(i) Every regular language has a regular proper subset (j) If L1 and L2 are nonregular languages, then L1 ∪ L2 is also not regular 4 Show that the language L
Home Pumping
If L1 ⊆ L2 and L2 is regular, then L1 must be regular There are subsets of a regular language which are not regular If L1 and L2 are nonregular, then L1 ∪ L2 must be nonregular If F is a finite language and L is some language, and L − F is a regular language, then L must be a regular language
lecture
For the true ones (if any) give a proof outline (a) Union of two non-regular languages cannot be regular Ans: False Let L1 = {ambn m ≥ n} and L2 = { ambn
Solution
4 1 Closure Properties of Regular Languages Closure under Simple Set Operators Thm 4 1: If L1 and L2 are regular languages, then so are L1 ∪L2,L1 ∩
Then, M = (Q,Σ, δ, q0,Q \ F) (i e , switch accept and non-accept states) accepts L Regular Languages are closed under intersection, i e , if L1 and L2 are
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This theorem is used in several proofs that certain languages are not context free The usual proof of this theorem is a cross product construction of a PDA and a
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You do not have to be a declared CS major and If L1 and L2 are languages over the alphabet Σ, the If L1 and L2 are regular languages, is L1 ∪ L2? start
Small
L2 = {0n1n n > 10} ○ Correct answer is (A): Only L1 ○ Why is L1 regular? – Easy: because it is a finite language (of size 10) ○ But why is L2 not regular?
Lect CSE ac
Theorem If a language L is accepted by a finite the suffix need not be proper} Union : Let L1 and L2 be two regular languages generated by two right
FSAcontd I
(In each case a fixed alphabet. ? is assumed.) (a) Every subset of a regular language is regular. (b) Let L? = L1 ? L2. If
There are subsets of a regular language which are not regular. 2. If L1 and L2 are nonregular then L1 ? L2 must be nonregular. 3.
For the true ones (if any) give a proof outline. (a) Union of two non-regular languages cannot be regular. Ans: False. Let L1 = {ambn
You may not leave the room during the examination even to go to the bathroom. If L1 and L1 ? L2 are regular languages
Hence ATM is not Turing-recognizable. 5. Let L1
17 oct. 2013 (f) If L1 and L2 are finite languages then
11 août 2000 n<N. This string is not in L a contradiction. ... (1) If L1 ? L2 is regular and L1 is finite
No late submissions will be accepted. 1. [20 points] If L1 and L2 are two b) If L1 = {1}? and L2 = {010} write down a regular expression that ...
L2 = decimal representations of nonnegative integers without leading 0's divisible by 2. L2 = L1 ? ?*{0 2
Thm 4 4: If L1 and L2 are regular languages then L1=L2 is regu-lar: The family of regular languages is closed under right quotient with a regular language Proof: 1 Assume that L1 and L2 are regular and let DFA M = (Q;?;–;q0;F) accept L1 2 We construct DFA Md = (Q;?;q0;Fc) as follows (a) For each qi 2 Q determine if there is a y 2
(i) Every regular language has a regular proper subset (j) If L1 and L2 are nonregular languages then L1 ? L2 is also not regular 4 Show that the language L = {anbm: n ? m} is not regular 5 Prove or disprove the following statement: If L1 and L2 are not regular languages then L1 ? L2 is not regular 6
Common Mistake: L1 and L2 are regular but that does not mean that L1 and L2 are finite (all finite languages are regular but not all regular languages are finite!) Problem 6 (10 points) Given the following state diagram of an NFA over the alphabet ? = {a b} convert it into the state diagram of its equivalent DFA
The set L = {0n1n n ? 0} is not regular Using the Pumping Lemma Claim: The set L = {0n1n n ? 0} is not regular Proof: Assume towards a contradiction that L is regular Therefore the Pumping Lemma applies to L and gives us some number p the pumping length of L
1 Closure properties of Regular Languages: By de nition of regular languages we can say: If L1 and L2 are two regular languages then L1[L2 L1L2 L1 are regular 2 How about L1L2 L1 L2 ? also regular 3 Regular languages are closed under union concata-nation Kleene star set intersection set di erence etc 4
L2 and L1 regularization for linear estimators A Bayesian interpretation of regularization If = 0 the solution is the same as in regular least-squares linear
What if L1 is regular?
(b) Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. FALSE. We know that the regular languages are closed under intersection. But it is important to keep in mind that this closure lemma (as well as all the others we will prove) only says exactly what it says and no more.
How to prove that L is not regular?
9. (a) L is not regular. We can prove this using the pumping lemma. Let w = aNbN. Since y must occur within the first N characters of w, y = ap for some p > 0. Thus when we pump y in, we will have more a’s than b’s, which produces strings that are not in L.
Which subset of a regular language is regular?
Every subset of a regular language is regular. Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. If L is regular, then so is L? = {xy : x ? and y ? L}.
Why is the intersection of L and LR regular?
is regular because the problem statement says so. LR is also regular because the regular languages are closed under reversal. The regular languages are closed under intersection. So the intersection of L and LR must be regular.
CS 341 Homework 9 Languages That Are and Are Not Regular