Build a TM over the alphabet Σ = {a,b} that accepts the language where each word has twice as many b’s as a’s.
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Build a TM over the alphabet Σ = {a,b} that accepts the language where each word has twice as
many b’s as a’s.
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- A uwuified sentence is sentence that has been transformed using a made-up Internet language in which some of the letters in the words are replaced by something else. The exact scheme is described below: Any uppercase/lowercase R or L is replaced by w/w, respectively. • If we encounter an o/o in a word, check if the previous letter (if it exists) is an M/m or N/n. If the previous letter is one of these, insert the lowercase letter y in between them, regardless of the capitalization of the other letters. • All other characters are left unchanged. Some examples: Professor will be converted to Pwofessow (There are two r's that are replaced by w's. Since the two o's aren't proceeded by an M/n or N/n, no y will be inserted.) LLunoacyo will be converted to wwunyoacyo (The two L's will be replaced with two ws according to the first rule. Then the first o will have a y inserted in front of it between then and the o according to the second rule. The last o won't have a y inserted in between…Consider the alphabet set Σ = {a b}; Write Regular Expressions for the followingdefinitions.1. Language of All those words which do not contains double a’s.2. Language of All those words which do not contains double letters.3. Language of All those words which has at most three letters.4. Language of All those words which do not contains substring ab.5. Language of All those words in which total number of a’s are divisible by 5 6. Language of All those words which do not contains substring baa.7. Language of All those words whose length is odd and they contain ab.8. Language of All those words which ends with bb and have odd length.9. Language of All those words which do not end with ba.10. Language of All those words which contains ODD a’s and ODD b’sLet L = { w | w cannot be written as st#ts with s, t {a, b}* }. Show that L is not regular.
- Finite language is a language with finite number of strings in it, i.e., there exist exactly k strings in this language such that k eNand k #00. For a finite language L, let |L| denote the number of elements of L. For example, |{A, a, ababb}| = 3. (Do not mix up with the length |x| of a string x.) The statement |L,L2| = |L1||L2| says that the number of strings in the concatenation LL2 is the same as the product of the two numbers |L1| and |L2|. Is this always true? If so, prove, and if not, find two finite languages L1, L2 S {a, b}* such that |L1L2| # |Li||L2l.IN HASKELL PROGRAMMING LANGUAGE PLEASE In case you do not know it: the game is played on a 3x3 grid that is initially empty. Two players are playing, by alternatingly making moves. A move by a player places their token (an X for player 1, an O for player 2) into a cell that was empty. We are using algebraic notations for indexing the positions in the board, with A,B,C indexing the columns and 1,2,3 the rows. Specifically, these coordinates would be used in the implementation for moves made by a human player. If the X X O Figure 1: Sample board position same token appears 3 times in any of the three columns, three rows or two main diagonals the game is over and that player wins. If the grid is filled without that happening the game is a draw. For the depicted board, we have Xs in positions C3 and A2, and an O in position B1. It would be O’s turn to make a move; a legal move would be C2, but it is not a good move, because X can force a win by responding A1. O cannot force a win, but…Consider the alphabet set Σ = {a b}; Write Regular Expressions for the following definitions. 7. Language of All those words whose length is odd and they contain ab.8. Language of All those words which ends with bb and have odd length.9. Language of All those words which do not end with ba.
- 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…L1={u €E•[u ends with aa}. 12 = {u €E•[u ends and begins with different letters }. L3 = {u €I•|u contains abba). L4 = {u €E•[u is of the form anbamfor n,m> 0}. Given the above languages: (a) Use the set operators 'union' and 'complement' to describe L5 = L1 n L2. (b) Prove that L5 is regular.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…Write a java code to implement four new functions DFA Union: Given Two DFAs, M1 and M2, create a DFA which accepts the union of the languages of M1 and M2. Intersection: Given Two DFAs, M1 and M2, create a DFA which accepts the intersection of the languages of M1 and M2. Difference: Given Two DFAs, M1 and M2, create a DFA which accepts the set difference of the languages of M1 and M2, that is, L(M1) - L(M2), the set of strings accepted by M1 but not M2. Complement: Given DFA M, create a DFA which accepts the complement of the language of M.A = {x| x = 3d + 2, d e N}, B = {x € Z]a? + 9 = 0}, C = {1,5}. Also assume the Universal set is N: Show the following: (a) (Cn A) – B (b) (AUC) – B (с) Р(P(C - В)nA)