BU CS 332 Theory of Computation Lecture 4: Reading: Non-regular - - PowerPoint PPT Presentation

BU CS 332 – Theory of Computation

Lecture 4:

• Non-regular languages
• Pumping Lemma

Reading: Sipser Ch 1.4

Mark Bun February 3, 2020

The Philosophical Question

• We’ve seen techniques for showing that languages are

regular

• Could it be the case that every language is regular?

2/3/2020 CS332 - Theory of Computation 2

Regular?

Construct an NFA for the following languages

2/3/2020 CS332 - Theory of Computation 3

{0𝑜1𝑜 | 0 < 𝑜 ≤ 2} 0𝑜1𝑜 0 < 𝑜 ≤ 𝑙} {0𝑜1𝑜 | 𝑜 > 0}

Proving a language is not regular

Theorem: 𝐵 = {0𝑜1𝑜 | 𝑜 > 0} is not regular Proof: (by contradiction) Let 𝑁 be a DFA with 𝑙 states recognizing 𝐵 Consider running 𝑁 on input 0𝑙1𝑙

2/3/2020 CS332 - Theory of Computation 4

Regular or not?

2/3/2020 CS332 - Theory of Computation 5

𝐷 = { 𝑥 | 𝑥 has equal number of 1s and 0s} 𝐸 = { 𝑥 | 𝑥 has equal number of 10s and 01s}

The Pumping Lemma

A systematic way to prove that a language is not regular

2/3/2020 CS332 - Theory of Computation 6

Why do we teach this?

Cons:

• The statement is difficult (5 quantifiers!)
• Some non-regular languages can still be pumped

Pros:

• Proof illuminates essential structure of finite automata
• Generalizes to other models of computation / classes of

languages (CFLs, self-assembly)

• Applying it can be fun!

2/3/2020 CS332 - Theory of Computation 7

Intuition for the Pumping Lemma

Imagine a DFA with 𝑞 states that recognizes strings of length > 𝑞

2/3/2020 CS332 - Theory of Computation 8

Idea: If you can go around the cycle once, you can go around 0 or 2,3,4… times

Pumping Lemma (Informal)

2/3/2020 CS332 - Theory of Computation 9

𝑟0 𝑟𝑗 𝑟𝑘 𝑟|𝑥| … 𝑦 𝑧 𝑨 Let 𝑀 be a regular language. Let 𝑥 be a “long enough” string in 𝑀. Then we can write 𝑥 = 𝑦𝑧𝑨 such that 𝑦𝑧𝑗𝑨 ∈ 𝑀 for every 𝑗 ≥ 0. 𝑗 = 0: 𝑗 = 2: 𝑗 = 1: 𝑗 = 3:

Pumping Lemma (Formal)

2/3/2020 CS332 - Theory of Computation 10

Let 𝑀 be a regular language. Then there exists a “pumping length” 𝑞 such that

• 1. |𝑧| > 0
• 2. |𝑦𝑧| ≤ 𝑞
• 3. 𝑦𝑧𝑗𝑨  𝑀 for all 𝑗 ≥ 0

For every 𝑥 ∈ 𝑀 where |𝑥| ≥ 𝑞, 𝑥 can be split into three parts 𝑥 = 𝑦𝑧𝑨 where: Example: Let 𝑀 = {𝑥 | all 𝑏’s in 𝑥 appear before all 𝑐’s} ; 𝑞 = 1

Using the Pumping Lemma

2/3/2020 CS332 - Theory of Computation 11

Theorem: 𝐵 = {0𝑜1𝑜 | 𝑜 > 0} is not regular Proof: (by contradiction) Assume instead that 𝐵 is regular. Then 𝐵 has a pumping length 𝑞. What happens if we try to pump 0𝑞1𝑞? If 𝐵 is regular, 𝑥 can be split into 𝑥 = 𝑦𝑧𝑨, where

• 1. |𝑧| > 0
• 2. |𝑦𝑧| ≤ 𝑞
• 3. 𝑦𝑧𝑗𝑨  𝐵 for all 𝑗 ≥ 0

General Strategy for proving 𝑀 is not regular

2/3/2020 CS332 - Theory of Computation 12

Proof by contradiction: assume 𝑀 is regular. Then there is a pumping length 𝑞.

