mapping reducible to ETM. 5.7 A ≤m ¬A implies ¬A ≤m A.

Why is ATM not Decidable?

Is ATM complement decidable?

Corollary 4.23: ATM is Turing-recognizable but not decidable, so its complement ATM is NOT Turing-recognizable.

Will halt on every input?

In computability theory, a machine that always halts, also called a decider or a total Turing machine, is a Turing machine that eventually halts for every input. Because it always halts, such a machine is able to decide whether a given string is a member of a formal language.