automata

How can I construct a context free grammar for the following language: L = {a^l b^m c^n d^p | l+n==m+p; l,m,n,p >=1} I started by attempting: S -> abcd | aAbBcd | abcCdD | aAbcdD | AabBcCd and then A = something else... but I couldn't get this working. . I was wondering how can we remember how many c's shud be increased for the no. of b's increased? For example: string : abbccd The grammar is : S1 -> a S1 d | S2 S2 -> S3 S4 S3 -> a S3 b | epsilon S4 -> S5 S6 S5 -> b S5 c | epsilon S6 -> c S6 d | epsilon Rule 1 adds equal number of a's and d's. Rule 3 adds equal number of a's and b's. Rule 5

I need some help with a pumping lemma problem. L = { {a,b,c}* | #a(L) < #b(L) < #c(L) } This is what I got so far: y = uvw is the string from the pumping lemma. I let y = abbc^n, n is the length from the pumping lemma. y is in L because the number of a:s is less than the number of b:s, and the number of b:s is less than the number of c:s. I let u = a, v = bb and w = c^n. |uv| < y, as stated in pumping lemma. If I "pump" (bb)^2 then i get y = abbbbc^n which violates the rule #b(L) < #c(L). Is this right ? Am I on the "right path" ? Thanks The main idea of the pumping lemma is to tell you that

I have a problem where I really need to be able to use finite automata as the keys to an associative container. Each key should actually represent an equivalence class of automata, so that when I search, I will find an equivalent automaton (if such a key exists), even if that automaton isn't structurally identical. An obvious last-resort approach is of course to use linear search with an equivalence test for each key checked. I'm hoping it's possible to do a lot better than this. I've been thinking in terms of trying to impose an arbitrary but consistent ordering, and deriving an ordered

Produce a PDA to recognise the following language : the language of strings containing more a's than b's I have been struggling with this question for several days now, I seem to have hit a complete mental block. Would any one be able to provide some guidance or direction to how I can solve this problem? Divyanjali Your problem of "more a's than b's" can be solved by PDA. All you have to do is: When input is a and the stack is either empty or has an a on the top, push a on the stack; pop b , if b is the top. When input is b and the stack is either empty or has an b on the top, push b on the

Good afternoon, Does anyone know of an "out-of-the-box" implementation of Levenshtein DFA ( deterministic finite automata ) in .NET (or easily translatable to it)? I have a very big dictionary with more than 160000 different words, and I want to, given an inicial word w , find all known words at Levenshtein distance at most 2 of w in an efficient way. Of course, having a function which computes all possible edits at edit distance one of a given word and applying it again to each of these edits solves the problem (and in a pretty straightforwad way). The problem is effiency --- given a 7 letter