PALINDROMES AND PSEUDO-PALINDROMES IN EPISTURMIAN AND
PSEUDO-PALINDROMIC INFINITE WORDS
VYOMA ANNE, LUCA Q. ZAMBONI, AND IOANA ZORCA
A BSTRACT. Let A be a finite set of cardinality greater or equal to 2. An infinite word ω ∈ AN is
called Episturmian if it is closed under mirror image (meaning if u = u1 u2 · · · uk is a subword of
ω, then so is ū = uk · · · u2 u1 ) and if for every n ≥ 1 there exists at most one subword u of ω of
length n which is right special. We show that if u is a subword of an Episturmian word ω which is a
palindrome, then every first return to u is also a palindrome. As a consequence, every Episturmian
word begins in an infinite number of distinct palindromes. Our methods extend to the context of
pseudo-palindromic infinite words: ω ∈ AN is called a pseudo-palindromic word if there exists a
bijection φ : A → A with φ2 the identity such that for each subword u of ω we have that φ(ū)
is also a subword of ω, and for every n ≥ 1 there exists at most one subword u of ω of length n
which is right special. These words arise naturally in the context of the Fine and Wilf Theorem on
k-periods. A factor u of ω is called a pseudo-palindrome if u = φ(ū). We deduce that if u is a
subword of a pseudo-palindromic word ω which is a pseudo-palindrome, then every first return to
u is also a pseudo-palindrome. In particular, every pseudo-palindromic infinite word begins in an
infinite number of distinct pseudo-palindromes.
1. I NTRODUCTION
Given an infinite word ω = ω0 ω1 ω2 . . . on a finite alphabet, denote by Ln (ω) the set of all
subwords of ω of length n, that is Ln (ω) = {ωj ωj+1 . . . ωj+n−1 | j ≥ 0}. The complexity function
p(n) = pω (n) is defined as the cardinality of Ln (ω). A celebrated result of Morse and Hedlund
states that an infinite word is ultimately periodic if and only if for some n the complexity p(n) ≤ n.
(See [30]). An infinite word ω is called Sturmian if p(n) = n + 1 for all n ≥ 1. Thus amongst
all aperiodic infinite words, Sturmian words are those having the smallest complexity. Perhaps the
most well known example is the Fibonacci word
01001010010010100101001001010010010100101001001010010 . . .
defined as the fixed point of the morphism 0 7→ 01 and 1 7→ 0.
The study of Sturmian words was originated by M. Morse and G.A. Hedlund in the 1930’s. They
showed that Sturmian words provide a symbolic coding of the orbit of a point on a circle under
an irrational rotation (c.f.[30]). Sturmian words have since been extensively studied from many
different points of view: (c.f. [3], [6], [4], [13], [16], [19], [27], [28], [31]). It is well known that
if ω is a Sturmian word, then for each factor u = u1 u2 · · · un with ui ∈ {0, 1} the mirror image
ū = un un−1 · · · u2 u1 is also a factor of ω, in other words the language of ω is closed under mirror
image. Also the condition p(n + 1) − p(n) = 1 implies that for each n there exists exactly one
word u ∈ Ln (ω) which is a prefix (respectively suffix) of more than one (in fact two) words in
Ln+1 (ω). Such a word is said to be right special (respectively left special). A word which is both
right and left special is called bispecial.
Date: October 18, 2003.
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VYOMA ANNE, LUCA Q. ZAMBONI, AND IOANA ZORCA
Let A be a finite set of cardinality greater or equal to two. An infinite word ω ∈ AN is called
Episturmian if the language of ω is closed under mirror image and for each n there exists at most
one right special factor of length n. It follows directly from the definition that ω contains at most
one left special factor of every length, and that each bispecial factor of ω is a palindrome. Also the
mirror image of every left special factor is right special, and a palindrome is left special if and only
if it is right special.
Episturmian words have been extensively studied by a number of people including Droubay,
Justin and Pirillo (cf. [17, 23, 24, 25, 26, 35]). Episturmian words are a generalization of Sturmian
words, in fact the set of Sturmian words is precisely the set of binary aperiodic Episturmian words.
Episturmian words are also a generalization of Arnoux-Rauzy infinite words (c.f. [3, 9, 14, 29, 32,
33, 36, 37]). In fact every Arnoux-Rauzy infinite word is Episturmian but not conversely. Although
every aperiodic Episturmian word is the morphic image of an Arnoux-Rauzy word.
Let ω be an aperiodic Episturmian word, and Xω the shift orbit closure of X. Then Xω contains
an infinite word ω̃ all of whose prefixes are left special factors. This word is called the associated standard Episturmian word (see [24, 25]). The bispecial prefixes of a standard Episturmian
word are palindromes, and hence every standard Episturmian word begins in an infinite number of
distinct palindromes.
Let ω be an Episturmian word on the alphabet A with |A| = k ≥ 2. Let u ∈ L(ω). A word
w ∈ L(ω) is called a first return word to u if w contains exactly two occurrences of u, one as a
prefix and one as a suffix. It is well known that every word u ∈ L(ω) has at most k distinct first
returns [26]. Our main result is the following:
Theorem 1. Let ω be an Episturmian infinite word, and x ∈ L(ω) a palindrome. Then every first
return to x is a palindrome.
As a consequence we deduce that every Episturmian infinite word begins in an infinite number
of distinct palindromes.
2. P ROOF OF T HEOREM 1
Let ω be anSaperiodic Episturmian word on the alphabet A and X = Xω the associated subshift.
Let L(ω) = n≥0 Ln (ω) be the set of all factors of ω. We denote the length of a word w ∈ L(ω)
by |w|. We regard the empty word, denoted ε, as the unique word in L(ω) of length zero.
If u and v are non-empty words in L(ω) we will write u ` uv to mean that for each word
w ∈ L(ω) with |w| = |u| + |v| if u is a prefix of w then w = uv. If it is not the case that u ` uv
then we will write ¬(u ` uv). Similarly we will write vu a u to mean that for each word w ∈ L(ω)
with |w| = |u| + |v| if u is a suffix of w then w = vu. Otherwise we write ¬(vu a u).
Lemma 1. Let a ∈ A and suppose axa ∈ L(ω) where x is a bispecial factor of ω. Then ax is a
right special factor of ω.
Proof. Suppose to the contrary that ax is not right special. Then ax ` axa. If no prefix of ax is
right special then a ` ax ` axa which would imply that X contains the periodic infinite word
axaxaxaxax . . . , a contradiction. Let v (possibly the empty word) be the longest prefix of x such
that av is right special. Since v is a prefix of x it follows that v is also left special, hence bispecial.
Whence v̄ = v. Since we are assuming that ax is not right special, it follows that |v| < |x|.
Equivalently, we can write ax = avu where u is not the empty word. Since av is right special, it
follows that v̄a = va is left special and hence the first letter of u must be a. Set u = au0 . It follows
PALINDROMES AND PSEUDO-PALINDROMES IN EPISTURMIAN AND PSEUDO-PALINDROMIC INFINITE WORDS3
by maximality of v that ava ` avau0 ` avau0 a = axa. Since av is a suffix of x we have that ava
is a suffix of axa. But this implies that ω is eventually periodic, a contradiction.
Lemma 2. Let w ∈ L(ω) be bispecial and a ∈ A.
(1) Suppose aw is right special and a ` aw. Then wa ` waw and waw is bispecial.
(2) Suppose wa is left special and wa a a. Then waw a aw and waw is bispecial.
Proof. Since L(ω) is closed under mirror image, it suffices to prove (1), and (2) will follow. Let us
assume that aw is right special and a ` aw. Clearly wa ` waw. We show that waw is bispecial.
Since wa is left special and wa ` waw, we have waw is also left special. Since waw is a palindrome (as w is a palindrome), it follows that waw is also right special and hence bispecial.
Lemma 3. Let w ∈ L(ω) be bispecial and a ∈ A.
(1) Suppose aw is right special and ¬(a ` aw). Let v (possibly empty) be the longest proper
prefix of w with the property that av is right special. Thus w = vau for some u ∈ L(ω).
Then wa ` wau and wau is bispecial.
(2) Suppose wa is left special and ¬(wa a a). Let v (possibly empty) be the longest proper
suffix of w with the property that va is left special. Thus w = uav for some u ∈ L(ω).
Then uaw a aw and uaw is bispecial.
Proof. Again it suffices to prove (1). We suppose aw is right special, and v (possibly empty) is the
longest proper prefix of w with the property that av is right special. Since v is bispecial it follows
that v is a palindrome and hence va is left special and hence a prefix of w. We write w = vau
for some u ∈ L(ω). The maximality of the length of v implies that ava ` avau. But since av is
right special and |av| ≤ |w| it follows that av is a suffix of w and hence ava a suffix of wa. Thus
wa ` wau. We now show wau is bispecial. Since wa is left special and wa ` wau we have wau
is also left special. But wau = (au)w = auw = auvau = (au)vau = vauau = wau = wau, and
hence wau is a palindrome. Thus wau is also right special.
Remark 1. It follows from the previous lemmas that if w ∈ L(ω) is bispecial and aw is right
special (with a ∈ A), then the shortest bispecial word w0 containing w as a proper prefix is of the
form wau for some suffix u (possibly empty) of w.
Lemma 4. Let å ∈ A be the unique bispecial factor of ω of length 1. Then the first returns to å are
contained in the set {åbå |b ∈ A, b 6= å} ∪ {åå}. In particular all first returns to å are palindromes.
Proof. Let b ∈ A with b 6= å. If åb ∈ L(ω) then it follows that bå ∈ L(ω) and since b 6= å we must
have that b ` bå. Thus åb ` åbå and hence åbå is a first return to å. It follows that a first return to å
which is not of the form åbå for some b 6= å must necessarily be of the form åå.
Lemma 5. Let å ∈ A be the unique bispecial factor of ω of length 1. Let b ∈ A with b 6= å. Then
each first return to b is a palindrome.
Proof. Let w be the shortest bispecial factor of ω containing b. Then by Remark 1 and Lemma 2
we deduce that w = ubu where u is the longest bispecial not containing b. We note that b ` bu and
ub a b. Thus every first return to b begins in bu.
Let c ∈ A and suppose that buc ∈ L(ω). If c = b then bub is a first return to b and is a
palindrome. Next suppose that c 6= b. If c does not occur in w then c ` cw and wc a c. In fact,
we know that c 6= å and that c ` cå and åc a c. If ¬(c ` cw) then there would be a bispecial
proper prefix x of w such that cx is right special. But then xc would be left special and hence a
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VYOMA ANNE, LUCA Q. ZAMBONI, AND IOANA ZORCA
prefix of w contradicting that c did not occur in w. Thus, in case c does not occur in w we have that
buc ` bucw = bucubu, and hence bucub is a first return to b. Since u is a palindrome, it follows
that bucub is also a palindrome.
Next suppose c occurs in w and hence in u. Let x be the longest bispecial proper prefix of w
such that cx is right special. Since b ` bu it follows that x is a prefix of u. We write u = xcy. Then
by maximality of the length of x we have that cxc ` cxcybu. Since cxc is a suffix of buc it follows
that buc ` bucybu and hence bucyb is a first return to b. To see that it is a palindrome it suffices
to show that ucy is a palindrome. Writing u = xcy with u and x both palindromes we see that
ucy = cy(u) = cyu = cyxcy = cy(x)cy = xcycy = ucy = ucy as required. Thus all first returns
to b are palindromes.
Lemma 6. Let a ∈ A and suppose that axa ∈ L(ω) where x is a bispecial factor of ω. Then there
exists a bispecial factor z containing x as a proper prefix and such that az is also right special.
Proof. Let w be the shortest bispecial factor containing axa as a subword. It follows from Remark 1 that w = zbz 0 where z is bispecial, b ∈ A, bz is right special, and z 0 is a suffix of z. By
minimality of the length of w we have that axa is not a subword of z nor a subword of z 0 . If axa
is a subword of bz 0 then a = b and hence az is right special and |z| ≥ |z 0 | ≥ |xa| > |x|. Otherwise
there is an occurrence of axa in w which begins in z and terminates in bz 0 . In this case there exists
u ∈ L(ω) such that ub is a prefix of x and au a suffix of z. Thus we deduce that au is right special
while ub is left special. Hence u is bispecial and a = b. Since za is a prefix of w we have that za
is left special and hence az is right special. Finally we claim that z 6= x. Since za ` w, if z were
equal to x we would have that xa ` w which contains axa implying that ω is eventually periodic,
a contradiction. Thus |z| > |x| and hence x is a proper prefix of z.
Lemma 7. Let a ∈ A and x be bispecial. Suppose that axa ∈ L(ω). Then each first return to axa
is a palindrome.
Proof. Let z be the shortest bispecial factor containing x as a proper prefix and such that az is right
special. The above lemma guarantees the existence of z.. Thus we can write z = xau for some
factor u ∈ L(ω). By minimality of the length of z we have that axa ` axau = az (and za a axa)
and hence axa is not a subword of z. Hence every first return to axa begins in axau. But axa is a
suffix of za. Hence za ` zau and zau is bispecial. Thus the only occurrence of axa in zau is as a
suffix of za.
Let b ∈ A and suppose that axaub ∈ L(ω). Since za a axa it follows that zaub ∈ L(ω). If
b = a, then axaua = aza is a first return to axa (recall that axa is both a prefix and a suffix of aza
and axa does not occur in z.) Since z is a palindrome, so is aza. Next suppose b 6= a. If b does
not occur in zau then b ` bzau, whence axaub ` axaubzau and hence axaubza is a first return to
axa. But this is a palindrome since axaubza = azbza and z is a palindrome.
Finally suppose that b occurs in zau. Let x0 be the longest bispecial prefix of zau such that bx0
is right special. Thus zau = x0 by 0 for some y 0 , and bx0 b ∈ L(ω) and bx0 b ` bx0 by 0 . Also, since x0
is actually a prefix of z we can write y 0 = y 00 au, in other words z = x0 by 00 . Since bx0 b is a suffix
of zaub it follows that zaub ` zauby 0 . Thus axauby 00 a is a first return to axa. To see that it is a
palindrome it suffices to show that xauby 00 = zby 00 = (x0 by 00 )by 00 is a palindrome where both z and
x0 are palindromes. But zby 00 = by 00 (z) = by 00 z = by 00 x0 by 00 = by 00 (x0 )by 00 = x0 by 00 by 00 = zby 00 =
zby 00 as required.
Proof of Theorem 1: It suffices to prove the result for aperiodic Episturmian infinite words; in
fact if ν is a periodic Episturmian word and u ∈ L(ν) then there exists an aperiodic Episturmian
PALINDROMES AND PSEUDO-PALINDROMES IN EPISTURMIAN AND PSEUDO-PALINDROMIC INFINITE WORDS5
word ω for which u ∈ L(ω). So suppose ω is an aperiodic Episturmian word and x ∈ L(ω) a
palindrome. We proceed by induction on |x|. If |x| = 1, then the result follows from lemmas 4
and 5. If |x| = 2, then the result follows from Lemma 7. So suppose x is a palindrome and |x| ≥ 3.
Let z be a first return to x. Set x = ax0 a with x0 a palindrome. If a−1 za−1 is a first return to x0 , then
by induction hypothesis a−1 za−1 is a palindrome and hence so is z. Otherwise a−1 za−1 contains
other occurrences of x0 other than the one at the beginning and at the end of the word. But since
a−1 za−1 does not contain any occurrences of x = ax0 a, it follows that there exists b ∈ A, with
b 6= a and a−1 za−1 contains either x0 b or bx0 as a subword. Thus x0 is either right special or left
special, and hence bispecial. In this case that z is a palindrome follows from Lemma 7.
It is well known that Episturmian words are linked to the so-called Fine and Wilf words: Given
{p1 , p2 , . . . , pk } with gcd(p1 , p2 , . . . , pk ) = 1, the longest non-constant word u having periods
p1 , p2 , . . . , pk on the maximum number of symbols is a bispecial factor of an Episturmian infinite
word (see [10, 23, 35]). In particular it is a palindrome. In [8], the authors define and study a
class of infinite words on a finite alphabet A they call pseudo-palindromic words: A word ω ∈ AN
is pseudo-palindromic if there exists a bijection φ : A → A with φ2 equal to the identity map
on A, such that a) if u ∈ L(ω) then so is φ(ū) and b) for each n ≥ 1, ω has at most one right
special factor of length n. If φ is the identity map, then the above definition coincides with that of
Episturmian words. Thus pseudo-palindromic words are a generalization of Episturmian words. In
[8] it is shown that given {p1 , p2 , . . . , pk } with gcd(p1 , p2 , . . . , pk ) = 1, and m ≥ 1, a non-constant
word u of length m having periods p1 , p2 , . . . , pk on the most number of symbols is unique up
to isomorphism and is a bispecial factor of a pseudo-palindromic word. So while the longest
non-constant word with periods p1 , p2 , . . . , pk (on the most number of symbols) is a palindrome,
shorter words having the requisite periods are only palindromes up to a map φ : A → A with
φ2 equal to the identity. Given {p1 , p2 , . . . , pk } with gcd(p1 , p2 , . . . , pk ) = 1, and m ≥ 1, there
exists an algorithm (arithmetic and combinatorial) similar to the one in [35] for generating a word
u of length m on the most number of symbols having periods p1 , p2 , . . . , pk (see [8]). Also in [8]
the authors investigate the sub-word complexity and pseudo-palindromic complexity of aperiodic
pseudo-palindromic infinite words. A subword u of a pseudo-palindromic infinite word ω is called
a pseudo-palindrome if u = φ(ū).
The results and proofs in this paper may be modified in a straightforward fashion to prove the
following:
Theorem 2. Let ω be a pseudo-palindromic infinite word on a finite alphabet A with |A| ≥ 2. Let
x ∈ L(ω) be a pseudo-palindrome. Then every first return to x is also a pseudo-palindrome.
As a consequence we deduce that
Corollary 1. Every pseudo-palindromic infinite word begins in an infinite number of distinct
pseudo-palindromes.
It should be noted that in this context a symbol a ∈ A is a pseudo-palindrome if and only if
φ(a) = a. However it is easily shown that every pseudo-palindromic infinite word begins in a
pseudo-palindrome, and hence by the above theorem, in an infinite number of pseudo-palindromes.
3. C ONCLUDING R EMARKS
3.1. Codings of 3-Interval Exchange Transformations. Some of the techniques used in the
proof of Theorem 1 may be extended to prove an analogous result for natural codings of orbits
6
VYOMA ANNE, LUCA Q. ZAMBONI, AND IOANA ZORCA
of points under a 3-interval exchange transformation. More precisely given two real numbers
α > 0 and β > 0, with α + β < 1, one considers the dynamical system I defined on the interval
[0, 1[ by
• Ix = x + 1 − α if x ∈ [0, α[,
• Ix = x + 1 − 2α − β if x ∈ [α, α + β[,
• Ix = x − α − β if x ∈ [α + β, 1[.
A natural coding of the orbit of a point x ∈ [0, 1[ is an infinite word (xn ) taking values {1, 2, 3}
according to whether the n-th iterate I n (x) lies in the first, second, or third interval. The subword
complexity of these words is known to be p(n) = 2n + 1. As is the case of Episturmian words,
they are also closed under mirror image. However, for each n ≥ 1 there are exactly two right
special factors of length n. Various combinatorial properties of the natural codings of 3-interval
exchange transformations were studied by Ferenczi, Holton and Zamboni in [21, 22]. In [22] a
combinatorial construction is given for generating the two standard orbits; a standard orbit is one
in which every prefix is left special. As in the Episturmian case, there is a notion of a directive
word (see [24, 25]). The directive word is obtained via an arithmetic division algorithm applied to
the lengths of the intervals. This arithmetic construction was originally introduced by the authors
in an earlier paper [20] and may be viewed as a two-dimensional generalization of the regular
continued fraction algorithm. Next a combinatorial construction is applied to the directive word
to generate the bispecial factors of the associated symbolic subshift as a function of the arithmetic
expansion. In [22] the authors obtain a complete characterization of those sequences of block
complexity 2n + 1 which are natural codings of orbits of 3-interval exchange transformations. In
[2] the authors show that every natural coding of a 3-interval exchange transformation begins in an
infinite number of distinct palindromes.
3.2. Avoidance of Large Palindromes. While every uniformly recurrent infinite word which contains arbitrarily large palindromes is necessarily closed under mirror image, we note that Theorem 1 is not merely a consequence of the fact that Episturmian words are closed under mirror
image. Consider the sequence ν = r1 r2 r3 r4 . . . ∈ {1, 2, 3}N defined by
ν = 22x˜1 33x˜2 22x˜3 33x˜4 22x˜5 33x˜6 22x˜7 33x˜8 . . .
where x1 x2 x3 x4 . . . ∈ {0, 1}N is a Sturmian word and 0̃ = 11 and 1̃ = 1. Let ω ∈ {0, 1}N be the
infinite word whose run length sequence is ν, that is ω = 0r1 1r2 0r3 1r4 0r5 1r6 . . . . Then ω is a uniformly recurrent infinite word, closed under mirror image, and does not contain any palindromes
of length greater than five1 [12]. While there are numerous examples of uniformly recurrent binary
infinite words closed under mirror image (eg. examples generated using paperfolding sequences),
the word ω given above is of lowest complexity amongst all such words [12]. The subword complexity of ω is of the form p(n) = 4n + k.
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