Lecture 6.1 - NFA to DFA toc and compiler design

compiler design

1. Non Deterministic Finite Automata
To
Deterministic Finite Automata Construction
1
Course Name: Compiler Design
Course Code: CSE331
Level:3, Term:3
Department of Computer Science and Engineering
Daffodil International University

2. There are three main cases of non-determinism in NFAs:
1.Transition to a state without consuming any input.
2.Multiple transitions on the same input symbol.
3.No transition on an input symbol.
To convert NFAs to DFAs we need to get rid of non-
determinism from NFAs.
Non Deterministic Features of NFA

3. Using Subset construction method to convert NFA to DFA
involves the following steps:
•For every state in the NFA, determine all reachable statesfor
every input symbol.
•The set of reachable states constitute a single statein the
converted DFA (Each state in the DFA corresponds to a subset of
states in the NFA).
•Find reachable statesfor each new DFAstate, until no more new
states can be found.
Subset Construction Method

4. Fig1. NFA without λ-transitions
1
3
2
4
5
a
a
a
b
a
b
b
a
a,b
a,b
Subset Construction Method

5. Fig1. NFA without λ-transitions
Step1
1
3
2
4
5
a
a
a
b
a
b
b
a
a,b
a,b
Construct a transition table showing
all reachable states for every state
for every input signal.
Subset Construction Method

6. Fig1. NFA without λ-transitions
Fig2. Transition table
1
3
2
4
5
a
a
a
b
a
b
b
a
a,b
a,b
Subset Construction Method

7. Fig1. NFA without λ-transitions
Fig2. Transition table
1
3
2
4
5
a
a
a
b
a
b
b
a
a,b
a,b
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3 ∅
{2}
4 {5} {4}
5 ∅ ∅
Subset Construction Method

8. Fig1. NFA without λ-transitions
Fig2. Transition table
1
3
2
4
5
a
a
a
a
a
b
b
b
a,b
a,b
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3 ∅
{2}
4 {5} {4}
5 ∅ ∅
Transition from state q with
input a
Transition from state q
with input b
Starts here
Subset Construction Method

9. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
Step2
Thesetofstatesresultingfromevery
transitionfunctionconstitutesanew
state.Calculateallreachablestates
foreverysuchstateforeveryinput
signal.
Subset Construction Method

10. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
Starts with
Initial state
Fig3. Subset Construction table

11. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}
{4,5}
Starts with
Initial state
Fig3. Subset Construction table

12. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}
{4,5}
Starts with
Initial state
Fig3. Subset Construction table
Step3
Repeat this process(step2) until no
more new states are reachable.

13. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5}
{2,4,5}
Fig3. Subset Construction table

14. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
5
4
Fig3. Subset Construction table

15. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5
4
{3,5}
Fig3. Subset Construction table

16. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4
{3,5}
∅
Fig3. Subset Construction table

17. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
We already got 4 and 5.
So we don’t add them again.
Fig3. Subset Construction table

18. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
∅
2
∅
2
Fig3. Subset Construction table

19. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
∅
2
∅ ∅ ∅
2
Fig3. Subset Construction table

20. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
∅
2
∅ ∅ ∅
2 3
5
3
Fig3. Subset Construction table

21. Fig2. Transition table
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
2 {3} {5}
3
∅
{2}
4 {5} {4}
5 ∅ ∅
q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
∅
2
∅ ∅ ∅
2 3
5
3
∅
2
Fig3. Subset Construction table
Stops here as there are
no more reachable states

22. q δ(q,a)δ(q,b)
1 {1,2,3,4,5}{4,5}
{1,2,3,4,5}{1,2,3,4,5}{2,4,5}
{4,5} 5 4
{2,4,5}
{3,5} {4,5}
5 ∅ ∅
4 5 4
{3,5}
∅
2
∅ ∅ ∅
2 3 5
3
∅
2
1
45
12345 245
35
5
4
∅
3
2
a
a
a
a
a
a
a
a
b
b
b
b
b
b
b
b
a,b
a,b
Fig3. Subset Construction table
Fig4. Resulting FA after applying
Subset Construction to fig1

23. NFA to DFA Conversion
LetX=(Q
x,∑,δ
x,q
0,F
x)beanNFAwhichacceptsthelanguageL(X).Wehaveto
designanequivalentDFAY=(Q
y,∑,δ
y,q
0,F
y)suchthatL(Y)=L(X).
ThefollowingprocedureconvertstheNFAtoitsequivalentDFA−
Algorithm:
Input−AnNFA
Output−AnequivalentDFA
Step1−CreatestatetablefromthegivenNFA.
Step2−Createablankstatetableunderpossibleinputalphabetsforthe
equivalentDFA.
Step3−MarkthestartstateoftheDFAbyq0(SameastheNFA).
Step4−FindoutthecombinationofStates{Q0,Q1,...,Qn}foreachpossibleinput
alphabet.
Step5−EachtimewegenerateanewDFAstateundertheinputalphabet
columns,wehavetoapplystep4again,otherwisegotostep6.
Step6−ThestateswhichcontainanyofthefinalstatesoftheNFAarethefinal
statesoftheequivalentDFA.
23

24. Example
Let us consider the NFA
shown in the figure:
q δ(q,0) δ(q,1)
a {a,b,c,d,e} {d,e}
b {c} {e}
c ∅ {b}
d {e} ∅
e ∅ ∅
The state transition table of
the NFA is:
24

25. q δ(q,0) δ(q,1)
a {a,b,c,d,e}{d,e}
b {c} {e}
c ∅ {b}
d {e} ∅
e ∅ ∅
Using the algorithm, we find its
equivalent DFA.
The state transition table of the
DFA is:
q δ(q,0) δ(q,1)
[a] [a,b,c,d,e] [d,e]
[a,b,c,d,e][a,b,c,d,e] [b,d,e]
[d,e] [e] ∅
[b,d,e] [c,e] [e]
[e] ∅ ∅
[c, e] ∅ [b]
[b] [c] [e]
[c] ∅ [b]
Example…
The state transition table
of the NFA is:
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26. Example…
The state transition table of
the DFA is:
q δ(q,0) δ(q,1)
[a] [a,b,c,d,e][d,e]
[a,b,c,d,e][a,b,c,d,e][b,d,e]
[d,e] [e] ∅
[b,d,e] [c,e] [e]
[e] ∅ ∅
[c, e] ∅ [b]
[b] [c] [e]
[c] ∅ [b]
The state diagram of the DFA is as follows:
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27. THANK YOU
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