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?) 2

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 3

Using 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 to build 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 Assume we have 𝑁 𝑏𝑑𝑑 𝑥 𝑵 𝒃𝒅𝒅 which computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 : Does ℳ 𝑥 𝑦 = 1? ℳ 𝑥 𝑦 == 1? 𝑦 We could then build 𝑁 𝑡𝑠 which computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 like this: 𝑵 𝒕𝒔 Does ℳ 𝑥 𝑥 = 0? 𝑵 𝒃𝒅𝒅 𝑥 ℳ 𝑥 1 == 0? Does ℳ 𝑥 𝑦 = 1? 4

Can 𝑁 𝑇𝑆 exist? Let 𝑥 𝑇𝑆 be the description of 𝑁 𝑇𝑆 • – ℳ 𝑥 𝑇𝑆 = 𝑁 𝑇𝑆 What should be 𝑁 𝑇𝑆 𝑥 𝑇𝑆 ? • – If 𝑁 𝑇𝑆 𝑥 𝑇𝑆 = 1 : • Since 𝑁 𝑇𝑆 computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 we conclude that 𝑥 𝑇𝑆 is rejected by whatever machine it describes. • Since 𝑥 𝑇𝑆 describes 𝑁 𝑇𝑆 , it must have been that 𝑁 𝑇𝑆 𝑥 𝑇𝑆 = 0 – If 𝑁 𝑇𝑆 𝑥 𝑇𝑆 = 0 : • Since 𝑁 𝑇𝑆 computes 𝑇𝑓𝑚𝑔𝑆𝑓𝑘𝑓𝑑𝑢 we conclude that 𝑥 𝑇𝑆 is accepted by whatever machine it describes. • Since 𝑥 𝑡𝑠 describes 𝑁 𝑇𝑆 , it must have been that 𝑁 𝑇𝑆 𝑥 𝑇𝑆 = 1 – There’s no answer that makes sense! Conclusion: 𝑁 𝑇𝑆 can’t be an always -halting Turing machine, so 𝑁 𝑏𝑑𝑑 • can’t exist 5

How to show things aren’t computable 1. Ask “can I have an always -halting Turing machine 𝑁 𝑞 for language/function/problem 𝑄 ?” 2. Show that, if 𝑁 𝑞 exists, it can be used to make an impossible machine 𝑁 𝑗𝑛𝑞 How do we know a machine is impossible? Option 1: It contradicts itself (e.g. 𝑁 𝑇𝑆 ) Option 2: Someone has done this before (e.g. 𝑁 𝑏𝑑𝑑 ) 6

Proving Other Problems are Uncomputable • Reduction – Convert some problem into a known uncomputable one (using only computable steps) 7

Proof by Reduction Shows how two different problems relate to each other

Reduction Proofs Opening a Lighting a door fire 𝐶 𝐵 reduces to Alcohol, wood, Keg cannon matches 𝑍 𝑌 battering ram can be used to make that can solve B that can solve A 𝑩 is not a harder problem than 𝑪 𝑩 ≤ 𝑪 The name “reduces” is confusing: it is in the opposite direction of the making

Proof of Impossibility by Reduction 1. X isn’t possible 𝑌 (e.g., X = some way to open the door) 2. Assume Y is possible 𝑍 ( Y = some way to light a fire) 𝑍 𝑌 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

Proof of Impossibility by Reduction 1. Take 𝑌 that does not exist. 𝑌 e.g., 𝑌 = Some TM that computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 2. Assume 𝑍 exists. 𝑍 𝑍 = Some TM that computes 𝐶 𝑍 3. Show how to use 𝑍 to perform 𝑌 . 𝑌 4. 𝑌 doesn’t exist, but 𝑍 could be used to make 𝑌 conclusion: 𝑍 must not exist either, so 𝐶 is • impossible

MacGyver’s Reduction Problem we think is impossible Problem know is impossible Lighting a fire Opening a door 𝐶 𝐵 Aim duct at door, insert keg If Solution for 𝑩 Solution for 𝑪 Keg cannon battering ram Alcohol, wood, matches Put fire under the Keg Reduction 12

Converse? 𝑩 reduces to 𝑪 𝑌 𝑍 can be used to make that can solve 𝑪 that can solve 𝑩 𝑩 is not harder to solve than 𝑪 (Lighting a fire) (opening a door) 𝑩 ≤ 𝑪 Does this mean 𝑪 is equally as hard as 𝑩 ? 𝑩 = 𝑪 No! Solving 𝑍 is only one way to solve 𝑌 There may be an easier way

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 𝐵 𝐶 ___ Reduces to ___ • 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?

