Consider the CFG: S -> XYX X -> aX | bX | λ Y -> bbb Prove that this generates the language of all strings with a triple b in them, which is the language defined by (a + b)*bbb(a + b)*
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Consider the CFG:
S -> XYX
X -> aX | bX | λ
Y -> bbb
Prove that this generates the language of all strings with a triple b in them, which is the language
defined by (a + b)*bbb(a + b)*
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- Write a CFG to generate the language of all strings that have more a's than b's (not necessarily only one more, as with the nonterminal A for the language EQUAL, but any number more a's than b's). {a aa aab aba baa aaaa aaab ... }Create a PDA that accepts the language L = {w ∈ {a.b}* : na(v) ≥ nb(v), where v is any prefix of w}Use the Pumping Lemma to show the following language L is NOT regular. L= {w E (a, b, c}* ; where w has equal number of a's, b's, and c's}
- Construct a PDA that matches all strings in the language over {x.y} such that each string has at least twice as many y's as x's. Editor Below, give a short description of the set of strings associated with each state of your PDA.Construct a CFG and PDA for the language B={w| w is a string over {a,b} with the same count of a's and b's}. For example, abba € B, abab e B,aabb € B, babbbaaaba € B, that is, there is no particular order in which a's and b's appear in w as long as there is the same amount of them.Implement the following two functions that allow breaking a string into non-empty tokens using a given delimiter in c language . For example, ● For a string “abc-EFG-hi”, and a delimiter ‘-’ : the list of tokens is [“abc”, “EFG”, “hi”] ● For a string “abc-EFG---hi-”, and a delimiter ‘-’ : the list of tokens is [“abc”, “EFG”, “hi”] ● For a string “abc”, and a delimiter ‘ ’ : the list of tokens is [“abc”] ● For a string “++abc++”, and a delimiter ‘+’ : the list of tokens is [“abc”] That is, we break the string using the given delimiter, and the tokens are only the non-empty substrings.
- Implement the following two functions that allow breaking a string into non-empty tokens using a given delimiter in c language . For example, ● For a string “abc-EFG-hi”, and a delimiter ‘-’ : the list of tokens is [“abc”, “EFG”, “hi”] ● For a string “abc-EFG---hi-”, and a delimiter ‘-’ : the list of tokens is [“abc”, “EFG”, “hi”] ● For a string “abc”, and a delimiter ‘ ’ : the list of tokens is [“abc”] ● For a string “++abc++”, and a delimiter ‘+’ : the list of tokens is [“abc”] That is, we break the string using the given delimiter, and the tokens are only the non-empty substrings. The function count_tokens gets a string str, and a char delim, and returns the number of tokens in the string separated by delim. int count_tokens(const char* str, char delim); For example● count_tokens("abc-EFG--",'-')needstoreturn2. ● count_tokens("++a+b+c",'+')needstoreturn3.● count_tokens("***",'*')needstoreturn0.The function get_tokens gets a string str, and a char delim, and returns the…Rewrite BNF to EBNF BNF: → + Kexpr> - → * / | → ** → () idDesign DPDA to accept language L= {xe [a, b]* N (x) > N,(x)). N(x) > N,(x) means number of a's are greater than number of b's in string x
- In C language, implement the following two functions that allow breaking a string into non-empty tokens using a given delimiter. For example, For a string "abc-EFG-hi", and a delimiter '-': the list of tokens is ["abc", "EFG", "hi"] For a string "abc-EFG---hi-", and a delimiter '-': the list of tokens is ["abc", "EFG", "hi"] For a string "abc", and a delimiter ' ': the list of tokens is ["abc"] For a string "++abc++", and a delimiter '+': the list of tokens is ["abc"] That is, we break the string using the given delimiter, and the tokens are only the non-empty substrings. 1. The function count_tokens gets a string str, and a char delim, and returns the number of tokens in the string separated by delim. int count_tokens(const char* str, char delim); For example count_tokens("abc-EFG--", '-') needs to return 2. count_tokens("++a+b+c", '+') needs to return 3. count_tokens("***", '*') needs to return 0. 2. The function get_tokens gets a string str, and a char delim, and returns the…Construct a PDA that matches all strings in the language over {x,y} such that each string has an odd length and its middle symbol is 'x'. Editor Below, give a short description of the set of strings associated with each state of your PDA.Question 2: Ʃ = {a, b} Give the regular expression for all strings in which any b’s that occur are in groups of an odd number at a time. Some sample strings in the language aabaabbbab, bbbbbabbbaab, aaaa, ababab Some sample strings NOT in the language abba, babbbb