Composing short 3-compressing words on a 2
letter alphabet⋆,⋆⋆
arXiv:1406.1413v1 [math.CO] 5 Jun 2014
Alessandra Cherubini1 , Achille Frigeri1 , and Zuhua Liu1,2
1
Dipartimento di Matematica “F.Brioschi”, Politecnico di Milano, 20133 Milano,
Italia {alessandra.cherubini,achille.frigeri}@polimi.it
2
Department of Mathematics, Kunming University, 650214 Kunming, China
[email protected]
Abstract. A finite deterministic (semi)automaton A = (Q, Σ, δ) is kcompressible if there is some word w ∈ Σ + such that the image of the
state set Q under the natural action of w is reduced by at least k states.
Such word, if it exists, is called a k-compressing word for A. A word
is k-collapsing if it is k-compressing for each k-compressible automaton.
We compute a set Γ of short words such that all 3-compressible automata on a two letter alphabet with a letter acting as a permutation
are 3-compressed by a word in Γ . Then we construct a shortest common
superstring of the words in Γ and we show that it is a 3-collapsing word
of length 53. Moreover, as previously announced in [4], we show that the
shortest 3-synchronizing word [1] is not 3-collapsing, illustrating the new
bounds 34 ≤ c(2, 3) ≤ 53 for the length c(2, 3) of the shortest 3-collapsing
word on a two letter alphabet.
1
Introduction
Let A = (Q, Σ, δ) be a finite deterministic complete (semi)automaton with state
set Q, input alphabet Σ, and transition function δ : Q × Σ → Q. For any word
w ∈ Σ + , the deficiency of w is the difference between the cardinality of Q and
the cardinality of the image of Q under the natural action of w. For a fixed k ≥ 1,
the word w is called k-compressing for A if its deficiency is greater or equal to k.
An automaton A is k-compressible, if there exists a k-compressing word for A.
A word w ∈ Σ + is k-collapsing, if it is k-compressing for every k-compressible
automaton with input alphabet Σ. A word w ∈ Σ + is called k-synchronizing if
it is k-compressing for all k-compressible automata with k + 1 states and input
alphabet Σ. Obviously each k-collapsing word is also k-synchronizing.
The concept of k-collapsing words arose (under a different name) in the
beginning of the 1990s with original motivations coming from combinatorics
⋆
⋆⋆
This work was completed during the third-named author’s visit to Politecnico di
Milano which was supported by the CSC (China Scholarship Council), and acknowledges support from National Natural Science Foundation of China, grant No.
11261066.
The other authors did this work in the framework of PRIN 2011-2012: “Automi e
linguaggi formali: aspetti matematici e applicativi”
2
and abstract algebra ([10,8]). In [10] it has been proved that k-collapsing words
always exist, for any Σ and any k ≥ 1, by means of a recursive construction which
k
gives a k-collapsing word whose length is O(22 ). Better lower and upper bounds
for c(k, t) and s(k, t), the length of the shortest k-collapsing and k-synchronizing
words respectively, on an alphabet of cardinality t were given in [6]. The bounds
for c(2, t) were slightly improved in [9] and [5], but the gaps from lower and upper
bounds are quite large even for small values of k and t. Exact values of s(k, t)
and c(k, t) are known for k = 2 and t = 2, 3 and are quite far from the theoretical
upper bounds ([2]). Moreover it is known that s(3, 2) = 33 and that the words
s3,2 = ab2 aba3 b2 a2 babab2 a2 b3 aba2 ba2 b2 a and its dual s̄3,2 are the unique shortest
synchronizing words on {a, b} ([1]). Observing that s(k, t) ≤ c(k, t), and applying
the construction in [6], one gets 33 ≤ c(3, 2) ≤ 154.
The reader is referred to [2,3,5,6] for references and connections to Theoretical
Computer Science and Language Theory. The paper is organized as follows: in
Section 2 we introduce some general concepts about 3-compressible automata, in
Section 3 we give a complete characterization of proper 3-compressible automata
on a two letter alphabet with a letter acting as a permutation. In Section 4 we
describe how to use the previous characterization to obtain a short 3-collapsing
word for automata with a permutation. In Section 5 we characterize all proper 3compressible automata without permutations on a two letter alphabet, showing
that the word found in the previous section also 3-compresses them, improving
the known upper bound for c(3, 2), as already announced in [4]. Section 6 ends
the paper with some considerations about the quest for short 3-collapsing words
in general and the relationship between 3-synchronizing and 3-collapsing words.
2
Background
Let A = (Q, Σ, δ) be a finite deterministic complete semiautomaton with state
set Q, input alphabet Σ = {a, b}, and transition function δ : Q × Σ → Q. The
action of Σ on Q given by δ extends naturally, by composition, to the action of
any word w ∈ Σ + on q ∈ Q; we denote it by qw = δ(q, w), while the action of w
on the entire state set Q is denoted by Qw = {qw|q ∈ Q}.
Definition 1. The difference |Q| − |Qw| is called the deficiency of the word w
with respect to A and denoted by dfA (w). For a fixed k ≥ 1, a word w ∈ Σ + is
called k-compressing for A, if dfA (w) ≥ k. An automaton A is k-compressible,
if there exists a k-compressing word for A. A word w ∈ Σ + is k-collapsing, if
it is k-compressing for every k-compressible automaton with input alphabet Σ.
A word w is called k-synchronizing if it is k-compressing for all k-compressible
automata with k + 1 states and input alphabet Σ. Obviously each k-collapsing
word is also k-synchronizing.
Actually, we view the automaton A as a set of transformations on Q induced
via δ and labeled by letters of Σ, rather than as a standard triple. Indeed, in
order to define an automaton, it is enough just to assign to every letter a ∈ Σ
the corresponding transformation τa : q → δ(q, a) on Q. Now, for a ∈ Σ, we get
3
dfA (a) = |Q| − | Im(τa )|, hence dfA (a) = 0 if and only if τa is a permutation on
Q. If dfA (a) = m ≥ 1, then there are exactly m different states y1 , y2 , . . . , ym ∈
/
Im(α), and there are some elements of Q whose images under τa are equal.
Definition 2. Let P = {{x11 , . . . , xj1 }, . . . , {x1r , . . . , xjr }} be a partition of Q
(where singleton sets are omitted), and y1 , . . . , ym ∈ Q, we say that τa is a transformation of type [x11 , . . . , xj1 ] . . . [x1r , . . . , xjr ]\y1 , . . . , ym if P is the partition
induced by the kernel of τa and the states y1 , y2 , . . . , ym do not belong to Im(τa ).
Remark 1. For instance, if A has at least three states denoted by 1, 2 and 3, the
transformation τ is of type [1, 2]\3, if and only if τ (1) = τ (2), the preimage of 3
is empty, and for any q, q ′ ∈
/ {1, 2}, τ (q) = τ (q ′ ) if and only if q = q ′ .
Remark 2. In the sequel we will identify each letter of the input alphabet with
its corresponding transformation.
Definition 3. Let a ∈ Σ, we say that a is a permutation letter if it induces
a permutation on the set of states, i.e., it has deficiency 0. We assume that
permutations on Q, viewed as elements of the symmetric group Sn with |Q| = n,
are written in the factorization in disjoint cycles where sometimes also cycles of
length 1 are explicitly written.
The notion of transformation induced by a letter naturally extends to words,
and then the semigroup generated by the transformations of A consists precisely
of the transformations corresponding to words in Σ + . If A is k-compressible at
least one letter of its input alphabet has deficiency greater than 0. It is well
known that each k-collapsing word over a fixed alphabet Σ is k-full ([10]), i.e.,
contains each word of length k on the alphabet Σ among its factors. Hence, to
characterize k-collapsing words it is enough to consider k-full words compressing
all proper k-compressible automata, i.e., k-compressible automata which are not
compressed by any word of length k.
Proposition 1. Let A be a finite complete automaton on the alphabet {a, b}: it
is 3-compressible and not proper if at least one letter, say a, fulfills one of the
following conditions:
1. it has deficiency greater than 2;
2. it has deficiency 2 and is of type [x, y, z]\u, v, with {u, v} * {x, y, z};
3. it has deficiency 2 and is of type [x, y][z, v]\u, w, with either {u, w} = {x, y},
or {u, w} = {z, v}, or {u, w} * {x, y, z, v};
4. it has deficiency 1 and is of type [x, y]\u, with u ∈
/ {x, y} and ua ∈
/ {x, y}.
The proof of the previous proposition is trivial, indeed if the letter a fulfills
one of the above conditions, then either a or a2 or a3 has deficiency 3. Since
we are looking for a 3-compressible proper automaton we may assume that each
letter of the alphabet either is a permutation or is of one of the following types
(we assume different letters represent different states):
1. [x, y, z]\x, y;
4
2. [x, y][z, v]\x, z;
3. [x, y]\x;
4. [x, y]\z with za ∈ {x, y}.
