Bounded number of squares in infinite
repetition-constrained binary words
Golnaz Badkobeh1 , Maxime Crochemore1,2
1 Kings
2 Université
College London
de Marne-la-Vallée, France
PSC 2010
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 1 / 19
Overview
Overview
Fraenkel and Simpson result,
Dejean’s Conjecture on the Repetitive threshold,
Overlap-free words and number of squares they contain,
Finite repetition Threshold, Shallit’s Theorem,
Result for alphabet sizes 2,3.
Generalizing the result for larger alphabet sizes?
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 2 / 19
Infinite Binary with few squares
Repetitions
Example:abaab
| {z } abaabab
Period=abaab
length
=
Exponent= periodlength
Square:exponent 2.
Cube:exponent 3.
Length=12
12
5
= 2.4
e.g abab
e.g aaa
Fraenkel and Simpson Theorem
There is an infinite Binary word containing 3 squares 00, 11 and 0101 only.
Example
01100011100101100111000101110010110001011100011001...
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 3 / 19
Avoidability on 2 letters
Thue-Morse morphism t
B = {0, 1}
(
t(0) = 01,
t(1) = 10.
t ∞ (0)= 0110100110010110....
t ∞ (0) is overlap-free.
Overlap is 2+ -repetition e.g |{z}
abc |{z}
abc a.
Squares are unavoidable
0
0
0
0
1
1
1
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 4 / 19
Avoidability on 3 letters
The morphism f is defined from A to itself
A = {a, b, c}


f (a) = abc,
f (b) = ac,


f (c) = b.
Iteration and Properties
f (a) = abc,
f 2 (a)= abcacb,
.
.
.
f ∞ (a)= abcacbabcbacabcacb....
It is known that f ∞ (a) is square-free.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 5 / 19
Avoidability on 3 letters
The Dejean’s morphism on 3 letters
The Dejean’s morphism d is defined from A to itself


d (a) = abcacbcabcbacbcacba,
d (b) = bcabacabcacbacabacb,


d (c) = cabcbabcabacbabcbac.
d ∞ (a)= abcacbcabcbacbcacbabcabacabcacbacabacb....
The d ∞ (a) is ( 47 )+ -free.
7
4
power is unavoidable
The longest word on 3 letters that can avoid ( 74 ) power has length 39.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 6 / 19
Repetitive Threshold
MaxExp(x)= max{ e ∈ Q|u e occurs in x for some u }
Threshold, tk for k-letter alphabet is t if
Exists infinite string x, MaxExp(x) = t
no infinite string y ,satisfies MaxExp(y ) < t
Example t2 = 2, t3 =
7
4
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 7 / 19
Dejean’s Conjecture on the Repetitive threshold
Dejean’s Conjecture
The repetition threshold for a k-letter alphabet (with k ≥ 2) is:

2



7
4
7



5k
k−1
for k = 2
for k = 3
for k = 4
otherwise.
Proofs
J.J. Pansiot. A Propos d’une Conjecture de F. Dejean sur les Répétitions dans les
Mots.( 1984)
J. Moulin Ollagnier. Proof of Dejean’s Conjecture for alphabets with 5, 6, 7, 8, 9, 10
and 11 letters. (1992),
A. Carpi. On Dejean’s conjecture over large alphabets.(2007)
J. Currie, N. Rampersad. Dejean’s Conjecture holds for n ≥ 27 (2009)
M. Rao. Last Cases of Dejean’s Conjecture. 8 ≤ k ≤ 34. (2009).
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 8 / 19
Overlap-free words and number of squares they contain
Observations
Maximum length of an overlap-free binary word with at most k squares.
k
=0 1
2
3
4
5
6
7
8
9
10 11
`(k) = 3 5
?
12 14 18 24 26 32 38 50 54
k
= 12 13 14 15 16 17 18 19 20 21 22 23
`(k) = 66 78 102 110 134 158 206 222 270 318 414 446
Theorem
[Shallit’s theorem (2008)]For all t ≥ 1 , there are no infinite binary words
that simultaneously avoid all squares yy with |y | ≥ t and 73 powers.
Consequently; There is no infinite overlap-free binary word with bounded
number of squares.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSC
binary
) 2010words 9 / 19
Extension of Shallit’s theorem (2008) and new Theorem
Shallit’s Theorem
There is an infinite binary word that simultaneously avoids all squares yy
+
with |y | ≥ 7 and 37 powers.
There are 20 squares in the example.
Finite Repetition Threshold
t is the Finite Repetition Threshold on k-letter alphabet if
Exist an infinite k-letter word x with MaxExp(x) = t and with finitely
many repetitions.
No infinite k-letter word y with MaxExp(y ) < t contains finitely
many repetitions
Example: k = 2 , t =
7
3
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words10 / 19
Extension of Shallit’s theorem (2008) and new Theorem
Theorem
There exists an infinite binary word whose factors have an exponent at
most 7/3 and that contains 12 squares, the fewest possible.
List of Squares:
{02 , 12 , (01)2 , (10)2 , (001)2 , (010)2 , (011)2 ,
(100)2 , (101)2 , (110)2 , (01101001)2 , (10010110)2 }
and the two words 0110110 and 1001001 of exponent 7/3.
12 is the minimum number.
Maximum length of (7/3)+ free binary words with at most k squares.
k
= 0 1 2 3 4 5 6 7 8 9 10 11
`(k) = 3 5 8 12 14 18 24 30 37 43 83 116
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words11 / 19
Sketch of the proof
Weakly square-free morphism g generating an infinite square-free word
g, Avoiding certain doublets and triplets.
Morphism h translating g to the binary word h with the desired
properties.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words12 / 19
A weakly square-free morphism on six letters
Morphism g defined from A∗6 to itself


