Intro to Theory of Computation LECTURE 16 Last time Designing deciders. Countable/uncountable sets. Diagonalization Today Undecidable/unrecognizable languages ATM is undecidable CS 464 Homework 6 due Homework 7 out Sofya Raskhodnikova 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

How to compare sizes of infinite sets? Two sets are the same size if there is a bijection between them. A set is countable if it is finite or it has the same size as ℕ, the set of natural numbers 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Theorem. There is no bijection from the positive integers to the real interval (0,1) Proof: Suppose f is such a function: n 1 2 3 4 5 : f(n) 0.28347279… 2 0.88388384… 8 0.77635284… 6 0.11111111… 1 0.12345678… 5 : Construct 𝒃∈(𝟎,𝟏) that does not appear in the table. b=0. 𝒅 𝟏 𝒅 𝟐 𝒅 𝟑 …, 𝒘𝒉𝒆𝒓𝒆 𝒅 𝒊 ≠𝒅𝒊𝒈𝒊𝒕 𝒊 𝒐𝒇 𝒇 𝒊 . 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Diagonalization The process of constructing a counterexample by “contradicting the diagonal” is called DIAGONALIZATION 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper What if we try this argument on ℚ instead of ℝ? Proof: Suppose f is such a function: n 1 2 3 4 5 : f(n) 0.28347279… 2 0.88388384… 8 0.77635284… 6 0.11111111… 1 0.12345678… 5 : Construct 𝒃∈(𝟎,𝟏) that does not appear in the table. b=0. 𝒅 𝟏 𝒅 𝟐 𝒅 𝟑 …, 𝒘𝒉𝒆𝒓𝒆 𝒅 𝒊 ≠𝒅𝒊𝒈𝒊𝒕 𝒊 𝒐𝒇 𝒇 𝒊 . 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

I-clicker question (frequency: AC) What if we try Cantor’s diagonalization argument on ℚ instead of ℝ? It works. It fails because there are some rational numbers that cannot be represented in decimal point notation. It fails because the 𝑖-th number might have no digit in the 𝑖-th position after the decimal point. It fails because the constructed number is not rational. None of the above. {madam, dad, but, tub, book}

Sofya Raskhodnikova; based on slides by Nick Hopper Let L be any set and P(L) be the power set of L Theorem: There is no bijection from L to P(L) Assume, for a contradiction, that there is bijection f : L  P(L) Proof: We construct a set S that cannot be the output, f(y), for any y L. Let S = { x  L | x  f(x) } Make less general If S = f(y) then y  S if and only if y  S 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper How is it diagonalization? x y1∈f(x)? y2∈f(x)? y3∈ f(x)? y4 ∈ f(x)? … y1 Y N y2 y3 y4   Y Y N Y (yi ∈ S) = Y iff (yi ∈ f(yi)) = N 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

EXAMPLE {0,1} Let L = {0,1,2}. Then P(L) = {Ø, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2}} Let f(0) = {1}, f(1) = Ø, f(2) = {0,2}. Then: x 0 ∈ f(x)? 1 ∈ f(x)? 2 ∈ f(x)? N Y 1 2 S = {0,1} L16.10

Sofya Raskhodnikova; based on slides by Nick Hopper For all sets L, P(L) has more elements than L 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Not all languages over {0,1} are decidable TM Deciders Languages over {0,1} Sets of strings of 0s and 1s Strings of 0s and 1s L P(L) 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Not all languages over {0,1} are recognizable Turing Machines Languages over {0,1} Sets of strings of 0s and 1s Strings of 0s and 1s L P(L) 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper A specific undecidable language ATM = { 𝑴, 𝒘 ∣ M is a TM, w is a string, and M accepts w } 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

𝑯 𝑴,𝒘 = 𝐚𝐜𝐜𝐞𝐩𝐭 𝐢𝐟 𝑴 𝐚𝐜𝐜𝐞𝐩𝐭𝐬 𝒘 𝐫𝐞𝐣𝐞𝐜𝐭 𝐢𝐟 𝑴 𝐝𝐨𝐞𝐬 𝐧 ′ 𝐭 𝐚𝐜𝐜𝐞𝐩𝐭 𝒘 Theorem. ATM is undecidable. Proof: For contradiction, suppose a TM H decides ATM. 𝑯 𝑴,𝒘 = 𝐚𝐜𝐜𝐞𝐩𝐭 𝐢𝐟 𝑴 𝐚𝐜𝐜𝐞𝐩𝐭𝐬 𝒘 𝐫𝐞𝐣𝐞𝐜𝐭 𝐢𝐟 𝑴 𝐝𝐨𝐞𝐬 𝐧 ′ 𝐭 𝐚𝐜𝐜𝐞𝐩𝐭 𝒘 Idea: Use 𝑯 to check what TM M does on its own description (and do the opposite). D is a decider. What does it do on 𝐷 ? TM D = `` On input 𝑴 where M is a TM: Run 𝐻 on input <M, 𝑴 >. Accept if it rejects. O.w. reject.’’ 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Is it diagonalization again? Does M accept 𝑴 ? TMs 〈𝑴 𝟏 〉 〈𝑴 𝟐 〉 〈𝑴 𝟑 〉 〈𝑴 𝟒 〉 … 𝑴 𝟏 Y N 𝑴 𝟐 𝑴 𝟑 𝑴 𝟒   Y Y N Y D accepts 𝑴 𝒊 iff entry (𝒊,𝒋) is N . 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Is it diagonalization again? Does M accept 𝑴 ? TMs 〈𝑴 𝟏 〉 〈𝑴 𝟐 〉 〈𝑴 𝟑 〉 〈𝑴 𝟒 〉 … 〈𝑫〉 𝑴 𝟏 Y N 𝑴 𝟐 𝑴 𝟑 𝑴 𝟒   D Y Y N Y ? 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper

Sofya Raskhodnikova; based on slides by Nick Hopper Classes of languages recognizable ATM decidable ATM CFL {0 𝑛 1 𝑛 0 𝑛 ∣𝑛≥0} {0 𝑛 1 𝑛 ∣𝑛≥0} regular 1* 2/23/2019 Sofya Raskhodnikova; based on slides by Nick Hopper