yeti_c wrote:I love this idea - I would be well up for coding the XML for this...
What we need is to decide on what consitutes 2 correct straights and 2 non correct straights?
i.e. This is 2 straights 2,3,4,5,6 & 7,8,9,10,J
Or This is 2 straights 2,3,4,5,6 & 6,7,8,9,10 (with one overlap)
I would prefer the first one...
Or alternatively to make it simpler then you could drop straights and only have flushes? (i.e straights of only one suit?) Or is that the rule anyway?
C.
PS Jokers will exponentially increase the XML... (i.e. don't have them!)
i was thinking of clean formations, no card should belong to 2 different straights.
for example if i have 2,3,4,5,6 and another 4,5,6,7,8 i want just 2 straights, no pairs no extra anything.
but it could be very difficult. even though i find it really logical.
if i have 2,3,4,5,6 and 4,5,6,7,8 plus another 6. this means i have 2 clean straights one overlapping straight 3,4,5,6,7 another 3 straights formed with using the third 6, another 6 straights for switching 4,5,6 between the first 2 straights, i have 3 one pairs 4,5,6 i have 3 two pairs (4-5,5-6,4-6) and another 4 two pairs by inserting the third 6 instead of one of the other sixes), i'd also have one 3 of a kind. and 2 full houses so i own 11 continents and i get 24 bonuses. this is really mad. could there be a way to allways implement the highest bonus obtainable without using a card for more than one formation?
another example: if i have QQKKK it's a full, 2 one pair, 1 3 of i kind, and 1 two pairs.
and i have the following bonuses.
full +6
3 of a kind +3
2 pair +2
1 pair +1
i want to receive only the full house bonus.
and if next turn i lose one K i want to receive the 2 pair bonus and not the 2 one pair bonuses.
and another problem.
i have QQ KK and i get the +2 bonus for havin a 2 pair formation.
and another guy has JJJJ and gets +7 bonus for having 4 of a kind.
that's a 5 armies difference for owning the same number of cards.
in the current format where all cards can attack in any direction this is a real problem, but it could be solved by adding impassable borders. what do you say?
“In the beginning God said, the four-dimensional divergence of an antisymmetric, second rank tensor equals zero, and there was light, and it was good. And on the seventh day he rested.”- Michio Kaku