(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
Then M = (Q
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