CIRCULAR WORDS AND AUTOMATA MINIMIZATION

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CIRCULAR WORDS AND AUTOMATA MINIMIZATION

• G. Castiglione, A. Restivo, M. Sciortino
• University of Palermo

WORDS 2007, Marseille 17-21 september 2007
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Contents

• Circular words
• Circular standard words
• Hopcrofts algorithm for the minimization of a
  finite states automaton
• Cyclic automaton
• Fibonacci finite words and tightness of the
  Hopcrofts algorithm.

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Circular words

• A0,1 binary alphabet.
• u,v ? A conjugate ? z,w in A s.t. vzw and
  uwz
• conjugation is an equivalence relation let w ?
  A,
• by (w) we denote the class of all the
  conjugates of the word w and we call it circular
  word
• a factor of (w) is a factor of ww of length not
  greater than w
• u is a special factor of (w) if both u0 and u1
  are factors of (w).

Example
00 is a special factor
(01001100)
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Circular standard words 1

• Recall the notion of standard word
• d1, d2,,dn, a sequence of natural integers
  d10, digt0
• snn0 sequence of words over A0,1 where
• s10, sn1 sndn sn-1 for n
  1
• Each finite word sn is called standard word.
• (d1, d2,,dn, ) directive sequence
• (1,1,,1,) corresponds to Fibonacci words.
• (w) is a circular standard word if w is a
  standard word.

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Circular standard words 2

• Proposition. Let v be a word of length n2. (v)
  is a standard
• circular word iff for k0,,n-1, (v) has exactly
  k1 factors of
• length k
• Proposition. Let v be a word of length n2. (v)
  is a standard
• circular word iff (v) has n-1 factors of length
  n-2 and v is
• primitive.

For k2, there are 3 distinct factors of length 2
(10010010)is standard
The factor 010010 has 2 occurrences.
(10010010)is standard
.
cf. Borel,Reutenauer, 2006
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Circular standard words 3

• Corollary. Let v be a word of length n.
• (v) is standard iff ? k0,,n-2, (v) has a unique
  special
• factor of length k.

Special factors k0 e k1 0 k2 10 k3 010 k4 0
010 k5 10010 k6 010010
(10010010)is standard
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Minimal Automaton

• A(Q,S,d,F) be a finite automaton. Given p in
  Q we define
• Lp(A)v? S d (p,v)? F
• For each regular language there exists a
  unique minimal deterministic automaton. It is
  computed by using the Nerode equivalence.
• pq if Lp(A) Lq(A)
• The Nerode equivalence
• is a congruence of A i.e.? a? S, pq ? d (p,a)
  d (q,a)
• is the coarsest congruence of A that saturates F
  i.e. such that F is union of classes of the
  congruence.

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Minimal Automaton

• Let A(Q,S,d,q0,F) automaton that recognizes the
  language L and ? Q1,Q2,,Qm the partition of
  Q corresponding to Nerode equivalence. The
  minimal automaton recognizing L is
  MA(QM,S,dM,q0M,FM) where
• QM Q1,Q2,,Qm
• q0Mq0
• dM(q,a)d(q,a) , ? q? Q and ? a? S
• FMq q ? F

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The Moore construction
The Nerode equivalence is computed by the Moore
construction. For any integer k0 Lpkv?Lp
vk We define k on Q as follows p k q if
Lpk Lqk
Theorem. Let Qn. The Nerode equivalence is
equal to n-2.
Let µ(A)mink k k1
By the theorem it follows that µ(A) ? n-2.
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Hopcrofts Algorithm
Hopcrofts Algorithm computes the minimal
automaton and it is based on splitting operation
of the equivalence classes.
Given a partition ? Q1,Q2,,Qm of Q, we say
that the pair (Qi,a) splits the class Qj if
da-1(Qi)? Qj??. In this case Qj is splitted in
Qj da-1(Qi)? Qj and Qj Qj \ da-1(Qi)
Q
Qi
Qj
Qj

• A partition ? Q1,Q2,,Qm is a congruence iff
• for any 1i,j m and any a ? S, the pair (Qi,a)
  does not splits Qj .

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Hopcrofts Algorithm

• The algorithm computes the coarsest congruence
  that saturates F. It starts from the partition ?
  F,Fc (that, trivially, saturates F) and
  refines it by splitting operations until it
  obtains the congruence i.e. until no split is
  possible.

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Hopcrofts Algorithm
The algorithm maintains the current partition ?
and the waiting set S containing the pairs for
which it has to check whether some classes are
splitted.