In the sequel we view the set Q of the states of A as a set of natural numbers:
Q = {1, 2, . . . , n}, so that, when no confusion arises, a letter a of types 1., 2., 3., 4.
is denoted respectively by [1, 2, 3]\1, 2, [1, 2][3, 4]\1, 3, [1, 2]\1, [1, 2]\3 with 3a ∈
{1, 2}.
Definition 4. Let w ∈ Σ + , we call M(w) = Q \ Qw the missing set of w. Let
Q1 ⊆ Q, we denote by M(Q1 , w) the set M(w) ∪ {qw | q ∈ Q1 and ∀q ′ ∈
Q \ Q1 , qw 6= q ′ w}, i.e., the missing set of a having already missed Q1 .
Remark 3. Observe that M(Q, w) = Q, M(∅, w) = M(w), and if a ∈ Σ is a
permutation, M(Q1 , a) = Q1 a. Moreover, |M(w)| ≥ |M(w1 )|, whenever w1 is a
factor of w.
Definition 5. For a permutation
and Q1 ⊆ Q, we call the orbit of a
S+∞ letter a S
over Q1 the set Orba (Q1 ) = n=0 Q1 an = q∈Q1 Orba (q).
We say that A is a (i., j.)-automaton, 1 ≤ i, j ≤ 4, if it is an automata on
a two letter alphabet {a, b} and one letter is of type i. and the other letter is
of type j. We say that A is a (i., p)-automaton, with 1 ≤ i ≤ 4, to denote
that a letter is of type i. and the other letter is a permutation. In the sequel,
without loss of generality, we will always suppose that in a (i., j.)-automaton
(resp. (i., p)-automaton) the letter a is of type i. and b is of type j. (resp. a
permutation).
3
3-compressible (i., p)-automata
In this section we characterize all proper 3-compressible automata over the alphabet {a, b} in which the letter b acts as a permutation on the set Q of states. In
particular in the following four lemmata we give a small set of short 3-collapsing
words when letter a is of type i., 1 ≤ i ≤ 4.
Lemma 1. Let A be a (1., p)-automaton with a = [1, 2, 3]\1, 2. A is not 3compressible if and only if Orbb (1, 2) ⊆ {1, 2, 3}, while it is 3-compressible but
not proper if and only if {1, 2}b * {1, 2, 3}. If Orbb (1, 2) * {1, 2, 3} and {1, 2}b ⊂
{1, 2, 3}, the word ab2 a 3-compresses A.
Proof. If Orbb (1, 2) ⊆ {1, 2, 3}, then A is not 3-compressible, indeed ∀n, m ∈
N, M(an ) = M({1, 2}, an) = M({1, 3}, an) = M({2, 3}, an) = {1, 2} and
{1, 2}bm ∈ {{1, 2}, {1, 3}, {2, 3}}. If {1, 2}b * {1, 2, 3}, i.e., there is x ∈ {1, 2}b \
{1, 2, 3}, then aba 3-compresses A which is not proper.
Conversely, if {1, 2}b ⊂ {1, 2, 3} and Orbb (1, 2) * {1, 2, 3}, then {1, 2}b ∈
{{1, 3}, {2, 3}} and it is easy to check that ∀n ∈ N, {1, 2}b2 ⊂ {1, 2, 3} implies
{1, 2}bn ⊂ {1, 2, 3} and so Orbb (1, 2) ⊆ {1, 2, 3}, whence {1, 2}b2 * {1, 2, 3} and
the word ab2 a 3-compresses A.
5
Lemma 2. Let A be a (2., p)-automaton with a = [1, 2][3, 4]\1, 3. A is not 3compressible if, and only if,
1. Orbb (1, 3) ⊆ {1, 2, 3, 4},
2. {1, 3}b ∈
/ {{1, 2}, {3, 4}}, and
3. if {1, 3}b ∈ {{1, 4}, {2, 3}} then |Orbb (1)| 6= 3 and |Orbb (3)| 6= 3.
A is 3 compressible but non proper if, and only if, {1b, 3b} ∈ {{1, 2}, {3, 4}}
or there is x ∈ {1b, 3b} \ {1, 2, 3, 4}. In all the other cases one of the words
ab2 a, ab3 a 3-compresses A.
Proof. Let Orbb (1, 3) ⊆ {1, 2, 3, 4} and m ∈ N. If {1, 3}b = {1, 3} then {1, 3}bm =
{1, 3}; if {1, 3}b = {2, 4} then {2, 4}b = {1, 3}, hence {1, 3}b2m+1 = {2, 4}
and {1, 3}b2m = {1, 3} and, in both cases, ∀n ∈ N, M(an ) = M({1, 3}, an) =
M({2, 4}, an) = {1, 3}, so A is not 3-compressible. If {1, 3}b = {1, 4}, |Orbb (1)| 6=
3 and |Orbb (3)| 6= 3, then b = (1)(34)(2)π, or b = (1423)π for some permutation π on Q \ {1, 2, 3, 4}. In the first case {1, 3}b2m = {1, 3}, and {1, 3}b2m+1 =
{1, 4}, in the latter {1, 3}b4m+1 = {1, 4}, {1, 3}b4m+2 = {2, 4}, {1, 3}b4m+3 =
{2, 3}, and {1, 3}b4m = {1, 3}, and since ∀n ∈ N, M(an ) = M({1, 3}, an) =
M({1, 4}, an) = M({2, 3}, an) = M({2, 4}, an) = {1, 3}, then A is not 3compressible. Similarly, if {1, 3}b = {2, 3}, |Orbb (1)| 6= 3 and |Orbb (3)| 6= 3, then
b = (12)(3)(4)π, or b = (1234)π for some permutation π on Q \ {1, 2, 3, 4}, and
again A is not 3-compressible. If {1, 3}b ∈ {{1, 2}, {3, 4}}, then M({1, 2}, a) =
{1, 3, 2a} and M({3, 4}, a) = {1, 3, 4a}, so aba 3-compresses A which is not
proper. If there is x ∈ {1, 3}b \ {1, 2, 3, 4}, then either {1, 3}b = {1b, x} or
{1, 3}b = {x, 3b} and again, since M({1b, x}, a) = {1, 3, xa} and M({x, 3b}, a) =
{1, 3, xa}, the word aba 3-compresses A which results not proper.
Conversely, if Orbb (1, 3) ⊆ {1, 2, 3, 4}, {1, 3}b ∈ {{1, 4}, {2, 3}} and |Orbb (1)| = 3
or |Orbb (3)| = 3, then b = (124)(3)π or b = (1)(342)π for some permutation
π on Q \ {1, 2, 3, 4}. In the first case {1, 3}b = {2, 3} and {1, 3}b2 = {3, 4},
in the latter {1, 3}b = {1, 4} and {1, 4}b2 = {1, 2}, and since M({2, 3}, a) =
M({1, 4}, a) = {1, 3}, M({1, 2}, a) = {1, 3, 2a}, and M({3, 4}, a) = {1, 3, 4a},
then aba does not 3-compresses A while ab2 a does. If Orbb (1, 3) * {1, 2, 3, 4}
and {1, 3}b ∈ {{1, 4}, {2, 3}, {2, 4}}, then if x ∈ {1, 3}b2 \ {1, 2, 3, 4} or {1, 3}b2 ∈
{{1, 2}, {3, 4}} the word ab2 a 3-compresses A while aba does not. Otherwise
{1, 3}b2 ∈ {{1, 4}, {2, 3}, {2, 4}}. If {1, 3}b = {2, 4}, then {1, 3}b2 = {2, 4}b
does not belong to {{1, 4}, {2, 3}, {2, 4}}, against the hypothesis. If {1, 3}b =
{1, 4}, then {1, 3}b2 = {1, 4}b = {2, 4} whence 1b = 4, 3b = 1, 4b = 2, while if
{1, 3}b = {2, 3} then {1, 3}b2 = {2, 3}b = {2, 4} whence 3b = 2, 1b = 3, 2b = 4.
As Orbb (1, 3) * {1, 2, 3, 4}, in the former case we have 2b = 1b3 ∈
/ {1, 2, 3, 4}, in
the latter 4b = 3b3 ∈
/ {1, 2, 3, 4}, and in both cases ab3 a 3-compresses A while
aba and ab2 a do not.
Remark 4. Each 3-compressible (i., p)-automaton A, with i ≥ 3 is proper. Namely
each 3-compressible word for A has to contain at least three occurrences of the
letter a and these occurrences cannot be all consecutive.
