Wednesday, July 03, 2013

No Triangles Allowed

Times out and quorums 6-0 / Skju

Adminned at 06 Jul 2013 07:09:33 UTC


Consider Atoms A, B, C where:
{A}:::{B} = 1,
{B}:::{C} = 1,
{A} does not equal {C},
and A, B, and C are all in the same Domain.

If {A}:::{B}, then A and B must each have two essential letters, and may not share any between themselves.

Thus, there are only 3! = 6 possible atoms that A can bond with, and B is one of them. The situation is the same for B and C, respectively.

Consider then A and C. A has occupied two of the Essential Letters, and B has occupied an additional two discreet from A’s. There is only one Letter left in the set of five; thus, C must take an Essential Letter already in use by A (not B because it has been previously stated that {B}:::{C}), and therefore {A}:::{C} = 0.

A more explicit form of the conclusion: {A}:::{B}&&{B}:::{C}&&{C}:::{A} = 0

Should anyone want to formally undertake the task, it can also be shown that the set of 20 Atoms with unique essence combinations bond into a structure resembling a dodecahedron. This realization of the triviality of the bonding structure is what inspired me to submit the Domains proposal in the first place.



07-05-2013 00:43:45 UTC

Looks correct.  for


07-05-2013 01:16:11 UTC



07-05-2013 12:01:16 UTC

for Good use of the pigenhole principle


07-05-2013 12:20:25 UTC



07-05-2013 15:48:15 UTC