FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

15-453 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA

MINIMIZING DFAs THURSDAY Jan 24

0 1 0 1 1 0 1 0 IS THIS MINIMAL? NO

0 1 1 0 IS THIS MINIMAL?

THEOREM For every regular language L, there exists a UNIQUE (up to re-labeling of the states) minimal DFA M such that L = L(M)

0 0 0 0 NOT TRUE FOR NFAs

EXTENDING  Given DFA M = (Q, Σ, , q0, F) extend  to  : Q Σ* → Q as follows: ^ ^ (q, ε) = q ^ (q, ) = (q, ) ^ ^ (q, 1 …k+1) = ( (q, 1 …k), k+1 ) ^ Note: (q0, w)  F  M accepts w String w  Σ* distinguishes states q1 and q2 iff ^ ^ (q1, w)  F  (q2, w)  F

EXTENDING  Given DFA M = (Q, Σ, , q0, F) extend  to  : Q Σ* → Q as follows: ^ ^ (q, ε) = q ^ (q, ) = (q, ) ^ ^ (q, 1 …k+1) = ( (q, 1 …k), k+1 ) ^ Note: (q0, w)  F  M accepts w String w  Σ* distinguishes states q1 and q2 iff ^ ^ exactly ONE of (q1, w), (q2, w) is a final state

p is distinguishable from q iff there is a w  Σ* that distinguishes p and q p is indistinguishable from q iff p is not distinguishable from q iff for all w  Σ*, (p, w)  F (q, w)  F Fix M = (Q, Σ, , q0, F) and let p, q  Q DEFINITION: ^ ^

q1 1 0 0,1 1 q0 q2 0 0 1 q3 ε distinguishes accept from non-accept states

p ~ q iff p is distinguishable from q / Fix M = (Q, Σ, , q0, F) and let p, q, r  Q Define relation ~ : p ~ q iff p is indistinguishable from q Proposition: ~ is an equivalence relation p ~ p (reflexive) p ~ q  q ~ p(symmetric) p ~ q and q ~ r  p ~ r (transitive) Proof (of transitivity): for all w, we have: (p, w)  F (q, w)  F  (r, w)  F ^ ^ ^

Fix M = (Q, Σ, , q0, F) and let p, q, r  Q so ~ partitions the set of states of M into disjoint equivalence classes Proposition: ~ is an equivalence relation [q] = { p | p ~ q } Q q q0

0 1 0 1 1 0 1 0

Algorithm MINIMIZE Input: DFA M Output: DFA MMIN such that: MMMIN(that is, L(M) = L(MMIN)) MMIN has no inaccessible states MMIN is irreducible || all states of MMIN are pairwise distinguishable Theorem: MMIN is the unique minimum DFA

Intuition: States of MMIN will be blocks of equivalent states of M We’ll find these equivalent states with a “Table-Filling” Algorithm

(1)DM = { (p,q) | p,q  Q and p ~ q } / TABLE-FILLING ALGORITHM Input: DFA M = (Q, Σ, , q0, F) Output: (2) EM = { [q] | q  Q } • We know how to find those pairs of states that ε distinguishes… • Use this and recursion to find those pairs distinguishable with longer strings • Pairs left over will be indistinguishable IDEA:

q0 Base Case: p accepts and q rejects  p ~ q q1 / (1) DM = { (p,q) | p,q  Q and p ~ q } / qi qn q0 q1 qi qn TABLE-FILLING ALGORITHM Input: DFA M = (Q, Σ, , q0, F) Output: (2) EM = { [q] | q  Q }

q0 Base Case: p accepts and q rejects  p ~ q q1 / (1) DM = { (p,q) | p,q  Q and p ~ q } / qi p ~ q / qn ~ / q0 q1 qi qn TABLE-FILLING ALGORITHM Input: DFA M = (Q, Σ, , q0, F) Output: (2) EM = { [q] | q  Q } Recursion: if there is σΣ and states p, q satisfying D D  (p, ) = p  D q  (q, ) = Repeat until no more new D’s

q0 q1 D D q2 D q3 D D D q0 q1 q2 q3 0,1 0 1 0 0 1 q0 q1 q2 q3 1

0 1 0 1 1 0 1 0 q1 q0 q0 q2 q3 q1 D D q2 q3 D D q0 q1 q2 q3

^ ^ (p, w)  F and (q, w)  F Claim: If p, q are distinguished by Table-Filling algorithm, then p ~ q / Proof: By induction on the length of the string distinguishing them. If (p, q) is marked D at the start, then one’s in F and one isn’t, so ε distinguishes p and q Suppose (p, q) is marked D at a later point. Then there are states p, q, and string w such that: 1. (p, q) are marked D p ~ q(by induction) /  2. p = (p,) and q = (q,), where   Σ The string w distinguishes p and q!

