1. DFA minimization The Chinese University of Hong Kong Fall 2011
CSCI 3130: Automata theory and formal languages
DFA minimization
Andrej Bogdanov

2. Example
L = {w: w ends in 111}
Isn’t there a smaller one?

3. Smaller DFA L = {w: w ends in 111} Can we do it with 3 states? 1 q0 q11 q0 q1 q2 q3 1
Can we do it with 3 states?

4. Even smaller DFA? L = {w: w ends in 111}
Suppose we had a 3 state DFA for L
There are 4 inputs but 3 states On two of these inputs M ends in the same state M
inputs:
e, 1, 11, 111

5. Pigeonhole principle
Suppose you are tossing 4 balls into 3 bins. Then two balls end up in the same bin.
Take 4 inputs and feed them into a 3 state DFA. Then two inputs end up in the same state.

6. ✘ A smaller DFA L = {w: w ends in 111}
Suppose inputs x = 1, y = 11 lead to same state
Then after reading one more 1
The state of x1 = 11 should be rejecting
The state of y1 = 111 should be accepting M
1, 11
inputs: 1
e, 1, 11, 111
11, 111
“ends in 111” ✘

7. ✘ A smaller DFA L = {w: w ends in 111}
Suppose inputs x = e, y = 1 lead to same state
Then after reading 11
The state of x1 = 11 should be rejecting
The state of y1 = 111 should be accepting M
e, 1
inputs: 1
e, 1, 11, 111
11, 111 ✘

8. No smaller DFA!
After looking at all possible pairs (x, y) we conclude that
So, this DFA is minimal
(e, 1)
(e, 11)
(e, 111)
(1, 11)
(1, 111)
(11, 111)
There is no DFA with 3 states for L 1 q0 q1 q2 q3

9. DFA minimization 1 … qe q0 q1
q00
q10
q01
q11
q000
q001
q101
q111 1 q0 q1 q2 q3
We now show how to turn any DFA for L into the minimal DFA for L

10. Minimal DFAs and distinguishable states
First, we have to understand minimal DFAs:
reject
accept 1 1 1 q0 q1 q2 q3 1
every pair of states is distinguishable
minimal DFA

11. Distinguishable states
Two states q and q’ are distinguishable if
accept w1 w2
wk-1 wk … q
reject w1 w2
wk-1 wk … q’
on the same continuation string w1w2...wk, one accepts, but the other rejects

12. Examples of distinguishable states1 1 1 1 q0 q1 q2 q3
(q0, q1)
distinguishable by 01
(q0, q2)
distinguishable by 1
(q0, q3)
distinguishable by e
DFA is minimal
(q1, q2)
distinguishable by 1
(q1, q3)
distinguishable by e
(q2, q3)
distinguishable by e

13. Examples of distinguishable statesq0 q1 q2 q3 1
0, 1
q01 q2 q3 1
0, 1
(q0, q3)
distinguishable by e
(q1, q3)
distinguishable by e
indistinguishable pairs can be merged
(q2, q3)
distinguishable by e
(q1, q2)
distinguishable by 0
(q0, q2)
distinguishable by 0
(q0, q1)
indistinguishable

14. Examples of distinguishable statesq0 q2
0, 1
q01
q23
0, 1 1 q1
0, 1 q3
0, 1
(q0, q2)
distinguishable by e
(q1, q2)
distinguishable by e
(q0, q3)
distinguishable by e
(q1, q3)
distinguishable by e
(q0, q1)
indistinguishable
(q2, q3)
indistinguishable

15. Finding (in)distinguishable states
If q is accepting and q’ is rejecting Mark (q, q’) as distinguishable (x)
Rule 1: q x q’ q1 x
q1’ q2
q2’ a
If (q1, q1’) are marked, Mark (q2, q2’) as distinguishable (x)
Rule 2: x
Rule 3:
Unmarked pairs are indistinguishable
Merge them into groups

16. Example of DFA minimizationq0
q00 q1 q0 1 1
q01
q00 qe 1
q01
q10 1 1
q10 q1 1
q11
q11 qe q0 q1
q00
q01
q10 1

17. Example of DFA minimizationq0
q00 q1 q0 1 1
q01
q00 qe 1
q01
q10 1 1
q10 q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 q11 is distinguishable from all other states

18. Example of DFA minimizationq0
q00 q1 1 q0 1 1
q01
q00 qe 1
q01
q10 1 1
q10 q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 Look at pair qe, q0
Neither (q0, q00) nor (q1, q01) are distinguishable

19. Example of DFA minimizationq0
q00 q1 x q0 1 1
q01
q00 qe 1 1
q01
q10 1 1
q10 q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 Look at pair qe, q1
(q1, q11) is distinguishable

20. Example of DFA minimizationq0
q00 q1 x x q0 1 1
q01
q00 x qe 1
q01 x x x
q10 1 1
q10 x x q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 After going thru the whole table once
Now we make another pass

21. Example of DFA minimizationq0
q00 q1 x x 1 q0 1 1
q01
q00 x qe 1
q01 x x x
q10 1 1
q10 x x q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 Look at pair qe, q0
Neither (q1, q00) nor (q1, q01) are distinguishable

22. Example of DFA minimizationq0
q00 q1 x x q0 1 1
q01
q00 x 1 qe 1
q01 x x x
q10 1 1
q10 x x q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 Look at pair qe, q00
Neither (q0, q00) nor (q1, q01) are distinguishable

23. Example of DFA minimizationq0
q00 q1 x x q0 1 1
q01
q00 x qe 1
q01 x x x
q10 1 1
q10 x x q1 1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10 1
 In the second pass, nothing changes
So we are ready to apply Rule 3

24. Example of DFA minimizationq0
q00 q1 x x q0
q01
q00 x qe
q01 x x x
q10
q10 x x q1
q11
q11 x x x x x x qe q0 q1
q00
q01
q10
 Merge unmarked pairs into groups

25. Example of DFA minimizationq0 A
q00 q1 x x q0 A
q01
q00 A A x qe
q01 x x B x
q10
q10 A A x A x q1 B
q11
q11 x x x x x x C qe q0 q1
q00
q01
q10
 Merge unmarked pairs into groups

26. Example of DFA minimizationq0 A
q00 q1 x x q0 A 1 1
q01
q00 A A x qe 1
q01 x x B x
q10 1 1
q10 A A x A x q1 1 B
q11
q11 x x x x x x C qe q0 q1
q00
q01
q10 1 1 qA qB qC
minimized DFA:

27. Example of DFA minimization1 qA qB qC
How do we know this DFA is minimal? 1 e qB qC qA
Answer: All pairs are distinguishable

28. Why it works Why do we end up finding all distinguishable pairs?
wk-1 wk … q x x x x w1 w2
wk-1 wk … q’
Because we work backwards

29. Why it works Why are there no inconsistencies when we merge?B w q5 a A q2 q3 q1 q7 a C q6
Because we only merge indistinguishable states

30. Why it works Why is there no smaller DFA? Suppose there is
By the pigeonhole principle this must happen: M q q’ v v’
smaller DFA M’ q”
v, v’

31. Why it works Why is there no smaller DFA? But then M smaller DFA M’ vq w q” w ?
v, v’ v’ q’ w
Every pair of states
is distinguishable
q” cannot exist!