我的朋友向我提出了这个问题；感觉很想在这里分享。
给定一副牌，我们将其分成两组，并“将它们交错排列”；让我们将此操作称为“拆分连接”。并在生成的牌组上重复相同的操作。
例如，{ 1, 2, 3, 4 } 变为 { 1, 2 } & { 3, 4 } （拆分） 我们得到 { 1, 3, 2, 4 } （合并） >
此外，如果我们有奇数张牌，即 { 1, 2, 3 }，我们可以像 { 1, 2 } & 那样将其拆分。 { 3 }（较大的一半在前）通向 { 1, 3, 2 }
（即，n 分为 Ceil[n/2] 和 n-Ceil[n/2]）
我朋友的问题问我的是：
需要多少这样的拆分连接才能恢复原始牌组？
这让我想知道：
，则需要多少分割连接
- 如果这副牌有 n 张牌，如果n 是偶数
- ？ n 是奇数吗？
- n 是 '2' 的幂吗？ [我发现我们需要 log (n) (base 2) 分割连接数...]
- （随意探索这样的不同场景。）
是否有一个简单的模式/公式/ n 与所需拆分连接数量相关的概念？
我相信，这是在 Mathematica 中探索的一件好事，特别是因为它提供了 Riffle[] 方法。
My friend posed this question to me; felt like sharing it here.
Given a deck of cards, we split it into 2 groups, and "interleave them"; let us call this operation a 'split-join'. And repeat the same operation on the resulting deck.
E.g., { 1, 2, 3, 4 } becomes { 1, 2 } & { 3, 4 } (split) and we get { 1, 3, 2, 4 } (join)
Also, if we have an odd number of cards i.e., { 1, 2, 3 } we can split it like { 1, 2 } & { 3 } (bigger-half first) leading to { 1, 3, 2 }
(i.e., n is split up as Ceil[n/2] & n-Ceil[n/2])
The question my friend asked me was:
HOW many such split-joins are needed to get the original deck back?
And that got me wondering:
If the deck has n cards, what is the number of split-joins needed if:
- n is even ?
- n is odd ?
- n is a power of '2' ? [I found that we then need log (n) (base 2) number of split-joins...]
- (Feel free to explore different scenarios like that.)
Is there a simple pattern/formula/concept correlating n and the number of split-joins required?
I believe, this is a good thing to explore in Mathematica, especially, since it provides the Riffle[] method.
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引用MathWorld：
未解决
n为奇数的情况。请注意，本文还包括一个具有探索 out-shuffle 功能的 Mathematica 笔记本。
To quote MathWorld:
The case when
nis odd isn't addressed.Note that the article also includes a Mathematica notebook with functions to explore out-shuffles.
如果我们有奇数张牌
n==2m-1，并且如果我们将牌分开，使得在每次洗牌期间第一组包含m张牌，第二组包含m张牌m-1张牌，并且将组连接起来，使得同一组中没有两张牌彼此相邻，则所需的洗牌次数等于MultiplicativeOrder[2, n ]。为了说明这一点，我们注意到，在洗牌一次后，位于位置
k的卡片已移动到位置2k，其中0<=k并为 2k-2m+1表示m<=k<2m-1，其中k是这样的0<=k<2m-1。写为模n==2m-1，这意味着对于所有0<=k，新位置为 Mod[2k, n]。因此，为了让每张牌返回到其原始位置，我们需要N次洗牌，其中N满足Mod[2^N k, n]==Mod[ k, n]对于所有0<=k ，其中 N是MultiplicativeOrder[2, n]< 的任意倍数/代码>。请注意，由于对称性，如果我们以相反的方式拆分牌组，结果将完全相同，即第一组始终包含
m-1卡，第二组始终包含m< /code> 卡。我不知道如果你交替的话会发生什么，即对于奇数洗牌，第一组包含m卡，对于偶数洗牌m-1卡。If we have an odd number of cards
n==2m-1, and if we split the cards such that during each shuffle the first group containsmcards, the second groupm-1cards, and the groups are joined such that no two cards of the same group end up next to each other, then the number of shuffles needed is equal toMultiplicativeOrder[2, n].To show this, we note that after one shuffle the card which was at position
khas moved to position2kfor0<=k<mand to2k-2m+1form<=k<2m-1, wherekis such that0<=k<2m-1. Written modulon==2m-1this means that the new position isMod[2k, n]for all0<=k<n. Therefore, for each card to return to its original position we needNshuffles whereNis such thatMod[2^N k, n]==Mod[k, n]for all0<=k<nfrom which is follows thatNis any multiple ofMultiplicativeOrder[2, n].Note that due to symmetry the result would have been exactly the same if we had split the deck the other way around, i.e. the first group always contains
m-1cards and the second groupmcards. I don't know what would happen if you alternate, i.e. for odd shuffles the first group containsmcards, and for even shufflesm-1cards.魔术师/数学家 Persi Diaconnis 的旧作讲述了如何通过完美的 riffle shuffle 来恢复顺序。 Ian Stewart 在他 1998 年《科学美国人》数学休闲专栏中写到了这项工作 - 请参阅，例如： http://www.whydomath.org/Reading_Room_Material/ian_stewart/shuffle/shuffle.html
There's old work by magician/mathematician Persi Diaconnis about restoring the order with perfect riffle shuffles. Ian Stewart wrote about that work in one of his 1998 Scientific American Mathematical Recreation columns -- see, e.g.: http://www.whydomath.org/Reading_Room_Material/ian_stewart/shuffle/shuffle.html
我知道老问题，但奇怪的是没有人提出实际的数学解决方案。
这处理偶数和奇数情况：
您可以很容易地证明，如果您将一张牌添加到奇数情况下，多余的牌将保留在底部并且不会改变顺序，因此奇数情况结果只是
n+1偶数结果..old question I know, but strange no one put up an actual mathematica solution..
This handles even and odd cases:
You can readily show if you add a card to the odd-case the extra card will stay on the bottom and not change the sequence, hence the odd case result is just the
n+1even result..