w W 2. Consider alphabet Σ = regular. {0,1} and language L = {wer : [w; # Σ Σ (1 - w;)). Prove or disprove that L is i=1 i=1
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- w |w Consider alphabet Σ = {0,1} and language L = = {wες : Σw # i=1 i=1 regular. Σ (1 - w;)}. Prove or disprove that L isFalse For each language L over a given alphabet E, let 1(L, E) denote the language over E obtained from L by generating all permutations of its strings - that is, r(L, E) -{we E*: there is a string v in L such that w is a permutation of v ). Specify which one of the following statements is true. Choose None of the other statements is true. The language r((0b1)*, (0, b,1}) is decidable. The language n((Ob1)*, (0, b,1}) is regular. There is no regular language L over an alphabet E such that the language n(L, E) fails to be context-free. Consider alnhaber E-Lo and context-free grammar G over E with precisely oneLet ∑ = {a, b, #} and L = { w | w cannot be written as t#s#t 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.∑ = {C,A,G,T}, L = { w : w = CAjGnTmC, m = j + n }. For example, CAGTTC ∈ L; CTAGTC ∉ L because the symbols are not in the order specified by the characteristic function; CAGTT ∉ L because it does not end with C; and CAGGTTC ∉ L because the number of T's do not equal the number of A's plus the number of G's. Prove that L ∉ RLs using the RL pumping theorem.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…
- Design in JFLAP a simulator of a finite deterministic automaton that (only) recognizes Even length binary numbers Example: = {00000000, 0100, 111110, etc.}Show that L = {an! : n ≥ 1} is not context-free.Give English language translations of the following wffs if L(x, y): x loves y H(x): x is handsome M(x): x is a man P(x): x is pretty W(x): x is a woman j: John k. Kathy L(x. y): x loves y а. Н() Л L(#) b. (Vx)[M(x) → H(x)] c. (Vx)(W(x) → (Vy)[L(x, y) → M(y) ^H(y)]) d. (3x)[M(x) ^ H(x) ^ L(x, k)] е. Вx) (W(x) Л Р() Л ()Le, y) — НО) Л М(У)) f. (Vx)[W(x) ^ P(x) →L(j, x)]
- TM M = (Q, E, I, 6, 90, 9a, qr), where Q = {90, 91, 92, 9a, 9r}, Σ = {0, 1}, r = {0, 1, L}, and 8 is: 8(qo, U) = (qr, U, R) 8(90, 0) = 8(go, 1) (91, 0, R) (qo, 1, R) = 8(g1, L) = (ga, U, R) 8(91,0) = (91, 0, R) 8(91, 1) = (92, 1, R) = (92, U, R) 8(92, U) 8(92, 0) = (90, 0, R) 8(92, 1) = (92, 1, R) i. Prove that M is NOT a decider. ii. Mathematically describe the language A that M recognises. Prove that A ≤ L(M). iii. Prove A = L(M). iv. Is A Turing-decidable? [Give clear reasons for your answer. No need for a formal proof.]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: FalseConstruct a CFG that generates L, where L is described below: L = {w|w€ {a, b}*, #a(x) = #b(x) }, Where #a(x) and #b(x) denote the number of a and b in x respectively. Provide arguments to convince the reader your solution is correct.