Short 3-Collapsing Words over a 2-Letter
Alphabet
Alessandra Cherubini1 , Achille Frigeri1 , and Brunetto Piochi2
1
Politecnico di Milano {alessandra.cherubini,achille.frigeri}@polimi.it
2
Università di Firenze
[email protected]
Let A = (Q, Σ, δ) be a finite deterministic complete automaton. A is called kcompressible if there is a word w ∈ Σ + such that the image of the state set Q
under the action of w has at most size |Q| − k, in such case the word w is called
k-compressing for A. A word w ∈ Σ + is k-collapsing if it is k-compressing for
each k-compressible automaton of the alphabet Σ and it is k-synchronizing if it
is k-compressing for each k-compressible automaton with k + 1 states (see [1,2]
for details).
For each alphabet Σ and k ≥ 1, k-collapsing words always exist [4], and
are k-full, i.e contain as factors each word of length k on Σ. Let |Σ| = t,
c(k, t) denotes the length of the shortest k-collapsing words on the alphabet
Σ. Exact values of c(k, t) are known only for k = 2 and t = 2, 3; moreover the shortest 3-synchronizing words for |Σ| = 2 (i.e., the word s3,2 =
ab2 aba3 b2 a2 babab2 a2 b3 aba2 ba2 b2 a and its dual s̄3,2 ) have length 33 [1], hence
by the construction in [3], one gets 33 ≤ c(3, 2) ≤ 154, and more precisely, since
we prove that neither s3,2 nor s̄3,2 are not 3-collapsing (see Fig. 1), 34 ≤ c(3, 2) ≤
154. Nevertheless, we observe that any 3-compressible 5-states automaton over
a two letter alphabet, is 3-compressed either by s3,2 or by s̄3,2 .
We are interested in 3-compressible
a
proper automata, i.e, 3-compressible
automata such that no word of length
b
3 is 3-compressing for them. So identi0
1
fying each letter of the alphabet with
b
a
its action on Q we can assume that
b
a b
b
each letter acts as a permutation or
as a transformation α of one of the
2
3
4
following types (different letters repa
resent different states):
a
1. [x, y, z]/x, y;
Fig. 1. An automaton which is not 32. [x, y][z, v]/x, z;
compressed by s3,2 : δ(Q, s3,2 ) = {0, 1, 3},
3. [x, y]/x;
δ(Q, s̄3,2 ) = {3}.
4. [x, y]/z with zα = x;
where this notation means that the states in the same square brackets are the
unique states identified by α, the state after the slash do not belong to Im(α).
G. Mauri and A. Leporati (Eds.): DLT 2011, LNCS 6795, pp. 469–471, 2011.
c Springer-Verlag Berlin Heidelberg 2011
470
A. Cherubini, A. Frigeri, and B. Piochi
We say that A is a (i., j.)-automaton, 1 ≤ i, j ≤ 4, if a letter is of type i.
and the other letter is of type j., and we say that A is a (i., p)-automaton, with
1 ≤ i ≤ 4, if a letter is of type i. and the other letter is a permutation. Then we
can prove that:
Lemma 1. A 3-compressible (i., j.)-automaton, with i ∈ {1, 2} and j ∈ {1, 2, 4}
on the alphabet {a, b}, is not proper.
Considering exhaustively the case of (i., p)-automata, we give a necessary and
sufficient condition for such automata to be not 3-compressible. In the proof of
such proposition, we produce for each proper 3-compressible (i., p)-automaton
a short 3-compressing word, leading to the following corollary:
Corollary 1. Let A be a 3-compressible proper automaton with input alphabet
Σ = {a, b} in which b acts as a permutation. Then:
1. if letter a is of type 1., then the word ab2 a 3-compress the automaton;
2. if letter a is of type 2., then either the words ab2 a or ab3 a 3-compress the
automaton;
3. if letter a is of type 3., then one of the following words 3-compresses the
automaton:
ababa, abab2 a, aba2 ba, abab2 aba, ab2 ab2 a, ab2 a2 b2 a,
ab2 abab2 a, ab2 aba, ab3 aba, abab3 a, ab3 ab3 a;
4. if letter a is of type 4., then one of the following words 3-compress the automaton:
a2 ba, a2 b2 a, aba2 , a2 b2 a2 , ababa, a2 ba2 , ab2 a2 , ab3 a2 , a2 b3 a, a2 b3 a2 .
One can also prove again via an exhaustive search, or proving that for each proper
3-compressible (i., j)-automaton A there exists an associated (i., p)-automaton
A such that a word 3-compressing A also 3-compresses A, that at least a
word in Corollary 1 or a dual of one of such words 3-compresses all proper
3-compressible automata of the remaining cases (i.e., (1., 3)-, (2., 3)-, (3., 3)-,
(3., 4)-, and (4., 4)-automata). It follows that the word
a2 b3 a3 b2 ab2 abab2 aba3 ba3 bab3 ab3 aba2 baba2 ba2 b2 a2 b2 a
which has length 55 and is a shortest word having all the above words and their
duals as factors, is 3-collapsing, improving the known upper bound. We also
recall that a short 3-collapsing word, can be used in the procedure arising from
([3], Theorem 3.5) to obtain shorter k-collapsing words for k ≥ 4, e.g., one have
c(4, 2) ≤ 1803 and c(5, 2) ≤ 113847, instead of the known upper bound of 4872
and 307194, respectively [3].
Short 3-Collapsing Words over a 2-Letter Alphabet
471
References
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(2007)
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