g (a) = abac,




g (b) = babd,



g (c) = eabdf,
g (d) = fbace,




g (e) = bace,



g (f) = abdf.
Properties
1
The set of doublets occurring in g is
D = {ab, ac, ba, bd, cb, ce, da, df, ea, fb}.
2
The set of triplets in g is
T = {aba, abd, acb, ace, bab, bac, bda, bdf, cba, cea, dab, dfb, eab, fba}.
3
The morphism g is weakly square-free; i.e., g = g ∞ (a) is an infinite square-free word.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words13 / 19
Proof of property 3
Gaps between consecutive occurrences of g(a)=abac in g.
g (b)
g (cb)
g (bd)
g (ce)
g (bdfb)
=
=
=
=
=
babd
eabdfbabd
babdfbace
eabdfbace
babdfbaceabdfbabd
4
9
9
9
17
Propositions
1
No square in g can contain less than four occurrences of abac.
2
No square in g can contain at least four occurrences of abac.
Proof of Proposition 2 by contradiction.
Let ww be a square occurring in g k+1 (a) so that g k (a) is square-free.
The square ww can be written as
v0 (abac · · · abac)u1 v1 (abac · · · abac)u2
|
{z
}|
{z
}
There are 6 cases for u1 v1 to prove existance of a square in g k (a), contradiction.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words14 / 19
Binary translation
Morphism h from A∗6 to B ∗


h(a) = 10011,




h(b) = 01100,



h(c) = 01001,

h(d) = 10110,





h(e) = 0110,



h(f) = 1001.
Parsing the word h(y ), according to the codwords, when y is a factor of g
is unique due to the absence of some doublets and triplets in g.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words15 / 19
Proof
Gaps between consecutive occurrences of 10011 in h.
h(b)
h(cb)
h(bd)
h(ce)
h(bdfb)
=
=
=
=
=
01100
0100101100
0110010110
010010110
0110010110100101100
5
10
10
9
19
Graph showing immediate successors of gaps in the word g: a suffix of it following an
occurrence of a is the label of an infinite path.
ba
cba
cea
bda
bdfba
1
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words16 / 19
Proof
Steps of the proof
occurrences of 10011 in h identify occurrences of a in g
If there is a large square in h it is in the form
w 2 = v0 (h(a) · · · h(a))u1 v1 (h(a) · · · h(a))v2 .
|
{z
}|
{z
}
Different cases of u1 v1 using the graph and the gap table.
Conclusion:
The square either comes from a square in g
Or it is a factor of image of a sequence that is not a factor of g.
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words17 / 19
Example
u1 v1 = 0110010110 corresponds to h(bd)
v0 (h(a) · · · h(a)) h(bd) (h(a) · · · h(a)) v2
|
{z
}
|
{z
}
v0 (h(abac) · · · h(ba)) h(bd) (h(abac) · · · h(ba)) v2
{z
}
|
{z
}
|
h(ce) (h(abac) · · · h(ba)) h(bd) (h(abac) · · · h(ba)) h(c).
|
|
{z
}
{z
}
bda
baX
bda
cba
cea
cba
bdaX
ba
cea XX
baXX
ba XX
bdfba
bda
cea
bdaXX
ba
cea X
bdfba...
ba X
bdfba
cea
bdfba...
If we continue these tries we will have image of the image of :
. . a} c
ce a
. . a} bd a
| .{z
| .{z
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words18 / 19
Conjecture and Questions
corollary
There exists an infinite (7/3)+ free binary word with only two
7/3-repetitions.
Conjecture
The Finite Repetition Threshold of 3-letter alphabet is 47 and computation
shows the number of repetitions is likely to be 2.
List of Repetitions:
7
7
{(1020) 4 , (2101) 4 },
Computation shows that the maximal length of (7/4)+ free ternary word
with only one 7/4-repetition is 102.
Conjecture tested up to length 20000.
Questions
Finite Repetition Threshold for larger alphabet sizes?
Golnaz Badkobeh, Maxime Crochemore (Bounded
Kings College
number
London
of squares
Université
in infinite
de Marne-la-Vallée,
repetition-constrained
France
PSCbinary
)2010 words19 / 19