To show that the set of all finite strings over the alphabet {0,1,2,3} is countable, you would define a bijection f: (0,1,2,3} - N, where f(A)=0 (remember that A is the empty string), listing strings in string order (with 0<1<2<3). Under this bijection, the f(313) = Your Answer: Answer
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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…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 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…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.Please help me fill in the blanks The reverse of a string x, denoted rev(x), is the string obtained by writing x backwards. For example, if x = abbaaab then rev(x) = baaabba. If A is a subset of Σ*, the reverse of A, denoted rev(A), is the subset of Σ* consisting of all reverses of strings in A: rev(A)={rev(x) | x∈A} For example, rev({a,ab,aab}) = {a,ba,baa}. Suppose you are given a deterministic finite automaton M accepting a set A. Show how to construct a DFA N accepting rev(A). (Kozen would suggest putting pebbles on the final states of M and moving them backwards along transition edges.) Describe N formally (i.e. in terms of Q, δ, etc.) including the definition of acceptance. Hint: the subset construction will be helpful here. You can start with M = (Q, Σ, δ, s, F) and define the components of N in terms of the components of M. 1. DFA N = (Q_N, Σ, δ_N, s_N, F_N) where Q_N = ___ 2. DFA N = (Q_N, Σ, δ_N, s_N, F_N) where δ_N = ___ 3. DFA N = (Q_N, Σ, δ_N, s_N, F_N)…
- Write Regular expression for the set of strings over {0, 1} that have atleast one 1.Question # 6 Make a suitable for machine for : L1={All strings that having prefix containing first 3 letters of your name separated by + or -, such that if there exits a vowel then there is + in between the letters and if there exists two consonants then there is - sign between them e.g. for the name ASM the string accepted will be A+S-M *Log There was a storm recently on Jolibi village. The storm was so strong that some treesfell. There are some logs of varied length lying on the ground. The village ground canbe represented by a string of length N, where the i-th character is either 1 or 0. A singlelog is represented by consecutive characters of 1, and two different logs are separated byone or more 0. For example, for the string 1100010111, there are 3 logs. The first one atposition 1 to 2 with length 2, the second one at position 6 with length 1, and the thirdone at position 8 to 10 with length 3.As a carpenter, you want to take one of these logs home. Because you are the seniorcarpenter, you may take the longest log home. Determine the length of the longest log! Format InputThe first line contains an integer N, the length of the string. The next line contains a string of length N, which represents thevillage ground. Format OutputOutput an integer X, the length of the longest log. Constraints• 1 ≤ N ≤ 104• the i-th…
- 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…Please do this in JAVA PROGRAMMING. Given: List L of pairs of charactersString SOutput: TRUE if S is a valid string, FALSE otherwise. A string is considered valid if each character in the string can be paired with another character in the string, where the pair belongs to the input list L. Furthermore, two pairs cannot cross each other. In other words, a pair most completely enclose another, or be completely separate. Design and implement an efficient dynamic programming solution to this problem. Examples: Input L: (a b) (b c) (c d) (a a) Input S: aaba Output: True (pairs shown color-coded: aaba, "aa" fully encloses "ab") Input L: (a b) (b c) (c d) (a a) Input S: abcaad Output: True (pairs shown color-coded: abcaad, "ab" is separate from the other pairs) Input L: (a b) (b c) (c d) (a a) Input S: acbd Output: False (acbd is not valid because the pairs cross each other.) Input L: (a b) (b c) (c d) (a a) Input S: aaac Output: FalseLet I be the set of strings (N, w) where N is an NFA that accepts the string w. Pick all that apply. L is decidable L recognisable. None of the above.