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

ℳ 𝑥 𝑦 Halts? It is infinite Language of 𝑁 𝑥𝑦 ℳ 𝑥 𝑦 Doesn’t Halt ? It isn’t infinite • If ℳ(𝑥)(𝑦) halts: Build this machine: 𝑁 𝑥𝑦 : – 𝑁 𝑥𝑦 always returns 1 1) run ℳ(𝑥)(𝑦) 2) return 1 – 𝑀 𝑁 𝑥𝑦 = Σ ∗ , which is infinite • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 1 and never returns – 𝑀 𝑁 𝑥𝑦 = ∅ , which is finite 2

Using 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐼𝐵𝑀𝑈 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 𝐼𝐵𝑀𝑈 which computes 𝐼𝐵𝑀𝑈 like this: 𝑵 𝑰𝑩𝑴𝑼 Does ℳ 𝑥 𝑦 halt? 𝑥 ℳ 𝑥 𝑦 𝑵 ∞ Build this machine: 𝑁 𝑥𝑦 : does/doesn’t Is L 𝑁 𝑥𝑦 infinite? 1) run ℳ(𝑥)(𝑦) 𝑦 halt 2) return 1 3

Reduction Problem we think is impossible Problem know is impossible 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 𝐼𝐵𝑀𝑈 𝐶 𝐵 Build (but don’t run) 𝑁 𝑥𝑦 Is L ℳ 𝑥 Does ℳ(𝑥) infinite? halt on input 𝑦 ? Assume Give 𝑁 𝑥𝑦 to the 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 solver, then 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 solver exists 𝑁 𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 answer the same Reduction 4

Using 𝐺𝐽𝑂𝐽𝑈𝐹 to build 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 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 ∞ which computes 𝐽𝑂𝐺𝐽𝑂𝐽𝑈𝐹 like this: 𝑵 ∞ Is L ℳ 𝑥 infinite? 𝑥 𝑵 𝑮𝑱𝑶𝑱𝑼𝑭 𝑀 ℳ 𝑥 Is L ℳ 𝑥 finite? is/isn’t infinite 5

Language 𝑂𝑝𝑜𝑆𝑓𝑕 • 𝑂𝑝𝑜𝑆𝑓𝑕 = 𝑥 𝑀 ℳ 𝑥 is non − regular} • We will show this is not computable by using 𝑂𝑝𝑜𝑆𝑓𝑕 to compute 𝐼𝐵𝑀𝑈 6

ℳ 𝑥 𝑦 Halts? It is non-regular ℳ 𝑥 𝑦 Doesn’t Halt ? It isn’t non-regular Language of 𝑁 𝑥𝑦 Build this machine: 𝑁 𝑥𝑦 (𝑧): 1) run ℳ(𝑥)(𝑦) 2) return MAJ(𝑧) • If ℳ(𝑥)(𝑦) halts: – 𝑁 𝑥𝑦 returns 1 if MAJ 𝑧 = 1 – 𝑀 𝑁 𝑥𝑦 = 𝑁𝐵𝐾(𝑧) , which is non-regular • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 1 and never returns 1 – 𝑀 𝑁 𝑥𝑦 = ∅ , which isn’t non-regular 7

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 ℳ 𝑥 non- 1) run ℳ(𝑥)(𝑦) 𝑦 halt regular? 2) return MAJ (𝑧) 8

Reduction 𝑂𝑝𝑜𝑆𝑓𝑕 𝐼𝐵𝑀𝑈 𝐶 𝐵 Describe that machine 𝑁 𝑥𝑦 Is L ℳ 𝑥 non- Does ℳ(𝑥) regular? halt on input 𝑦 ? Assume Give 𝑁 𝑥𝑦 to the 𝑂𝑝𝑜𝑆𝑓𝑕 solver, then 𝑂𝑝𝑜𝑆𝑓𝑕 solver exists 𝑁 𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 answer the same Reduction 9

Using 𝑆𝑓𝑕 to build 𝑂𝑝𝑜𝑆𝑓𝑕 Assume we have 𝑁 𝑆𝑓𝑕 𝑵 𝑆𝑓𝑕 which computes 𝑂𝑝𝑜𝑆𝑓𝑕 : 𝑀 ℳ 𝑥 is/isn’t 𝑥 Is L ℳ 𝑥 regular? Regular We could then build 𝑁 𝑂𝑆 which computes 𝑂𝑝𝑜𝑆𝑓𝑕 like this: 𝑵 𝑶𝑺 Is L ℳ 𝑥 non-regular? 𝑥 𝑵 𝑆𝑓𝑕 𝑀 ℳ 𝑥 is/isn’t non- Is L ℳ 𝑥 regular? regular 10

