BU CS 332 – Theory of Computation Lecture 4: Reading: • Non-regular languages Sipser Ch 1.4 • Pumping Lemma 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 {0 𝑜 1 𝑜 | 0 < 𝑜 ≤ 2} 0 𝑜 1 𝑜 0 < 𝑜 ≤ 𝑙} {0 𝑜 1 𝑜 | 𝑜 > 0} 2/3/2020 CS332 - Theory of Computation 3
Proving a language is not regular Theorem: 𝐵 = {0 𝑜 1 𝑜 | 𝑜 > 0} is not regular Proof: (by contradiction) Let 𝑁 be a DFA with 𝑙 states recognizing 𝐵 Consider running 𝑁 on input 0 𝑙 1 𝑙 2/3/2020 CS332 - Theory of Computation 4
Regular or not? 𝐷 = { 𝑥 | 𝑥 has equal number of 1 s and 0 s } 𝐸 = { 𝑥 | 𝑥 has equal number of 10 s and 01 s } 2/3/2020 CS332 - Theory of Computation 5
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 Idea: If you can go around the cycle once, you can go around 0 or 2,3,4… times 2/3/2020 CS332 - Theory of Computation 8
Pumping Lemma (Informal) Let 𝑀 be a regular language. Let 𝑥 be a “long enough” string in 𝑀 . 𝑧 𝑦 𝑨 … 𝑟 0 𝑟 𝑗 𝑟 𝑘 𝑟 |𝑥| Then we can write 𝑥 = 𝑦𝑧𝑨 such that 𝑦𝑧 𝑗 𝑨 ∈ 𝑀 for every 𝑗 ≥ 0. 𝑗 = 0: 𝑗 = 2: 𝑗 = 1: 𝑗 = 3: 2/3/2020 CS332 - Theory of Computation 9
Pumping Lemma (Formal) Let 𝑀 be a regular language. Then there exists a “pumping length” 𝑞 such that 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 1. |𝑧| > 0 before all 𝑐’ s } ; 𝑞 = 1 2. |𝑦𝑧| ≤ 𝑞 3. 𝑦𝑧 𝑗 𝑨 𝑀 for all 𝑗 ≥ 0 2/3/2020 CS332 - Theory of Computation 10
Using the Pumping Lemma Theorem: 𝐵 = {0 𝑜 1 𝑜 | 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 0 𝑞 1 𝑞 ? If 𝐵 is regular, 𝑥 can be split into 𝑥 = 𝑦𝑧𝑨 , where 1. |𝑧| > 0 2. |𝑦𝑧| ≤ 𝑞 3. 𝑦𝑧 𝑗 𝑨 𝐵 for all 𝑗 ≥ 0 2/3/2020 CS332 - Theory of Computation 11
General Strategy for proving 𝑀 is not regular Proof by contradiction: assume 𝑀 is regular. Then there is a pumping length 𝑞 . 2/3/2020 CS332 - Theory of Computation 12
Pumping Lemma as a game 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 𝑦𝑧 𝑗 𝑨 is not in 𝑀 . If regardless of how the ADVERSARY plays this game, you can always win, then 𝑀 is nonregular 2/3/2020 CS332 - Theory of Computation 13
Example: Palindromes Claim: 𝑀 = 𝑥𝑥 𝑆 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 2/3/2020 CS332 - Theory of Computation 14
Example: Palindromes Claim: 𝑀 = 𝑥𝑥 𝑆 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 If 𝑥 = 𝑦𝑧𝑨 with | 𝑦𝑧 | ≤ 𝑞 , then… Formally 2/3/2020 CS332 - Theory of Computation 15
Now you try! Claim: 𝑀 = 0 𝑗 1 𝑘 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 2/3/2020 CS332 - Theory of Computation 16
Now you try! Claim: 𝑀 = 0 𝑗 1 𝑘 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 If 𝑥 = 𝑦𝑧𝑨 with | 𝑦𝑧 | ≤ 𝑞 , then… Formally 2/3/2020 CS332 - Theory of Computation 17
Choosing wisely Claim: 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 = 𝑥 𝑥 has an equal # of 0 s and 1 s } 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 If 𝑥 = 𝑦𝑧𝑨 with | 𝑦𝑧 | ≤ 𝑞 , then… Formally 2/3/2020 CS332 - Theory of Computation 18
Reusing a Proof Pumping a language can be lots of work… Let’s try to reuse that work! How else might we show that 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 is regular? 0 𝑜 1 𝑜 𝑜 ≥ 0} = 𝐶𝐵𝑀𝐵𝑂𝐷𝐹𝐸 ∩ 𝑥 all 0 s in 𝑥 appear before all 1 s } 2/3/2020 CS332 - Theory of Computation 19
Using Closure Properties If 𝐵 is not regular, we can show a related language 𝐶 is not regular ∩ 𝐶 𝐷 𝐵 = (Not regular) (regular) any of {∘ , ∪ , ∩} or, for one language, {¬ , R , * } By contradiction: If 𝐶 is regular, then 𝐶 ∩ 𝐷 (= 𝐵) is regular. But 𝐵 is not regular so neither is 𝐶 ! 2/3/2020 CS332 - Theory of Computation 20
Example Prove 𝐶 = {0 𝑗 1 𝑘 |𝑗 ≠ 𝑘} is not regular using nonregular language 𝐵 = 0 𝑜 1 𝑜 𝑜 ≥ 0 and regular language 𝐷 = 𝑥 all 0 s in 𝑥 appear before all 1 s } 2/3/2020 CS332 - Theory of Computation 21
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