DFA minimization The Chinese University of Hong Kong Fall 2011 CSCI 3130: Automata theory and formal languages DFA minimization Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

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

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

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

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.

✘ 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” ✘

✘ 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 ✘

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

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

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

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

Examples of distinguishable states 1 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

Examples of distinguishable states q0 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

Examples of distinguishable states q0 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

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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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

Example of DFA minimization q0 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:

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

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

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

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’

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