superprismatic

Perhaps going down to a smaller alphabet will help shed light on this problem. I wrote a little program to generate all possible ways of making adjacencies on an alphabet of size 8 with each letter having 3 adjacencies. The number of these I got was 19,355.

Curr3nt and brifri238 both found good adjacency hookups. I think phil1882 got into a bit of trouble trying to keep the dimension down to two or three. Also, curr3nt's solution makes for an easy second part solution. I had hoped that someone would go a bit further and hint at some algorithm for generating these things. I know of no way to generate a random one -- i.e. an algorithm which can choose one with probability 1/N where N is the total number of possible adjacencies. I don't even know how to find N. Thanks for working on it.

There is only one A which has eight neighbours. There is only one of each of the 26 letters from A to Z, and each letter has to be connected (adjacent) to exactly 8 others. The arrangement can be on any manifold with any number of dimensions -- Whence my last paragraph. All that is required is that each of the 26 letters is adjacent to 8 others and "adjacency" is commutative. How this plays out in your head is of no concern. I hope this clarifies things.

A Boggle-like Challenge In the game, Boggle, a letter may have at most 8 adjacent letters. That fact inspired this challenge. This first part of this challenge is to place letters in such a way that each letter of the alphabet has precisely eight other different letters adjacent to it. You must use all 26 letters and, of course, "adjacent" is a commutative relation. To specify your placement, all you need to do is list the eight letters adjacent to A, the eight letters adjacent to B, the eight letters adjacent to C,...,etc. But remember that, if Q is on A's adjacency list, then A must be on Q's adjacency list, and this is true for every pair of letters -- not just A and Q. The second part of the challenge is to create a cycle of all 26 letters, such that each adjacent pair of letters in the cycle are adjacent in the sense of the first part of the challenge. Note that there is no requirement that the graph of adjacent letters is realizable in a small number of dimensions. So, trying to visualize such a graph may be hazardous to your mental health!

Nice! I just checked your 30-face solution and it works just fine. I hope you can go lower.

Nice solution for 3 dice! I hope you can beat my 4 dice solution of 48 faces. I suspect a solution with fewer than 48 faces exists, but I haven't found one.

I don't know, but I can modify my program to count those things. I have an appointment soon. When I return from that, I'll fix up the program. Perhaps knowing such things may make it easier to analyse the problem analytically.

No, that can't work because there are 24 permutations of 4 things and your dice can produce 256 different results, but 256 results can't be split evenly amongst 24 permutations. Some permutations would have to get more dice results than others.

If we have any 4-tuple of distinct real numbers, there is a simple way to have this determine a permutation on 4 things: Just replace each number with its ranking amongst the 4 numbers. For example, suppose I had the 4-tuple, (36,95,1,18). By replacing each number with its rank, I get the permutation (2,1,4,3). I would like to be able to use 4 fair dice to generate a permutation in this way. The dice would give me the 4-tuple (each die would have its own spot in the tuple) and I would use the ranks to determine a permutation. Of course, In order to insure that I get 4 distinct numbers, no die can have a number which is on any other die. But it may be the case that a paricular die has two or more faces having the same value. The number of faces on the dice may be any positive integer (I'm assuming that fair dice can always be made this way). It is not necessary that all of the dice have the same number of faces. Can you construct a set of 4 dice which can produce all 24 permutations of 4 things, each with probability 1/24 ? I have several such sets with 12 faces on each die. Can you find a set with fewer total faces?

You can't delete posts, only moderators can. But, I think you may want to leave it here so that others may enjoy working on it. You might even offer hints and/or answer questions for people who may have trouble with it. Thanks for posting.

I used the Wikipedia article entitled "Anno Domini" as the source of my information. There's a section of the page about going from 1BC to 1AD with no year 0 between them.

Yep. I didn't think things through enough.

Find 11 common English words, each of which is a subsequence of the string PLUFIEMDATNGORPAPNLCHGE. The words are all related by their meanings. As an example, IMPALE is a subsequence since its letters occur, in sequence, from left to right in the string PLUFIEMDATNGORPAPNLCHGE. Unfortunately, IMPALE isn't one of the words in the answer to this puzzle.

Arrr, mesmells a rat................Methinks Simpson's Paradox is rearing its ugly head!

I had to write a little program to get:

Nice, but do you have any reason to believe there is no 8-long? If you look at the end of my 9-long, you'll see that it ends in what are almost "wasted" moves:

You guys are much better than me at doing this. I fooled around with 311 for some time, and I can't do it with an 8-long program (including the first 1). But eventually I got a 9-long which ended in 311. Can you guys do it in an 8-long, like you did with the previous numbers?