**Edited by brifri238, 16 November 2012 - 12:07 PM.**

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# A Boggle-like Challenge

Started by superprismatic, Nov 13 2012 09:05 PM

18 replies to this topic

### #11

Posted 16 November 2012 - 11:59 AM

Spoiler for Possible solution for the first part:

### #12

Posted 16 November 2012 - 09:11 PM

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.

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.

### #13

Posted 17 November 2012 - 02:04 PM

Spoiler for assuming curr3nts

### #14

Posted 18 November 2012 - 07:41 AM

Spoiler for assuming curr3nts

N is not going to be that simple to calculate. Any given adjacency you find could have multiple possible cycles that could be found within it (So it may incorrectly be counted multiple times). Adjacencies don't need to be as neatly organized as curr3nt's answer was.

Spoiler for Large range for N

Spoiler for Anticlimactic way to generate with 1/N probability

### #15

Posted 19 November 2012 - 07:58 PM

Spoiler for Why wouldn't phil's approach for N work?

### #16

Posted 19 November 2012 - 11:06 PM

Spoiler for Why wouldn't phil's approach for N work?

Spoiler for can you "cube" this?

### #17

Posted 19 November 2012 - 11:39 PM

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.

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.

### #18

Posted 20 November 2012 - 01:47 AM

Yay for integer sequences (found using your 19355 )...

Time to look at 3-regular graphs... which aren't as simple and quite possibly don't have this property.

Spoiler for Not the answer, but some pretty numbers anyway

Spoiler for Estimation of N based on list 3 + connected=cycle assumption

Spoiler for Simple proof for connected=cycle in 2-regular graphs

Time to look at 3-regular graphs... which aren't as simple and quite possibly don't have this property.

Spoiler for Useful article...

### #19

Posted 20 November 2012 - 04:21 PM

Spoiler for can you "cube" this?

Spoiler for You had me at...

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