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Nearlydouble Chess. A 2d variant with the whole of two identical sets minus the second King. (11x11, Cells: 121) [All Comments] [Add Comment or Rating]
Jeremy Good wrote on Sat, Mar 31, 2007 12:35 AM UTC:
Charles, I really enjoyed reading your description, at the bottom of this page, of all possible ways pawns could make their first doublemove. That sort of systematic approach is one of the things I really like about your work.

Charles Gilman wrote on Sat, Mar 17, 2007 07:41 AM UTC:
132 is not 'nearly' twice 64 in the sense of being just short of it. It is surely more logical to have just under twice the number of both squares and pieces than to have just over twice one and just under twice the other. The latter would exacerbate the change in piece density (in the case of your suggestion to 46.97%) rather than partially cancel it out (in the case of this variant to 51.24%).

Abdul-Rahman Sibahi wrote on Thu, Mar 15, 2007 10:41 PM UTC:
In fact, the 11x12 (132 squares) board is 'nearer' to double boards (128) than 11x11 (121). The simple calculations show that 132-128 is 4 , while 128-121 is 7 .

So it would still be nearly double.

Charles Gilman wrote on Thu, Mar 15, 2007 07:20 PM UTC:

I've only just read the comments, as for some reason the first one didn't get e-mailed to me. My first thoughts are as follows.

1: I'd be reluctant to go to 11x12 as that would over twice the FIDE number of squares and so no longer 'nearly' double in that aspect.

2: You're right about the point in pieces not being demoted or lost due to another piece being captured. I'll try to rephrase it to make this clear. As for promoting if you have the full quota, yes, the first two must become Queens.

3: I am indeed planning a Nearlydouble Wildebeest Chess, on a board of 14 ranks by 15 files, and decreasingly powerful double Pawn moves from successive ranks.

Abdul-Rahman Sibahi wrote on Thu, Mar 15, 2007 09:14 AM UTC:
Another question regarding promotion:

Assuming the player has a full set of pieces, and a pawn reaches the last rank; I understand from the rules that the pawn can only promote to a queen because he has less than four queens. Am I right ?

Abdul-Rahman Sibahi wrote on Wed, Mar 14, 2007 09:23 PM UTC:Good ★★★★
I would suggest enlarging the board a little, just a little, to 11 files x 12 ranks, for the following reasons:

1. dividing the board to three equal areas (The warfield, the white camp, and the black camp.)

2. More importantly, the white squared bishops, in both camps, have quite a field in the opening, which might lead to quick exchanges. Adding a rank would let the white squared bishops from one camp oppose the dark squared ones in the other.

3. The white squared bishop is slightly stronger than the dark squared bishop. (In the current form, a dark colored bishop and a knight can't force mate.)

I wonder how the game would be like played with Extinction or Suicide rules.

(It would also be interesting to have a Wildebeest variant.)

'the restriction that only if they have at least four of every type can they have more than four of any type.'

I assume that this applies to promotion only ? I mean, if a player has five rooks and one of his knights was captured, surely a rook wouldn't demote ?

Joe Joyce wrote on Wed, Mar 14, 2007 07:09 PM UTC:
Very interesting game. Ten unlimited FIDE sliders of 15 total pieces, on an 11x11 board, with knights and pawns slightly augmented. Nice setup, a classic type: the FIDE standards, lots of power, just a little over 50% starting density. Be interesting to see how it plays, and very likely instructive. I'd expect it to often be a rather bloody tactical game with a somewhat extended opening phase and later phases that would feature chains of trades. [I do wonder how many pawns would be left at the end.] 
I also wonder how scalable this game is; how would a 12x20, with about double the squares and 20 sliders/side, play? Or a 16x16, with 3 sets of FIDES? Do you think it's more or less directly scalable, Charles? Or anybody?

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