• At the first step ? F,Fc and
  S(min(F,Fc),a)
• The main loop of the algorithm takes and deletes
  a pair of S.
• If a class B is splitted, it is replaced in ? by
  the two new classes obtained from the split.
  Furthermore, if (B,a) is in S it is replaced by
  (B,a) and (B,a) otherwise the pair
  (min(B,B),a) is added to S.
• When S is empty, no split is possible and the
  current partition is the coarsest congruence.

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Cyclic Automata

• (w) a circular word wa1a2an over A0,1.
• Aw(Q,S,d,F) automaton associated to w
• Q1,2,,n
• Sa
• d(i,a)(i1), iltn
• d(n,a)1
• F i ?Q ai1

w 11101000
Aw
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Hopcrofts Algorithm
w 11101000 de Brujin word
Notation Qu the set of positions of cyclic
occurrence of u in w.
Aw
S1,2,3,5
S1,2, 4,8
S4,8, 1, 8
S1, 8, 3, 7
S8, 3, 7
S3, 7
S7
S
Q
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Hopcrofts Algorithm

• The algorithm has some degrees of freedom
  because at each step it allows a free choice of
  the pair of S to process.
• As regards to the running time, the splitting
  with respect a pair (C,a) takes a time
  proportional to the cardinality of C. Hence, the
  running time of the algorithm is proportional to
  the sum of the cardinality of all sets processed
  of S.
• The complexity of the algorithm is O(nlogn)
  Hopcroft
• The complexity is tight for the cyclic automata
  associated to de Brujin words Berstel-Carton.
• For the same automata with n states there would
  be executions that run in linear time and other
  executions that run in nlogn.
• Are there automata on which all the executions
  of the algorithm do not run in linear time?

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Hopcrofts Algorithm
w 11101000 de Brujin word
We say Qu to be a splitting set if ? Qv that
splits Qu.
Proposition. If Qu is a splitting set of states
of Aw then u is a prefix of a special factor of
w.
Corollary. If Qv that splits Qu and uv then
u is a cyclic special factor of w and the
resulting sets are Qu0 and Qu1 .
Q
FcQ04,6,7,8
FQ11,2,3,5
Q103,5
Q111,2
Q006,7
Q014,8
Q1005
Q1013
Q1102
Q1101
Q0006
Q0017
Q0104
Q0118
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Hopcrofts Algorithm
w 11101000 de Brujin word
Aw
µ(Aw)logn-1
Q
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The algorithm and Standard Words
Recall Corollary. Let v be a word of length n.
(v) is standard iff ? k0,,n-2, (v) has a
unique special factor of length k.

• Proposition. If w is a standard word the
  executions of the Hopcrofts
• algorithm are uniquely determined. In particular,
  at each step 1kn-2
• ?k Qv v is a special factor of length k and
  Sk1.

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Hopcrofts Algorithm
Proposition. If Qu is a splitting set of states
of Aw then u is a prefix of a special factor of
w.
µ(Aw)max u u is a special factor of w
Corollary. Let v be a word of length n. (v) is
standard iff ? k0,,n-2, (v) has a unique
special factor of length k.
(w) is standard iff µ(Aw)n-2
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Fibonacci words

• Recall the notion of Fibonacci finite word
• fnn0 sequence of words over A0,1 where
• f01, f11,
• fn1 fn fn-1 for n 2
• Each finite word fn is called finite Fibonacci
  word.

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The Algorithm and Fibonacci Words 1
f50 10 0 10 10
a conjugate of f6
f60100101001001
f4 0 1 0 0 1
Q0
Q10
Q10010
Q1010010
Q0010
Q010
Q010010
Q01010010
Q001010010
Q1001010010
Q01001010010
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The algorithm and Fibonacci words 2
Notation Fnfn, Fibonacci numbers C(Fn) the
size of S in the minimization of the automata Afn
Proposition. For each n2, C(Fn)
C(Fn-1)C(Fn-2) Fn-2 C(F0) C(F1)0.
(recurrence relation of Fibonacci convolution
sequence)
C(Fn)1/5((n-1) Fn2nFn-1) FnFn/ sqrt(5)
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The algorithm and Fibonacci words 3
Theorem. Hopcrofts algorithm on cyclic automata
associated to finite Fibonacci words runs in
time O(nlogn).
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Standard words ?
w 0000001 wn
Q
1,2,3,4,5,6
7
1,2,3,4,5
6
C(n)n-1
5
1,2,3,4
µ(Aw)n-2
1,2
2
1
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Thanks for your attention! ?