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 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 ) 2

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 3

Non-Computable Problems 4

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 5

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 6

𝐺𝐽𝑂𝐽𝑈𝐹 Reduction Problem we think is impossible Problem know is impossible 𝐺𝐽𝑂𝐽𝑈𝐹 𝐼𝐵𝑀𝑈 𝐶 𝐵 Build (but don’t run) 𝑁 𝑥𝑦 Is L ℳ 𝑥 finite? Does ℳ(𝑥) halt on input 𝑦 ? Assume Give 𝑁 𝑥𝑦 as input to 𝑁 𝑔𝑗𝑜𝑗𝑢𝑓 computes 𝐺𝐽𝑂𝐽𝑈𝐹 𝑁 𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 𝑁 𝑔𝑗𝑜𝑗𝑢𝑓 answer opposite Reduction 7

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 ∗ (all strings) 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 8

Using 𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐼𝐵𝑀𝑈 Assume we have 𝑁 𝐺𝐽𝑂𝐽𝑈𝐹 𝑵 𝑮𝑱𝑶𝑱𝑼𝑭 which computes 𝐺𝐽𝑂𝐽𝑈𝐹 : 𝑀 ℳ 𝑥 is/isn’t 𝑥 Is L ℳ 𝑥 finite? finite We could then build 𝑁 𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? 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 9

Showing 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 is not computable • Use _____ to build _____. 10

𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 Reduction 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 𝐺𝐽𝑂𝐽𝑈𝐹 𝐶 𝐵 Is L ℳ 𝑥 Is L ℳ 𝑥 infinite? finite? Assume 𝑁 ∞ computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 𝑁 𝐺𝐽𝑂𝐽𝑈𝐹 computes 𝐺𝐽𝑂𝐽𝑈𝐹 Reduction 11

𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 Reduction 𝐺𝐽𝑂𝐽𝑈𝐹 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 𝐶 𝐵 Is L ℳ 𝑥 finite? Is L ℳ 𝑥 infinite? Assume 𝑁 𝐺𝐽𝑂𝐽𝑈𝐹 computes 𝐺𝐽𝑂𝐽𝑈𝐹 𝑁 ∞ computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 Reduction 12

Using 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐺𝐽𝑂𝐽𝑈𝐹 Assume we have 𝑁 ∞ which 𝑵 ∞ computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 : 𝑀 ℳ 𝑥 is/isn’t 𝑥 Is L ℳ 𝑥 infinite? infinite We could then build 𝑁 𝐺𝐽𝑂𝐽𝑈𝐹 which computes 𝐺𝐽𝑂𝐽𝑈𝐹 like this: 𝑵 𝑮𝑱𝑶𝑱𝑼𝑭 Is L ℳ 𝑥 finite? 𝑥 𝑵 ∞ 𝑀 ℳ 𝑥 Is L ℳ 𝑥 infinite? is/isn’t finite 𝑦 13

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 𝑁 𝑥𝑦 such that 𝑀(𝑁 𝑥𝑦 ) is regular if and only if ℳ(𝑥)(𝑦) runs forever. 14

Building 𝑁 𝑥𝑦 Build this machine: 𝑁 𝑥𝑦 (𝑧): 1) run ℳ(𝑥)(𝑦) • If ℳ(𝑥)(𝑦) halts: 2) return 𝑌𝑃𝑆(𝑧) – 𝑁 𝑥𝑦 returns 1 if 𝑌𝑃𝑆 𝑧 = 1 – 𝑀 𝑁 𝑥𝑦 = 𝑌𝑃𝑆(𝑧) , which is not regular • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 2 and never returns 1 – 𝑀 𝑁 𝑥𝑦 = ∅ , which is regular 15

