Use the construction given in Theorem 1.39 to convert the NFA to equivalent DFA. Show that for each n >= 1 the language B n is regular. For each n>=1
Answer all questions on test paper. (b) {w : every odd position of w is a 1}. Solutions ... Show that for n ? 1 the language Bn is. [20] regular.
i.e. M accepts all of the strings in the language
Answer: A CFG is in Chomsky normal form if each of its rules has one of 3 forms: A Next show that a known NP-Complete language B can be reduced to C in ...
integer n > 1 . Denote the set of all binary relations on N by Bn ... Birget uses the following n regular languages in his proof in [1] : Lqi) = {RI .
Show that for each n > 1 the language Bn
1. Use the procedure described in Lemma 1.55 to convert the regular expression (((00)?(11))? Prove that the following languages are not regular.
19 avr. 2019 Key words and phrases: Difference hierarchy regular language
An analogous argument shows that every bounded decreasing sequence converges. Example 2. Let (an)n=12
As for regular limits to establish that lim{an} = ?
Show that for each n >= 1 the language Cn is regular By simulating binary division we create a DFA M with n states that recognizes Cn M has n states which keep track of the n possible remainders of the division process The start state is the only accept state and corresponds to remainder 0
1w 2 ···w n the reverse of w written as wR is the string w in reverse order w n ···w 2w 1 For any language A let AR = {wR w ? A} Show that if A is regular so is AR [20 points] Solution: One solution is recursively (or inductively) de?ne a reversing operation on regular expressions and apply that operation on the regular
To show that L is not regular we find one that isn’t To use the Pumping Theorem to show that a language L is not regular we must: 1 Choose a string w where w k Since we do not know what k is we must state w in terms of k 2 Divide the possibilities for y into a set of equivalence classes that can be considered together 3
n = f0k: where k is a multiple of n g Show that for n 1 the language B n is regular Solution: For any xed n B n is the language of strings of the form 0ni where i is any non-negative integer Note that in particular i = 0 gives " 2B n (for any n) Let A be a DFA designed as follows: / q 0 0/ q 1 / 2 0 /::: 0 / q n 1 0 h Clearly L(A) = B n
Regular languages are closed under complementation Proof Let A be a regular language We will show that A is regular Since A is regular a DFA M accepts it By turning all the accept states of M into non-accept and all non-accept states into accept every input in A ends up in a non-accept state and every input not in A ends up in an
(i) Every regular language has a regular proper subset (j) If L1 and L2 are nonregular languages then L1 ? L2 is also not regular 4 Show that the language L = {anbm: n ? m} is not regular 5 Prove or disprove the following statement: If L1 and L2 are not regular languages then L1 ? L2 is not regular 6 Show that the language L = {x
Show that for each n >= 1, the language Cn is regular. By simulating binary division, we create a DFA M with n states that recognizes Cn. M has n states which keep track of the n possible remainders of the division process.
Every subset of a regular language is regular. Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. If L is regular, then so is L? = {xy : x ? and y ? L}.
(b) Let L? = L1 ? L2. If L? is regular and L2 is regular, L1 must be regular. FALSE. We know that the regular languages are closed under intersection. But it is important to keep in mind that this closure lemma (as well as all the others we will prove) only says exactly what it says and no more.
7. First, let L' = L ? a*b*, which must be regular if L is. We observe that L' = anbn+2: n ? 0. Now use the pumping lemma to show that L' is not regular in the same way we used it to show that anbn is not regular.