CS3102 Theory of Computation Warm up: = { 0,1 | has an odd - PowerPoint PPT Presentation

CS3102 Theory of Computation Warm up: 𝑌𝑃𝑆 = {𝑦 ∈ 0,1 ∗ |𝑦 has an odd number of 1s} Write a regex for 𝑌𝑃𝑆 𝑑 (i.e. 𝑌𝑃𝑆 , i.e. the complement of 𝑌𝑃𝑆 )

AND to NAND • AND: – 𝑅 = 𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡, 𝑇𝑝𝑛𝑓0𝑡 0,1 – 𝑟 0 = 𝑡𝑢𝑏𝑠𝑢 Some 0s 0 – 𝐺 = {𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡} – 𝜀 defined as the arrows 0 start • NAND: – 𝑅, 𝑟 0 , 𝜀 don’t change – 𝐺 = 𝑅 − 𝐺 1 1 • In general, If we can compute a language 𝑀 No 0s with a FSA, we can compute 𝑀 𝑑 as well 2

Logistics • Homework due Tonight and Friday • You’ll have an assignment due the Friday you return from the break (no early deadline) • Quiz due Tuesday 3

Last Time • Regular Expressions – Equivalent to FSAs (but we haven’t shown that yet) 4

Regular Expressions Name Decision Problem Function Language 𝑐 ∈ Σ ∗ 𝑐 matches the pattern} Regex Does this string 𝑔 𝑐 = ቊ 0 the string matches the string doesn ′ t match this 1 pattern? • A way of describing a language • Give a “pattern” of the strings, every string matching that pattern is in the language • Examples: – (𝑏|𝑐)𝑑 matches : 𝑏𝑑 and 𝑐𝑑 – 𝑏|𝑐 ∗ 𝑑 matches : 𝑑 , 𝑏𝑑 , 𝑐𝑑 , 𝑏𝑏𝑑 , 𝑏𝑐𝑑 , 𝑐𝑏𝑑 , 𝑐𝑐𝑑 , … 5

FSA = Regex • Finite state Automata and Regular Expressions are equivalent models of computing • Any language I can represent as a FSA I can also represent as a Regex (and vice versa) • How would I show this? 6

Showing FSA ≤ Regex • Show how to convert any FSA into a Regex for the same language • We’re going to skip this: – It’s tedious, and people virtually never go this direction in practice, but you can do it (see textbook theorem 9.12) 7

Showing Regex ≤ FSA • Show how to convert any regex into a FSA for the same language • Idea: show how to build each “piece” of a regex using FSA 8

“Pieces” of a Regex • Empty String: – Matches just the string of length 0 – Notation: 𝜁 or “” • Literal Character – Matches a specific string of length 1 – Example: the regex 𝑏 will match just the string 𝑏 • Alternation/Union – Matches strings that match at least one of the two parts – Example: the regex 𝑏|𝑐 will match 𝑏 and 𝑐 • Concatenation – Matches strings that can be dividing into 2 parts to match the things concatenated – Example: the regex 𝑏 𝑐 𝑑 will match the strings 𝑏𝑑 and 𝑐𝑑 • Kleene Star – Matches strings that are 0 or more copies of the thing starred – Example: 𝑏 𝑐 𝑑 ∗ will match 𝑏 , 𝑐 , or either followed by any number of 𝑑 ’s 9

FSA for the empty string 10

FSA for a literal character 11

FSA for Alternation/Union • Tricky… • What does it need to do? 12

Recall: AND to NAND • AND: – 𝑅 = 𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡, 𝑇𝑝𝑛𝑓0𝑡 0,1 – 𝑟 0 = 𝑡𝑢𝑏𝑠𝑢 Some 0s 0 – 𝐺 = {𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡} – 𝜀 defined as the arrows 0 start • NAND: – 𝑅, 𝑟 0 , 𝜀 don’t change – 𝐺 = 𝑅 − 𝐺 1 1 • In general, If we can compute a language 𝑀 No 0s with a FSA, we can compute 𝑀 𝑑 as well 13

Computing Complement • If FSA 𝑁 = (𝑅, Σ, 𝜀, 𝑟 0 , 𝐺) computes 𝑀 • Then FSA 𝑁 ′ = (𝑅, Σ, 𝜀, 𝑟 0 , 𝑅 − 𝐺) computes ത 𝑀 • Why? – Consider string 𝑥 ∈ Σ ∗ – 𝑥 ∈ 𝑀 means it ends at some state 𝑔 ∈ 𝐺 , which will be non-final in 𝑁 ′ and therefore it will return False – 𝑥 ∉ 𝑀 means it ends at some state 𝑟 ∉ 𝐺 , which will be final in 𝑁 ′ and therefore it will return True 14