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 𝑦 runs forever 0 if ℳ 𝑥 – 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 16

𝐼𝐵𝑀𝑈 Reduction Problem we think is impossible Problem know is impossible 𝐼𝐵𝑀𝑈 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 𝐶 Does ℳ(𝑥) 𝐵 Does ℳ(𝑥) halt on input 𝑦 ? accept 𝑦 ? Assume 𝑁 ℎ𝑏𝑚𝑢 computes 𝐼𝐵𝑀𝑈 𝑁 𝑏𝑑𝑑 computes 𝐵𝐷𝐷𝐹𝑄𝑈 Reduction 17

Using 𝐼𝐵𝑀𝑈 to build 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 Assume we have 𝑁 𝐼𝐵𝑀𝑈 𝑥 𝑵 𝑰𝑩𝑴𝑼 which computes 𝐼𝐵𝑀𝑈 : ℳ 𝑥 𝑦 Does ℳ 𝑥 𝑦 halt? does/doesn’t halt 𝑦 We could then build 𝑁 𝑏𝑑𝑑 which computes 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 like this: 𝑵 𝒃𝒅𝒅 Does ℳ 𝑥 𝑥 = 1? If it halts: Return ℳ(𝑥) 𝑥 𝑵 𝑰𝑩𝑴𝑼 ℳ 𝑥 1 == 1 Does ℳ 𝑥 𝑦 halt? 𝑦 If it doesn’t halt: Return 0 18

𝐼𝐵𝑀𝑈 Reduction Problem we think is impossible Problem know is impossible 𝐼𝐵𝑀𝑈 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 𝐶 Does ℳ(𝑥) 𝐵 Give 𝑥, 𝑦 as inputs to 𝐼𝐵𝑀𝑈 Does ℳ(𝑥) halt on input 𝑦 ? accept 𝑦 ? Assume If ℳ 𝑥 𝑦 halts, return ℳ 𝑥 𝑦 𝑁 ℎ𝑏𝑚𝑢 computes 𝐼𝐵𝑀𝑈 𝑁 𝑏𝑑𝑑 computes 𝐵𝐷𝐷𝐹𝑄𝑈 Else return 0 Reduction 19

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 20

Example: 𝐺𝐽𝑂𝐽𝑈𝐹 • 𝐺𝐽𝑂𝐽𝑈𝐹 𝑥 = 1 if 𝑀 ℳ 𝑥 is finite is infinite 0 if 𝑀 ℳ 𝑥 • To show 𝐺𝐽𝑂𝐽𝑈𝐹 is uncomputable – Show how to use a TM for 𝐺𝐽𝑂𝐽𝑈𝐹 to solve 𝐼𝐵𝑀𝑈 • 𝐺𝐽𝑂𝐽𝑈𝐹 ≥ 𝐼𝐵𝑀𝑈 • 𝐼𝐵𝑀𝑈 reduces to 𝐺𝐽𝑂𝐽𝑈𝐹 21

𝐺𝐽𝑂𝐽𝑈𝐹 Reduction Problem we think is impossible Problem know is impossible 𝐺𝐽𝑂𝐽𝑈𝐹 𝐼𝐵𝑀𝑈 𝐶 𝐵 Is L ℳ 𝑥 finite? Does ℳ(𝑥) halt on input 𝑦 ? Assume 𝑁 𝑔𝑗𝑜𝑗𝑢𝑓 computes 𝐺𝐽𝑂𝐽𝑈𝐹 𝑁 𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 Reduction 22

Using 𝐼𝐵𝑀𝑈 to build 𝐵𝐷𝐷𝐹𝑄𝑈𝑇 Assume we have 𝑁 𝐺𝐽𝑂𝐽𝑈𝐹 𝑵 𝑮𝑱𝑶𝑱𝑼𝑭 which computes 𝐺𝐽𝑂𝐽𝑈𝐹 : 𝑀 ℳ 𝑥 is/isn’t 𝑥 Is L ℳ 𝑥 finite? We could then build 𝑁 𝐼𝐵𝑀𝑈 finite which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑥 = 1? 𝑥 ℳ 𝑥 1 == 1 𝑵 𝑮𝑱𝑶𝑱𝑼𝑭 Build this machine: 𝑁 𝑥𝑦 : Is L 𝑁 𝑥𝑦 finite? 1) run ℳ(𝑥)(𝑦) 𝑦 2) return 1 23

What’s the Language of 𝑁 𝑥𝑦 ? • If ℳ(𝑥)(𝑦) halts: Build this machine: 𝑁 𝑥𝑦 : – 𝑁 𝑥𝑦 always returns 1 1) run ℳ(𝑥)(𝑦) – 𝑀 𝑁 𝑥𝑦 = Σ ∗ 2) return 1 – 𝑀 𝑁 𝑥𝑦 is infinite • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 1 and never returns – 𝑀 𝑁 𝑥𝑦 = ∅ – 𝑀 𝑁 𝑥𝑦 = 0 24