e)answer)
regular expression is a(a+b)*b
f)answer)
regular expression is (ab)* where n start from 0
g)answer)
regular expression is (a+b)*aaaa*(a+b)*
h)answer)
regular expression is (a+b)*bab(a+b)*
i)answer)
regular expression is aaaa*(a+b)*
e)answer)
regular expression is a(a+b)*b
f)answer)
regular expression is (ab)* where n start from 0
g)answer)
regular expression is (a+b)*aaaa*(a+b)*
h)answer)
regular expression is (a+b)*bab(a+b)*
i)answer)
regular expression is aaaa*(a+b)*
Give a regular expression for each of the following languages. e. {axb | x∈{a, b}*} f....
please tell me how to do (p), (s), (t). 85 Exercises EXERCISE 1 on for each of the following languages. Give a regular expression for each of the follow ke the machine from 0 back, a. label rip from 0 back co state 0 on an input b. {abc, xyz] c. a, b, d. {ax | x € {a,b]"} e axb | x € {a,b}} [ {(ab)"} assing through 0. bo a piece we already have a input string. So...
Give the regular expressions of the following languages (alphabet is ab): a. {w | w has a length of at least three and its second symbol is a b} b. {w | w begins with an a and ends with a b} c. {w | w contains a single b} d. {w | w contains at least three a's} e. {w | w contains the substring baba} d. {w | w is a string of even length} e. The empty...
L = {w|w contains the substring bab} give the regular expression that describes L are the 2 languages L and L* the same language? Is L(aba)* a regular language?
3) Construct a regular expression defining each of the following languages over the alphabet {a, b}. (a) L = {aab, ba, bb, baab}; (b) The language of all strings containing exactly two b's. (c) The language of all strings containing at least one a and at least one b. (d) The language of all strings that do not end with ba. (e) The language of all strings that do not containing the substring bb. (f) The language of all strings...
1. Design an NFA (Not DFA) of the following languages. a) Lw E a, b) lw contain substring abbaab) b) L- [w E 10,1,2) lsum of digits in w are divisible by three) c) L-(w E {0,1,2)' |The number is divisible by three} d) The language of all strings in which every a (if there are any) is followed immediately by bb. e) The language of all strings containing both aba and bab as substrings. f L w E 0,1every...
For each of the languages listed below, give a regular expression that generates the lan- guage. Briefly justify your answer. (a) The set of strings over (a, b such that any a in the string is followed by an odd number of b's. Examples: bbbab E L, but abb f L. (b) The set of strings over fa, b in which there is an a in every even position and the total number of b's is odd, where the first...
Give a regular expression for these languages i) {w| w is a word of the alphabet = {0,1} that represents an integer in a binary form that is a multiple of 4} ii) {w belongs to {0,1,2}* | w contains the string ab exactly 2 times but not at the end} iii) { w belongs to {0,1,2}* | w=uxvx that x belongs to {0,1,2} u,v belongs to {0,1,2}* and there isn't any string y in the sequence v that x<y}
given ∑ = {a,b}: 1. describe in English the languages denoted by the regular expression: (a+b)*b(a+b)* 2. Write a regular expression: L(w) = {w | w has exactly a single substring abaa or exactly a single substring babb} 3. Write a regular expression for the following language: L(w) = {w | w ends in bb and does contain the substring aba}
Provide a regular expression for the following languages: (a) the set of all strings over {a, b} that start with ab and end with ba, (b) the set of strings over {a, b} where four consecutive occurrences of both letters occur in every word.
Automata Theory - Finding a regular expression for each of the following languages over {a,b} or {0,1}: I've written the solution . Please show steps on how to approach the problems that I mentioned in parentheses. The ones where I put my own regular expression check and see if it's still right. Thanks Strings with .... odd # of a's ---> (b*ab*ab*)b*ab* even # of 1's ---> 0*(10*10*)* ---> my answer was 0*10*10* (is this still right?) start & end...