Transition Diagrams Lecture 3 Fri, Jan 21, 2005

Building Transition Diagrams from Regular Expressions A regular expression consists of symbols a, b, c, …, operators, parentheses, and . We describe a recursive method of building a transition diagram from a regular expression.

Building Transition Diagrams The basic cases. For , build :  For each symbol a  , build a: a

Building Transition Diagrams The recursive cases. For the expression r | s, build r | s:  r s

Building Transition Diagrams For the expression rs, build rs: r s  For the expression r*, build r*:  r

Building Transition Diagrams Applying these rules builds an NFA representing the regular expression. Note that each diagram has unique start and accepting states. Note also that generous use was made of -moves. This facilitates joining them together without any complications.

Example: Building a Transition Diagram Build a transition diagram from the regular expression ab*(a | ). Applying the rules rigorously produces the following. a  b

Converting an NFA to a DFA Let Q be the states of the NFA. The -closure of a state q in the NFA is the set of all states that are reachable from q through sequences of -moves (including q itself). Define the states of the DFA to be (Q), i.e., sets of states in the NFA.

Converting an NFA to a DFA For every state A  (Q) and every symbol x  , the transition (A, x) is the -closure of all states in the NFA that are reached from states in A by reading x. That is, First find all states reached from A by following x-moves. Then find the -closure of that set of states.

Example: A DFA from an NFA Consider the NFA of the regular expression ab*(a | ). Number the states 1 through 12. 1 3 4 5 6 8 9 12 10 11 a  b 2 7

Example: A DFA from an NFA Find the -closure of each state. -cl(1) = {1}. -cl(2) = {2, 3, 4, 6, 7, 8, 10, 11, 12}. -cl(3) = {3, 4, 6, 7, 8, 10, 11, 12}. -cl(4) = {4}. -cl(5) = {4, 5, 6, 7, 8, 10, 11, 12}. -cl(6) = {6, 7, 8, 10, 11, 12}. -cl(7) = {7, 8, 10, 11, 12}.

Example: A DFA from an NFA -cl(8) = {8}. -cl(9) = {9, 12}. -cl(10) = {10, 11, 12}. -cl(11) = {11, 12}. -cl(12) = {12}. The start state of the DFA is -cl(1). From there, follow the rule for the transitions of the DFA.

Example: A DFA from an NFA The result is a b 4, 5, 6, 7, 8, 10, 11, 12 2, 3, 6, 7, 8, 10, 11, 12 9, 12 1

Minimizing a DFA To minimize a DFA is to reduce the number of states to a minimum without changing the language accepted by the DFA. Two states p and q are equivalent if for every string w  *, (p, w) and (q, w) are either both accepting states or both rejecting states.

Example: Minimizing a DFA Minimize the DFA of regular expression ab*(a | ). First, add a dead state to make the DFA fully defined. 1 5 2 3 4 a a | b b

Example: Minimizing a DFA The initial partition is {1, 5}, {2, 3, 4}. Apply the transitions by a and b: 1 5 2 3 4 a b a distinguishes 1 from 5, and 2 and 4 from 3. b distinguishes 2 and 4 from 3.

Example: Minimizing a DFA The second partition is {1}, {2, 4}, {3}, {5}. a and b do not distinguish 2 and 4. Therefore, this is the final partition. States 2 and 4 are equivalent and should be merged. Also, remove the dead state.

Example: Minimizing a DFA The minimized DFA is 1 2 3 a b

Programming a DFA There are two basic methods of programming a DFA. Use switch statements. Use a transition table.

Using Switch Statements The main function contains a switch statement whose cases are the different states, including the dead state. Each case contains a switch statement whose cases are the different symbols.

Using a Transition Table The program uses a 2-dimensional array to store the transitions. Rows represent states. Columns represent symbols. The programs lex, flex, and JLex all use the tables.