6
Lemma 3. Let A be a (3., p)-automaton with a = [1, 2]\1. Then A is not 3compressible if, and only if, one of the following conditions holds:
1. b fixes 1 or the set {1, 2},
2. b = (13)π for some permutation π on Q \ {1, 3} and 3a = 3,
3. b = (13)(2)π, b = (132)π or b = (123)π for some permutation π on Q \
{1, 2, 3} and {2, 3}a = {2, 3},
4. b = (13)(24)π or b = (1324)π for some permutation π on Q \ {1, 2, 3, 4} and
{3, 4}a = {3, 4}.
In all other cases A is a proper 3-compressible automaton and one of the words
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, 3-compresses A.
Proof. First observe that ∀n ∈ N, M(an ) = M({1}, an ) = M({2}, an) = {1},
then if b fixes 1 or the set {1, 2}, then A is not 3-compressible. Let b = (13)π
for some permutation π on Q \ {1, 3}. Remark that M({3}, a) = M({1, 3}, a) =
{1, 3a}, so if 3a = 3, then ∀n ∈ N, M({3}, an ) = M({1, 3}, an) = {1, 3}; if
b = (13)(2), 3a = 2 and 2a = 3, then {1, 2}b = {2, 3}, {2, 3}b = {1, 3} and
M({3}, a2n ) = M({1, 3}, a2n) = {1, 3}, M({3}, a2n+1 ) = M({1, 3}, a2n+1) =
{1, 2} and M({2, 3}an = {2, 3}; if b = (13)(24)π with 3a = 4 and 4a =
3 then M({3}, a2n+1 ) = M({1, 3}, a2n+1) = M({2, 3}, a2n+1) = {1, 4} and
M({3}, a2n ) = M({1, 3}, a2n) = M({2, 3}, a2n) = {1, 4}, hence in all three
cases A is not 3-compressible.
If b = (123)π (or b = (132)π) for some permutation π on Q \ {1, 2, 3} with
{2, 3}a = {2, 3}, then M({3}, a) = M({1, 3}, a) = M({2, 3}, a) = {1, 3a} ∈
{{1, 2}, {1, 3}}, hence
– if 3a = 3 then M({3}, an) = M({1, 3}, an) = M({2, 3}, an) = {1, 3} and
M({1, 2}, an) = {1, 2},
– if 3a = 2 then M({3}, a2n) = M({1, 3}, a2n) = M({2, 3}, a2n) = {1, 3} and
M({3}, a2n+1 ) = M({1, 3}, a2n+1) = M({2, 3}, a2n+1) = {1, 2}, M({1, 2}, a2n) =
{1, 2} and M({1, 2}, a2n+1) = {1, 3}
and then, being {1, 2}b = {2, 3}, {1, 3}b = {1, 2} and {2, 3}b = {1, 3} (resp.
{1, 2}b = {1, 3}, {1, 3}b = {2, 3} and {2, 3}b = {1, 2}), A is not 3-compressible.
Lastly, if b = (1324)π for some permutation π on Q \ {1, 2, 3, 4} and {3, 4}a =
{3, 4}, then M({3}, a) = M({1, 3}, a) = M({2, 3}, a) = {1, 3a} ∈ {{1, 3}, {1, 4}},
M({4}, a) = M({1, 4}, a) = {1, 4a} ∈ {{1, 3}, {1, 4}} \ {{1, 3a}}, so if 3a = 3
then M({3}, an) = M({1, 3}, an) = M({2, 3}, an) = {1, 3} and M({4}, an ) =
M({1, 4}, an) = {1, 4}, if 3a = 4 then M({3}, a2n ) = M({1, 3}, a2n) = M({2, 3}, a2n) =
{1, 3}, M({3}, a2n+1 ) = M({1, 3}, a2n+1) = M({2, 3}, a2n+1) = {1, 4}, M({4}, a2n) =
M({1, 4}, a2n) = {1, 4} and M({4}, a2n+1 ) = M({1, 4}, a2n+1) = {1, 3} so,
since {1, 3}b = {2, 3} and {1, 4}b = {1, 3}, again A is not 3-compressible.
Conversely assume that b does not satisfy any of the above conditions. Since
b fixes neither 1 nor {1, 2}, Orbb (1) * {1, 2}.
If Orbb (1) \ {1, 2} = {3}, then there are three cases: b = (13)π for some permutation π on Q \ {1, 3} with 3a 6= 3, or b = (132)π, or b = (123)π for some
permutation π on Q \ {1, 2, 3} and in both cases {2, 3}a 6= {2, 3}.
7
a) Let b = (13)π with 3a 6= 3.
Then 3a2 6= 3a, 3ab 6= 1 and 3a2 b 6= 3ab. If 3ab 6= 2, then dfA (ababa) = 3. So
assume 3ab = 2, hence 3a 6= 1 and 3a2 b 6= 2. If 3a = 2, then b = (13)(2)π ′
with π = (2)π ′ and 2a 6= 3, whence 2ab = 3a2 b ∈
/ {1, 2} and dfA (aba2 ba) = 3.
2
2
Then let 3a = 4, if 3a b 6= 1 again dfA (aba ba) = 3, otherwise 3a2 = 3 and
b = (13)(24)π ′ with π = (2)π ′ and {3a, 4a} 6= {3, 4}, whence 4a 6= 3 and
4ab ∈
/ {1, 2} so dfA (ab3 aba) = 3.
b) Let b = (132)π and {2a, 3a} 6= {2, 3}.
If 3a ∈
/ {2, 3}, then 3ab ∈
/ {1, 2} and dfA (ababa) = 3. Else, if 3a ∈ {2, 3},
then 2a ∈
/ {2, 3} and 2ab ∈
/ {1, 2}. If 3a = 2, then dfA (aba2 ba) = 3, else if
2
3a = 3, then dfA (abab aba) = 3.
c) Let b = (123)π and {2a, 3a} =
6 {2, 3}.
If 3a ∈
/ {2, 3}, then 3ab2 ∈
/ {1, 2} and dfA (ab2 ab2 a) = 3. Else, if 3a ∈ {2, 3},
then 2a ∈
/ {2, 3} and 2ab2 ∈
/ {1, 2}. If 3a = 2, then dfA (ab2 a2 b2 a) = 3, else if
2
2
3a = 3, then dfA (ab abab a) = 3.
Let now |Orbb (1) \ {1, 2}| ≥ 2, we have again three cases: b = (134 . . .)π, or
b = (1324 . . .)π, or b = (1234 . . .)π.
a’) Let b = (134 . . .)π.
If 3ab ∈
/ {1, 2}, then dfA (ababa) = 3, else if 4ab ∈
/ {1, 2}, then dfA (ab2 aba) =
3. So suppose {3ab, 4ab} = {1, 2}; if 3ab = 1, then 3ab2 = 3 and dfA (abab2 a) =
3, else if 4ab = 1, then 4ab2 = 3 and dfA (ab2 ab2 a) = 3.
b’) Let b = (1324 . . .)π.
If 3ab ∈
/ {1, 2}, then dfA (ababa) = 3, if 3ab3 ∈
/ {1, 2}, then dfA (abab3 a) = 3, if
3
4ab ∈
/ {1, 2}, then dfA (ab aba) = 3, else if 4ab3 ∈
/ {1, 2}, then dfA (ab3 ab3 a) =
3
3
3. So suppose {3ab, 3ab , 4ab, 4ab } = {1, 2}, as 3ab 6= 4ab, it follows 3ab =
4ab3 and 4ab = 3ab3 , and then 3a = 4ab2 and 4a = 3ab2 . So 3a = 3ab4 and
b = (1324)π, and from {3ab, 4ab} = {1, 2}, it follows {3a, 4a} = {3, 4}, that
is a contradiction because in such case A satisfies condition 4).
c’) Let b = (1234 . . .)π.
Suppose {3ab2 , 4ab3 , 3abab2, 3a2 b2 } ⊆ {1, 2}.
First assume 3ab2 = 1. If 4ab3 = 2, then 4ab2 = 1 = 3ab2 and 3a = 4a, a
contradiction, so 4ab3 = 1, whence 3a = 4ab. If 3abab2 = 1, then 3aba = 3a
whence 3ab = 3 and 3ab2 = 4, a contradiction, so 3abab2 = 2. Then 3aba =
3ab = 4ab2 . If 3a2 b2 = 1, then 3 = 3a, whence 3b2 = 1 and b = (1234)π,
so 4ab2 = 3b = 4, but 3abab2 = 2 implies 4ab2 = 3bab2 = 3abab2 = 2, a
contradiction. If 3a2 b2 = 2, then 3aba = 3a2 , that implies either 3ab = 1 or
3ab = 3a, a contradiction, as 3a = 4ab.
So assume 3ab2 = 2, whence 3ab = 1. If 3abab2 = 1, then 1ab2 = 1.
