Zadání
DS(n)
DS(n<sup>3</sup>)
DT(2<sup>(n^3)</sup>)
NS(n*log(n))
NT(n*log(n))
Řešení
Bez záruky
DS(n) a DS(n<sup>3</sup>)
Z prostorové hiearchie DS(n) ⊂ DS(n<sup>3</sup>).
ドは髑髏のド、レはレレレのレェェ、ミはみんなの髑髏のミ、ファはファファファのファァ、ソはそれ行け髑髏のソ、ラはラララのラァァ、シは死神の・・・。表示テストをしています。表示テストをしています。ul.tabs { margin: 0; paddnig: 0; float: left; list-style: none; height: 32px; /--Set height of tabs--/ border-bottom: 1px solid #999; border-left: 1px solid #999; width: 100%;}
DS(n) a NS(n*log(n))
DS(n) ⊂ DS(nlog(n)) ⊆ NS(nlog(n))
DS(n) a NT(n*log(n))
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DS(n<sup>3</sup>) a NS(n*log(n))
NS(nlog(n)) ⊆ DS((nlog(n))<sup>2</sup>) ⊂ DS(n<sup>3</sup>)
XDADsW <a href="http://ornfldligjpu.com/">ornfldligjpu</a>
DT(2<sup>(n^3)</sup>) a NS(n*log(n))
Savičova věta: NS(nlog(n)) ⊆ DS(n<sup>2</sup>(log(n))<sup>2</sup>)<BR> Věta o prostorové hierarchii: DS(n<sup>2</sup>*(log(n))<sup>2</sup> ⊂ DS(n<sup>5/2</sup>)<BR>
L ∈ NS(n*log(n)) ⊂ DS(n<sup>5/2</sup>) => L ∈ DT(C<sub>L</sub><sup>n^(5/2)</sup>) = DT(2<sup>log(CL)*n^(5/2)</sup>) ⊂ DT(2<sup>(n^3)</sup>)
DT(2<sup>(n^3)</sup>) a NT(n*log(n))
NT(nlog(n)) ⊆ NS(nlog(n)) a zřetězit s předchozím (tranzitivita)
NS(nlog(n)) a NT(nlog(n))
NS(nlog(n)) ⊇ NT(nlog(n))