29 Dec 08
Originally posted by progprinterHi
the board is invalid since both white bishops are on white squares.
You are allowed to do these things in problems of this nature
just so long as it can be seen that an under promtion has legally taken place.
If you could prove that White never under-promoted to a white
squared Bishop ("for Christmas giggles" I like that phrase)
then it would be invalid.
Sometimes the beauty behind such a problem is not the solution itself,
But the idea behind it and the proof that the position is legal.
Composing such a position can be just as rewarding as playing a
good OTB combination. Even better in this case because I asked
the boys at The Problemist Forum to come up with a unique
Christmas Tree problem just for here, so all Heinzkat had was a blank
canvas and a set task.
29 Dec 08
Originally posted by progprinterWhat can I say, "I'm dreaming of a white Christmas". 🙂 For me that was justification enough to place them...
the board is invalid since both white bishops are on white squares.
The board is "invalid" from a composer's perspective indeed, as normally one does not use promoted force (i.e. three Knights, two Queens, two light-square Bishops). But it is a legal position - look, White has seven pawns on the board, the eighth one promoted to a LSB. I'd rather have used nine pawns to finish the construction, but hey then you would not be satisfied either.
😉