Question

Please circle final answer

20 points) Com , contains at least three strings such that the length of the strings is od facts that everyings function atle ast 30ddorstartBab show below bab; or both. Assume that ΣG-{a,b). Recall the lab wh which decides if strings st art ; or the ite language must contain a string whose length is less tha an also important to hich nhnse lengths are greater than bn and less than or equal to bi ca if a language is infinite. These are the only decidecFLusingPDA (p, w) where p is a PDA and w is string. Returns true bn, and that strings can also important for deciding procedures you may use, other than the ones shown: if w e L(p), and false otherwise. regEXTOFSM (regex)-where regex is a regular expression. language described by regex. For example, L(p) n L(f) P,where p is a PDA and f is an FSM. Returns a PDA accepting boolean atLeast3oddorstartBab (CFG g) int b getBranchingFactor (g) int n getNumNonterminals (g)i PDA Pg = CFGtoPDA (g);

Answer 1

rightarrow In the regular expression given below: - FSM accepting only odd length string over and starting with bad on both rightarrow we assume that E_G = {a, b} so that R = bab{a(a + b) + b(a + b) } * Deterministic finite state machine: - * In the Regular language , we know that Regular language (REL) is on set of constant free language. * In the flow of a PDA accpets CFL (constant free language) and FSM accepts REL regular language. Therefore intersection of these two language is always regular accepted by FSM. than RE = bab[a(a + b) + b(a + b)]^* (2) rightarrow starts with bab rightarrow [Q(a + b) = b(a + b)] is after bab these are called be only combination of a's and b's with a total string length rightarrow which is odd including Null * In the solution path following: - Rightarrow CFG g rightarrow call CFG to PDA (g) to get the PDA Rightarrow accepting g rightarrow consider the alone mentioned regular expression RE