Language 𝐵𝑑𝑑𝑓𝑞𝑢101 • 𝐵𝑑𝑑𝑓𝑞𝑢101 = 𝑥 ℳ 𝑥 101 = 1} • We will show this is not computable by using 𝐵𝑑𝑑𝑓𝑞𝑢101 to compute 𝐼𝐵𝑀𝑈 11

ℳ 𝑥 𝑦 Halts? It does accept 101 ℳ 𝑥 𝑦 Doesn’t Halt ? It doesn’t accept101 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 accept 101 • If ℳ(𝑥)(𝑦) doesn’t halt: – 𝑁 𝑥𝑦 gets “stuck” in step 1 and never returns 1 – 𝑀 𝑁 𝑥𝑦 = ∅ , so it doesn’t accept 101 12

Using 𝐵𝑑𝑑𝑓𝑞𝑢101 to build 𝐼𝐵𝑀𝑈 Assume we have 𝑁 𝐵101 𝑵 𝑩𝟐𝟏𝟐 which computes 𝐵𝑑𝑑𝑓𝑞𝑢101 : ℳ(𝑥) does/doesn’t 𝑥 Does ℳ(𝑥) accept accept 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 𝑩𝟐𝟏𝟐 Build this machine: 𝑁 𝑥𝑦 (𝑧): does/doesn’t Does ℳ(𝑥) accept 1) run ℳ(𝑥)(𝑦) 𝑦 halt 101? 2) return 𝑧 == 101 13

𝐵𝑑𝑑𝑓𝑞𝑢101 Reduction 𝐵𝑑𝑑𝑓𝑞𝑢101 𝐼𝐵𝑀𝑈 𝐶 𝐵 Describe that machine 𝑁 𝑥𝑦 Is 101 ∈ L ℳ 𝑥 ? Does ℳ(𝑥) halt on input 𝑦 ? Assume Give 𝑁 𝑥𝑦 to the 𝐵𝑑𝑑𝑓𝑞𝑢101 solver, then 𝐵𝑑𝑑𝑓𝑞𝑢101 solver exists 𝑁 𝐼𝐵𝑀𝑈 computes 𝐼𝐵𝑀𝑈 answer the same Reduction 14

Using 𝑆𝑓𝑘𝑓𝑑𝑢101 to build 𝐵𝑑𝑑𝑓𝑞𝑢101 Assume we have 𝑁 𝑆101 𝑵 𝑆101 which computes 𝑆𝑓𝑘𝑓𝑑𝑢 101: 101 isn’t/is in 𝑥 Is 101 ∉ L ℳ 𝑥 𝑀 ℳ 𝑥 ? We could then build 𝑁 𝐵101 which computes 𝐵𝑑𝑑𝑓𝑞𝑢101 like this: 𝑵 𝐵101 Is 101 ∈ L ℳ 𝑥 ? 𝑥 𝑵 𝑆101 101 is/isn’t in 𝑀 ℳ 𝑥 Is 101 ∉ L ℳ 𝑥 ? 15

Computability and non-computability closed under complement • If 𝑀 is computable, then 𝑀 𝑑 is computable Assume we have 𝑁 𝑀 which 𝑵 𝑴 computes 𝑀 : 𝑥 is/isn’t in 𝑀 𝑥 Is 𝑥 ∈ L ? We could then build 𝑁 𝑀 𝑑 which computes 𝑀 𝑑 like this: 𝑵 𝑀 𝑑 Is 𝑥 ∉ L ? 𝑥 𝑵 𝑴 𝑥 isn’t/is in 𝑀 Is 𝑥 ∈ L ? 16

Complement Reduction 𝑀 𝑀 𝑑 𝐶 𝑥 ∈ 𝑀 ? 𝐵 Don’t change 𝑥 𝑥 ∉ 𝑀 ? If Give 𝑥 to the 𝑀 solver, 𝑀 solver exists 𝑀 𝑑 solver exists then answer the opposite 𝑀 solver doesn’t exist 𝑀 𝑑 solver doesn’t exist Reduction 17

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 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 20

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 𝑁 𝑄 be a machine that does have property 𝑄 • • Idea: – If ℳ(𝑥)(𝑦) halts, 𝑀 𝑁 𝑥𝑦 = 𝑀 𝑁 ℎ𝑏𝑡 𝑄 – 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 halts ℳ 𝑥 𝑦 Halts? It has property 𝑄 Build this machine: 𝑁 𝑥𝑦 (𝑧): ℳ 𝑥 𝑦 Doesn’t Halt ? It doesn’t have property 𝑄 1) run ℳ(𝑥)(𝑦) 2) return 𝑁 ℎ𝑏𝑡 𝑄 (𝑧) 21

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 ℎ𝑏𝑡 𝑄 (𝑧) 22