Transition Diagrams

1. Transition Diagrams Lecture 3 Fri, Jan 21, 2005

2. 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.

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

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

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

6. 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.

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

8. 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.

9. 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.

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

11. 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}.

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.

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

14. 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.

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

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

17. 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.

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

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

20. 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.

21. 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.