Examples for Finite Automata CSC3130 Tutorial One Bridge Zhao bridger.zhao@gmail.com Department of Computer Science & Engineering

Strings With Common Prefix Construct a DFA that accepts a language L over  = {0, 1} such that L is the set of all strings starting with “101”. 0,1 q1 1 q2 q3 1 q0 start q4 1 absorbing state dead state CSC3130 Tutorial One 9/22/2018

Strings With Common Suffix Construct a DFA that accepts a language L over  = {0, 1} such that L is the set of all strings ending with “101”. 1 q0 start q1 1 q2 q3 1 1 CSC3130 Tutorial One 9/22/2018

NFA for Common Suffix We can have a simpler representation for common suffix language using NFA: Use subset construction to convert it to a DFA. q0 1,0 q1 1 q2 q3 1 start compare with previous DFA 1 q0 start q0q1 1 q0q2 q0q1q3 1 1 CSC3130 Tutorial One 9/22/2018

NFA Example Construct NFA for the language over the alphabet {a, b, …, z} such that every string doesn’t contain “fool”. ‘f’ dead state not ‘f’ ‘f’ ‘f’ q1 ‘f’ ‘o’ q2 q3 ‘l’ start q0 q2 ‘o’ not ‘f’,‘l’ not ‘f’,‘o’ not ‘f’,‘o’ CSC3130 Tutorial One 9/22/2018

State Upper Bound By subset construction, an NFA with n states can be expressed by a DFA with no more than 2n states. Is this upper bound tight? b δ(i, a)={i+1}, (0<i<n) δ(i, b)={i+1}, (1<i<n) δ(1, b)={1} δ(n, b)=Ø δ(n, a)={1, 2} 1 start a a n i 2 3 a,b CSC3130 Tutorial One 9/22/2018

the DFA has 16=24 states in total n=4 Case a 2 a 3 a 4 b Ø 1 start a 1,2 a 1,2,4 a 3,4 a 2,3 b 1,3 a 1,2,3,4 a 2,3,4 a 1,2,3 a 2,4 b 1,3,4 b 1,4 δ(i, a)={i+1}, (i=1,2,3) δ(i, b)={i+1}, (i=2,3) δ(1, b)={1} δ(4, b)=Ø δ(4, a)={1, 2} the DFA has 16=24 states in total upper bound is tight CSC3130 Tutorial One 9/22/2018

Regular Expressions Construct a RE over d={0,1} such that It contains all strings that have two consecutive “0”s. It contains all strings except those with two consecutive “0”s. It contains all strings with an even number of “0”s. CSC3130 Tutorial One 9/22/2018

Tips for RE Construction (1+0)*00(1+0*) We can divide the string into segments: Use similar idea as 2: ((1*0)(1*0))*1* 01110101111 (ε +0) (11*0)* 1* CSC3130 Tutorial One 9/22/2018