CS3102 Theory of Computation - - PowerPoint PPT Presentation

CS3102 Theory of Computation

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

Warm up: Recall:

• 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 𝑥, 𝑦 = 1 if 𝑥 (as a TM description, call it

ℳ 𝑥 ) accepts 𝑦, 0 otherwise Why would 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 be a useful function to implement?

Last Time

• An uncomputable problem

– 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 𝑥, 𝑦 = 1 if 𝑥 (as a TM description, call it ℳ 𝑥 ) accepts 𝑦, 0 otherwise – First attempt to solve:

• Try running ℳ(𝑥), see what happens

– Challenge: ℳ(𝑥) could not accept for two reasons

• ℳ 𝑥 halts and returns 0
• ℳ 𝑥 runs forever (how long do we wait to see?)

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Proving 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 is uncomputable

• Proof by contradiction:

– Assume (toward contradiction) that there is an always- halting Turing machine to compute 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 – Show that we could use that Turing machine to build an impossible Turing machine

• What’s the impossible machine?

– 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢= The set of all Turing machine descriptions that don’t accept themselves – 𝑦 ∈ 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 if and only if ℳ 𝑦 𝑦 = 0

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Using 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 to build 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢

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Does ℳ 𝑥 𝑦 = 1? 𝑵𝒃𝒅𝒅 𝑥 𝑦 ℳ 𝑥 𝑦 == 1? 𝑵𝒕𝒔 Does ℳ 𝑥 𝑥 = 0? Does ℳ 𝑥 𝑦 = 1? 𝑵𝒃𝒅𝒅 𝑥 ℳ 𝑥 1 == 0? Assume we have 𝑁𝑏𝑑𝑑 which computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇: We could then build 𝑁𝑡𝑠 which computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 like this:

Can 𝑁𝑇𝑆 exist?

• Let 𝑥𝑇𝑆 be the description of 𝑁𝑇𝑆

– ℳ 𝑥𝑇𝑆 = 𝑁𝑇𝑆

• What should be 𝑁𝑇𝑆 𝑥𝑇𝑆 ?

– If 𝑁𝑇𝑆 𝑥𝑇𝑆 = 1:

• Since 𝑁𝑇𝑆 computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 we conclude that 𝑥𝑇𝑆 is rejected by whatever

machine it describes.

• Since 𝑥𝑇𝑆 describes 𝑁𝑇𝑆, it must have been that 𝑁𝑇𝑆 𝑥𝑇𝑆 = 0

– If 𝑁𝑇𝑆 𝑥𝑇𝑆 = 0:

• Since 𝑁𝑇𝑆 computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 we conclude that 𝑥𝑇𝑆 is accepted by whatever

machine it describes.

• Since 𝑥𝑡𝑠 describes 𝑁𝑇𝑆, it must have been that 𝑁𝑇𝑆 𝑥𝑇𝑆 = 1

– There’s no answer that makes sense!

• Conclusion: 𝑁𝑇𝑆 can’t be an always-halting Turing machine, so 𝑁𝑏𝑑𝑑

can’t exist

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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)

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Proof by Reduction

Shows how two different problems relate to each other

Reduction Proofs

reduces to

that can solve B

can be used to make

that can solve A

The name “reduces” is confusing: it is in the opposite direction of the making 𝑩 is not a harder problem than 𝑪 𝑩 ≤ 𝑪

Opening a door Lighting a fire Alcohol, wood, matches Keg cannon battering ram

𝐶 𝑌 𝑍 𝐵

Proof of Impossibility by Reduction

• 1. X isn’t possible

(e.g., X = some way to open the door)

• 2. Assume Y is possible

(Y = some way to light a fire)

• 3. Show how to use Y to perform X.
• 4. X isn’t possible, but Y could be used to perform X

conclusion: Y must not be possible either

𝑌 𝑍 𝑌 𝑍

Proof of Impossibility by Reduction

• 1. Take 𝑌 that does not exist.

e.g., 𝑌 = Some TM that computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇

• 2. Assume 𝑍 exists.

𝑍 = Some TM that computes 𝐶

• 3. Show how to use 𝑍 to perform 𝑌.
• 4. 𝑌 doesn’t exist, but 𝑍 could be used to make 𝑌
• conclusion: 𝑍 must not exist either, so 𝐶 is

impossible

𝑌 𝑍 𝑌 𝑍

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

Converse?

that can solve 𝑪 (Lighting a fire)

can be used to make

that can solve 𝑩 (opening a door)

𝑩 is not harder to solve than 𝑪 𝑩 reduces to 𝑪 Does this mean 𝑪 is equally as hard as 𝑩? No! Solving 𝑍 is only one way to solve 𝑌 There may be an easier way

𝑍 𝑌

𝑩 ≤ 𝑪 𝑩 = 𝑪

Common Reduction Traps

• Be careful: the direction matters a great deal

– Using a solver for 𝐶 to solve 𝐵 shows 𝐵 is not harder than 𝐶

• The transformation must use only things you can do:

– Otherwise it may be that 𝐶 exists, but some other step doesn’t! – Example:

• A witch/wizard could open the door by waving a wand and casting

a magic spell

• MacGyver can’t do magic, and is in a room that cannot be opened
• Can we conclude that MacGyver can’t wave a wand?