If 3a2 b2 = 1, then 3a2 = 1a whence 3a ∈ {1, 2} and 3ab ∈ {2, 3}, a
contradiction; otherwise if 3a2 b2 = 2, then 3a2 = 3a and 3a = 3 that
implies 3b = 1, a contradiction. If 3abab2 = 2, then 3ab = 3, a contradiction. So {3ab2 , 4ab3 , 3abab2, 3a2 b2 } * {1, 2}. If 3ab2 ∈
/ {1, 2} then
dfA (ab2 ab2 a) = 3, if 4ab3 ∈
/ {1, 2}, then dfA (ab3 ab3 a) = 3, if 3abab2 ∈
/ {1, 2},
then dfA (ab2 abab2 a) = 3, if 3a2 b2 ∈
/ {1, 2}, then dfA (ab2 a2 b2 a) = 3.
8
So each (3., p)-automaton with a = [1, 2]\1, that fulfils no condition given in the
statement is 3-compressed by one of the following words: 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.
Lemma 4. Let A be a (4., p)-automaton with a = [1, 2]\3, 3a = 1. Then A is
not 3-compressible if and only if one of the following conditions holds:
1. b fixes {1, 3};
2. b = (3)(12)π for some permutation π on Q \ {1, 2, 3};
3. b = (1)(23)π, or b = (123)π, or b = (132)π for some permutation π on
Q \ {1, 2, 3} and in any case 2a = 2;
4. b = (1)(2)(34)π, or b = (12)(34)π, or b = (14)(23)π, or b = (1423)π, or
b = (1324)π for some permutation π on Q \ {1, 2, 3, 4} and in any case and
4a = 2.
In all other cases A is a proper 3-compressible automaton and it is 3-compressed
by one of the following words: ababa, a2 ba, a2 b2 a, ab2 a2 , a2 b2 a2 , a2 ba2 , a2 b3 a,
ab3 ab3 a and ab2 ab2 a or a2 bab2 a
Proof. Let A be a (4., p)-automaton with a = [1, 2]\3, 3a = 1, then M(a) =
M({1}, a) = {3}, M(a2) = M({3}, an ) = M({1, 3}, an) = {1, 3} for each positive integer n. So if {1, 3}b = {1, 3}, A is not 3-compressible. If b = (3)(12)π
for some permutation π on Q \ {1, 2, 3}, then, since M({2, 3}, an) = {1, 3}
for each positive integer n, A is again not 3-compressible. If b = (1)(23)π,
or b = (123)π, or b = (132)π for some permutation π on Q \ {1, 2, 3} with
2a = 2 , since M({1, 2}, a) = {2, 3} and M({2, 3}, an) = {1, 3} for each positive integer n, A is not 3-compressible. If b = (1)(2)(34)π, or b = (12)(34)π,
or b = (14)(23)π, or b = (1423)π, or b = (1324)π for some permutation π on
Q \ {1, 2, 3, 4} with 4a = 2, then, since M({1, 4}, a) = M({2, 4}, a) = {2, 3} and
M({2, 3}, an) = {1, 3} for each positive integer n, A is not 3-compressible.
Conversely, A is a (4., p)-automaton which does not satisfy any condition of
the statement. First assume that Orbb (1, 3) ⊆ {1, 2, 3}, then since b does not
fix {1, 3}, then Orbb (1, 3) = {1, 2, 3}, hence b = (1)(23)π, or b = (123)π, or b =
(132)π for some permutation π on Q \ {1, 2, 3} and 2a 6= 2, then M({1, 2}, a) =
{3, 2a} and M({1, 2}, a2) = {1, 3, 2a2} and in the first two cases {1, 3}b = {1, 2},
in the latter {1, 3}b2 = {1, 2} , whence either a2 ba2 or a2 b2 a2 3-compresses
A. So let Orbb (1, 3) * {1, 2, 3}. Let Orbb (3) ⊆ {1, 2, 3}, then 1 ∈
/ Orbb (3),
hence b = (14 . . .)(3)π, or b = (124 . . .)(3)π, or b = (14)(23)π with 4a 6= 2, or
b = (145 . . .)(23)π for some permutation π on Q \ Orbb (1, 3). In the first case
a2 ba 3-compresses A, in the third a2 ba2 3-compresses A and in the remaining
two cases a2 b2 a does that. So let Orbb (3) * {1, 2, 3}. If 3b = 4 and 4a 6= 2
then aba2 3-compresses A. So assume 3b = 4 and 4a = 2. If |Orbb (3)| = 2 then
b = (15 . . .)(34)π, b = (125 . . .)(34)π for some permutation π on Q \ Orbb (1, 3).
In the first case a2 ba 3-compresses A, and in the latter a2 b2 a does that. If
|Orbb (3)| ≥ 3 then b = (341 . . .)π, or b = (342 . . .)π, or b = (345 . . .)π for some
permutation π on Q \ Orbb (3). In the first case if 1b 6= 2 then a2 ba 3-compresses
A, if 1b = 2 then ababa 3-compresses A, in the second case if 2b 6= 1 then ababa
9
does that otherwise if 2b = 1 a2 ba is a 3-compressing word for A. In the third
case, since 5a 6= 2, the word ab2 a2 is 3-compressing for A. So let 3b ∈ {1, 2} and
3b2 = 4. If 4a 6= 2 then ab2 a2 3-compresses A. So assume 3b ∈ {1, 2}, 3b2 = 4
and 4a = 2, whence b = (314 . . .)π or b = (324 . . .)π for some permutation
π on Q \ Orbb (3). In the first case a2 b2 a 3-compresses A if 4b 6= 2, otherwise
b = (31425 . . .)π and both the words ab2 ab2 a, a2 bab2 a 3-compress A. In the
latter M(a2 b) = {1b, 2} and if 1b 6= 3 then a2 ba2 is a 3-compressible word
for A, if 1b = 3 then b = (13245 . . .)π and a2 b3 a 3-compresses A. Finally let
{3b, 3b2} = {1, 2} and 3b3 = 4. Then b = (3124 . . .)π or b = (3214 . . .)π for some
permutation π on Q \ Orbb (3). In the first case a2 b3 a 3-compresses A, in the
second ab3 ab3 a does that.
4
A short 3-compressing word automata for
(i., p)-automata
From lemmata 1,2,3 and 4, taking into account that the roles of letters a and b
are interchangeable, the following corollary immediately follows:
Corollary 1. Each 3-full word, containing as factors the following words is a
3-compressible word for each 3-compressible proper automaton on a 2 letter alphabet and a permutation:
abab2 aba, ab2 ab2 a, ab2 a2 b2 a, ab2 abab2 a, ab3 aba, abab3 a, ab3 ab3 a, a2 b3 a
baba2 bab, ba2 ba2 b, ba2 b2 a2 b, ba2 baba2 b, ba3 bab, baba3 b, ba3 ba3 b, b2 a3 b.
We find out that the length of a shortest common 3-full superstring for these
words is 53, such word is for example
b2 a3 ba3 b3 aba2 baba2 ba2 b2 a2 b2 ab2 abab2 aba3 bab3 ab3 a.
5
3-compressible automata without permutations
In this section we characterize proper 3-compressible automata on 2-letter alphabet where no letter acts as permutation.
Lemma 5. All 3-compressible (i., j.)-automata with i ∈ {1, 2} and j ∈ {1, 2, 4}
are not proper.
Proof. We have to consider five different cases.
Let A be a 3-compressible (1., 1.)-automaton on the alphabet {a, b} where
a and b have respectively types [1, 2, 3]\1, 2 and [x, y, z]\x, y. Then {1, 2} *
{x, y, z} or {x, y} * {1, 2, 3}, otherwise A is not 3-compressible. In the first case
ab and in the latter ba are 3-compressible words for A.
Let A be a 3-compressible (1., 2.)-automaton on the alphabet {a, b} where
a and b have respectively types [1, 2, 3]\1, 2 and [x, y][z, v]\x, z. Then {1, 2} ∈
/
10
{{x, z}, {x, v}, {y, z}, {y, v}}, or {x, z} * {1, 2, 3}, otherwise A is not 3-compressible.
In the first case ab and in the latter ba are 3-compressible words for A.
Let A be a 3-compressible (2., 2.)-automaton on {a, b} where a and b have
respectively types [1, 2][3, 4]\1, 3 and [x, y][z, v]\x, z.
Then {1, 3} ∈
/ {{x, z}, {x, v}, {y, z}, {y, v}}, or {x, z} ∈
/ {{1, 3}, {1, 4}, {2, 3}, {2, 4}},
otherwise A is not 3-compressible. In the first case ab and in the latter ba are
3-compressible words for A.
