Language IN = is - - PowerPoint PPT Presentation

Language IN𝐺𝐽𝑂𝐽𝑈𝐹

• 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 = 𝑥 𝑀 ℳ 𝑥

is infinite}

• We will show this is not computable by using

𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 to compute 𝐼𝐵𝑀𝑈

1

Language of 𝑁𝑥𝑦

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 always returns 1 – 𝑀 𝑁𝑥𝑦 = Σ∗, which is infinite

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 1 and never returns – 𝑀 𝑁𝑥𝑦 = ∅, which is finite

2

Build this machine: 𝑁𝑥𝑦: 1) run ℳ(𝑥)(𝑦) 2) return 1 ℳ 𝑥 𝑦 Halts? It is infinite ℳ 𝑥 𝑦 Doesn’t Halt? It isn’t infinite

Using 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐼𝐵𝑀𝑈

3

Is L ℳ 𝑥 infinite? 𝑵∞ 𝑥 𝑀 ℳ 𝑥 is/isn’t infinite 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ 𝑥 𝑦 does/doesn’t halt Assume we have 𝑁∞ which computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹: We could then build 𝑁𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈like this: 𝑥 𝑦

Build this machine: 𝑁𝑥𝑦: 1) run ℳ(𝑥)(𝑦) 2) return 1

Is L 𝑁𝑥𝑦 infinite? 𝑵∞

Reduction

4

𝐼𝐵𝑀𝑈

Problem know is impossible

𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹

Problem we think is impossible 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 solver exists

𝐵 𝐶

Build (but don’t run) 𝑁𝑥𝑦

Assume Reduction Does ℳ(𝑥) halt on input 𝑦? Is L ℳ 𝑥 infinite? 𝑁𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 Give 𝑁𝑥𝑦 to the 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 solver, then answer the same

Using 𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹

5

Is L ℳ 𝑥 finite? 𝑵𝑮𝑱𝑶𝑱𝑼𝑭 𝑥 𝑀 ℳ 𝑥 is/isn’t finite 𝑵∞ Is L ℳ 𝑥 infinite? 𝑀 ℳ 𝑥 is/isn’t infinite Assume we have 𝑁𝐺𝐽𝑂𝐽𝑈𝐹 which computes 𝐺𝐽𝑂𝐽𝑈𝐹: We could then build 𝑁∞ which computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 like this: 𝑥 Is L ℳ 𝑥 finite? 𝑵𝑮𝑱𝑶𝑱𝑼𝑭

Language 𝑂𝑝𝑜𝑆𝑓𝑕

• 𝑂𝑝𝑜𝑆𝑓𝑕 = 𝑥 𝑀 ℳ 𝑥

is non − regular}

• We will show this is not computable by using

𝑂𝑝𝑜𝑆𝑓𝑕 to compute 𝐼𝐵𝑀𝑈

6

Language of 𝑁𝑥𝑦

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 returns 1 if MAJ 𝑧 = 1 – 𝑀 𝑁𝑥𝑦 = 𝑁𝐵𝐾(𝑧), which is non-regular

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 1 and never returns 1 – 𝑀 𝑁𝑥𝑦 = ∅, which isn’t non-regular

7

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return MAJ(𝑧) ℳ 𝑥 𝑦 Halts? It is non-regular ℳ 𝑥 𝑦 Doesn’t Halt? It isn’t non-regular

Using 𝑂𝑝𝑜𝑆𝑓𝑕 to build 𝐼𝐵𝑀𝑈

8

Is L ℳ 𝑥 non- regular? 𝑵𝑶𝑺 𝑥 𝑀 ℳ 𝑥 is/isn’t non-regular 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ(𝑥)(𝑦) does/doesn’t halt Assume we have 𝑁𝑂𝑆 which computes 𝑂𝑝𝑜𝑆𝑓𝑕: We could then build 𝑁𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈like this: 𝑥 𝑦

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return MAJ(𝑧)

Is L ℳ 𝑥 non- regular? 𝑵𝑶𝑺

Reduction

9

𝐼𝐵𝑀𝑈 𝑂𝑝𝑜𝑆𝑓𝑕

𝐵 𝐶

Describe that machine 𝑁𝑥𝑦

Assume Reduction 𝑁𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 Give 𝑁𝑥𝑦 to the 𝑂𝑝𝑜𝑆𝑓𝑕 solver, then answer the same 𝑂𝑝𝑜𝑆𝑓𝑕 solver exists Does ℳ(𝑥) halt on input 𝑦? Is L ℳ 𝑥 non- regular?

