. Give a CFG G that generates all odd length strings over {a,b}. B. Give a Push-down Automaton that accepts strings generated by G in Part a.
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A. Give a CFG G that generates all odd length strings over {a,b}.
B. Give a Push-down Automaton that accepts strings generated by G in Part a.
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- 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.Write a regular expression for the set of strings over the alphabet {a, b} that have at least one a andat least one b. For example, aaaba and bbaba are strings in the set, but aaaa and bbb are not.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…
- 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…Draw a DFA over the alphabet {a,b,c}that only accepts strings with no c’s and an even number of a’s, or an odd number of c’s and an odd number of b’s. For example, it should accept aa, aba, abbabaa, cbacabcb, cbcca, and abc but not aca, bacc, or aaa.Code in C++ only. Correct answer will upvoted else downvoted. framework of size n×m, with the end goal that every cell of it contains either 0 or 1, is considered lovely if the total in each adjoining submatrix of size 2×2 is actually 2, i. e. each "square" of size 2×2 contains precisely two 1's and precisely two 0's. You are given a network of size n×m. At first every cell of this network is unfilled. How about we indicate the cell on the crossing point of the x-th line and the y-th segment as (x,y). You need to handle the inquiries of three sorts: x y −1 — clear the cell (x,y), in case there was a number in it; x y 0 — compose the number 0 in the cell (x,y), overwriting the number that was there already (assuming any); x y 1 — compose the number 1 in the cell (x,y), overwriting the number that was there beforehand (assuming any). After each question, print the number of ways of filling the unfilled cells of the grid so the subsequent network is delightful. Since the appropriate…
- Must be new solution and run on GNU Common Lisp! Using Lisp, write a program that solves the Missionaries and Cannibals problem that uses a DFS( depth first search). It should use (mac start end). Start is the current state (which can be (3 3 l) and End is the goal state (which can be (0 0 r). This should output the sequences of moves needed to reach the end state from the start state. This should print nil if there is no solution. For example, the call should be something like this! Call: (mac '(3 3 l) '(0 0 r)) Output: ((3 3 l) (2 2 r) (3 2 l) (3 0 r) (3 1 l) (1 1 r) (2 2 l) (0 2 r) (0 3 l) (0 1 r) (1 1 l) (0 0 r))Give a regular expression for the language of strings over (a,b,c) in which the substring ab occurs no more than once. ({submit:RE-abc-not-more-than-one-ab)}}solve in C please. Implement the following two functions that get a string, and compute an array of non-emptytokens of the string containing only lower-case letters. For example:● For a string "abc EFaG hi", the list of tokens with only lower-case letters is ["abc", "hi"].● For a string "ab 12 ef hi ", the list of such tokens is ["ab","ef","hi"].● For a string "abc 12EFG hi ", the list of such tokens is ["abc","hi"].● For a string " abc ", the list of such tokens is ["abc"].● For a string "+*abc!! B" the list of such tokens is empty.That is, we break the string using the spaces as delimiters (ascii value 32), and look only at thetokens with lower-case letters only .1. The function count_tokens gets a string str, and returns the number ofsuch tokens.int count_tokens(const char* str);For example● count_tokens("abc EFaG hi") needs to return 2.● count_tokens("ab 12 ef hi") needs to return 3.● count_tokens("ab12ef+") needs to return 0.2. The function get_tokens gets a string str, and…
- of the form 1m0n1k, where m, n, k are odd. 3. Build a Finite Automaton to generate all string: of the form 1m0n1k, where m, n and k are greater than 0 and divisible by 3.1. Build a Finite Automaton to generate all strings with any combinations of 0's and 1's except for an empty string, i.e. empty string is not generated. 2. Build a Finite Automaton to generate all strings of the form 1m0n1k, where m, n, k are odd. 3. Build a Finite Automaton to generate all strings of the form 1m0n1k, where m, n and k are greater than 0 and divisible by 3. 4. Build a Finite Automaton with four states generating just one string 5. Build a Finite Automaton with four states generating just two strings 6. Build a Finite Automaton with four states generating Infinitely many strings. 7. Please convert the following Finite Automaton into DFA (a is a start state and d is a final state) ▪a,1 => b ▪a,0=>d ▪b,0=>c ▪c, 1=>c - C,0=>d ▪d,0=>a ■ 8. Please convert the owing Finite Automaton into DFA (a is a start state, while d and e are final states) 1. a,1 => b 2. a, 1=>c 3. b,0=>d 4. C,0=>e 9. Limitation Question. Using the idea of Pumping Lemma, please explain, why we can't build…Write more-.