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# A Boggle-like Challenge

18 replies to this topic

### #1 superprismatic

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Posted 13 November 2012 - 09:05 PM

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!

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### #2 BobbyGo

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Posted 13 November 2012 - 10:56 PM

Spoiler for Questions

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### #3 superprismatic

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Posted 13 November 2012 - 11:25 PM

Spoiler for Questions

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.
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### #4 CaptainEd

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Posted 14 November 2012 - 12:11 AM

I think I just learned something from this clarification. In Boggle, the tableau
ABC
DEF
GHI

has E adjacent to each of the others, and each of the others adjacent to two others, arranged in a ring.

But I believe you are willing to relax the ring requirement on the neighbors of E; your only requirement is that AE <-> EA.
All that is required is the graph.

Thanks! (Another interesting puzzle from superprismatic!)
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### #5 TheChad08

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Posted 14 November 2012 - 03:44 AM

Seems as though there is a near infinite amount of answers.

EDIT:
Nevermind, did it in a less graphical way

Edited by TheChad08, 14 November 2012 - 03:46 AM.

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### #6 curr3nt

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Posted 14 November 2012 - 03:21 PM

Spoiler for Is this what you wanted?

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### #7 curr3nt

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Posted 14 November 2012 - 04:01 PM

Spoiler for am I missing something?

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### #8 phil1882

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Posted 14 November 2012 - 04:03 PM

Spoiler for

Edited by phil1882, 14 November 2012 - 04:08 PM.

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### #9 curr3nt

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Posted 14 November 2012 - 04:11 PM

Spoiler for example...

Edited by curr3nt, 14 November 2012 - 04:12 PM.

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### #10 phil1882

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Posted 14 November 2012 - 04:35 PM

edit of above: still not quite right.
[spoiler]
```   Z  Y  X  W
V  A  B  C  D  Q
E  F  G  H  I  J
K  L  M  N  O  P K
Q  R  S  T  U  V
J  W  X  Y  Z  E
D  C  B  A
```

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