Using 𝑆𝑓𝑕 to build𝑂𝑝𝑜𝑆𝑓𝑕

10

Is L ℳ 𝑥 regular? 𝑵𝑆𝑓𝑕 𝑥 𝑀 ℳ 𝑥 is/isn’t Regular 𝑵𝑶𝑺 Is L ℳ 𝑥 non-regular? 𝑀 ℳ 𝑥 is/isn’t non- regular Assume we have 𝑁𝑆𝑓𝑕 which computes 𝑂𝑝𝑜𝑆𝑓𝑕: We could then build 𝑁𝑂𝑆 which computes𝑂𝑝𝑜𝑆𝑓𝑕 like this: 𝑥 Is L ℳ 𝑥 regular? 𝑵𝑆𝑓𝑕

Language 𝐵𝑑𝑑𝑓𝑞𝑢101

• 𝐵𝑑𝑑𝑓𝑞𝑢101 = 𝑥 ℳ 𝑥

101 = 1}

• We will show this is not computable by

using𝐵𝑑𝑑𝑓𝑞𝑢101 to compute 𝐼𝐵𝑀𝑈

11

Building 𝑁𝑥𝑦

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 returns 1 if 𝑧 == 101 – 𝑀 𝑁𝑥𝑦 = {101}, so it does accept 101

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 1 and never returns 1 – 𝑀 𝑁𝑥𝑦 = ∅, so it doesn’t accept 101

12

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑧 == 101 ℳ 𝑥 𝑦 Halts? It does accept 101 ℳ 𝑥 𝑦 Doesn’t Halt? It doesn’t accept101

Using 𝐵𝑑𝑑𝑓𝑞𝑢101 to build 𝐼𝐵𝑀𝑈

13

Does ℳ(𝑥) accept 101? 𝑵𝑩𝟐𝟏𝟐 𝑥 ℳ(𝑥) does/doesn’t accept 101 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ(𝑥)(𝑦) does/doesn’t halt Assume we have 𝑁

𝐵101

which computes𝐵𝑑𝑑𝑓𝑞𝑢101: We could then build 𝑁𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈like this: 𝑥 𝑦

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑧 == 101

Does ℳ(𝑥) accept 101? 𝑵𝑩𝟐𝟏𝟐

𝐵𝑑𝑑𝑓𝑞𝑢101 Reduction

14

𝐼𝐵𝑀𝑈 𝐵𝑑𝑑𝑓𝑞𝑢101

𝐵 𝐶

Describe that machine 𝑁𝑥𝑦

Assume Reduction 𝑁𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 Give 𝑁𝑥𝑦 to the 𝐵𝑑𝑑𝑓𝑞𝑢101 solver, then answer the same 𝐵𝑑𝑑𝑓𝑞𝑢101 solver exists Does ℳ(𝑥) halt on input 𝑦? Is 101 ∈ L ℳ 𝑥 ?