Computing Union • Let FSA 𝑁 1 = (𝑅 1 , Σ, 𝜀 1 , 𝑟 01 , 𝐺 1 ) compute 𝑀 1 • Let 𝑁 2 = (𝑅 2 , Σ, 𝜀 2 , 𝑟 02 , 𝐺 2 ) compute 𝑀 1 • Will there always be some automaton 𝑁 ∪ to compute 𝑀 1 ∪ 𝑀 2 • What must 𝑁 ∪ do? – Somehow end up in a final state if either 𝑁 1 or 𝑁 2 did – Idea: build 𝑁 ∪ to “simulate” both 𝑁 1 and 𝑁 2 15

Example • 𝐵𝑂𝐸 ∪ 𝑌𝑃𝑆 – What is the resulting language? 𝑁 𝐵𝑂𝐸 0,1 Some 0s 0 𝑁 𝑌𝑃𝑆 1 0 0 start Even 0 Odd 1 1 No 0s 1 16

Cross-Product Construction • 2 machines at once! Start Start Odd Even 𝑁 𝐵𝑂𝐸 0,1 Some 0s 0 start 0 Some0s Some0s Odd Even 1 No 0s 1 𝑁 𝑌𝑃𝑆 1 No0s No0s 0 Even Odd Odd Even 0 17 1

Cross-Product Construction • 2 machines at once! Start Start Odd Even 𝑁 1 𝐵𝑂𝐸 0,1 Some 0s 0 0 1 start 0 0 0 Some0s Some0s Odd Even 1 1 No 0s 1 0 0 𝑁 𝑌𝑃𝑆 1 1 No0s No0s 0 Even Odd Odd Even 1 0 18 1

Cross Product Construction • Let FSA 𝑁 1 = (𝑅 1 , Σ, 𝜀 1 , 𝑟 01 , 𝐺 1 ) compute 𝑀 1 • Let 𝑁 2 = (𝑅 2 , Σ, 𝜀 2 , 𝑟 02 , 𝐺 2 ) compute 𝑀 1 • 𝑁 ∪ = (𝑅 1 × 𝑅 2 , Σ, 𝜀 ∪ , (𝑟 01 , 𝑟 02 ), 𝐺 ∪ ) computes 𝑀 1 ∪ 𝑀 2 – 𝜀 ∪ 𝑟 1 , 𝑟 2 , 𝜏 = 𝜀 1 𝑟 1 , 𝜏 , 𝜀 2 𝑟 2 , 𝜏 – 𝐺 ∪ = 𝑟 1 , 𝑟 2 ∈ 𝑅 1 × 𝑅 2 𝑟 1 ∈ 𝐺 1 or 𝑟 2 ∈ 𝐺 2 } • How could we do intersection? 19

Non-determinism • Things could get easier if we “relax” our automata • So far: – Must have exactly one transition per character per state – Can only be in one state at a time • Non-deterministic Finite Automata: – Allowed to be in multiple (or zero) states! – Can have multiple or zero transitions for a character – Can take transitions without using a character – Models parallel computing 20

Nondeterminism Driving to a friend’s house Friend forgets to mention a fork in the directions Which way do you go? Why not both? ? 21

Example Non-deterministic Finite Automaton • 𝑈ℎ𝑗𝑠𝑒𝑀𝑏𝑡𝑢1 = { 𝑥 ∈ 0,1 ∗ the third from last character is a 1} | 0,1 𝑜𝑓𝑦𝑢2 start 𝑝𝑜𝑓 𝑜𝑓𝑦𝑢1 1 0,1 0,1 22

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Non-Deterministic Finite State Automaton • Implementation: – Finite number of states – One start state – “Final” states – Transitions: (partial) function mapping state-character (or epsilon) pairs to sets of states • Execution: – Start in the initial “state” – Enter every state reachable without consuming input ( 𝜻 -transitions) – Read each character once, in order (no looking back) – Transition to new states once per character (based on current states and character) – Enter every state reachable without consuming input ( 𝜻 -transitions) – Return True if any state you end is final • Return False if every state you end in is non-final 24

Union Using Non-Determinism 0,1 Some 0s 0 New start 0 start 1 1 No 0s 1 0 Even 0 Odd 25 1

Union Using Non-Determinism 0,1 Some 0s 0 𝜁 new 0 start 𝜁 1 1 No 0s 1 0 𝑁 ∪ = 𝑅 1 ∪ 𝑅 2 ∪ 𝑜𝑓𝑥 , Σ, 𝜀 ∪ , 𝑜𝑓𝑥, 𝐺 1 ∪ 𝐺 2 Even 0 Odd 𝜀 ∪ (𝑟, 𝜏) = ቊ 𝜀 1 𝑟, 𝜏 if 𝑟 ∈ 𝑅 1 {𝜀 2 𝑟, 𝜏 } if 𝑟 ∈ 𝑅 2 26 𝜀 ∪ 𝑜𝑓𝑥, 𝜁 = {𝑡𝑢𝑏𝑠𝑢, 𝑓𝑤𝑓𝑜} 1

What’s the language? 0, 𝜁 final start 27

NFA Example {𝑥 ∈ 0,1 ∗ |𝑥 contains 0101} 28