WebLet L = { w ∈ ( 0 + 1 ) * w has even number of 1s } , i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expressions below WebStep: 02 For L1, the strings can be read as: w is a word such that it contains any combination of 0's or 1's (including NULL value), where the number of 0's is equal to number of 1's. Sample strings: {, Ʌ 01, 10, 1010, 0011, 1001, 1100, . . .}For L2, the strings can be read as: w is a word such that it contains any combination of 0's or 1's, where it contains odd number …
CSE 105, Fall 2024 - Homework 2 Solutions - University of …
WebJul 20, 2024 · DFA that accepts strings where there are odd number of 1's, and any number of 0's. The alphabet Σ = { 0, 1 } Well since it's odd 1 's, then there must be at least one 1. So I think the regex is the following: R = 0 ∗ 10 ∗ This is correct because I can have any number of 0 ′ s, but also incorrect because I do not offer a way to add more 1's. WebAug 28, 2016 · L1= {€, 00,0000,.....} FA1 The language accepted for numbers of 1's divisible by 3 is L2= {€,111, 111111,...} FA2 The finite automata that accepts the string having even numbers of 0's and numbers of 1's divisible by 3 is shown below: Transition diagram Share Follow edited Jul 3, 2024 at 20:01 flyingfishcattle 1,525 3 11 25 osx touchscreen
[Solved] Let L = {w ∈ (0 +1) * w has even number of - Testbook
WebHint: L0 can be read as: w is a word such that w contains any possible combination of 0 and 1 including null value (Kleene star closure property applied here) where the number of 0s is equal to the number of 1s. Sample Strings: {Ʌ, 01, 10, 1010, 0011, 1001, 1100, ...} L1 = {w ∈ {0,1} ∗: the number of 0s in w equals the number of 1s in w} WebJun 13, 2024 · After this we need to see an even number of $(ab+ba)$ blocks (remember zero is an even number), possibly interspersed with zero or more $(aa+bb)$ blocks. I'll let … WebOct 28, 2013 · 1 L = { w w is of even length and begins with 01 } Ans: 01 ( (0 + 1) (0 + 1))* Explanation: 01 itself of even length to, we can suffix any even length string consist of 0 s and 1 s. L = { w the numbers of 1's in w is multiple of 3 } Ans: (0*10*10*10*)* rock creek saddlebred horse show