Using𝑆𝑓𝑘𝑓𝑑𝑢101 to build 𝐵𝑑𝑑𝑓𝑞𝑢101

15

Is 101 ∉ L ℳ 𝑥 ? 𝑵𝑆101 𝑥 101 isn’t/is in 𝑀 ℳ 𝑥 𝑵𝐵101 Is 101 ∈ L ℳ 𝑥 ? 101 is/isn’t in 𝑀 ℳ 𝑥 Assume we have 𝑁𝑆101 which computes 𝑆𝑓𝑘𝑓𝑑𝑢101: We could then build 𝑁

𝐵101

which computes 𝐵𝑑𝑑𝑓𝑞𝑢101 like this: 𝑥 Is 101 ∉ L ℳ 𝑥 ? 𝑵𝑆101

Computability and non-computability closed under complement

• If 𝑀 is computable, then 𝑀𝑑 is computable

16

Is 𝑥 ∈ L? 𝑵𝑴 𝑥 𝑥 is/isn’t in 𝑀 Assume we have 𝑁𝑀 which computes 𝑀: 𝑵𝑀𝑑 Is 𝑥 ∉ L? We could then build 𝑁𝑀𝑑 which computes 𝑀𝑑 like this: 𝑥 Is 𝑥 ∈ L? 𝑵𝑴 𝑥 isn’t/is in 𝑀

Complement Reduction

17

𝑀𝑑 𝑀

𝐵 𝐶

Don’t change 𝑥

If Reduction 𝑀𝑑 solver exists Give 𝑥 to the 𝑀 solver, then answer the

• pposite

𝑀 solver exists 𝑥 ∉ 𝑀? 𝑥 ∈ 𝑀? 𝑀 solver doesn’t exist 𝑀𝑑 solver doesn’t exist

Sematic Property

• Turing machines 𝑁, 𝑁′ are Functionally Equivalent if

∀𝑦 ∈ Σ∗, 𝑁 𝑦 == 𝑁 𝑦′

– i.e. they compute the same function/language

• A Semantic Property of a Turing machine is one that

depends only on the input/output behavior of the machine

– Formally, if 𝑄 is semantic, then for machine 𝑁, 𝑁′ that are functionally equivalent, 𝑄 𝑁 == 𝑄 𝑁′ – If 𝑁, 𝑁′ have the same input/output behavior, and 𝑄 is a semantic property, then either bother 𝑁 and 𝑁′ have property 𝑄, or neither of them do.

18

Examples

• These properties are Semantic:

– Is the language of this machine finite? – Is the language of this machine Regular? – Does this machine reject 101? – Does this machine return 1001 for input 001? – Does this machine only ever return odd numbers? – Is the language of this machine computable?

• These properties are not Semantic:

– Does this machine ever overwrite cell 204 of its tape? – Does this machine use more than 3102 cells of its tape on input 101? – Does this machine take at least 2020 transitions for input 𝜁? – Does this machine ever overwrite the 𝛼 symbol?

19

Rice’s Theorem

• For any Semantic property 𝑄 of Turing Machines,

either:

– Every Turing machine has property 𝑄 – No Turing machines have property 𝑄 – 𝑄 is uncomputable

• In other words:

– If 𝑄 is semantic, and computable, then one of these two machines computes it:

20

𝑄 is “trivial” Return 1 Return 0

Proof of Rice’s Theorem

• Let 𝑄 be a semantic property of a Turing machine
• Assume 𝑁∅ (a machine whose language is ∅) doesn’t have property 𝑄

(otherwise substitute ¬𝑄, then answer opposite)

• Let 𝑁𝑄 be a machine that does have property 𝑄
• Idea:

– If ℳ(𝑥)(𝑦) halts, 𝑀 𝑁𝑥𝑦 = 𝑀 𝑁ℎ𝑏𝑡 𝑄 – If ℳ(𝑥)(𝑦) doesn’t halt, 𝑀 𝑁𝑥𝑦 = L 𝑁∅ = ∅ – 𝑀(𝑁𝑥𝑦) has property 𝑄 if and only if ℳ 𝑥 𝑦 halts

21

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑁ℎ𝑏𝑡 𝑄(𝑧) ℳ 𝑥 𝑦 Halts? It has property 𝑄 ℳ 𝑥 𝑦 Doesn’t Halt? It doesn’t have property 𝑄

Using 𝑄 to build 𝐼𝐵𝑀𝑈

22

Does ℳ(𝑥) have property 𝑄? 𝑵𝑸 𝑥 ℳ(𝑥) does/doesn’t have property 𝑄 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ(𝑥) does/doesn’t have property 𝑄 Assume we have 𝑁𝑄 which computes 𝑄: We could then build 𝑁𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈like this: 𝑥 𝑦

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑁ℎ𝑏𝑡 𝑄(𝑧)

Does ℳ(𝑥) have property 𝑄? 𝑵𝑸

What if 𝑄 is “trivial”?

