Theory of Computation FSM Conversions and Problems
- 1. Theory of Computation By Rushabh Wadkar
- 2. Topics to be covered DFA & NFA Problems Contd. NFA to DFA Extended transition functions(NFA) eNFA to NFA Day 4
- 3. DFA Problems Draw a DFA to accept decimal strings divisible by 3. M=(Q, ∑, δ, q0 , F)
- 4. DFA Problems Draw a DFA to accept decimal strings divisible by 3. ● To begin, we identify the radix, input alphabet and divisor. r=10 | d={0,1,2,3,4,5,6,7,8,9} | k=3 ● 𝛿(qi ,0) = qj Where, j =( r * i + d) mod k. ● j =( 10 * i + d) mod 3. ● Since mod3 has remainders 0,1,2 ; i = 0,1,2.
- 5. DFA Problems Draw a DFA to accept decimal strings divisible by 3. M=(Q, ∑, δ, q0 , F)
- 6. DFA Problems Draw a DFA to accept na (w) and nb (w) are both even. ∑ = (a,b)
- 7. DFA Problems Draw a DFA to accept na (w)mod3=0 and nb (w)mod2=0. ∑ = (a,b)
- 8. DFA Problems Draw a DFA to accept na (w)mod5=0 and nb (w)mod3=0. ∑ = (a,b)
- 9. DFA Problems Draw a DFA to accept na (w)mod5 ≠ nb (w)mod3. ∑ = (a,b)
- 10. DFA Problems Draw a DFA to accept na (w)mod5 > nb (w)mod3. ∑ = (a,b)
- 11. NFA Problems Draw a NFA that satisfies the language: L= {an ∪bn : n>=0}
- 12. NFA Problems Draw a NFA that satisfies the language: L= {0101n ∪0100 : n>=0}
- 13. NFA Problems Draw a NFA that contains either ‘01’ or ‘10’ on Σ = {0,1}
- 14. NFA Problems Draw a NFA that according to the transition graph 𝛿 0 1 q0 q0 ,q1 q0 ,q2 q1 q3 - q2 q2 ,q3 q3 q3 q3 q3
- 15. NFA to DFA Step 1 − Create state table from the given NFA Step 2 − Mark the start state of the DFA by q0 (Same as the NDFA). Step 3 − Find out the combination of States {Q0 , Q1 ,... , Qn } for each possible input alphabet. Step 4 − Each time we generate a new DFA state under the input alphabet columns, we have to apply step 3 again, otherwise go to step 5. Step 5 − The states which contain any of the final states of the NDFA are the final states of the equivalent DFA. Let X = (Qx, ∑, δx, q0, Fx) be a NFA which accepts the language L(X). We have to design an equivalent DFA Y = (Qy, ∑, δy, q0, Fy) such that L(Y) = L(X).
- 16. NFA to DFA Q.Consider a NFA with the following transitions: 𝛿(q0 ,0) = {q0 q1 } 𝛿(q1 ,0) = Φ 𝛿(q0 ,1) = q1 𝛿(q1 ,1) = {q0 q1 }
- 17. NFA to DFA 𝛿 0 1 q0 q0 q1 q1 q1 - q0 q1 Required transition graph :
- 18. NFA to DFA We identify the required states : We already have the transitions for q0 ,q1 on 0 & 1. The new found state is {q0 q1 }. Let us find transition for this.
- 19. NFA to DFA Required transition graph for DFA : Notice the final states.
- 20. NFA to DFA Required DFA :
- 21. NFA to DFA Q.Consider a NFA with the following transitions:
- 22. NFA to DFA We identify the required states : We already have the transitions for q0 ,q1 on 0 & 1. The new found state is {q0 q1 }. Let us find transition for this.
- 23. NFA to DFA Similarly for the newly generated q0 q1 q2
- 24. Required transition graph for DFA : NFA to DFA *notice that q1 was not marked as final in slide 22. Here we can either consider it as a final state. Or remove finals that contain q1
- 25. NFA to DFA Q.Consider the following NFA
- 26. NFA to DFA Required transition graph :
- 27. Required transition graph for DFA : NFA to DFA
- 28. ε-NFA An example of ε-NFA to understand it better.
- 29. ε-NFA States: It consists of 3 states, q0 , q1 , q2 ∈ Q Input Alphabets: There are 2 input symbols in the alphabet ∑ = {a, b} Transitions: δ(qi ,a) = qj δ(qi ,a) = {qi ,qj } δ(qi ,ε) = qj Let us take a closer look at this ε-NFA now.
- 30. ε-NFA It is defined using the quintuple M=(Q, Σ, 𝛿, q0 , F) Where, Q = Set of internal states in the ε-NFA Σ = Finite set of symbols (input alphabet) 𝛿 = Transition function defined by 𝛿= QxΣ∪ε--->2Q q0 = Initial state ∈ Q F = Set of final states ∈ Q
- 31. ε-NFA Epsilon closure(ECLOSE) is finding all the states which can be reached from the present state on one or more epsilon transitions. ECLOSE(A) = ABD ECLOSE(B) = BD ECLOSE(C) = C ECLOSE(D) = D
- 32. ε-NFA Epsilon closure(ECLOSE) follows the properties of extended transitions
- 33. ε-NFA Ref: Finite Automata and Formal Languages (Padma Reddy)
- 34. ε-NFA Example Problems Draw a ε-NFA that accepts zero or more a’s or zero OR more b’s or zero OR more c’s
- 35. ε-NFA to DFA Step 1 : Take ∈ closure for the beginning state of NFA as beginning state of DFA. Step 2 : Find the states that can be traversed from the present for each input symbol(union of transition value and their closures for each states of NFA). Step 3 : If any new state is found take it as current state and repeat step 2. Step 4 : Do repeat Step 2 and Step 3 until no new state present in DFA transition table. Step 5 : Mark the states of DFA which contains final state of NFA as final states of DFA.
- 36. ε-NFA to DFA Find the DFA for the following NFA.
- 37. ε-NFA to DFA To find the start state of the DFA we will find the eclosure of the current start state. . . . the start state for DFA is ECLOSE(q0 ) = {q0 q1 q2 }
- 38. ε-NFA to DFA From the transitions of {q0 q1 q2 }, we arrive on state {q1 q2 }
- 39. ε-NFA to DFA And then finally for state {q2 },
- 40. ε-NFA to DFA From the following we can draw the transition table
- 42. End of Day 4 www.linkedin.com/in/wadkar-rushabh @RushabhWadkar Thank you...