Fraenkel word: Difference between revisions

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<math>\displaystyle{
<math>\displaystyle{
\begin{align*}
\begin{aligned}
F_0 &= \epsilon, \\
F_0 &= \epsilon, \\
F_1 &= \mathbf{0}, \\
F_1 &= \mathbf{0}, \\
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&\ \ \vdots \\
&\ \ \vdots \\
F_{n} &= F_{n-1}(\mathbf{n-1})F_{n-1},
F_{n} &= F_{n-1}(\mathbf{n-1})F_{n-1},
\end{align*}}
\end{aligned}}
</math>
</math>


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== 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 ===
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{{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}}.
}}
}}


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# 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
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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}}.
}}
}}


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<math>\displaystyle{
<math>\displaystyle{
\begin{align*}
\begin{aligned}
G_0 &= \epsilon, \\
G_0 &= \epsilon, \\
G_1 &= \mathbf{0}, \\
G_1 &= \mathbf{0}, \\
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&\ \ \vdots \\
&\ \ \vdots \\
G_{n} &= G_{n-1}(\mathbf{n-1})G_{n-1}(\mathbf{n-1})G_{n-1}.
G_{n} &= G_{n-1}(\mathbf{n-1})G_{n-1}(\mathbf{n-1})G_{n-1}.
\end{align*}}
\end{aligned}}
</math>
</math>


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[[Category:Terms]]
[[Category:Terms]]
[[Category:Combinatorics on words]]
[[Category:Combinatorics on words]]
[[Category:Math]]
[[Category:Pages with open problems]]
[[Category:Pages with open problems]]