Comments on: Maximal Partitions http://yaniv.leviathanonline.com/blog/math/maximal-partitions/ Strain your Brain Tue, 13 Sep 2011 13:21:47 -0700 hourly 1 http://wordpress.org/?v=3.1 By: yaniv http://yaniv.leviathanonline.com/blog/math/maximal-partitions/comment-page-1/#comment-83 yaniv Sat, 25 Aug 2007 15:02:00 +0000 http://yaniv.leviathanonline.com/blog/math/maximal-partitions/#comment-83 Nadav, A very interesting riddle indeed. A more standard way to phrase it (using group theory notations) is to ask: what is the maximal order of an element of Sn. Although this is a very interesting question, as far as I know no complete answer exists. There are some (rather poor) bounds available with Cauchy's theorem, stating that for each prime p dividing n there is a member of Sn of order p, but since Sn is not commutative, we cannot conclude much about the existance of elements of composite order. Indeed, a simple O(n^2) algorithm for calculating this exists. It is similar to the one I originally used to solve the Maximal Partitions riddle. Nadav,

A very interesting riddle indeed. A more standard way to phrase it (using group theory notations) is to ask: what is the maximal order of an element of Sn.

Although this is a very interesting question, as far as I know no complete answer exists. There are some (rather poor) bounds available with Cauchy’s theorem, stating that for each prime p dividing n there is a member of Sn of order p, but since Sn is not commutative, we cannot conclude much about the existance of elements of composite order.

Indeed, a simple O(n^2) algorithm for calculating this exists. It is similar to the one I originally used to solve the Maximal Partitions riddle.

]]> By: Nadav http://yaniv.leviathanonline.com/blog/math/maximal-partitions/comment-page-1/#comment-80 Nadav Sat, 25 Aug 2007 13:02:55 +0000 http://yaniv.leviathanonline.com/blog/math/maximal-partitions/#comment-80 This reminds me of another riddle I once invented. First part: Let there be a 52-cards pack, currently sorted. I will shuffle the cards as follows: split it to two equal parts. Merge the two parts by putting one card from each part at the time. (Note that we keep each part in the same order, and that the topmost card, stays the topmost card after the shuffling) Repeating this procedure several times, causes an intresting phenomena: After 8 times, the pack is sorted again. This happens because of a certain property of the used permutaion. The permutation created by this procedure is: P = [0, 26, 1, 27, 2, 28, 3, 29, 4, 30, 5, 31, 6, 32, 7, 33, 8, 34, 9, 35, 10, 36, 11, 37, 12, 38, 13, 39, 14, 40, 15, 41, 16, 42, 17, 43, 18, 44, 19, 45, 20, 46, 21, 47, 22, 48, 23, 49, 24, 50, 25, 51] But P is formed from several circles: [[0], [1, 26, 13, 32, 16, 8, 4, 2], [2, 1, 26, 13, 32, 16, 8, 4], [3, 27, 39, 45, 48, 24, 12, 6], [5, 28, 14, 7, 29, 40, 20, 10], [6, 3, 27, 39, 45, 48, 24, 12], [9, 30, 15, 33, 42, 21, 36, 18], [10, 5, 28, 14, 7, 29, 40, 20], [11, 31, 41, 46, 23, 37, 44, 22], [17, 34], [18, 9, 30, 15, 33, 42, 21, 36], [19, 35, 43, 47, 49, 50, 25, 38], [22, 11, 31, 41, 46, 23, 37, 44], [34, 17], [38, 19, 35, 43, 47, 49, 50, 25], [51]] The i-th card lifecycle, is to move through the all locations in its circle, again and again. In that case, the pack will be sorted again, if all the circles are fully completed. For that reason, the number of times needed to resort the pack, is LCM(circle_lengths), which in that case is 8. The riddle I invented, is as follows: Given the number of cards N, find the worst permutaion (IE. Find a permutation with the longest order). Hint - so you can be sure your answer is correct: for N=52, the worst permutation have an order of 180180. So in fact, this riddle can be translated to: Find a partition of N, with maximal LCM(P) (instead of the original demand of maximal MUL(P)) BTW - I don't have a formula for calculation it, but I do have an algorithm for finding it in polynomal time. This reminds me of another riddle I once invented.
First part:
Let there be a 52-cards pack, currently sorted.
I will shuffle the cards as follows:
split it to two equal parts.
Merge the two parts by putting one card from each part at the time.
(Note that we keep each part in the same order, and that the topmost card, stays the topmost card after the shuffling)

Repeating this procedure several times, causes an intresting phenomena:
After 8 times, the pack is sorted again.
This happens because of a certain property of the used permutaion.
The permutation created by this procedure is:
P = [0, 26, 1, 27, 2, 28, 3, 29, 4, 30, 5, 31, 6, 32, 7, 33, 8, 34, 9, 35, 10, 36, 11, 37, 12, 38, 13, 39, 14, 40, 15, 41, 16, 42, 17, 43, 18, 44, 19, 45, 20, 46, 21, 47, 22, 48, 23, 49, 24, 50, 25, 51]
But P is formed from several circles:
[[0], [1, 26, 13, 32, 16, 8, 4, 2], [2, 1, 26, 13, 32, 16, 8, 4], [3, 27, 39, 45, 48, 24, 12, 6], [5, 28, 14, 7, 29, 40, 20, 10], [6, 3, 27, 39, 45, 48, 24, 12], [9, 30, 15, 33, 42, 21, 36, 18], [10, 5, 28, 14, 7, 29, 40, 20], [11, 31, 41, 46, 23, 37, 44, 22], [17, 34], [18, 9, 30, 15, 33, 42, 21, 36], [19, 35, 43, 47, 49, 50, 25, 38], [22, 11, 31, 41, 46, 23, 37, 44], [34, 17], [38, 19, 35, 43, 47, 49, 50, 25], [51]]
The i-th card lifecycle, is to move through the all locations in its circle, again and again.
In that case, the pack will be sorted again, if all the circles are fully completed.
For that reason, the number of times needed to resort the pack, is LCM(circle_lengths), which in that case is 8.

The riddle I invented, is as follows:
Given the number of cards N, find the worst permutaion (IE. Find a permutation with the longest order).
Hint – so you can be sure your answer is correct:
for N=52, the worst permutation have an order of 180180.

So in fact, this riddle can be translated to:
Find a partition of N, with maximal LCM(P) (instead of the original demand of maximal MUL(P))

BTW – I don’t have a formula for calculation it, but I do have an algorithm for finding it in polynomal time.

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