CS3102 Theory of Computation - - PowerPoint PPT Presentation

CS3102 Theory of Computation

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

Warm up: How might we try to show that this language is not computable: 𝐽𝑜𝑔𝑗𝑜𝑗𝑢𝑓 = 𝑥 𝑀 ℳ 𝑥 is infinite}

How to show things aren’t computable

• 1. Ask “can I have an always-halting Turing

machine 𝑁𝑞 for language/function/problem 𝑄?”

• 2. Show that, if 𝑁𝑞 exists, it can be used to

make an impossible machine 𝑁𝑗𝑛𝑞

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How do we know a machine is impossible? Option 1: It contradicts itself (e.g. 𝑁𝑇𝑆) Option 2: Someone has done this before (e.g. 𝑁𝑏𝑑𝑑)

Proving Other Problems are Uncomputable

• Reduction

– Convert some problem into a known uncomputable one (using only computable steps) – Show how you can use a solution to one problem to help you to solve another

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Non-Computable Problems

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MacGyver’s Reduction

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Opening a door Problem know is impossible Lighting a fire Problem we think is impossible Alcohol, wood, matches Solution for 𝑪

𝐵 𝐶

Keg cannon battering ram Solution for 𝑩

Aim duct at door, insert keg

If

Put fire under the Keg

Reduction

Example: 𝐺𝐽𝑂𝐽𝑈𝐹

• 𝐺𝐽𝑂𝐽𝑈𝐹 𝑥 = 1 if 𝑀 ℳ 𝑥

is finite 0 if 𝑀 ℳ 𝑥 is infinite

• To show 𝐺𝐽𝑂𝐽𝑈𝐹 is uncomputable

– Show how to use a TM for 𝐺𝐽𝑂𝐽𝑈𝐹 to solve 𝐼𝐵𝑀𝑈

• 𝐺𝐽𝑂𝐽𝑈𝐹 ≥ 𝐼𝐵𝑀𝑈
• 𝐼𝐵𝑀𝑈 reduces to 𝐺𝐽𝑂𝐽𝑈𝐹

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𝐺𝐽𝑂𝐽𝑈𝐹 Reduction

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𝐼𝐵𝑀𝑈

Problem know is impossible

𝐺𝐽𝑂𝐽𝑈𝐹

Problem we think is impossible 𝑁

𝑔𝑗𝑜𝑗𝑢𝑓 computes 𝐺𝐽𝑂𝐽𝑈𝐹

𝐵 𝐶

Build (but don’t run) 𝑁𝑥𝑦

Assume Reduction Does ℳ(𝑥) halt on input 𝑦? Is L ℳ 𝑥 finite? 𝑁𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 Give 𝑁𝑥𝑦 as input to 𝑁

𝑔𝑗𝑜𝑗𝑢𝑓 answer opposite

What’s the Language of 𝑁𝑥𝑦?

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 always returns 1 – 𝑀 𝑁𝑥𝑦 = Σ∗ (all strings) – 𝑀 𝑁𝑥𝑦 is infinite

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 1 and never returns – 𝑀 𝑁𝑥𝑦 = ∅ – 𝑀 𝑁𝑥𝑦 = 0

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

Using 𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐼𝐵𝑀𝑈

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Is L ℳ 𝑥 finite? 𝑵𝑮𝑱𝑶𝑱𝑼𝑭 𝑥 𝑀 ℳ 𝑥 is/isn’t finite 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ 𝑥 1 ! = 1 Assume we have 𝑁𝐺𝐽𝑂𝐽𝑈𝐹 which computes 𝐺𝐽𝑂𝐽𝑈𝐹: 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 finite? 𝑵𝑮𝑱𝑶𝑱𝑼𝑭

Showing 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 is not computable

• Use _____ to build _____.

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𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 Reduction

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𝐺𝐽𝑂𝐽𝑈𝐹 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹

𝑁∞ computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹

𝐵 𝐶

Assume Reduction Is L ℳ 𝑥 finite? Is L ℳ 𝑥 infinite? 𝑁𝐺𝐽𝑂𝐽𝑈𝐹 computes 𝐺𝐽𝑂𝐽𝑈𝐹

𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 Reduction

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𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 𝐺𝐽𝑂𝐽𝑈𝐹

𝑁𝐺𝐽𝑂𝐽𝑈𝐹 computes 𝐺𝐽𝑂𝐽𝑈𝐹

𝐵 𝐶

Assume Reduction Is L ℳ 𝑥 infinite? Is L ℳ 𝑥 finite? 𝑁∞ computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹

Using 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐺𝐽𝑂𝐽𝑈𝐹

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

Language 𝑂𝑝𝑜𝑆𝑓𝑕

• 𝑂𝑝𝑜𝑆𝑓𝑕 = 𝑥 𝑀 ℳ 𝑥

is not regular}

• We will show this is not computable by using

𝑂𝑝𝑜𝑆𝑓𝑕 to compute 𝐼𝐵𝑀𝑈

• Idea: given an input for 𝐼𝐵𝑀𝑈, 𝑥 and 𝑦, build

a machine 𝑁𝑥𝑦 such that 𝑀(𝑁𝑥𝑦) is regular if and only if ℳ(𝑥)(𝑦) runs forever.

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Building 𝑁𝑥𝑦

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 returns 1 if 𝑌𝑃𝑆 𝑧 = 1 – 𝑀 𝑁𝑥𝑦 = 𝑌𝑃𝑆(𝑧), which is not regular

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 2 and never returns 1 – 𝑀 𝑁𝑥𝑦 = ∅, which is regular

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

Using 𝑂𝑝𝑜𝑆𝑓𝑕 to build 𝐼𝐵𝑀𝑈

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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 𝑌𝑃𝑆(𝑧)

Is L 𝑁𝑥𝑦 non- regular? 𝑵𝑶𝑺

Language 𝑆𝑓𝑘𝑓𝑑𝑢𝑡101

• 𝑆𝑓𝑘𝑓𝑑𝑢𝑡101 = 𝑥 ℳ 𝑥

101 = 0}

• We will show this is not computable by using

𝑆𝑓𝑘𝑓𝑑𝑢𝑡101 to compute 𝐼𝐵𝑀𝑈

• Idea: given an input for 𝐼𝐵𝑀𝑈, 𝑥 and 𝑦, build

a machine 𝑁𝑥𝑦 such that 101 ∈ 𝑀(𝑛𝑥𝑦) if and only if ℳ(𝑥)(𝑦) runs forever.

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Building 𝑁𝑥𝑦

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 returns 1 if 𝑧 == 101 – 𝑀 𝑁𝑥𝑦 = {101}, so it does not reject 101

• If ℳ(𝑥)(𝑦) doesn’t halt:

– 𝑁𝑥𝑦 gets “stuck” in step 2 and never returns 1 – 𝑀 𝑁𝑥𝑦 = ∅, so it does reject 101

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

Using 𝑆𝑓𝑘𝑓𝑑𝑢𝑡101 to build 𝐼𝐵𝑀𝑈

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Does ℳ(𝑥) reject 101? 𝑵𝑺𝟐𝟏𝟐 𝑥 ℳ(𝑥) does/doesn’t reject 101 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? ℳ(𝑥)(𝑦) does/doesn’t halt Assume we have 𝑁𝑆101 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 𝑧 == 101

Does ℳ(𝑥) reject 101? 𝑵𝑺𝟐𝟏𝟐

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.

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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?

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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:

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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 ∅) has property 𝑄 (otherwise

substitute ¬𝑄, then answer opposite)

• Let 𝑁𝐺𝐵𝑀𝑇𝐹 be a machine that doesn’t have property 𝑄
• Idea:

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

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

Using 𝑄 to build 𝐼𝐵𝑀𝑈

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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 𝑁𝐺𝑏𝑚𝑡𝑓(𝑧)

Does ℳ(𝑥) have property 𝑄? 𝑵𝑸

What if 𝑄 is “trivial”?

• If 𝑄 applies to no Turing machines:

– 𝑁𝐺𝑏𝑚𝑡𝑓 can’t exist

• If 𝑄 applies to all Turing machines:

– It applies to 𝑁∅, ¬𝑄 applies to all machines

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Build this machine: 𝑁𝑥𝑦(𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑁𝐺𝑏𝑚𝑡𝑓(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 ∇ symbol on input 𝜁?

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𝑇𝑢𝑓𝑞𝑡2020

• 𝑇𝑢𝑓𝑞𝑡2020 =

{𝑥|ℳ 𝑥 takes at least 2020 steps}

• Is 𝑇𝑢𝑓𝑞𝑡2020 computable?

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𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓∇

• 𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓𝛼 =

w ℳ 𝑥 overwrites ∇ on input 𝜁}

• Is 𝑃𝑤𝑓𝑠𝑥𝑠𝑗𝑢𝑓∇ computable?

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