superprismatic

What is the code for § ? I don't see it in your extended ASCII set.

Find a set of fifteen distinct positive integers such that each number in the set divides the sum of all of the other fourteen numbers in the set. Can you find more than one such set?

There's an explanation under "100 Prisoners" at http://en.wikipedia.org/wiki/Random_permutation_statistics but it is pretty terse.

Consider the following list of alphabetic sequences: ABCDEFG ABCEGFD ACGFEBD ADEBFGC AEGFBCD AGBCEDF AGFCEBD BAFECDG BCFEDAG BDACFEG BDAGCEF BFDACGE BFEACDG BGCDEFA CAEBFGD CAEFDGB CBFAGDE CDAEFBG CDGEFAB CEDGFAB CFDGAEB DAECGBF DBECAGF DECBAGF DEGBAFC DFAGCEB DGBAECF DGBFACE EAGCDBF EBCFGAD EBDFGCA ECFAGDB ECFGBDA EFBDGCA EGBDCFA FABGDEC FAEGDCB FCDEBAG FDABEGC FEGCDBA FEGDABC FGABDEC GBCAFDE GCADBFE GCFDBEA GDCABFE GEBFACD GFDBCAE GFDEBAC [/code]They are listed in alphabetic order -- [b]NOT[/b] in the order in which they were generated. These were generated using a pair of 7-cycle permutations, P and Q, on 7 objects. Each of these 49 sequences was produces by applying P[sup]i[/sup]Q[sup]j[/sup] to ABCDEFG for all pairs i and j such that 0 < i,j < 8. Your task is to: 1. find such a P and Q which will produce all the sequences. 2. determine how many equivalent pairs of such permutations there are.

Thanks, Phillip1882, Anza Power, and Tuckleton. It makes perfect sense now. I guess that the idea was so simple and elegant that it flew right over my head! This is really a very nice proof!

Please expand your proof. I can't see how the first paragraph relates to the second. The proof is just too terse for me. I would like to understand it. Thanks.

Let P be a permutation on 12 objects which has cycle length 12. As an example, suppose P were given by 1 2 3 4 5 6 7 8 9 10 11 12 2 8 6 10 4 9 11 7 12 3 5 1 [/code] which tells us to move the object in the 1st position to the 2nd position, the object in the 2nd position to the 8th position, ..., and the object in the 12th position to the 1st position. We will use this to permute the 12 letters ABCDEFGHIJKL. Applying P to this yields LAJEKCHBFDGI. For this we write P(ABCDEFGHIJKL)=LAJEKCHBFDGI. We can continue to apply P so that P[sup]2[/sup](ABCDEFGHIJKL)=P(LAJEKCHBFDGI) and we can compute P(LAJEKCHBFDGI). Since P has cycle length 12, applying P over and over again 12 times will get us back to ABCDEFGHIJKL, written P[sup]12[/sup](ABCDEFGHIJKL)=ABCDEFGHIJKL. So, I can easily produce all 12 sequences of letters which powers of P can make when applied to ABCDEFGHIJKL. Now, suppose I do this with another permutation Q on 12 objects with cycle length 12. The 12 letter sequences which were produced (in alphabetical order -- [b]NOT[/b] the order of their generation) is: [code] ABCDEFGHIJKL BDGKAILFJECH CGFLKAIEBDHJ DKLCBJHIEAGF EAKBJHCLFIDG FIAJHCBKGLED GLIHCBJADKFE HFEILKADCGJB IJBEFGDCLHAK JEDAILKGHFBC KCHGDEFJABLI LHJFGDEBKCIA Find all the possible permutations which Q could be?

It's supposed to have ten of each number from 1 to 10. So far, Tuckleton has the record. I'd like to know what the maximum really is.

You may have noticed that the example I used in the OP has 3 possible ending positions. I tried very hard to find one which had 4 or more. I failed. Can you construct an array which has 4 or more possible ending positions?

