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Smaller DFA qq qq qq qq 11 Can we do it with 3 states?
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CSC 3130: Automata theory and formal languages
Andrej Bogdanovhttp://www.cse.cuhk.edu.hk/~andrejb/
csc3130
The Chinese University of Hong Kong
DFA minimization
Fall 2009
Example• Construct a DFA over alphabet {0, 1} that
accepts those strings that end in 111
• This is big, isn’t there a smaller DFA for this?
0
1
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……
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Smaller DFA
1
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00
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1q q q q1 1
Can we do it with 3 states?
Even smaller DFA?• Suppose we had a 3 state DFA M for L
… let’s imagine what happens when:
• By the pigeonhole principle, on two of these inputs M ends in the same state
Minputs:, 1, 11, 111
Pigeonhole principle
• Here, balls are inputs, bins are states:
Suppose you are tossing m balls into n bins, and m > n. Then two balls end up in the same bin.
If you have a DFA with n states and you run it on m inputs, and m > n, then two inputs end up in same state.
A smaller DFA• Suppose M ends up in the same state after
reading inputs x = 1 and y = 11
• Then after reading one more 1– The state of x1 = 11 should be rejecting– The state of y1 = 111 should be accepting
… but they are both the same state!
Minputs:, 1, 11, 111
1, 11
11, 111“ends in 111”
A smaller DFA• Suppose M ends up in the same state after
reading inputs x = and y = 1
• Then after reading 11– The state of x1 = 11 should be rejecting– The state of y1 = 111 should be accepting
… but they are both the same state!
Minputs:, 1, 11, 111
, 1
11, 111“ends in 111”
No smaller DFA!• After looking at all possible pairs for x, y, x ≠ y
we conclude that
• So, this DFA is minimal
There is no DFA with 3 states for L
1q q q q1 1 1
0
00
0
(, 1) (, 11) (, 111) (1, 11) (1, 111)(11, 111)
DFA minimization
We will show how to turn any DFA for L into the minimal DFA for L
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00 0
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Minimal DFAs and distinguishable states• First, we have to understand minimal DFAs:
1
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1q q q q1 1
reject accept
minimal DFA
every pair of states is
distinguishable
Distinguishable states• Two states q and q’ are distinguishable if
q
q’
w1
w1
w2
w2
wk
wk
wk-1
wk-1
…
…
on the same continuation string w1w2...wk, one accepts, but the other rejects
reject
accept
Examples of distinguishable states
distinguishable by 01(q0, q2)(q0, q3)
(q0, q1)
(q1, q3)(q2, q3)
(q1, q2)
distinguishable by 1distinguishable by distinguishable by 1distinguishable by distinguishable by
0 0
1q q q q0 1
0 1 1
DFA is minimal
Examples of distinguishable states
distinguishable by
q
q
q
q
0
1
0
1
0, 1
0, 1
q0
q
q
0
1
0, 1
0, 1
(q1, q3)(q2, q3)
(q0, q3)
(q0, q2)(q0, q1)
(q1, q2)
distinguishable by distinguishable by distinguishable by 0distinguishable by 0indistinguishable
indistinguishable pairs can be merged
Examples of distinguishable states
distinguishable by (q1, q2)(q0, q3)
(q0, q2)
(q0, q1)(q2, q3)
(q1, q3)
distinguishable by distinguishable by distinguishable by indistinguishableindistinguishable
q
q
q
q
1
0, 1
0, 1
0, 1
0
qq0, 1
0, 1
Finding (in)distinguishable states
Rule 1:
Rule 2:
q q’If q is accepting and q’ is rejectingMark (q, q’) as distinguishable (x)
x
q1 q1’x
xq2 q2’
a a If (q1, q1’) are marked,Mark (q2, q2’) as distinguishable (x)
Rule 3: Unmarked pairs are indistinguishableMerge them together
Example of DFA minimization
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1q
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qq q q qq
Example of DFA minimization
0
1q
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01
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x x x x x x
q
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qq q q qq
q11 is distinguishable from all other states
Example of DFA minimization
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1q
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01
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x
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x
xx x x x
q
q
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qq q q qq
q1 is distinguishable from q, q0, q00, q10
On transition 1, they go to distinguishable states
Example of DFA minimization
0
1q
q
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q
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01
1
0
0
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x
x
x
x
x
x
x
xx
x
xxx x
q
q
q
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q
qq q q qq
q01 is distinguishable from q, q0, q00, q10
On transition 1, they go to distinguishable states
Example of DFA minimization
0
1q
q
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q
q
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AxAxAx
xAxAx
xBxx
xAx
xx x
q
q
q
q
q
q
qq q q qq
Merge states not marked distinguishable q, q0, q00, q10 are equivalent → group A q1, q01 are equivalent → group B q11 cannot be merged → group C
A
BC
Example of DFA minimization
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AxAxAx
xAxAx
xBxx
xAx
xx x
q
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qq q q qq
1q q qC1 1
0
00
minimized DFA:
A
BC
Food for thought• Why does method find all distinguishable pairs?
q
q’
w1
w1
w2
w2
wk
wk
wk-1
wk-1
…
…xxxx
Because we work backwards
Food for thought• Why are there no inconsistencies when we merge?
A q3
q7
q1
Bq5
q2
Cq6
a
a
Because we only merge indistinguishable states
w
w
Food for thought• Why is there no smaller DFA?
Mq
q’
v
v’
smaller DFA M’
q’’v, v’
Suppose there is
By the pigeonhole principle this must happen:
Food for thought• Why is there no smaller DFA?
But then
M smaller DFA M’q
q’
v
v’
q’’v, v’
w
w?w
Every pair of statesis distinguishable
q’’ cannot exist!
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