Ndryshimi mes inspektimeve të "Përdoruesi:Armend/nënfaqe"

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Lemma: <math>|t^2(c)|\geq |c|\,</math> if and only if either <math>|{c}|=1\,</math>, or <math>|{c}|=2\,</math> and <math>c=(a,b^{|c|-1})\,</math> (up to permutation) where <math>a\ne b\,</math>.
 
Proof: We have <math>|t^2(c)|=\max\{t_0,\dots,t_p}t_{p+1}\}\,</math>, so <math>|t^2(c)|\ge |c|\,</math> if and only if <math>|t^2(c)|\in \{|c|,|c|+1\}\iff \{|c|-1,|c|\}\cap \{t_0,\dots,t_p\}\ne \emptyset\,</math>, which is equivalent to the conditions above.
 
Claim: If <math>m=|c|\ge 3\,</math> then <math>|t^n(c)|<|c|\,</math> for some <math>n\ge 1\,</math>.
Proof of the claim: this is a case-by-case analysis.
#If <math>|{c}|>2\,</math> then <math>n=2\,</math> will do by the lemma.
#If <math>|{c}|=1\,</math>, either <math>c=(0^m)\implies t(c)=(m)\,</math>, or <math>c=(1^m)\implies t(c)=(0,m)\,</math>, or <math>c=(2^3)\implies t^6(c)=(0,3)\,</math>, or <math>c=(2^m)\,</math> where <math>m\ge4\,</math>, so that <math>t(c)=(0,0,m)\,</math>, or <math>c=(x^m)\,</math> where <math>x\ge3\,</math>. In this case, <math>t(c)=(0^x,m)\,</math>, <math>t^2(c)=x,0^{m-1},1), t^3(c)=(m-1,1,0^{x-2},1)\,</math>, <math>t^4(c)=(x-2,2,0^{m-3},1)\,</math>.
#If <math>m\ge4\,</math> or <math>x\ge5\,</math> then <math>|{t^4(c)}|>2\,</math>, so <math>|t^4(c)|=m\,</math> and $|t^6(c)|
#If <math>|{c}|=2\,</math> and <math>c\,</math> is not of the form <math>(x,y^{m-1})\,</math> (up to permutation) then <math>|t^2(c)|<|c|\,</math> by the lemma.
#If <math>|{c}|=2\,</math> and <math>c=(x,y^{m-1})\,</math>, suppose first that <math>x 3\,</math> then <math>|t^4(c)|=\max\{m-3,2\}+1=\max\{m-2,3\} y\,</math> then <math>t(c)\,</math> is a permutation of <math>t(y,x^{m-1})\,</math>, so the previous argument applies.
 
So we prove that, the assumption is true for <math>m \in \{1,2\}\,</math>, so it appears to be true in general.
 
So we prove that, the assumption is true for <math>m \in {1,2}\,</math>, so it appears to be true in general.
===1===
If <math>m=1\,</math>, <math>c=(x)\,</math> for some <math>x \in \mathbb{N}\,</math>. If <math>x>0\,</math>, <math>t(c) = (0^x,1)\,</math>, and either <math>x=1\,</math>, in which case <math>t^2(c) = (0,2), t^3(c) = (1,0,1)\,</math>, and <math>t^4(c) = (1,2) \in R\,</math>. If <math>x=0\,</math>, <math>t(c) = (1)\,</math>, and we’re in the first case.

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