Let A be a 3-compressible (1., 4.)-automaton with a = [1, 2, 3]\1, 2 and
b = [x, y]\z, zb = x. Then {x, z} 6= {1, 2} otherwise M(a) = M(a2 ) = M(ab) =
M(ba) = {1, 2} and A is not 3-compressible. Similarly b 6= [3, 2]\1 (b 6= [3, 1]\2),
otherwise M(ab) = M(b2 ) = M(ab2 ) = M(b3 ) = {1, 3}, M(a2) = M({1, 3}, a) =
{1, 2} (M(ab) = M(b2 ) = M(ab2 ) = M(b3 ) = {2, 3}, M(a2) = M({2, 3}, a) =
{1, 2}), and A is not 3-compressible. If {x, z} * {1, 2, 3} then b2 a 3-compresses
A and if {1, 2} ∩ {x, y} = ∅ then ab does that. So let {x, z} ∈ {{1, 3}, {2, 3}}. If
{x, z} = {1, 3}({x, z} = {2, 3}) then b = [1, y]\3 (b = [2, y]\3). Moreover 2b 6= 2
(1b 6= 1), otherwise A is not 3-compressible, whence aba 3-compresses A
Let A be a 3-compressible (2., 4.)-automaton with a = [1, 2][3, 4]\1, 3 and
b = [x, y]\z, zb = x. If {x, z} = {1, 3} then M(a) = M(a2 ) = M(ab) = M(ba) =
{1, 3} and A is not 3-compressible. Similarly if b = [4, 3]\1 or b = [2, 1]\3, A
is not 3-compressible. If {x, z} ∈
/ {{1, 4}, {2, 3}, {2, 4}} then b2 a 3-compresses A
and if {1, 3} ∩ {x, y} = ∅ then ab does that. So let {x, z} ∈ {{1, 4}, {2, 3}, {2, 4}},
b 6= [4, 3]\1 and b 6= [2, 1]\3. If {x, z} = {1, 4}({x, z} = {2, 3}) then b = [1, y]\4
(b = [3, y]\2). If 3b 6= 2 (1b 6= 4), then aba 3-compresses A, if 3b 6= y (1b 6= y),
then ab2 3-compresses A, in the remaining case , 3b = 2 = y (1b = y = 4), since
M(a) = M({1, 3}, a) = M({1, 4}, a) = M({2, 3}, a) = {1, 3}, M({1, 3}, b) =
{2, 4}, M({2, 4}, b) = M({1, 4}, b) =, (M({2, 4}, b) = M({2, 3}, b) = {2, 3}),
we get the contradiction that A is not 3-compressible. Lastly let {x, z} = {2, 4}.
If b = [2, 3]\4 or b = [4, 1]\2 either ab2 or aba 3-compresses A, so let b = [2, 1]\4
(b = [4, 3]\2), then 3b 6= 1 (1b 6= 3), otherwise A is not 3-compressible, and aba
3-compresses A.
Lemma 6. Let A be a (1., 3.)-automaton with a = [1, 2, 3]\1, 2, and b = [x, y]\x.
A is a 3-compressible proper automaton if and only if x ∈ {1, 2, 3}, {1, 2} ∩
{x, y} 6= ∅ and z ∈ {1, 2} \ {x} , Orbb (z) * {1, 2, 3} and zb ∈ {1, 2, 3}. In such
case ab2 a 3-compresses A. Moreover each word w ∈ {a, b}+ 3-compressing all
proper 3-compressible (1., p.)-automata, 3-compresses also A.
Proof. If {1, 2} ∩ {x, y} = ∅ then ab 3-compresses A, if x ∈
/ {1, 2, 3} then ba
3-compresses A so in both cases A is not proper. So let {1, 2} ∩ {x, y} =
6 ∅
and x ∈ {1, 2, 3}, then M(ab) = {x, zb} with z ∈ {1, 2}, z 6= x. If Orbb (z) ⊆
{1, 2, 3}, then A is not 3−compressible. Therefore Orbb (z) * {1, 2, 3}. Moreover
zb ∈ {1, 2, 3} otherwise aba 3−compresses A and A is not proper. If zb2 ∈
{1, 2, 3} then {z, zb, zb2} ⊆ {1, 2, 3} \ {1} and by Dirichlet drawer principle, two
of z, zb, zb2 are equal, hence we get the contradiction Orbb (z) ⊆ {1, 2, 3}. So
zb2 ∈
/ {1, 2, 3} which implies that ab2 a 3−compresses A. Furthemore for each
3-compressible proper (1., 3.)-automaton A one can build a (1., 3.)-automaton
11
A′ = (Q, {a, b}, δ ′) with δ ′ (q, a) = δ(q, a) for all q ∈ Q, δ ′ (q, b) = δ(q, b) for all
q ∈ Q \ {x}, and δ ′ (x, b) = x. Then in A′ zb2 ∈
/ {1, 2, 3} and zb ∈ {1, 2, 3}. Hence
w 3-compresses A′ if and only if it contains a factor abj a with 2bj ∈
/ {1, 2, 3},
but in such case w 3-compresses also A.
Lemma 7. Let A be a (2., 3.)-automaton with a = [1, 2][3, 4]\1, 3, and b =
[x, y]\x. Then A is a proper 3-compressible automaton if and only if the following
conditions hold
1. {1, 3} ∩ {x, y} 6= ∅,
2. x ∈ {1, 2, 3, 4},
3. Orbb (z) * {z, v} where za = va and {z} = {1, 3} \ {x} if x ∈ {1, 3} and
{z} = {1, 3} \ {y} if y ∈ {1, 3}, and zb = v.
Moreover one of the words ab2 a, ab3 a 3-compresses each proper 3-compressible
(2., 3.)-automaton.
Proof. If x ∈
/ {1, 2, 3, 4}, then ba 3-compresses A, if {1, 3} ∩ {x, y} = ∅, then
ab 3-compresses A. So let x ∈ {1, 2, 3, 4} and {1, 3} ∩ {x, y} =
6 ∅. If x ∈ {1, 3}
then we can write b = [x, f ][z, v]\x, z with {z} = {1, 3} \ {x}, otherwise y ∈
{1, 3} and b = [y, f ][z, v]\y, z with {z} = {1, 3} \ {y}. Hence M(ab) = {x, zb},
where x 6= zb. If {zb, zb2, zb3 } ⊆ {z, v}, by Dirichlet drawer principle, two words
of zb, zb2, zb3 are equal, and this implies that zbn ∈ {z, v} for any positive
integer n, then A is not 3-compressible, since M(a2 ) = M(ba) = M({x, z}, a) =
M({x, v}, a){1, 3}, M(b2) = {x}, M({x, z}, b) ∈ {{x, z}, {x, v}}, M({x, v}, b) ∈
{{x, z}, {x, v}}. Therefore {zb, zb2, zb3 } * {z, v}. Now zb 6= v implies aba 3compresses A and A is not proper; so zb2 ∈
/ {z, v} or zb3 ∈
/ {z, v} which give
2
3
respectively that ab a or ab a are 3−compressing words for A.
Lemma 8. Let A be a 3-compressible (3., 3.)-automaton with a = [1, 2]\1, b =
[2, y]\2 then {1ba, 1b2a, 1b3 a} * {2, y} or Orba (1ba) * {2, y}.
Proof. Assume by contradiction that {1ba, 1b2a, 1b3 a} ⊆ {2, y} and Orba (1ba) ⊆
{2, y}. If 1b = 1 then 1ba = 1b2 a = 1b3 a = 1a = 2a and Orba (1ba) ⊆ {2, y}
gives 1ba = 2a = 2 or 1ba = 2a = y and = 1ba2 = ya = 2. In the first
case M(a) = M({1}, a) = M({2}, a) = {1}, M({1, 2}, a) = {1, 2}, M(b) =
M({2}, b) = {2}, M({1}, b) = {1}, M({1, 2}, b) = {1, 2}. In the latter M(a) =
M({1}, a) = M({2}, a) = {1}, M({1, 2}, a) = {1, y}, M({1, y}, a) = {1, 2},
M(b) = M({2}, b) = M({y}, b) = {2}, M({1}, b) = {1}, M({1, 2}, b) =
M({1, y}, b) = {1, 2}. Hence A is not 3-compressible, a contradiction.
So let 1b 6= 1, hence 1ba 6= 1b2 a, 1b2 a 6= 1b3 a and so 1ba = 1b3 a and 1b2 = 1.
