1. 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

2. 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

3. 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) … 2 … 8 … 6 … 1 … 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

4. 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

5. 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) … 2 … 8 … 6 … 1 … 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

6. 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}

7. 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

8. 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

9. 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

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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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