superprismatic

I took the statement " Each glass is either right-side-up or up side down" to mean that all the glasses were all in the same orientation to start with. I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid. But, that makes it too easy, I suppose. I haven't yet considered the case where of an arbitrary beginning setup. Thanks for pointing that out, phil. If all the glasses were already in the same orientation (all up or all down), the game would be over before it started. But I assumed that the bartender had to change at least one glass on his first move. Then the spin would cause a problem for him.

I took the statement " Each glass is either right-side-up or up side down" to mean that all the glasses were all in the same orientation to start with. I thought it must have been there to emphasize that no glasses where placed on their sides, which would, of course, be stupid. But, that makes it too easy, I suppose. I haven't yet considered the case where of an arbitrary beginning setup. Thanks for pointing that out, phil.

Thats a nice picture, phil

I would like to see what you mean. I wish I could make a picture of it, but I have never done anything like that. I've only used text.

Very impressive! Did you just devise this yourself?

Suppose coins 1,2, and 3 are gold, silver, and bronze respectively. - Put coin 1 in pile A. - Weigh coin 2 against coin 1, not equal. (Both coins have been weighed once) - Put coin 2 in pile B. - Re-order piles so coin 2 is in pile A, coin 1 in pile B. - Weigh coin 3 against coin 2, not equal. - Weigh coin 3 against coin 1, not equal. (All coins have been weighed twice) I think this will only work if you can see the result of the third weighing of a coin before the coin magically disappears. I now see that I have much more of a problem with the algorithm than can be fixed in the simple way I've tried. I must be getting tired.

I think you have a problem, say the coin types were as follows: 0 1 2 3 A B C A Coin 0 will go to pile A, coin 1 to B and 2 to C, but in the process you have compared coin 0 with 2 coins so you can't compare it against coin 3. You're right. I had to modify step 2 in my post to fix that problem.

I've exhausted all triangles whose two legs differ by 1 where the length of the shortest leg is less than or equal to 1010. I found only 9: 3 4 5 20 21 29 119 120 169 696 697 985 4059 4060 5741 23660 23661 33461 137903 137904 195025 803760 803761 1136689 4684659 4684660 6625109

I don't understand your example. It seems to me that C is joined to E is joined to A is joined to F is joined to D is joined to B, but B and C aren't joined. So, it seems that player 2 didn't win because a loop wasn't made! Can you clear up my misunderstanding?

I hope you're not intimating that -1 is also a solution!

If a deck of 52 cards is cut exactly in half (top half and bottom half), then the top half can be partitioned into T segments (keeping the order of the cards) and, similarly, the bottom half can be partitioned into B segments. If T=B then there are two ways to interleave these partitions of cards -- dropping segments of cards alternately from T and B, beginning with T (the first way) or beginning with B (the second way). If T=B+1, then there is only one way to interleave the segments -- one must begin with T. If T+1=B, one must begin with B. Any other relationship between T and B make them impossible to interleave. We can count the number of partitions of 26 in all possible arrangements, they are: #partitions of 26; #of ways to make partitions of this number respecting order 1 1 2 25 3 300 4 2300 5 12650 6 53130 7 177100 8 480700 9 1081575 10 2042975 11 3268760 12 4457400 13 5200300 14 5200300 15 4457400 16 3268760 17 2042975 18 1081575 19 480700 20 177100 21 53130 22 12650 23 2300 24 300 25 25 26 1 We can now count all possible T,B pairs where T and B are at most one apart. This gives us a grand total of 374,369,872,911,804 possible riffle shuffles of 52 cards. Considering that 52! is the enormous 80658175170943878571660636856403766975289505440883277824000000000000, I can't see how to show that there is some number, N, of riffle shuffles such that one can always produce any particular permutation of the 52! by applying at most N riffle shuffles to the identity permutation.

That answer works. Is there a shorter answer, one that uses less numbers? I don't think so. 5,1,1,1 also works. I think these are the only solutions There are two others: 1,5,1,1 and 1,1,5,1. But is there a three number solution? No

That answer works. Is there a shorter answer, one that uses less numbers? I don't think so. 5,1,1,1 also works. I think these are the only solutions There are two others: 1,5,1,1 and 1,1,5,1.

What's a "random shuffle"? If it's the same as a random permutation, then phil1882 must be correct. So, it's not that. Please define "random shuffle".