Mapping Reducibility Sipser 5.3 (pages 206-210)

Computable functions Definition 5.17: A function f:Σ*→Σ* is a computable function if some Turing machine M, on every input w, halts with just f(w) on its tape. Example: The increment function inc++:{1}* → {1}* is Turing-computable

Incrementing inc++:{1}* → {1}* Infinite tape 1 1 1 1 ⨆ ⨆ ⨆ ⨆ Finite control

Transforming machines F = "On input <M>: Construct the machine M∞ = "On input x: Run M on x. If M accepts, accept. If M rejects, loop." Output <M∞>."

Mapping reducibility Definition 5.20: Language A is mapping reducible to language B, written A ≤m B, if there is a computable function f:Σ* → Σ*, where for every w, w ∈ A ⇔ f(w) ∈ B A B f f

Problem reduction Theorem 5.22: If A ≤m B and B is decidable, then A is decidable. A B f f

And… the contrapositive Theorem 5.22: If A ≤m B and B is decidable, then A is decidable. Corollary 5.23: If A ≤m B and A is undecidable, then B is undecidable.

A familiar mapping reduction… ATM = {<M,w> | M is a TM and M accepts w} ≤m HALTTM = {<M,w> | M is a TM & M halts on input w} ATM HALTTM f <M,w> <M∞,w> f <M,w> <M∞,w>

ATM ≤m HALTTM F = "On input <M>: M∞ = "On input x: Construct the machine M∞ = "On input x: Run M on x. If M accepts, accept. If M rejects, loop." Output <M∞>." ATM HALTTM f <M,w> <M∞,w> f <M,w> <M∞,w>

Similarly… Theorem 5.28: If A ≤m B and B is Turing-recognizable, then A is Turing-recognizable. Theorem 5.29: If A ≤m B and A is not Turing-recognizable, then B is not Turing-recognizable.

Solvable, half-solvable, hopeless ATM ADFA ETM co-Turing- recognizable Turing- recognizable Turing-decidable

EQTM = {<M1,M2>| L(M1)=L(M2)} is hopeless Theorem 5.30: EQTM is neither Turing-recognizable nor co-Turing-recognizable Proof: What if we show ATM ≤m EQTM? ATM EQTM f <M,w> <M1,M2> f EQTM ATM <M,w> <M1,M2>

ATM ≤m EQTM G = "On input <M,w>: Construction the following two machines: M1 = "On any input: Accept.” M2 = "On any input: Run M on w. If it accepts, accept." Output <M1,M2>.” ATM EQTM f <M,w> <M1,M2>

EQTM is not Turing-recognizable Theorem 5.30: EQTM is neither Turing-recognizable nor co-Turing-recognizable Proof: Show ATM ≤m EQTM ATM EQTM f <M,w> <M1,M2> f EQTM ATM <M,w> <M1,M2>

ATM ≤m EQTM G = "On input <M,w>: Construction the following two machines: M1 = "On any input: Reject.” M2 = "On any input: Run M on w. If it accepts, accept." Output <M1,M2>.” ATM EQTM f <M,w> <M1,M2> f EQTM ATM <M,w> <M1,M2>

Solvable, half-solvable, hopeless EQTM ATM ADFA ETM co-Turing- recognizable Turing- recognizable Turing-decidable