___ Reduces to ___ 𝐶 𝐵

What “Can Do” Means

• Tools used in a reduction are limited by what you are

proving

• Undecidability:

– You are proving something about all TMs: – The transformation “can do” things a terminating TM “can do”

Spoiler alert!

• Complexity:

– You are proving something about time required: – The time it takes to do the transformation is limited

Example

• 𝐼𝐵𝑀𝑈 𝑥, 𝑦 = 1 if ℳ 𝑥

𝑦 halts 0 if ℳ 𝑥 𝑦 runs forever

– Does the machine halt on this input?

• To show 𝐼𝐵𝑀𝑈 is uncomputable:

– Show how to use a TM for 𝐼𝐵𝑀𝑈 to solve an uncomputable problem

• Show 𝐼𝐵𝑀𝑈 ≥ 𝐵𝐷𝐷𝐹𝑄𝑈𝑇
• Show 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 reduces to 𝐼𝐵𝑀𝑈

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𝐼𝐵𝑀𝑈 Reduction

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𝐵𝐷𝐷𝐹𝑄𝑈𝑇

Problem know is impossible

𝐼𝐵𝑀𝑈

Problem we think is impossible 𝑁ℎ𝑏𝑚𝑢 computes 𝐼𝐵𝑀𝑈

𝐵 𝐶

Assume Reduction Does ℳ(𝑥) accept 𝑦? Does ℳ(𝑥) halt on input 𝑦? 𝑁𝑏𝑑𝑑 computes 𝐵𝐷𝐷𝐹𝑄𝑈

Using 𝐼𝐵𝑀𝑈 to build 𝐵𝐷𝐷𝐹𝑄𝑈𝑇

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Does ℳ 𝑥 𝑦 halt? 𝑵𝑰𝑩𝑴𝑼 𝑥 𝑦 ℳ 𝑥 𝑦 does/doesn’t halt 𝑵𝒃𝒅𝒅 Does ℳ 𝑥 𝑥 = 1? ℳ 𝑥 1 == 1 Assume we have 𝑁𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈: We could then build 𝑁𝑏𝑑𝑑 which computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 like this: Does ℳ 𝑥 𝑦 halt? 𝑵𝑰𝑩𝑴𝑼 𝑥 𝑦

If it halts: Return ℳ(𝑥) If it doesn’t halt: Return 0

𝐼𝐵𝑀𝑈 Reduction

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𝐵𝐷𝐷𝐹𝑄𝑈𝑇

Problem know is impossible

𝐼𝐵𝑀𝑈

Problem we think is impossible 𝑁ℎ𝑏𝑚𝑢 computes 𝐼𝐵𝑀𝑈

𝐵 𝐶

Give 𝑥, 𝑦 as inputs to 𝐼𝐵𝑀𝑈

Assume Reduction Does ℳ(𝑥) accept 𝑦? Does ℳ(𝑥) halt on input 𝑦? 𝑁𝑏𝑑𝑑 computes 𝐵𝐷𝐷𝐹𝑄𝑈 If ℳ 𝑥 𝑦 halts, return ℳ 𝑥 𝑦 Else return 0

Conclusion

• 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 is not computable
• If 𝐼𝐵𝑀𝑈 was computable, an implementation

could be used to compute 𝐵𝐷𝐷𝐹𝑄𝑈𝑇

• So it must be that 𝐼𝐵𝑀𝑈 is not computable

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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

Assume Reduction Does ℳ(𝑥) halt on input 𝑦? Is L ℳ 𝑥 finite? 𝑁𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈

Using 𝐼𝐵𝑀𝑈 to build 𝐵𝐷𝐷𝐹𝑄𝑈𝑇

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Is L ℳ 𝑥 finite? 𝑵𝑮𝑱𝑶𝑱𝑼𝑭 𝑥 𝑀 ℳ 𝑥 is/isn’t finite 𝑵𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑥 = 1? 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

What’s the Language of 𝑁𝑥𝑦?

• If ℳ(𝑥)(𝑦) halts:

– 𝑁𝑥𝑦 always returns 1 – 𝑀 𝑁𝑥𝑦 = Σ∗ – 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

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