Since Orba (1ba) ⊆ {2, y} then either 1ba = 1b3 a = 2 and 1b2 a = 1a = 2a = y =
1ba2 or 1ba = 1b3 a = y and 1b2 a = 1a = 2a = 2. In the both cases y ∈ Orba (1ba)
hence in the first case ya = 2 whence ya = 1ba and so 1b = y, yb = 1 and 2a = y,
in the latter case 2a = 2 and ya = y, hence ya = 1ba which implies 1b = y,
yb = 1. In both cases M(a) = M({1}, a) = M({2}, a) = {1}, M(b) = {2},
M({1}, b) = M({1, 2}, b) = M({1, y}, b) = {2, y}, M({2, y}, b) = {1, 2}, moreover in the first case M({2, y}, a) = M({1, y}, a) = {1, 2}, M({1, 2}, a) = {1, y},
12
in the second one M({2, y}, a) = M({1, y}, a) = {1, y}, M({1, 2}, a) = {1, 2}.
Hence A is not 3-compressible against the hypothesis.
Lemma 9. Let A be a (3., 3.)-automaton with a = [1, 2]\1, and b = [x, y]\x.
Then A is a proper 3-compressible automaton if and only if one of the following
conditions holds
1. x ∈
/ {1, 2}, y 6= 1 with 1b ∈ {1, 2}, xa ∈ {x, y} and Orbb (1) * {1, 2} or
Orba (x) * {x, y};
2. x = 2, y 6= 1 and {1ba, 1b2a, 1b3 a} * {2, y} or Orba (1ba) * {2, y};
3. y = 1, x 6= 2 and {xab, xa2 b, xa3 b} * {1, 2} or Orbb (xab) * {1, 2}.
Moreover each proper 3-compressible (3., 3.)-automaton is 3-compressed by a
word in the set
{abab, ab2ab, aba2 b, ab3 ab, aba3 b, baba, ba2 ba, bab2a, ba3 ba, bab3 a},
or equivalently by a word in the set
{abab, ab2ab, aba2 b, ab2 a2 b, baba, ba2 ba, bab2a, ba2 b2 a}.
Proof. Let A be a (3., 3.)-automaton with a = [1, 2]\1, and b = [x, y]\x. If
x = 1, or {1, 2} = {x, y}, then it is easy to see that A is not 3−compressible.
Then either {1, 2} ∩ {x, y} = ∅, or y ∈ {1, 2} and x ∈
/ {1, 2} , or x = 2 and
y 6= 1. If x 6= 2, y 6= 1 and 1b ∈
/ {1, 2} then M(aba) = {1ba, xa, 1} and A is a
non proper 3-compressible automaton. Similarly if x 6= 2, y 6= 1 and xa ∈
/ {x, y}
then bab 3-compresses A which is a non proper 3-compressible automaton. So
let 1b ∈ {1, 2} and {xa ∈ {x, y}. If Orbb (1) ⊆ {1, 2} and Orba (x) ⊆ {x, y},
then we have the following cases: 1b = 1, xa = x, or 1b = 2, 2b = 1, xa = x, or
1b = 1, xa = y, ya = x, or 1b = 2, 2b = 1, xa = y, ya = x.
– If 1b = 1, xa = x, then M(a) = M ({1}, a) = {1}, M(b) = M({x}, b) =
{x}, M({x}, a) = M({1, x}, a) = M({1}, b) = M({1, x}, b) = {1, x}.
– If 1b = 2, 2b = 1, xa = x, then M(a) = M ({1}, a) = {1}, M(b) =
M({x}, b) = {x}, M({x}, a) = M({1, x}, a) = M({2, x}, a) = {1, x},
M({1}, b) = M({1, x}, b) = {2, x}, M({2, x}, b) = {1, x}.
– If 1b = 1, xa = y, ya = x, then M(a) = M ({1}, a) = {1}, M(b) =
M({x}, b) = {x}, M({x}, a) = M({1, x}, a) = {1, y}, M({1}, b) = M({1, x}, b) =
{1, x}.
– If 1b = 2, 2b = 1, xa = x, then M(a) = M ({1}, a) = {1}, M(b) =
M({x}, b) = {x}, M({x}, a) = M({1, x}, a) = M({2, x}, a) = {1, x},
M({1}, b) = M({1, x}, b) = {2, x}, M({2, x}, b) = {1, x}.
– Lastly if 1b = 2, 2b = 1, xa = y, ya = x, then M(a) = M ({1}, a) =
{1}, M(b) = M({x}, b) = {x}, M({x}, a) = M({1, x}, a) = M({2, x}, a) =
{1, y}, M({1, y}, a) = M({2, y}, a) = {1, x}, M({1}, b) = M({1, x}, b) =
M({1, y}, b) = {2, x}, M({2, x}, b) = M({2, y}, b) = {1, x}.
13
So in all cases A is not 3-compressible.
If x = 2, y 6= 1, {1ba, 1b2a, 1b3 a} ⊆ {2, y} and Orba (1ba) ⊆ {2, y}, then by
lemma 8 A is not 3-compressible. If y = 1, x 6= 2, {xab, xa2 b, xa3 b} ⊆ {1, 2} and
Orbb (xab) ⊆ {1, 2}, then, exchanging a and b, A is not 3-compressible by lemma
8 . Conversely let A be a (3., 3.)-automaton satisfying one of the conditions of
the statement. Suppose that A satisfies condition (1) with Orbb (1) * {1, 2},
then 1b2 ∈
/ {1, 2} because {1b, 1b2} ⊆ {1, 2} implies 1bm ∈ {1, 2} for each positive integer m, hence ab2 a 3-compresses A. Similarly if A satisfies condition
(1) with Orba (x) * {x, y}, ba2 b 3-compresses A. So one of the words ab2 a, ba2 b
3-compresses each (3., 3.)-automaton satisfying condition (1).
Condition (3) is obtained by condition (2) exchanging the roles of a and b. So assume that A satisfies condition (2). If {1ba, 1b2a, 1b3 a} * {2, y} then one of the
words abab, ab2 ab, ab3 ab 3 compresses A. So assume that {1ba, 1b2a, 1b3 a} ⊆
{2, y}, hence Orba (1ba) * {2, y} and one of the words abab, aba2b, aba3 b 3compresses A. So each (3., 3.)-automaton satisfying condition (2) is 3 compressible and it is 3 compressed by a word in the set {abab, ab2ab, aba2 b, ab3 ab, aba3 b}.
Moreover if A satisfies condition (2) then {1ba, 1b2a, 1ba2 , 1b2 a2 } * {2, y}. Assume by contradiction {1ba, 1b2a, 1ba2 , 1b2 a2 } ⊆ {2, y}. We consider the following two cases
Case 1: 1b = 1. Then 1ba = 1b2 a = 1b3 a ∈ {2, y}. If 1ba = 2 = 1a = 2a then
Orba (1ba) = {2}, otherwise if 1ba = y = 1a = 2a, then 1ba2 = ya = 2 and
Orba (1ba) = {2, y}, so A does not satisfies condition (2).
Case 2: 1b 6= 1 whence 1ba 6= 1b2 a and {1ba, 1b2a} = {2, y} that implies
{2a, ya} = {1ba2, 1b2 a2 } = {2, y}. First assume 1ba = 2, 1b2a = y. Then
1ba2 = 2a = 2 gives the contradiction 1b = 1. So let 1ba2 = 2a = 1a = y = 1b2 a
and ya = 1a2 = 1b2 a2 = 2. The equalities 1b2 a = 1a, 1ba = ya give 1b2 = 1
and 1b = y whence 1b3 = 1b = y, so we get the contradiction {1ba, 1b2a, 1b3 a} =
{1ba, 1b2a,1b3a} = {2, y}. Then let 1ba = y, 1b2 a = 2. If ya = y and 2a = 2 then
1b = y, 1b2 = 1, 1b3 = y and {1ba, 1b2a, 1b3 a} = {2, y}, {1ba, 1ba2, 1ba3 } = {y},
a contradiction. So let ya = 2 and 2a = y, hence 1ba = 2a gives the contradiction
1b = 1.
Then each (3., 3.)-automaton satisfying condition (2) is 3 compressed also by
a word in the set {abab, ab2ab, aba2 b, ab2 a2 }. So we can conclude that each
(3., 3.)-automaton satisfying one of the conditions of the statement is a proper
3-compressible automaton and is compressed by a word in the set
{abab, ab2 ab, aba2 b, ab3 ab, aba3 b, baba, ba2ba, bab2 a, ba3 ba, bab3 a}
or equivalently by a word in the set
{abab, ab2ab, aba2 b, ab2 a2 b, baba, ba2 ba, bab2a, ba2 b2 a}.
Lemma 10. Let A be a (3., 4.)-automaton with a = [1, 2]\1 and b = [x, y]\z
with zb = x. Then A is a proper 3-compressible automaton if and only if one of
the following conditions holds
1. b = [1, y]\z with 2 ∈
/ {z, y} and za 6= z;
14
2.
3.
4.
5.
