Suppose L is a subset of {a, b}". If xo,x1,... is a sequence of distinct strings in {a, b}* such that for every n> 0, xn and xn+1 are L-distinguishable, does it follow that the strings xo, x1,... are pairwise L- distinguishable?
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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.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…4. For E = {a, b}, construct dfa's that accept the sets consisting of (d) all strings with at least one b and exactly two a's. (e) all the strings with exactly two a's and more than three b's.
- For E= {a,b}, onstruct dfa's that accept the sets consisting of (a) all strings with exactly one a, (b) all strings with at least one a, (c) all strings with no more than three a's, (d) all strings with at least one a and exactly two b’s, (e) all the strings with exactly two a's and more than two b’s.L1 = {u ∈ Σ∗| u ends with aa}.L2 = {u ∈ Σ∗| u ends and begins with different letters }.L3 = {u ∈ Σ∗| u contains abba}.L4 = {u ∈ Σ∗| 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 ∩ L2.(b) Prove that L5 is regular.(c) Construct an NFA M that accepts L5, and prove its correctness.(d) Convert such NFA (from question above) to a DFA using the algorithm from class.(e) Given the DFA you’ve designed in Q3(d), what is δ(s,au), for u ∈ Σ∗. What is δ(s,buaa), foru ∈ Σ∗. Justify your answers. Note that s is the starting state of the given DFA.Let A ={a, b, c} and B ={a, b} a. Is A a subset of B? b. Is Ba subset of A? c. What is A U B? d. What is A n B? e. What is A n B ? f. What is A x B?
- Construct an NFA for the set of all strings w *epsilon* {a,b}*; such that (1) w contains both aa and bb, or (2)w does not contain aa and w does not contain bb.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…2. Let n be a positive integer, and let A be a list of positive integers. We say that the integer n can be factorized by A if there exists a sequence of integers 11, 12,..., ik (with 0 ≤i; < len (A), for j = 1,..., k) such that n is the product of the integers A[i], A[i2],..., A[ik]. Write an efficient algorithm factorizable (n, A) that returns True if n can be factorized by A, and False otherwise. Prove that your algorithm is correct, and bound its running time. Larger scores will be awarded to faster solutions. Example 1: if n = 6 and A [4,2,3], then factorizable (n, A) should return True, L[1] * L[2]. since n == Example 2: if n = 8 and A = L[0] * L[0] * L[0]. [2,5], then factorizable (n, A) should return True, since ,2,3,5,7], then factorizable (n, A) should return Example 3: if n = 13 and A = False since n cannot be factorized by A.
- A substring of a string is a contiguous sequence of characters from the string. For example, BC is a substringof ABCD which starts from the second character of ABCD. Another example, ABC is a substring of ABCD whichstarts from the first character of ABCD. Note that ABCD itself is also a substring of ABCD.In this problem, we define a special substring as a non-empty substring that contains only the same character.For example, B and CC are special substrings of ABBCCC, while ABBC and BC are not special substrings.You are given a string S of length N and an integer K. Your task is to determine the minimum number ofcharacters of S that need to be changed such that there exists a special substring of length K in S.For example, let N = 6, K = 4, and S = ABBCCC. In this example, we only need to change the third characterof S to C (i.e. ABBCCC → ABCCCC) so that we have a special substring CCCC of length 4.InputInput begins with a line containing two integers: N K (1 ≤ K ≤ N ≤ 100 000)…The reverse of a string x denoted by rev(x) and is defined by the following recursiverule: •rev(ε) = ε•rev(xa) = a.rev(x) Prove that if A is a regular set then so is the following set:rev(A) = {rev(x) : x ∈A}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.