Why start from a fixed pick? That fixed pick can be chosen to maximize versatility too.
The set you describe could be represented as a cycle:
, 
, 
, 
(alphabetical)
, , , (Archetypes
http://elementscommunity.org/forum/index.php/topic,14931.12.html)
I count (1)(11)(10)/6 x (9)(8 )(7)/6 = 1540 cycles although that is probably an overestimate again.
What cycle defined by a set of 6 non opposite elements divided into two sets of 3 would create the most versatile and balanced set?
I'm not sure how picking a cycle determines what we are looking for (four 6-element subsets such that each pair of is either disjoint or has exactly 3
elements in common... if i said that right). Would we take the first six for one, the last six for the second, and alternating choices for the other two?
So if we start with the cycle
our four sets would be
|||
|||
|||
( 1 2 3 4 5 6 ; 7 8 9 10 11 12 ; 1 3 5 7 9 11 ; 2 4 6 8 10 12 ) ? This would also allow for two additional pillars based on the circle of fifths (1 6 11 4 9 2 ; 7 12 5 10 3 8 ), if we were to allow 4-element overlaps. Or six beyond that if we add in the other cyclic subgroups of Z/12Z (the group of integers modulo 12), although this would create pillars with 5-element overlaps. Some of these "facts" are unverified and based on extrapolation from example and intuition, so please take with grain of salt.
The number of cycles can be found by considering the number of permutations of the twelve elements and dividing by 12 (there are 12 permutations in each
equivalency class, one for each way we can start the cycle.) This is a whopping 11! = 39,916,800 possible (distinct) cycles.
If we start with SnoWeb's criterion,
How do you create four types of quantum producer which produce 2 quanta out of 6 in maximizing the versatility? you create two couples of opposite arrangement which have 3 commune elements with each non-partner.
how many possibilities are there?
First we choose one of our subsets of six. There are 924 possibilities --combinations of 12 taken 6 at a time = 12!/[6!(12-6)!]. The second subset of six
has to include the remaining elements.
The third subset has to have exactly 3 elements in common with the first. We need to choose 3 out of 6 elements = 6!/[3!(6-3)!] = 20 ways to choose. And 3
in common with the second subset. Again 20 ways to choose. The fourth subset must be the complement of the third.
So there are 924*20*20= 369600 ways to choose our 4 subsets.
We must choose the most versatile, balanced, and symmetric set. Why?
We must choose a versatile set in order to satisfy our mechanic. We want to address the difficulty in creating a deck that utilizes a "middle number" of
elements, especially four or five. If our set lacks in versatility, we may be able to, for example, create decks that utilitze some sets of five elements
but not other 5-element combinations.
We must choose a balanced set to avoid overpowering synergies. If one of our pillars allows for elements that work especially well together then that pillar
could become overused in the meta-game.
We must choose a symmetric set to maintain the aesthetic of Elements and keep it as pleasing as possible to the mind's eye. This also makes it easier to
remember which elements are together on a single pillar.
In fact we chose the subsets in this way (SnoWeb's criterion) exactly in order to maximize versatility. If we define versatility in terms of "If we choose
two of the pillars how can we guarantee the smallest overlap?" then all our choices have equal versatility. The overlap will always be exactly 3 elements. If
we wish to distinguish further among these possibilities based on versatility, we will need a different definition for versatility. Perhaps, "If we choose
two pillars how can we guarantee the largest number of elements will be represented?" or "When choosing 2 pillars how can we maximize the number of 4/5/6/...
element sets that we have available".
To calculate this directly sounds doable but i last studied group theory 11 years ago so i would have to brush up. (Hopefully someone who still studies maths will stroll by :p) For now i will concentrate on an example:
Example: Chromatophore's groupings |||
and their opposites
First criterion: How many elements can we guarantee are represented when we are allowed to choose two (non-opposite) pillars? There is a set of four that
does not appear.
darkness. Every set of three appears. It seems like this should be easy to prove but i am running out of steam for now. I'll have to pick up on this later.
Second criterion: If we set out to make a deck using just one of the four types we can account for 4*
C(6,4)]=60 out of the 495 possible 4-combinations and 24 of the possible 792 5-combinations. With two pillar types we can account for more but again i'll have to investigate further at a later time.
A reasonable way to create the most balanced set, would be to examine synergies amongst the elements. If we can count the synergies between the elements,
then we can choose pillars with roughly equal numbers of synergies amongst their component elements. Of course this is difficult to count. If a card
requires two elements to use (like lava golem) then it should be counted. But what about less obvious synergies (say otyugh/archangel)? Which synergies do
we count, how do we weight them and where do we draw the line?
Another way we could choose is based on symmetries. We could insist, for example, that one pair out of our four satisfy the oondition that opposing elements
are kept together while the other pair have opposing elements separated. (Like in the original proposal and in SnoWeb's list of 32). This would cut the
number of possiblities down significantly (i haven't calculated how much as of yet).
The question of how do we choose is rarely an easy one but it just as rarely proves unintriguing. OldTrees and SnoWeb thank you for continuing down this
path of discovery. I hope we like what we find!