6.
b = [x, y]\z with {x, y} = {1, 2} and |Orba (z)| > 2 or zab ∈
/ {x, y};
b = [1, y]\2 with |Orba (2)| > 2 or 2ab ∈
/ {1, y};
b = [x, 1]\2 with Orba (2, x) 6= {2, x} or {xab, 2ab} =
6 {x, 1};
b = [x, y]\2 with 1 ∈
/ {x, y}, 1b = y and {xa, ya} 6= {x, y};
b = [x, y]\1 with Orba (x) * {x, y}.
Moreover one of the words in {b2 ab2 , b2 a2 b2 , b2 a3 b2 , b2 abab2 } 3-compresses A.
Proof. Obviously M(a) = M({1}, a) = M({2}, a) = {1}, M(b) = {z}, M(b2 ) =
M({x, z}, b) = M({y, z}, b) = {x, z}. If {x, z} ∩ {1, 2} = ∅ then M(b2 a) =
{xa, za, 1}, if z ∈
/ {1, 2} and {1, za} ∩ {x, y} = ∅ then M(bab) = {zab, 1b, z},
if {1, 1b} ∩ {x, y} = ∅ then M(ab2 ) = {1b2 , x, z}, if 1 ∈
/ {x, y} and {z, 1b} ∩
{1, 2} = ∅ then M(aba) = {1ba, za, 1}. So in order that a 3-compressible (3., 4.)automaton is proper, the following conditions have to be satisfied ({x, z} ∩
{1, 2} 6= ∅) ∧ (z ∈ {1, 2} ∨ {1, za} ∩ {x, y} =
6 ∅) ∧ ({1, 1b} ∩ {x, y} =
6 ∅) ∧ (1 ∈
{x, y} ∨ {z, 1b} ∩ {1, 2} 6= ∅).
Assume that A is a proper 3-compressible (3., 4.)-automaton with z ∈
/ {1, 2},
hence M({1, z}, a) = M({2, z}, a) = {1, za}. The above condition for A being
proper gives x ∈ {1, 2} and 1 ∈ {x, y} or 1b ∈ {x, y} ∩ {1, 2}, moreover 1 ∈
/ {x, y}
implies x = 2 and 1b ∈ {x, y} ∩ {1, 2} = {2} whence the contradiction z = 1,
so x ∈ {1, 2} and 1 ∈ {x, y} hence x = 1, y 6= 2 or {x, y} = {1, 2}. If za = z
then M({z}, a) = M ({x, z}, a) = {x, z}, M(b2 ) = M({z}, b) = M({x, z}, b) =
{x, z} and A is not 3-compressible. Then let za 6= z. If x = 1, y 6= 2 then
M({1, za}, b) = M({y, za}, b) = {zab, z}, and if y 6= 2 then M({zab, z}, b) =
{zab2 , 1, z} or M({zab, z}, a) = {zaba, za, 1}, hence one of the words baba, bab2
3-compresses A. So let {x, y} = {1, 2}, |Orba (z)| ≤ 2 and zab ∈ {x, y}. Since we
assumed za 6= z then |Orba (z)| = 2 hence za2 = z that implies za2 b = zb = x
and zab = y. Then M({1, za}, b) = {y, z}, M({1, za}, a) = {1, z} and A is not
3-compressible. Conversely if b = [x, y]\z with {x, y} = {1, 2} and zab ∈
/ {x, y}
then bab2 3-compresses A, if b = [x, y]\z with {x, y} = {1, 2}, zab ∈ {x, y} and
|Orba (z)| > 2 then za, za2 , za3 are pairwise distinct and either za2 b or za3 b does
not belong to {x, y}, hence either ba2 b2 or ba3 b2 3-compresses A.
So it remains to assume that A is a proper 3-compressible (3., 4.)-automaton
with z ∈ {1, 2}, then the fact that A is proper implies that 1 ∈ {x, y} and z = 2
or 1 ∈
/ {x, y}, 1b ∈ {x, y} i.e z = 2 and 1b = y or z = 1.
If b = [1, y]\2 with |Orba (2)| ≤ 2 or 2ab ∈ {1, y}, then either 2a = 2 and 2ab =
2b = 1 or 2a = y, ya = 2 and 2ab = yb = y, in the first case M({1, 2}, a) = {1, 2}
in the latter M({1, 2}, a) = {1, y}, M({1, y}, a) = M({2, y}, a) = {1, 2}, M({1, 2}, b) =
{1, 2}, M({1, y}, b) = {2, y}, M({2, y}, b) = {1, 2} and in both cases A is not 3compressible. Conversely if b = [1, y]\2 and 2ab ∈
/ {1, y} then b2 ab2 3-compresses
A, if 2ab ∈ {1, y} and |Orba (2)| > 2, then 2a, 2a2 , 2a3 are pairwise distinct and
either 2a2 b or 2a3 b does not belong to {1, y}, hence either b2 a2 b2 or b2 a3 b2 3compresses A.
If b = [x, 1]\2 with Orba (2, x) = {2, x} and {xab, 2ab} = {x, 1}, then either
2a = 2, xa = x and xab = 1 = xb, 2ab = x or 2a = x, xa = 2 and
xab = x, 2ab = 1 = xb. In the first case M({1, x}, a) = M({2, x}, a) =
{1, x} and M({1, x}, b) = M({2, x}, b) = {2, x}, in the latter M({2, x}, a) =
15
M({1, x}, a) = {1, 2}, M({1, 2}, a) = {x, 1} and M({1, 2}, b) = {2, x}, M({ x}, b) =
{1, 2}, so in both cases A is not 3-compressible. Conversely let b = [x, 1]\2 with
{xab, 2ab} 6= {x, 1}, if xab ∈
/ {x, 1} then b2 ab2 3-compresses A, so let xab ∈ {x, 1}
and 2ab ∈
/ {x, 1}. If xab = x then xa = 2 and M(b2 a) = {1, 2}, so b2 a2 b2 3compresses A, if xab = 1, then M(b2 ab) = {1, 2}, and b2 abab2 3-compresses A.
So let b = [x, 1]\2 with {xab, 2ab} = {x, 1} and Orba (2, x) 6= {2, x}. We claim
that if 2a ∈
/ {2, x} then 2a2 b ∈
/ {1, x}. Namely 2a ∈
/ {2, x} implies 2ab = 1 and
xab = x whence xa = 2, and 2a2 b = 1 gives the contradiction 2a = 2, 2a2 b = x
gives 2a2 = 2 = xa that implies the contradiction 2a = x. Then b2 a3 b2 is a
3-compressible word for A. Similarly if xa ∈
/ {2, x} then xab = 1 and 2ab = x
whence 2a = 2 and again xa2 b = 1 gives the contradiction xa = x while xa2 b = x
gives the contradiction xa = 2, so xa2 b ∈
/ {1, x} and b2 a2 b2 3-compresses A.
If b = [x, y]\2 with 1 ∈
/ {x, y}, 1b = y with {xa, yz} = {x, y}, then M({1}, b) =
{2, y}, M({1, x}, a) = M({2, x}, a) = {1, xa} ∈ {{1, x}, {1, y}}, M({1, y}, a) =
M({2, y}, a) = {1, ya} ∈ {{1, x}, {1, y}}, M({1, x}, b) = M({1, y}, b) = {2, y},
M({2, x}, b) = M({2, y}, b) = {2, x}, hence A is not 3-compressible. Conversely
if b = [x, y]\2 with 1 ∈
/ {x, y}, 1b = y with {xa, yz} 6= {x, y}, then if xa ∈
/
{x, y}, b2 ab is a 3-compressible word for A, if xa ∈ {x, y}, then M({1, x}, a) ∈
{{1, x}, {1, y}}, M({1, x}, b) = M({1, y}, b) = {2, y}, M({2, y}, a) = {1, ya}
with ya ∈
/ {x, y}, hence b2 abab 3-compresses A.
Lastly assume b = [x, y]\1 with Orba (x) ⊆ {x, y}, then either M({1, x}, a) =
{1, x} and A is not 3-compressible, or M({1, x}, a) = {1, y}, whence M({1, y}, a) =
{1, x}, M({1, x}, b) = M({1, y}, b) = {1, x} and A is not 3-compressible. Conversely let b = [x, y]\1 with Orba (x) * {x, y}. Then either xa ∈
/ {x, y} and
b2 ab 3-compresses A, or xa = y and ya ∈
/ {x, y} and b2 a2 b 3-compresses A. In
conclusion each 3-compressible proper automaton is 3-compressed by a word in
the following set {b2 ab2 , b2 a2 b2 , b2 a3 b2 , b2 abab2 }.
.
Lemma 11. Let A be a (4., 4.)-automaton with a = [1, 2]\3, 3a = 1, b =
[x, y]\z, zb = x. Then A is a proper 3-compressible automaton if and only if
one of the following conditions holds
1.
2.
3.
4.