You may want to try making up such an array. It's pretty easy to to -- just scatter ten 10s in a blank array, then, scatter in ten 9's, etc. until you're done. Then pick a few starting points and go through the process to see what kinds of things are happening. Watch how two or more such paths interact. That might give you some ideas.

Consider a 100-long array of numbers containing 10 of each of the numbers 1,2,3,4,5,6,7,8,9,10 in random order. As an example, consider the array 3 3 1 7 1 5 8 10 4 7 10 3 1 6 9 3 9 8 1 4 9 5 1 1 7 5 5 5 2 9 2 8 1 3 7 5 7 8 10 6 10 9 6 9 3 1 8 9 10 7 2 2 1 8 5 1 4 7 4 6 4 8 10 4 6 8 7 6 4 5 5 3 10 4 7 6 6 10 2 2 2 9 2 10 7 9 10 2 3 4 8 3 6 3 9 8 6 4 2 5 [/code] Consider the following: First, choose an element in a random position from the first 10 elements of the array. That number will determine how far to advance in the array to get the next number. This continues until one gets to a number such that one cannot advance by that amount and still stay inside the array. The final number together with its position in the array is our result. For example, suppose one were to start with the number in the 7[sup]th[/sup] position. This number is an 8. So we move eight positions to a 9. From there, we move nine positions to a 1, then to a 7, then 8,6,1,8,5,6,8,4,10, 2,4,3, and finally to a 6 in position 97. Thus, the pair (6,97) is our result. If a random array such as is described in the first sentence of this puzzle were presented to two people, and each were to pick a random starting position in the first 10 positions, what is the probability that they would both end end up with identical results after completing the process described above? Identical results would mean that each would arrive at the same number in the same position in the array.

Ha! Aren't you playing a little fast and loose with definitions? I guess that's OK, but it does make the problem more of a riddle than a math puzzle! Thanks, I enjoyed thinking about this especially while I was doing my boring exercises. And I came up with an alternate, albeit complicated, solution to yours!

I don't believe that. If you cut out a quadrilateral and throw away what's left, you've got a quadrilateral -- not a triangle. So, what you throw away matters.

I have a larger cycle length than yours but I haven't proven that it is the best. My idea was to find a partition of 100 such that the LCM of the numbers was large. So, I didn't just stick with primes. My best was 232,792,560 using cycles of lengths 19,17,16,13,11,9,7,5,and 3. Note that the 3 adds nothing to the magnitude of the cycle length, it just makes it all add up to 100. I threw the problem out there without thinking too much about it. I know of no theory behind this kind of thing. I don't think that Group theory addresses this type of maximization problem.

This is not the largest cycle possible. Can anyone find a larger one?

I don't quite see how that reference applies to this problem. I intentionally didn't use the term 'cycle' because that term is used in the way permutations can be written. But your comment about disorder may be a way of looking at it, if I understand you correctly. I was going to post the following before I saw your response -- it may help clarify things: Bushindo's answer is way too small and phillip1882's is way too big! Suppose, I had a permutation on 5 things. Let P be the permutation that takes ABCDE to EABCD. Then ABCDE -> EABCD -> DEABC -> CDEAB -> BCDEA -> ABCDE, and so, P5=P. But what if P took ABCDE to CABED. Then, ABCDE -> CABED -> BCADE -> ABCED -> CABDE -> BCAED -> ABCDE shows that P6=P. So, 5 is too small because I found a permutation on 5 things which requires six steps to return to the start. However, 5! = 120. I'd like to see someone find a permutation on 5 things which requires 120 steps to get back to the start. I can't.

Let P be a permutation on 100 things. For that P, let N be the smallest number such that PN=P. What is the largest possible value of N over some such P? In other words, how large a cycle can be made by applying a 100-long permutation over and over again?

I was talking about Bayes' treatment of prior probabilities especially when nothing is known about them. My point was that these are quite subjective.