Suppose the pair (p, q) is not marked D by the algorithm, yet p ~ q (a “bad pair”) / ^ ^ (p, w)  F and (q, w)  F Claim: If p, q are not distinguished by Table-Filling algorithm, then p ~ q Proof (by contradiction): Then there is a string w such that: (Why is |w| >0 ?) So, w = w, where   Σ Of all such bad pairs, let p, q be a pair with the shortest w Let p = (p,) and q = (q,) Then (p, q) is also a bad pair, but with a SHORTER w !

Algorithm MINIMIZE Input: DFA M Output: DFA MMIN (1) Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: EM = { [q] | q is an accessible state of M } Define: MMIN = (QMIN, Σ, MIN, q0 MIN, FMIN) QMIN = EM, q0 MIN = [q0], FMIN = { [q] | q  F } MIN( [q],  ) = [ ( q,  ) ] Must show MIN is well defined!

Algorithm MINIMIZE Input: DFA M Output: DFA MMIN (1) Remove all inaccessible states from M (2) Apply Table-Filling algorithm to get: EM = { [q] | q is an accessible state of M } Define: MMIN = (QMIN, Σ, MIN, q0 MIN, FMIN) QMIN = EM, q0 MIN = [q0], FMIN = { [q] | q  F } MIN( [q],  ) = [ ( q,  ) ] Claim: MMIN M

0 0 1 1 0 0 1 0 1 1 MINIMIZE q0 q1 q2

1 0 0,1 0 q4 q0 q1 1 1 0 q5 0 0,1 q0 1 q2 q1 D q3 q3 D D q4 D D q5 D D D D q0 q1 q3 q4 q5

1 0 0,1 0 q4 q0 q1 1 1 0 q5 0 q0 q1 D q3 q3 D D q4 D D q5 D D D D q0 q1 q3 q4 q5

*MMIN is the unique minimal DFA equivalent to M Claim: Suppose MMMIN, and M has no inaccessible states and is irreducible. Then there is a bijection that preserves transitions between M and MMIN

*MMIN is the unique minimal DFA equivalent to M Claim: Suppose MMMIN, and M has no inaccessible states and is irreducible. Then there is a bijection that preserves transitions between M and MMIN Note: If Mis minimal, then M has no inaccessible states and is irreducible. (So the Claim implies *) Corollary to the Claim: If M has no inaccessible states and is irreducible, then M is minimal. Proof: Let Mmin  M be minimal. Then Mmin  MMIN So, by Claim, both Mmin and M are isomorphic to MMIN NOT TRUE for NFAs !

  *MMIN is the unique minimal DFA equivalent to M Claim: Suppose MMMIN, and M has no inaccessible states and is irreducible. Then there is a bijection that preserves transitions between M and MMIN Proof: We construct a map from MMIN to M recursively Base Case: q0 MIN→ q0 Recursive Step: If p → p Then q → q q q

Base Case: q0 MIN→ q0 Recursive Step: If p → p Then q → q q q   We need to prove: That is, for all q  MMIN there is a q  M such that q → q The map is defined everywhere The map is well defined If q  MMIN, there is a string w such that MIN(q0 MIN,w) = q The map is a bijection ^ ^ Let q = (q0,w). q will map to q

Base Case: q0 MIN→ q0 Recursive Step: If p → p Then q → q q q   The map is well defined Suppose there exist q and q such that q → q and q → q We show that q and q are indistinguishable, so it must be that q = q

Suppose there exist q and q such that q → q and q → q Suppose q and q are distinguishable MMIN M u u w w Accept Accept q q q0 q0 MIN Contradiction! v w v w Reject Reject q q q0 q0 MIN

Base Case: q0 MIN→ q0 Recursive Step: If p → p Then q → q q q   The map is onto For all q Mthere is a q  MMIN such that q → q If q M, there is w such that (q0,w) = q ^ ^ Let q = MIN(q0 MIN,w)

The map is 1-1 Suppose there are distinct p and q such that p → q and q → q Suppose p and q are distinguishable MMIN M u u w w Accept Accept q p q0 q0 MIN Contradiction! v w v w Reject Reject q q q0 q0 MIN

How can we prove that two regular expressions are equivalent?

WWW.FLAC.WS Finish reviewing Chapter 1 of the book and read 2.1 & 2.2 for next time

FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY