Fraenkel word: Difference between revisions

From Xenharmonic Wiki
Jump to navigation Jump to search
ArrowHead294 (talk | contribs)
mNo edit summary
ArrowHead294 (talk | contribs)
mNo edit summary
 
Line 17: Line 17:


== Facts ==
== Facts ==
Below we denote the length of a word ''w'' by {{abs|''w''}} and the number of occurrences of the letter '''i''' in ''w'' as {{abs|''w''}}<sub>'''i'''</sub>, as is standard notation in combinatorics on words. The notation ''w''(''u''<sub>0</sub>, ..., {{nowrap|''u''<sub>''r'' − 1</sub>}}) represents the word ''w'' in '''0''', '''1''', ..., {{nowrap|'''r − 1'''}} but with '''i''' replaced by the word ''u''<sub>''i''</sub>.
Below we denote the length of a word ''w'' by {{abs|''w''}} and the number of occurrences of the letter '''i''' in ''w'' as {{abs|''w''}}<sub>'''i'''</sub>, as is standard notation in combinatorics on words. The notation ''w''(''u''<sub>0</sub>, ..., {{nowrap|''u''<sub>''r'' − 1</sub>}}) represents the word ''w'' in '''0''', '''1''', ..., {{nowrap|'''''r'' − 1'''}} but with '''i''' replaced by the word ''u''<sub>''i''</sub>.


=== Fraenkel words are balanced ===
=== Fraenkel words are balanced ===
Line 39: Line 39:
{{theorem|name=Lemma|contents=For all {{nowrap|''n'' &ge; 1|0 &le; ''i'' &le; ''n'' − 1}}, and {{nowrap|1 &le; {{abs|''w''}} &le; 2<sup>''n''/2</sup> − 2}}, the following holds for any subword ''w'' of ''F''<sub>''n''</sub>:
{{theorem|name=Lemma|contents=For all {{nowrap|''n'' &ge; 1|0 &le; ''i'' &le; ''n'' − 1}}, and {{nowrap|1 &le; {{abs|''w''}} &le; 2<sup>''n''/2</sup> − 2}}, the following holds for any subword ''w'' of ''F''<sub>''n''</sub>:
# If {{nowrap|{{abs|''w''}} ≡ 0 (mod 2<sup>''i'' + 1</sup>)}}, then {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{abs|''w''}}/2<sup>''i'' + 1</sup>}}.
# If {{nowrap|{{abs|''w''}} ≡ 0 (mod 2<sup>''i'' + 1</sup>)}}, then {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{abs|''w''}}/2<sup>''i'' + 1</sup>}}.
# If {{nowrap|{{abs|''w''}} ≢ 0 (mod 2<sup>''i'' + 1</sup>)}}, then {{abs|''w''}}<sub>'''i'''</sub> {{=}} either {{nowrap|{{floor|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}} or {{nowrap|{{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}}.
# If {{nowrap|{{abs|''w''}} ≢ 0 (mod 2<sup>''i'' + 1</sup>)}}, then {{abs|''w''}}<sub>'''i'''</sub> {{=}} either {{nowrap|{{floor|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}} or {{nowrap|{{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}}.
#* More precisely, if for a given ''i'' we have {{nowrap|''w'' {{=}} ''uv''}} or ''vu'' where ''u'' is a possibly empty word whose length is {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}, and ''v'' is a nonempty word intersecting the middle of an {{nowrap|''F''<sub>''i'' + 1</sub>}}, then {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}}. Otherwise, {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{floor|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}}.
#* More precisely, if for a given ''i'' we have {{nowrap|''w'' {{=}} ''uv''}} or ''vu'' where ''u'' is a possibly empty word whose length is {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}, and ''v'' is a nonempty word intersecting the middle of an {{nowrap|''F''<sub>''i'' + 1</sub>}}, then {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}}. Otherwise, {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{floor|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}}.
}}
}}


{{proof|contents=We use the previous lemma. In the first case, ''w'' is guaranteed to have exactly ''k''-many '''i'''s where {{abs|''w''}} = ''k''2<sup>''i'' + 1</sup>. In the second case, if ''k''2<sup>''i'' + 1</sup> < {{abs|''w''}} < (''k'' + 1)2<sup>''i'' + 1</sup> and ''w'' = ''uv'' or ''vu'' where {{abs|''u''}} ≡ 0 mod 2<sup>''i'' + 1</sup>,  then ''u'' satisfies {{abs|''u''}}<sub>'''i'''</sub> = ''k''2<sup>''i'' + 1</sup>/2<sup>''i'' + 1</sup> = ''k'' by the previous case. Thus {{abs|''w''}}<sub>'''i'''</sub>  is determined by {{abs|''v''}}<sub>'''i'''</sub>, which is 1 if ''v'' contains the '''i''' in the middle of ''F''<sub>''i''</sub>, implying {{abs|''w''}}<sub>'''i'''</sub> = {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}, and 0 otherwise, implying {{abs|''w''}}<sub>'''i'''</sub> = floor({{abs|''w''}}/2<sup>''i'' + 1</sup>).}}
{{proof|contents=We use the previous lemma. In the first case, ''w'' is guaranteed to have exactly ''k''-many '''i'''s where {{abs|''w''}} = ''k''2<sup>''i'' + 1</sup>. In the second case, if ''k''2<sup>''i'' + 1</sup> < {{abs|''w''}} < (''k'' + 1)2<sup>''i'' + 1</sup> and ''w'' = ''uv'' or ''vu'' where {{abs|''u''}} ≡ 0 mod 2<sup>''i'' + 1</sup>,  then ''u'' satisfies {{abs|''u''}}<sub>'''i'''</sub> = ''k''2<sup>''i'' + 1</sup>/2<sup>''i'' + 1</sup> = ''k'' by the previous case. Thus {{abs|''w''}}<sub>'''i'''</sub>  is determined by {{abs|''v''}}<sub>'''i'''</sub>, which is 1 if ''v'' contains the '''i''' in the middle of ''F''<sub>''i''</sub>, implying {{abs|''w''}}<sub>'''i'''</sub> = {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}, and 0 otherwise, implying {{abs|''w''}}<sub>'''i'''</sub> = floor({{abs|''w''}}/2<sup>''i'' + 1</sup>).}}


{{theorem|name=Lemma|contents=Let [''F''<sub>''n''</sub>] denote the circular Fraenkel word on ''n'' letters. Suppose ''w'' is a proper subword of [''F''<sub>''n''</sub>] such that {{nowrap|''w'' {{=}} ''uv''}} where ''u'' is a nonempty suffix of ''F''<sub>''n''</sub> and ''v'' is a nonempty prefix of ''F''<sub>''n''</sub>. For {{nowrap|1 &le; {{abs|''w''}} &le; 2<sup>''n''/2</sup> − 2}}, either {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}} or {{nowrap|{{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}} − 1}}.
{{theorem|name=Lemma|contents=Let [''F''<sub>''n''</sub>] denote the circular Fraenkel word on ''n'' letters. Suppose ''w'' is a proper subword of [''F''<sub>''n''</sub>] such that {{nowrap|''w'' {{=}} ''uv''}} where ''u'' is a nonempty suffix of ''F''<sub>''n''</sub> and ''v'' is a nonempty prefix of ''F''<sub>''n''</sub>. For {{nowrap|1 &le; {{abs|''w''}} &le; 2<sup>''n''/2</sup> − 2}}, either {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}} or {{nowrap|{{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}} − 1}}.
}}
}}


Line 52: Line 52:
# Both {{abs|''u''}} and {{abs|''v''}} are {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}.
# Both {{abs|''u''}} and {{abs|''v''}} are {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}.
# At least one of {{abs|''u''}} and {{abs|''v''}} is not {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}.
# At least one of {{abs|''u''}} and {{abs|''v''}} is not {{nowrap|0 (mod 2<sup>''i'' + 1</sup>)}}.
In case 1, by the preceding lemma {{nowrap|{{abs|''u''}}<sub>'''i'''</sub> {{=}} {{abs|''u''}}/2<sup>''i'' + 1</sup>}} and {{nowrap|{{abs|''v''}}<sub>'''i'''</sub> {{=}} {{abs|''v''}}/2<sup>''i'' + 1</sup>}}, and hence {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{abs|''w''}}/2<sup>''i'' + 1</sup>}} {{nowrap|{{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}}.
In case 1, by the preceding lemma {{nowrap|{{abs|''u''}}<sub>'''i'''</sub> {{=}} {{abs|''u''}}/2<sup>''i'' + 1</sup>}} and {{nowrap|{{abs|''v''}}<sub>'''i'''</sub> {{=}} {{abs|''v''}}/2<sup>''i'' + 1</sup>}}, and hence {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{abs|''w''}}/2<sup>''i'' + 1</sup>}} {{nowrap|{{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}}.


In case 2, suppose {{nowrap|''w'' {{=}} ''ustv''}} where ''st'' is as in case 1 and {{abs|''u''}} and {{abs|''v''}} are less than {{nowrap|2<sup>''i'' + 1</sup>}}. Neither ''u'' nor ''v'' can contain an '''i''', as they are subwords of ''F''<sub>''i''</sub>. Using {{nowrap|{{abs|''st''}} {{=}} {{abs|''st''}}<sub>'''i'''</sub>2<sup>''i'' + 1</sup>}}, we have
In case 2, suppose {{nowrap|''w'' {{=}} ''ustv''}} where ''st'' is as in case 1 and {{abs|''u''}} and {{abs|''v''}} are less than {{nowrap|2<sup>''i'' + 1</sup>}}. Neither ''u'' nor ''v'' can contain an '''i''', as they are subwords of ''F''<sub>''i''</sub>. Using {{nowrap|{{abs|''st''}} {{=}} {{abs|''st''}}<sub>'''i'''</sub>2<sup>''i'' + 1</sup>}}, we have
Line 60: Line 60:
thus
thus


{{nowrap|{{abs|''w''}}<sub>'''i'''</sub> &ge; {{abs|''st''}}<sub>'''i'''</sub>}} {{nowrap|{{=}} {{abs|''st''}}/2<sup>''i'' + 1</sup>}} {{nowrap|{{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}} − 1}}.  
{{nowrap|{{abs|''w''}}<sub>'''i'''</sub> &ge; {{abs|''st''}}<sub>'''i'''</sub>}} {{nowrap|{{=}} {{abs|''st''}}/2<sup>''i'' + 1</sup>}} {{nowrap|{{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}} − 1}}.  


On the other hand, {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> &lt; {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}}}}, lest ''u'' or ''v'' have an '''i'''. Therefore {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>}} − 1}}.
On the other hand, {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> &lt; {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}}}}, lest ''u'' or ''v'' have an '''i'''. Therefore {{nowrap|{{abs|''w''}}<sub>'''i'''</sub> {{=}} {{ceil|{{abs|''w''}}/2<sup>''i'' + 1</sup>|1.75}} − 1}}.
}}
}}



Latest revision as of 15:49, 17 April 2025

In combinatorics on words, the Fraenkel word over n letters [math]\displaystyle{ \mathbf{0}, \mathbf{1}, ..., (\mathbf{n-1}) }[/math] is defined recursively by

[math]\displaystyle{ \displaystyle{ \begin{align*} F_0 &= \epsilon, \\ F_1 &= \mathbf{0}, \\ F_2 &= \mathbf{010}, \\ F_3 &= \mathbf{0102010}, \\ &\ \ \vdots \\ F_{n} &= F_{n-1}(\mathbf{n-1})F_{n-1}, \end{align*}} }[/math]

where ε is the empty word. Fraenkel words may be encountered as exceptional examples of scale properties alongside larger families, e.g. for the maximum/strict variety 3 and balance properties.

Fraenkel words are named after mathematician Aviezri S. Fraenkel.

Facts

Below we denote the length of a word w by |w| and the number of occurrences of the letter i in w as |w|i, as is standard notation in combinatorics on words. The notation w(u0, ..., ur − 1) represents the word w in 0, 1, ..., r − 1 but with i replaced by the word ui.

Fraenkel words are balanced

Theorem — As circular words, Fraenkel words are balanced.

The theorem will be a consequence of the following lemmas:

Lemma — Let Fn denote the non-circular Fraenkel word on n letters. For n ≥ 1 and 0 ≤ in − 1, the letter i appears once every 2i + 1 letters in Fn; i.e. in every subword of the form iwi, |w| = 2i + 1 − 1.

Proof
We prove this by induction on n. The n = 1 case being trivial, for n > 1 we start by observing that i always occurs as the greatest letter of a subword that is the Fraenkel word Fi + 1, and Fi + 1 is always flanked (on at least one side) by letters that are greater. This subword iFi is not a suffix of Fn, and we thus have

iFik...

where k > i, and F0 is the empty word. Since k occurs as the middle letter of Fk + 1, there is a copy of Fk that follows k; Fk has Fi as a prefix. Thus Fn has a subword

iFikFii,

as desired, since |Fi| = 2i − 1. [math]\displaystyle{ \square }[/math]

Lemma — For all n ≥ 1, 0 ≤ in − 1, and 1 ≤ |w| ≤ 2n/2 − 2, the following holds for any subword w of Fn:

  1. If |w| ≡ 0 (mod 2i + 1), then |w|i = |w|/2i + 1.
  2. If |w| ≢ 0 (mod 2i + 1), then |w|i = either |w|/2i + 1 or |w|/2i + 1.
    • More precisely, if for a given i we have w = uv or vu where u is a possibly empty word whose length is 0 (mod 2i + 1), and v is a nonempty word intersecting the middle of an Fi + 1, then |w|i = |w|/2i + 1. Otherwise, |w|i = |w|/2i + 1.
Proof
We use the previous lemma. In the first case, w is guaranteed to have exactly k-many is where |w| = k2i + 1. In the second case, if k2i + 1 < |w| < (k + 1)2i + 1 and w = uv or vu where |u| ≡ 0 mod 2i + 1, then u satisfies |u|i = k2i + 1/2i + 1 = k by the previous case. Thus |w|i is determined by |v|i, which is 1 if v contains the i in the middle of Fi, implying |w|i = |w|/2i + 1, and 0 otherwise, implying |w|i = floor(|w|/2i + 1). [math]\displaystyle{ \square }[/math]

Lemma — Let [Fn] denote the circular Fraenkel word on n letters. Suppose w is a proper subword of [Fn] such that w = uv where u is a nonempty suffix of Fn and v is a nonempty prefix of Fn. For 1 ≤ |w| ≤ 2n/2 − 2, either |w|i = |w|/2i + 1 or |w|/2i + 1 − 1.

Proof
There are 2 cases:
  1. Both |u| and |v| are 0 (mod 2i + 1).
  2. At least one of |u| and |v| is not 0 (mod 2i + 1).

In case 1, by the preceding lemma |u|i = |u|/2i + 1 and |v|i = |v|/2i + 1, and hence |w|i = |w|/2i + 1 = |w|/2i + 1.

In case 2, suppose w = ustv where st is as in case 1 and |u| and |v| are less than 2i + 1. Neither u nor v can contain an i, as they are subwords of Fi. Using |st| = |st|i2i + 1, we have

|st|i2i + 1 < |w| = |st|i2i + 1 + |u| + |v| ≤ |st|i2i + 1 + 2i + 1 − 2,

thus

|w|i ≥ |st|i = |st|/2i + 1 = |w|/2i + 1 − 1.

On the other hand, |w|i < |w|/2i + 1, lest u or v have an i. Therefore |w|i = |w|/2i + 1 − 1. [math]\displaystyle{ \square }[/math]

Open problems

Fraenkel's conjecture

For circular words (equivalently, infinite periodic words), Fraenkel's conjecture asserts that the only balanced circular words over n ≥ 3 letters with letter occurrences pairwise distinct are (letter reassignments of) [math]\displaystyle{ F_n. }[/math][1] The conjecture is known to be true for 3 ≤ n ≤ 7.

Other conjectures

Conjecture: Let MV(s) denote the maximum variety of the circular word s. Then {MV(F2k − 1), MV(F2k), MV(F2k + 1)} is an arithmetic progression with common difference f2k (the 2k-th Fibonacci number: 1, 3, 8, 21, ...) for every k ≥ 1.

Conjecture: For all k > 0, MV(Fkn) = fk + 1.

Let Gk be a modified Fraenkel word, defined by

[math]\displaystyle{ \displaystyle{ \begin{align*} G_0 &= \epsilon, \\ G_1 &= \mathbf{0}, \\ G_2 &= \mathbf{01010}, \\ G_3 &= \mathbf{01010201010201010}, \\ &\ \ \vdots \\ G_{n} &= G_{n-1}(\mathbf{n-1})G_{n-1}(\mathbf{n-1})G_{n-1}. \end{align*}} }[/math]

Conjecture: For all k > 1, MV(Gk) = 3 ⋅ 2k − 1 − 1.

Conjecture: For all k > 1, MV(Gkn) = 2k.

See also

References

  1. Bulgakova, D. V., Buzhinsky, N., & Goncharov, Y. O. (2023). On balanced and abelian properties of circular words over a ternary alphabet. Theoretical Computer Science, 939, 227-236.