• If 𝑄 applies to all Turing machines:

– It applies to 𝑁∅, so we’ll consider ¬𝑄 which applies to no Turing machines

• If 𝑄 applies to no Turing machines:

– 𝑁𝐼𝑏𝑡 𝑄 can’t exist

23

Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑁𝐼𝑏𝑡 𝑄(𝑧)

Using Rice’s Theorem

• These properties are Semantic:

– Is the language of this machine finite? – Is the language of this machine Regular? – Does this machine reject 101? – Does this machine return 1001 for input 001? – Does this machine only ever return odd numbers? – Is the language of this machine computable?

• These properties are not Semantic:

– Does this machine ever overwrite cell 204 of its tape? – Does this machine use more than 3102 cells of its tape on input 101? – Does this machine take at least 2020 transitions for input 𝜁? – Does this machine ever overwrite the 𝛼 symbol on input 𝜁?

24

𝑇𝑢𝑓𝑞𝑡2020

• 𝑇𝑢𝑓𝑞𝑡2020 =

{𝑥|ℳ 𝑥 takes at least 2020 steps}

• Is 𝑇𝑢𝑓𝑞𝑡2020 computable?

25

𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓𝛼

• 𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓𝛼 =

w ℳ 𝑥 overwrites 𝛼 on input 𝜁}

• Is 𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓𝛼 computable?

26

CS3102 Theory of Computation

www.cs.virginia.edu/~njb2b/cstheory/s2020

Warm up: To measure the “cost” of computing something, what would units should we use?

Units

28

Computing Cost

• What do we actually care about with

computing cost?

29

Notions of function “difficulty”

• Can an algorithm for function 𝑔 be implemented using this

computing model?

– NAND-CIRC: answer is YES iff 𝑔 is finite – FSA: answer is YES if function doesn’t require “memory” – TM: Answer is NO for 𝐼𝐵𝑀𝑈𝑈𝑁, 𝐺𝐽𝑂𝐽𝑈𝐹, … (and many other things)

• How efficient is an algorithm for function 𝑔 implemented using

this computing model?

– NAND-CIRC: How many gates? – FSA: (we never talked about this) – TM: How many transitions are required (time)? How many tape cells are required (space)?

30

Larger inputs = More time

• Run time is not measured by a number, but a

function.

• Running time: 𝑈 𝑜 is a function mapping naturals to
• naturals. We say 𝐺: 0,1 ∗ → 0,1 ∗ is computable in

𝑈 𝑜 time if there exists a TM 𝑁 s.t. for every large 𝑜 and ever input 𝑦 ∈ 0,1 𝑜, 𝑁 halts after at most 𝑈 𝑜 steps and outputs 𝐺 𝑦 .

• 𝑈𝐽𝑁𝐹 𝑈 𝑜

represents the set of boolean functions computable within 𝑈 𝑜 time

31

Examples

• How long will 𝑌𝑃𝑆 take on a Turing Machine?

– We have an even state and an odd state – For each bit, move right, switch states if 1 – Halt when you get to end of input

• How long will 𝑁𝐵𝐾 take on a Turing Machine?

– Find a zero, cross it off – Go to beginning – Find a one, cross it off – Go to beginning – Halt when no more 0s or no more 1s

32

More time gives more functions

33

𝑈𝐽𝑁𝐹 𝑜 𝑈𝐽𝑁𝐹 10𝑜 𝑈𝐽𝑁𝐹 𝑜3 𝑈𝐽𝑁𝐹 2𝑜

Different computing Models may have Different Running Times

• So far: a TM uses a tape. Can only visit a

neighboring cell from the current one.