Pumping Lemma as a game

2/3/2020 CS332 - Theory of Computation 13

• 1. YOU pick the language 𝑀 to be proved nonregular.
• 2. ADVERSARY picks a possible pumping length 𝑞.
• 3. YOU pick 𝑥 of length at least 𝑞.
• 4. ADVERSARY divides 𝑥 into 𝑦, 𝑧, 𝑨, obeying rules of the Pumping

Lemma: |𝑧| > 0 and |𝑦𝑧| ≤ 𝑞.

• 5. YOU win by finding 𝑗 ≥ 0, for which 𝑦𝑧𝑗 𝑨 is not in 𝑀.

If regardless of how the ADVERSARY plays this game, you can always win, then 𝑀 is nonregular

Example: Palindromes

2/3/2020 CS332 - Theory of Computation 14

Claim: 𝑀 = 𝑥𝑥𝑆 𝑥 ∈ 0,1 ∗} is not regular Proof: Assume 𝑀 is regular with pumping length 𝑞

• 1. Find 𝑥 ∈ 𝑀 with 𝑥 > 𝑞
• 2. Show that 𝑥 cannot be pumped

Intuitively

Example: Palindromes

2/3/2020 CS332 - Theory of Computation 15

Claim: 𝑀 = 𝑥𝑥𝑆 𝑥 ∈ 0,1 ∗} is not regular Proof: Assume 𝑀 is regular with pumping length 𝑞

• 1. Find 𝑥 ∈ 𝑀 with 𝑥 > 𝑞
• 2. Show that 𝑥 cannot be pumped

Formally If 𝑥 = 𝑦𝑧𝑨 with |𝑦𝑧| ≤ 𝑞, then…

Now you try!

2/3/2020 CS332 - Theory of Computation 16

Claim: 𝑀 = 0𝑗1𝑘 𝑗 > 𝑘 ≥ 0} is not regular Proof: Assume 𝑀 is regular with pumping length 𝑞

• 1. Find 𝑥 ∈ 𝑀 with 𝑥 > 𝑞
• 2. Show that 𝑥 cannot be pumped

Intuitively

Now you try!

2/3/2020 CS332 - Theory of Computation 17

Claim: 𝑀 = 0𝑗1𝑘 𝑗 > 𝑘 ≥ 0} is not regular Proof: Assume 𝑀 is regular with pumping length 𝑞

• 1. Find 𝑥 ∈ 𝑀 with 𝑥 > 𝑞
• 2. Show that 𝑥 cannot be pumped

Formally If 𝑥 = 𝑦𝑧𝑨 with |𝑦𝑧| ≤ 𝑞, then…

Choosing wisely

2/3/2020 CS332 - Theory of Computation 18

Claim: 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 = 𝑥 𝑥 has an equal # of 0s and 1s} is not regular Proof: Assume 𝑀 is regular with pumping length 𝑞

• 1. Find 𝑥 ∈ 𝑀 with 𝑥 > 𝑞
• 2. Show that 𝑥 cannot be pumped

Formally If 𝑥 = 𝑦𝑧𝑨 with |𝑦𝑧| ≤ 𝑞, then…

Reusing a Proof

2/3/2020 CS332 - Theory of Computation 19

Pumping a language can be lots of work… Let’s try to reuse that work!

0𝑜1𝑜 𝑜 ≥ 0} = 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 ∩ 𝑥 all 0s in 𝑥 appear before all 1s}

How else might we show that 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 is regular?

Using Closure Properties

2/3/2020 CS332 - Theory of Computation 20

𝐶 𝐷 𝐵 ∩ = (Not regular) If 𝐵 is not regular, we can show a related language 𝐶 is not regular any of {∘, ∪, ∩} or, for one language, {¬, R, *} By contradiction: If 𝐶 is regular, then 𝐶 ∩ 𝐷 (= 𝐵) is regular. (regular) But 𝐵 is not regular so neither is 𝐶!

Example

2/3/2020 CS332 - Theory of Computation 21

Prove 𝐶 = {0𝑗1𝑘|𝑗 ≠ 𝑘} is not regular using nonregular language 𝐵 = 0𝑜1𝑜 𝑜 ≥ 0 and regular language 𝐷 = 𝑥 all 0s in 𝑥 appear before all 1s}