Using 𝑂𝑝𝑜𝑆𝑓𝑕 to build 𝐼𝐵𝑀𝑈 Assume we have 𝑁 𝑂𝑆 which 𝑵 𝑶𝑺 computes 𝑂𝑝𝑜𝑆𝑓𝑕 : 𝑀 ℳ 𝑥 is/isn’t 𝑥 Is L ℳ 𝑥 non- non-regular regular? We could then build 𝑁 𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? 𝑥 ℳ(𝑥)(𝑦) 𝑵 𝑶𝑺 Build this machine: 𝑁 𝑥𝑦 (𝑧): does/doesn’t Is L 𝑁 𝑥𝑦 non- 1) run ℳ(𝑥)(𝑦) 𝑦 halt regular? 2) return 𝑌𝑃𝑆(𝑧) 16

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 𝑁 𝑥𝑦 such that 101 ∈ 𝑀(𝑛 𝑥𝑦 ) if and only if ℳ(𝑥)(𝑦) runs forever. 17

Building 𝑁 𝑥𝑦 Build this machine: 𝑁 𝑥𝑦 (𝑧): 1) run ℳ(𝑥)(𝑦) • If ℳ(𝑥)(𝑦) halts: 2) return 𝑧 == 101 – 𝑁 𝑥𝑦 returns 1 if 𝑧 == 101 – 𝑀 𝑁 𝑥𝑦 = {101} , so it does not reject 101 • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 2 and never returns 1 – 𝑀 𝑁 𝑥𝑦 = ∅ , so it does reject 101 18

Using 𝑆𝑓𝑘𝑓𝑑𝑢𝑡101 to build 𝐼𝐵𝑀𝑈 Assume we have 𝑁 𝑆101 𝑵 𝑺𝟐𝟏𝟐 which computes 𝑂𝑝𝑜𝑆𝑓𝑕 : ℳ(𝑥) does/doesn’t 𝑥 Does ℳ(𝑥) reject reject 101 101? We could then build 𝑁 𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? 𝑥 ℳ(𝑥)(𝑦) 𝑵 𝑺𝟐𝟏𝟐 Build this machine: 𝑁 𝑥𝑦 (𝑧): does/doesn’t Does ℳ(𝑥) reject 1) run ℳ(𝑥)(𝑦) 𝑦 halt 101? 2) return 𝑧 == 101 19

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. 20

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? 21

Rice’s Theorem • For any Semantic property 𝑄 of Turing Machines, either: – Every Turing machine has property 𝑄 𝑄 is “trivial” – 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: Return 1 Return 0 22

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 𝑁 𝐺𝐵𝑀𝑇𝐹 be a machine that doesn’t have property 𝑄 • • Idea: – If ℳ(𝑥)(𝑦) halts, 𝑀 𝑁 𝑥𝑦 = 𝑀 𝑁 𝐺𝐵𝑀𝑇𝐹 – If ℳ(𝑥)(𝑦) doesn’t halt, 𝑀 𝑁 𝑥𝑦 = L 𝑁 ∅ = ∅ – 𝑀(𝑁 𝑥𝑦 ) has property 𝑄 if and only if ℳ 𝑥 𝑦 runs forever Build this machine: 𝑁 𝑥𝑦 (𝑧): 1) run ℳ(𝑥)(𝑦) 2) return 𝑁 𝐺𝑏𝑚𝑡𝑓 (𝑧) 23

Using 𝑄 to build 𝐼𝐵𝑀𝑈 Assume we have 𝑁 𝑄 which 𝑵 𝑸 computes 𝑄 : ℳ(𝑥) does/doesn’t 𝑥 Does ℳ(𝑥) have have property 𝑄 property 𝑄 ? We could then build 𝑁 𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? 𝑥 ℳ(𝑥) 𝑵 𝑸 Build this machine: 𝑁 𝑥𝑦 (𝑧): does/doesn’t Does ℳ(𝑥) have 1) run ℳ(𝑥)(𝑦) 𝑦 have property 𝑄 property 𝑄 ? 2) return 𝑁 𝐺𝑏𝑚𝑡𝑓 (𝑧) 24