こんにちは。前に質問させてもらい、
内容として
「21 枚のカードがあり、絵が見えるように上にして、三枚ずつ七行に並べていく。七行・三列の配列で考える。
そして、観客に一枚を選ばせ、三列のうちのどの列なのか言わせる。カードは手渡さないで、指示された列が真ん
中になるよう各列のカードを集めたあと、また繰り返す。このイカサマを 3ラウンド繰り返せば、イカサマ師は観
客の選んだ札を言い当てられる。」
ということで、理解できました。そして、そのあとの英文のないようで、数式が出てきているのですが、
どのような意味なのかが、読みとれません＞＜
その内容は、
The convergence of the process
We consider a pack with PQ cards which will be arranged in an array of P columns, each of Q cards or, equivalently, Q rows, each of P cards. The case described above is with P=3 and Q=7. We shall only analyse the case when P and Q are odd and greater than 1 (leaving the other cases to the interested reader), and we write P=2p+1 and Q=2q+1, where p>=1 and q>=1. We are interested in the central card in the pack; this has exactly 1/2(P×Q-1) cards above and below it and, labelling the cards 1,2,3,... from the top of the pack, the central card is labelled C, where
C=1/2(P×Q+1)=p×Q+q+1=q×P+p+1.
Suppose now that the chosen card lies in the Xn th position in the pack after n rounds have been played. During the play in the next round the first Xn cards will be spread out over the first few rows, and the chosen card will occur in the r th row, where r is the smallest integer greater than or equal to Xn/P. We write <Y> for the smallest integer m satisfying m>=Y. It follows that, after reforming the pack by ensuring that the column containing the chosen card is the
central column, the position Xn+1 of the chosen card will satisfy the recurrence relation
Xn+1=pQ+<Xn/P>.
This is a recurrence relation, but it is non-linear, and we do not know the initial value X0 of the chosen card in the pack; all we know is that 1=<X0=<PQ. Nevertheless, this information is sufficient to give us a complete solution of our problem, and we shall see that the dynamics of the trick is reminiscent of the convergence of iterates of a function towards an attracting fixed point of the function.
We begin by noting that there is a symmetry in the process, so that we may confine our attention to those cases where the chosen card lies in the top half of the pack; thus we may assume that 1=<X0=<C. An induction argument now shows that, for all n, Xn=<C. For suppose that Xn=<C; then
Xn+1=pQ+<Xn/P>=<pQ+<C/P>
=pQ+<(qP+p+1)/P>
=pQ+q+1=C
This argument also shows that if Xn=C, then Xn+1=C.
Observe next that X1>pQ. We shall now show that there is some integer N with
pO<X1<X2<....<Xn-1<Xn=C=Xn+1=Xn+2=.... ・・・(1)
いつも質問させてもらい、本当に助かってます。時間がある方、アドバイスお願いします＞＜
