The time complexity for e-closure(T). Push all states in T onto stack; initialize e-closure(T) to T; while stack is not empty do begin pop t, the top element, off the stack for each state u with an edge from t to u labeled e do if u is not in e-closure(T) do begin add u to e-closure(T) push u onto stack end if end do end while (page 119, Fig. 3.26) Time complexity?

The complexity for the algorithm that recognizes the language accepted by NFA(revisit). Input: an NFA (transition table) and a string x (terminated by eof). output “yes” if accepted, “no” otherwise. S = e-closure({s0}); a = nextchar; while a != eof do begin S = e-closure(move(S, a)); a := next char; end if (intersect (S, F) != empty) then return “yes” else return “no” Time complexity ?? Space complexity ??

Algorithm to convert an NFA to a DFA that accepts the same language (algorithm 3.2, page 118) initially e-closure(s0) is the only state in Dstates and it is marked while there is an unmarked state T in Dstates do begin mark T; for each input symbol a do begin U := e-closure(move(T, a)); if (U is not in Dstates) then add U as an unmarked state to Dstates; Dtran[T, a] := U; end end; Initial state = e-closure(s0), Final state = ?

Question: for a NFA with |S| states, at most how many states can its corresponding DFA have? Using DFA or NFA?? Trade-off between space and time!!

The number of states determines the space complexity. A DFA can potentially have a large number of states. Converting an NFA to a DFA may not result in the minimum-state DFA. In the final product, we would like to construct a DFA with the minimum number of states (while still recognizing the same language). Basic idea: assuming all states have a transition on every input symbol (what if this is not the case??), find all groups of states that can be distinguished by some input strings. An input string w distinguishes two states s and t, if starting from s and feeding w, we end up in a nonaccepting state while starting from t and feeding w, we end up in an accepting state, or vice versa.

Algorithm (3.6, page 142): Input: a DFA M output: a minimum state DFA M’ If some states in M ignore some inputs, add transitions to a “dead” state. Let P = {All accepting states, All nonaccepting states} Let P’ = {} Loop: for each group G in P do Partition G into subgroups so that s and t (in G) belong to the same subgroup if and only if each input a moves s and t to the same state of the same P-groups put the new subgroups in P’ if (P != P’) {P = P’; goto loop} Remove any dead states and unreachable states.

Example: minimize the DFA for (ab|ba)a* Example: minimize the DFA for Fig 3.29 (pages 121) Questions: How can we implement Lex? %% BEGIN {return(BEGINNUMBER);} END {return(ENDNUMBER);} IF {return(IFNUMBER);}

Lex internal: construct an NFA to recognize the sum of all patterns convert the NFA to a DFA (record all accepting states for each individual pattern). Minimize the DFA (separate distinct accepting states for the initial pattern). Simulate the DFA to termination (that is, no further transitions) Find the last DFA state entered that holds an accepting NFA state (this picks the longest match). If no such state, then it is an invalid token.