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Utente:Grasso Luigi/sandbox4/Permutazioni: differenze tra le versioni

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Grasso Luigi (discussione | contributi)
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Grasso Luigi (discussione | contributi)
17 501 modifiche
Riga 953:
ESEMPI: obviously, for <math>\lambda_2</math> one has <math>\lambda_2=(12)</math>; to build <math>\lambda_3</math> there are only two possibilities for the coset beginnings satisfying the no repeat condition; the choice <math> p_1=p_2=1</math> leads to <math>\lambda_3=\lambda_2(13)\lambda_2(13)\lambda_2=(13)</math>. To continue generating <math>S_4</math> one needs appropriate coset beginnings (satisfying the no repeat condition): there is a convenient choice: <math>p_1=1, p_2=2, p_3=3</math>, leading to <math>\lambda_4=(13)(1234)(13)=(1432)</math>. Then, to build <math>\lambda_5</math> a convenient choice for the coset beginnings (satisfying the no repeat condition) is <math> p_1=p_2=p_3=p_4=1</math>, leading to <math>\lambda_5=(15)</math>.
 
FromDagli examplesesempi aboveprecedenti onesi canpuò inductivelyinduttivamente goandare toa highervalori di <math>k</math> in a similar wayelevati, choosing coset beginnings of <math>S_{k}</math> in <math>S_{k+1}</math>, as follows: for <math>k</math> even choosing all coset beginnings equal to 1 and for <math>k</math> odd choosing coset beginnings equal to <math>(1, 2,\dots , k)</math>. With such choices the "last" permutation is <math>\lambda_k=(1k)</math> for <math>k</math> odd and <math>\lambda_k=(1k_-)(12\cdots k)(1k_-)</math> for <math>k</math> even (<math>k_-=k-1</math>). Using these explicit formulae one can easily compute the permutation of certain index in the counting/generation steps with minimum computation. ForPer this,questo writingè theutile indexscrivere l’indice in factorial base is usefulfattoriale. Consideriamo l'indice
Ad esempio, consideriamo l'indice
:<math>699699_{(10)}\ =\ 5(5!)\ +\ 4(4!)\ +\ 0(3!)\ +\ 1(2!)\ +\ 1(1!)\ =\ 54011_{(!)} </math>
e la corrispondente permutazione è:diventa
:<math>\begin{align}
\sigma &= \lambda_2(13)\lambda_2(15)\lambda_4(15)\lambda_4(15)\lambda_4(15)\lambda_4(56)\lambda_5(46)\lambda_5(36)\lambda_5(26)\lambda_5(16)\lambda_5 = \\
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