Exercises for Chapter 2

Exercises for Chapter 2

Summary for the Quiz • Some numbers • 199  less than 100  55 • Conclusion • There are students who did read textbook at least for review; • Most submitted answers are actually correct; • Rethinks on teaching method

Explanations on RE • A regular expression defines a set of strings(language); • A regular expression states the common pattern(structure) of the strings that belong to its language; • For the same set of strings, there could be several RE definitions for it; • Explanation of three operations • (顺序;选择;重复);

Example of RE • {a, b, c} • Defining a set of strings, where each string should starts with “a” and ends with “c”; • Defining a set of strings, if “a” appears in one string, “c” should appear right after “a”;

From RE to NFA • The generated NFA could be different, but they are equivalent; • The only criteria for judging whether the generated NFA is correct: L(RE) = L(NFA) • You can follow the general rules; • There are some shortcuts;

S0 S0 S S Rules ■ is a regular expression，L()={ } ■  is a regular expression，L()={  } ■ for any c , c is a regular expression, L(c)={c} c

Rules ■ ( A ), L( (A) ) = L(A), no change; ■ A B， L( A B )= L(A)L(B) NFA(B) NFA(A) 

Rules ■ ( A ), L( (A) ) = L(A), no change; ■ A | B，L( A | B )=L(A)L(B) NFA(A)    NFA(B) 

Rules ■ A* ，L( A*) = L(A)*  NFA(A)   

Attention • The rules introduced above are effective for those NFAs that have one start state and one terminal state; • Any NFA can be extended to meet this requirement; …… NFA ……    

Quiz • (a|b)*abb(a|b)* • Follow the rules  a     b   b b a

Quiz • (a|b)*abb(a|b)* • The NFA without  edge; a a b a b 3 4 2 1 b b

{1} --- S0; {1,2} --- S1; {1,3} --- S2; {1,4} --- S3; {1,2,4} --- S4; {1,3,4} --- S5;

Minimizing DFA • Current groups:{S0,S1,S2}, {S3,S4,S5} Splitting {S0, S1,S2}  {S0,S1},{S2} • Current groups:{S0,S1},{S2} , {S3,S4,S5} Splitting {S0, S1}  {S0},{S1}

Minimizing DFA • Current groups:{S0},{S1},{S2}, {S3,S4,S5} • Splitting {S3, S4,S5}  {S3,S4,S5}

Minimized DFA • {S0},{S1},{S2}, {S3,S4,S5} • {S0} ---- 0; {S1} --- 1 ; • {S2} ---- 2; {S3,S4,S5} --- 3; a b 00 1 00 1 2 1 2 1 3 3 3 3*

You should know how to do these problems!!!!

Exercises for Chapter 2