34

1960s

35

A tape was probably a reasonable memory model

Today

36

Can look up two locations without visiting all locations between. “Random” access (RAM)

RAM Machine

• We can go directly to a certain index in the tape

To transition: 1. Have a second tape to keep track of current location (increment each time we move right, decrement for left) 2. Have a third tape to record the target location 3. Move until the two tapes match

1. Maybe we need another tape to do this?

(details not important, but if you want them, see 7.2 in text) Important observation: Tape-machine takes more steps than a RAM machine (if a RAM-TM computes 𝑔 in 𝑈(𝑜) time, a tape TM can compute 𝑔 in 𝑈 𝑜

4

time, see theorem 12.5 for details)

37

Finding Running Times

• We will find running times for the following:

– Shortest Path in a graph – Longest Path in a graph – 3SAT – 2SAT

38

Graphs

39

10 2 6 11 9 5 8 3 7 3 1 8 12 9 A B C D E F G I H

Definition: 𝐻 = (𝑊, 𝐹)

𝑥 𝑓 = weight of edge 𝑓

𝑊 = {𝐵, 𝐶, 𝐷, 𝐸, 𝐹, 𝐺, 𝐻, 𝐼, 𝐽} 𝐹 = { 𝐵, 𝐶 , 𝐵, 𝐷 , 𝐶, 𝐷 , … }

Adjacency List Representation

40

10 2 6 11 9 5 8 3 7 3 1 8 12 9 A B C D E F G I H A B C D E F G H I B C A C E A B D F C E F B D G H C D G E F H I E G I G H

Tradeoffs Space: Time to list neighbors: Time to check edge (𝐵, 𝐶):

𝑊 + 𝐹 𝐸𝑓𝑕𝑠𝑓𝑓(𝐵) 𝐸𝑓𝑕𝑠𝑓𝑓(𝐵)

Adjacency Matrix Representation

41

10 2 6 11 9 5 8 3 7 3 1 8 12 9 A B C D E F G I H A B C D E F G H I

Tradeoffs Space: Time to list neighbors: Time to check edge (𝐵, 𝐶):

𝑊2 𝑊 𝑃(1) A B C D E F G H I 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Definition: Path

42

10 2 6 11 9 5 8 3 7 3 1 8 12 9 A B C D E F G I H

A sequence of nodes (𝑤1, 𝑤2, … , 𝑤𝑙) s.t. ∀1 ≤ 𝑗 ≤ 𝑙 − 1, 𝑤𝑗, 𝑤𝑗+1 ∈ 𝐹 Simple Path: A path in which each node appears at most once Cycle: A path of > 2 nodes in which 𝑤1 = 𝑤𝑙

Shortest Path

• Given an unweighted graph, start node 𝑡 and

an end node 𝑢, how long is shortest path from 𝑡 to 𝑢?

43

A B C D E F G I H Shortest path from A to E has length 2

Breadth First Search

44

Find a path from 𝑡 to 𝑢 Keep a queue 𝑅 of nodes ℎ𝑝𝑞𝑡 = 0 𝑅. 𝑓𝑜𝑟𝑣𝑓𝑣𝑓((𝑡, ℎ𝑝𝑞𝑡)) While 𝑅 is not empty and 𝑤 ! = 𝑢: 𝑤, ℎ𝑝𝑞𝑡 = 𝑅. 𝑒𝑓𝑟𝑣𝑓𝑣𝑓() for each “unvisited” 𝑣 ∈ 𝑊 s.t. 𝑤, 𝑣 ∈ 𝐹: 𝑅. 𝑓𝑜𝑟𝑣𝑓𝑣𝑓((𝑣, ℎ𝑝𝑞𝑡 + 1))

Running time: 𝑃 𝑊 + 𝐹

Longest Path

• Given a start node 𝑡 and an end node 𝑢, how

long is longest path from 𝑡 to 𝑢?

45

Longest Path

Enumerate all possible sequences of nodes check if it’s a path print the length of the longest one Running time: 𝑜!

46