5.
b = [x, 3]\1 with xa 6= 2 or 2b 6= 3;
b = [x, 3]\2 with xa 6= 2 or 1b 6= 3;
b = [3, y]\2 with ya 6= 2 or 1b 6= y;
b = [x, 3]\z with z ∈
/ {1, 2}, x ∈ {1, 2}, za ∈ {1, 2};
b = [x, y]\2 with 3 ∈
/ {x, y}, 1 ∈ {x, y}, 3b ∈ {x, y};
Moreover one of the words in the set {a2 ba2 , a2 b2 , b2 ab2 , b2 a2 } 3-compresses A.
Proof. Let A be a 3-compressible (4., 4.)-automaton with a = [1, 2]\3, 3a = 1,
b = [x, y]\z, zb = x. If {1, 2} ∩ {x, z} = ∅ then dfA (b2 a) = 3, if z ∈
/ {1, 2} and
{3, za} ∩ {x, y} = ∅ then bab 3-compresses A, if {3, 3b} ∩ {x, y} = ∅ then ab2
3-compresses A, if {1, 3} ∩ {x, y} = ∅ then a2 b 3-compresses A, if 3 ∈
/ {x, y} and
{z, 3b} ∩ {1, 2} = ∅ then aba 3-compresses A, if {z, za} ∩ {1, 2} = ∅ then ba2
16
3-compresses A. Hence in order that A is proper the following condition holds
({1, 2} ∩ {x, z} =
6 ∅) ∧ (z ∈ {1, 2} ∨ {3, za} ∩ {x, y} 6= ∅) ∧ ({3, 3b} ∩ {x, y} =
6 ∅) ∧
({1, 3}∩{x, y} 6= ∅)∧(3 ∈ {x, y}∨{z, 3b}∩{1, 2} =
6 ∅)∧({z, za}∩{1, 2} =
6 ∅), i.e.
(z ∈ {1, 2} ∧ 3 ∈ {x, y}) ∨ (z ∈ {1, 2} ∧ 1 ∈ {x, y} ∧ 3b ∈ {x, y}) ∨ (3 ∈ {x, y} ∧ x ∈
{1, 2} ∧ za ∈ {1, 2}) ∨ (x ∈ {1, 2} ∧ 1 ∈ {x, y} ∧ {za, 3b} ⊆ {x, y} ∩ {1, 2}).
Let z ∈ {1, 2} ∧ 3 ∈ {x, y}. If z = 1 and x = 3, then A is not 3-compressible.
If z = 1,y = 3, xa = 2 and 2b = 3 then M(b2 ) = {1, x}, M(a2 ) = {1, 3},
M({1, x}, a) = {2, 3}, M({2, 3}, a) = M({2, 3}, b) = {1, 3}, M({1, 3}, b) =
{1, x} and A is not 3-compressible. If z = 2, x = 3, ya = 2 and 1b = y then
M(b2 ) = {2, 3}, M(a2 ) = M({2, 3}, a) = {1, 3}, M({1, 3}, b) = {2, y}, M({2, y}, b) =
M({2, y}, a) = {2, 3} and A is not 3-compressible. If z = 2,y = 3, xa = 2 and
1b = 3 then M(b2 ) = {2, x}, M(a2 ) = {1, 3}, M({2, x}, a) = {2, 3}, M({2, 3}, a) =
{1, 3}, M({1, 3}, b) = {2, 3}, M({2, 3}, b) = {2, x} and again A is not 3compressible. If z ∈
/ {1, 2} ∧ 3 ∈
/ {x, y}, then in order that A is proper 1 ∈
{x, y}, 3b ∈ {1, 2} ∩ {x, y}, za ∈ {1, 2} ∩ {x, y}, x ∈ {1, 2} whence either
x = 1, y = 2, 3b ∈ {1, 2}, za ∈ {1, 2}, or x = 1, y 6= 2, 3b = za = 1, or
x = 2, y = 1, 3b ∈ 1, 2}, za ∈ {1, 2.
In the first case M(b2 ) = {z, 1}, M(a2 ) = {3, 1}, M({z, 1}, a) = {za, 3},
M({za, 3}, a) = {3, 1}, M({za, 3}, b) = {z, 1}, M({3, 1}, b) = {3b, z}, M({3b, z}, a) =
{za, 3}, M({3b, z}, b) = {z, 1} and A is not 3-compressible. The third case is
symmetric to the previous one and again A is not 3-compressible. Finally if
b = [1, y]\z with y 6= 2, z 6= 2, 3b = za = 1, then M(b2 ) = M({1, 3}, b) =
{1, z}, M(a2 ) = M({1, z}, a) = {1, 3} and A is not 3-compressible.
Conversely assume that A has one of the forms 1,2,3 in the Statement. If
b = [x, 3]\z with z ∈ {1, 2}, xa 6= 2, then xa 6= 1 because x 6= 3, M(b2 ) =
{z, x}, M({z, x}, a) = {xa, 3}, M({xa, 3}, a) = {xa2 , 1, 3} and b2 a2 3-compresses
A. If b = [x, 3]\1 with xa = 2 and 2b 6= 3, then 2b 6= x because 1 6= 2, M(b2 ) =
{1, x}, M({1, x}, a) = {2, 3}, M({2, 3}, b) = {2b, 1}, M({2b, 1}, b) = {2b2 , x, 1}
and b2 ab2 3-compresses A. Similarly if b = [x, 3]\2 with xa = 2 and 1b 6= 3, then
1b 6= x because 1 6= 2, M(b2 ) = {2, x}, M({2, x}, a) = {2, 3}, M({2, 3}, b) =
{2b, 2}, M({2b, 1}, b) = {2b2, x, 2} and b2 ab2 3-compresses A. If b = [3, y]\2
with 1b 6= y then 1b 6= 3 because 1 6= 2 so M(a2 ) = {1, 3}, M({1, 3}, b) =
{1b, 2}, M ({1b, 2}, b) = {1b2 , 3, 2} and a2 b2 3-compresses A. If 1b = y and
ya 6= 2, then ya 6= 1 because y 6= 3 and M(a2 b) = {y, 2}, M({y, 2}, a) =
{ya, 3}, M({ya, 3}, a) = {ya2 , 1, 3} so a2 ba2 is a 3-compressible word for A .
Now let z ∈
/ {1, 2} ∧ 3 ∈ {x, y}, then in order that A is proper x ∈ {1, 2} whence
y = 3 and za ∈ {1, 2}, so b = [x, 3]\z with z ∈
/ {1, 2}, x ∈ {1, 2}, za ∈ {1, 2}
, then M(b2 ab) = {2b, z}, M(a2b) = {1b, z} with 1b 6= 3, 2b 6= 3 and 1b 6= 2b,
hence 1b or 2b is not x and b2 ab2 or a2 b2 3-compresses A. The case z ∈ {1, 2}∧3 ∈
/
{x, y} is symmetric of the previous one and gives 3b ∈ {x, y}, 1 ∈ {x, y} hence
b = [x, y]\2 with 3 ∈
/ {x, y}, 1 ∈ {x, y}, 3b ∈ {x, y} which is 3-compressed by
a2 ba2 or by b2 a2 .
Therefore, the word w given in Section 4 is a 3−collapsing word.
17
6
Lower and upper bounds
a
0
b
1
b
b
2
a
a b
3
a
b
4
a
Fig. 1. An automaton which is not 3-compressed by s3,2 : Qs̄3,2 = {3}, Qs3,2 = {0, 1, 3}.
The 3−collapsing word above can be used to improve the procedure arising
from ([6], Theorem 3.5) to obtain shorter k-collapsing words for k ≥ 4. In particular it follows that c(4, 2) ≤ 1741 and c(5, 2) ≤ 109941. Though very lengthy,
they can be effectively used in testing the compressibility of an automaton. In
particular this can accelerate the algorithm presented in [1,7] to find short (possibly shortest) 4- and 5-synchronizing words.
It is known that in general the language Ck,t of k-collapsing words on an
alphabet of t letters differs from the language Sk,t of k-synchronizing words on
the same alphabet. However, this not excludes that in some cases c(k, t) = s(k, t).
Up to now it was only know that c(2, 2) = s(2, 2) ([10]) and that c(2, 3) 6=
s(2, 3) ([1]). We find a counterexample proving that also c(3, 2) 6= s(3, 2) (and
so c(3, 2) ≥ 34). In fact, the automaton in Fig. 1 is 3-compressible (and also
3-synchronizing), but the word s3,2 do not compresses it. On the other hand
its dual s̄3,2 synchronizes it. We check with a computer calculation that any
3-compressible 5-states automaton over a two letter alphabet, is 3-compressed
either by s3,2 or by s̄3,2 , so we believe interesting to understand if it is always
the case for every 3-compressible automata.
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