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 𝑌𝑃𝑆𝑑 (i.e. 𝑌𝑃𝑆, i.e. the complement of 𝑌𝑃𝑆)

AND to NAND

• AND:

– 𝑅 = 𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡, 𝑇𝑝𝑛𝑓0𝑡 – 𝑟0 = 𝑡𝑢𝑏𝑠𝑢 – 𝐺 = {𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡} – 𝜀 defined as the arrows

• NAND:

– 𝑅, 𝑟0, 𝜀 don’t change – 𝐺 = 𝑅 − 𝐺

• In general, If we can compute a language 𝑀

with a FSA, we can compute 𝑀𝑑 as well

2 start

No 0s

0,1 1

Some 0s

1

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)

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Regular Expressions

• 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, …

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Name Decision Problem Function Language Regex Does this string match this pattern? 𝑔 𝑐 = ቊ0 the string matches 1 the string doesn′t 𝑐 ∈ Σ∗ 𝑐 matches the pattern}

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?

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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)

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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

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“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

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FSA for the empty string

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FSA for a literal character

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FSA for Alternation/Union

• Tricky…
• What does it need to do?

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Recall: AND to NAND

• AND:

– 𝑅 = 𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡, 𝑇𝑝𝑛𝑓0𝑡 – 𝑟0 = 𝑡𝑢𝑏𝑠𝑢 – 𝐺 = {𝑡𝑢𝑏𝑠𝑢, 𝑂𝑝0𝑡} – 𝜀 defined as the arrows

• NAND:

– 𝑅, 𝑟0, 𝜀 don’t change – 𝐺 = 𝑅 − 𝐺

• In general, If we can compute a language 𝑀

with a FSA, we can compute 𝑀𝑑 as well

13 start

No 0s

0,1 1

Some 0s

1

Computing Complement

• If FSA 𝑁 = (𝑅, Σ, 𝜀, 𝑟0, 𝐺) computes 𝑀
• Then FSA 𝑁′ = (𝑅, Σ, 𝜀, 𝑟0, 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

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Computing Union

• Let FSA 𝑁1 = (𝑅1, Σ, 𝜀1, 𝑟01, 𝐺

1) compute 𝑀1

• Let 𝑁2 = (𝑅2, Σ, 𝜀2, 𝑟02, 𝐺

2) compute 𝑀1

• Will there always be some automaton 𝑁∪ to

compute 𝑀1 ∪ 𝑀2

• What must 𝑁∪ do?

– Somehow end up in a final state if either 𝑁1 or 𝑁2 did – Idea: build 𝑁∪ to “simulate” both 𝑁1 and 𝑁2

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Example

• 𝐵𝑂𝐸 ∪ 𝑌𝑃𝑆

– What is the resulting language?

16 start

0,1 1

Some 0s No 0s

1 𝑁

𝐵𝑂𝐸

Even 1 1

Odd

𝑁𝑌𝑃𝑆

Cross-Product Construction

• 2 machines at once!

17 start

0,1 1

Some 0s

No 0s

1 𝑁

𝐵𝑂𝐸

Even

1 1

Odd

𝑁𝑌𝑃𝑆

Start Even Some0s Even No0s Even Start Odd Some0s Odd No0s Odd

Cross-Product Construction

• 2 machines at once!

18 start

0,1 1

Some 0s

No 0s

1 𝑁

𝐵𝑂𝐸

Even

1 1

Odd

𝑁𝑌𝑃𝑆

Start Even Some0s Even No0s Even Some0s Odd No0s Odd

1 1 1 1 1

Start Odd

Cross Product Construction

• Let FSA 𝑁1 = (𝑅1, Σ, 𝜀1, 𝑟01, 𝐺

1) compute 𝑀1

• Let 𝑁2 = (𝑅2, Σ, 𝜀2, 𝑟02, 𝐺

2) compute 𝑀1

• 𝑁∪ = (𝑅1 × 𝑅2, Σ, 𝜀∪, (𝑟01, 𝑟02), 𝐺

∪) computes

𝑀1 ∪ 𝑀2

– 𝜀∪ 𝑟1, 𝑟2 , 𝜏 = 𝜀1 𝑟1, 𝜏 , 𝜀2 𝑟2, 𝜏 – 𝐺∪ = 𝑟1, 𝑟2 ∈ 𝑅1 × 𝑅2 𝑟1 ∈ 𝐺

1 or 𝑟2 ∈ 𝐺2}

• How could we do intersection?

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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

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Nondeterminism

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?

Why not both?

Driving to a friend’s house Friend forgets to mention a fork in the directions Which way do you go?

Example Non-deterministic Finite Automaton

• 𝑈ℎ𝑗𝑠𝑒𝑀𝑏𝑡𝑢1 = {

| 𝑥 ∈ 0,1 ∗ the third from last character is a 1}

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𝑝𝑜𝑓 0,1 start 𝑜𝑓𝑦𝑢1 𝑜𝑓𝑦𝑢2 0,1 0,1 1

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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

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Union Using Non-Determinism

25 start

0,1 1

Some 0s No 0s

1 Even 1 1

Odd

New start

Union Using Non-Determinism

26 start

0,1 1

Some 0s No 0s

1 Even 1 1

Odd

new 𝜁 𝜁 𝑁∪ = 𝑅1 ∪ 𝑅2 ∪ 𝑜𝑓𝑥 , Σ, 𝜀∪, 𝑜𝑓𝑥, 𝐺

1 ∪ 𝐺 2

𝜀∪(𝑟, 𝜏) = ቊ 𝜀1 𝑟, 𝜏 if 𝑟 ∈ 𝑅1 {𝜀2 𝑟, 𝜏 } if 𝑟 ∈ 𝑅2 𝜀∪ 𝑜𝑓𝑥, 𝜁 = {𝑡𝑢𝑏𝑠𝑢, 𝑓𝑤𝑓𝑜}

What’s the language?

27

start

final

0, 𝜁

NFA Example

{𝑥 ∈ 0,1 ∗|𝑥 contains 0101}

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