I have not been following this thread and just noticed it today.
What are you doing about the legal positions which cannot be reached in play, are you counting them?
I'm really sorry about my late reply but due to some rather unfortunate events I wasn't able to post one earlier.
"P(64,2) = 4,032 where we assume two kings can be next to each other."
Yes, I got so far.
"if we follow the rule that no two kings are next to each other. That is P(64,2)-420"
And I do get why that is correct.
What I don't get, however, is how you reached the 420.
Or to put it differently, how do you know that cardinality? (That I didn't know the answer to that question was the very reason I calculated the valid positions for a two king game in that intricate a way.)
"It shouldn't matter yet which diagonal the bishop travels on because combined the bishops travel on all the squares."
Right. Why do such easy things elude me? -.-
What he said."Meaning you're allowing the possibility for one of the kings to be in check?"
Yes, 1 king in check is possible and we are also counting the checkmates for example, a white king at A1 with a black queen on B2 and the black king on C3 would be checkmate.
Where we could have a problem, however, would be in preventing both kings' being checked/mated at the same time (later).
"Given the constraints of the problem are 20 rooks on the chess board possible?"
No. For 20 rooks, all pawns would have to be promoted to rooks, which is impossible as they block each other (a black pawn must leave to enable a white promotion and v.v.).
"We start with 2 rooks how many rooks on a chess board are possible?"
"How many pawns can make it to their last row?"
Why not start with a simple problem which you can actually solve?
I think a list of "rules" for this problem might prove invaluable.
Same idea differently put? Gee, as usual, I don't see the obvious. (or maybe I'm simply dumb)There are four squares where the second king cannot stand on 3 adjacent squares, and 24 squares where the king cannot stand on 5 adjacent squares. On any of the other 36 squares, the opposing king cannot stand on 8 adjacent squares.
Given enough time (like really really long), you could set up each position on a board and count them, no? Then is it a fallacy to assume the problem must therefore be solvable?If you're immediately set against the notion that we can have 20 rooks at once, then you must, in effect, solve all possible games - which is beyond current mathematical capabilities.
Then tell me one thing, please: How can a position that isn't playable be legal?No, it's not really a contradiction, as such.
After all, it's considered legal IF you can reach it by playing valid moves.