- 11 Nov '08 21:25 / 1 editIf anyone wants to know the answer, this hopefully shouldn't spoil it too much. Here's a possibility for 13 with all the pieces represented by white pawns:

edit: and a proof showing that there isn't going to be a higher answer than 14 (even though I currently still think the maximum is 13)

The two edges have to have 4 blank squares. This is very easily seen from the fact that the edge can only be occupied by a piece extremity, and then the bit that juts out from it prevents the square above/below being used. So if 4 edge squares are used on one edge, another 4 must be blank. This leaves us with 56 squares remaining. 56/4=14

Though I doubt any tile combination would slot together this perfectly. - 14 Nov '08 07:41I think I understand the quesiton now.

This figure:

Darn attempts at explaining it

Pieces are either all black or all white, I tried to make it easy enough to see.

It is the only way to fill a 4x8

Which leads me to suspect that you can only have two of them for an 8x8, leaving us with 6 pieces facing the same way (up), and 12 pieces with the 3 bar in the same direction (up and down) - 16 Nov '08 07:50but one can fill a 4x4 section with 4 "tetris pieces" where 2 are placed vertically and 2 horizontally. for example look at the top left quadrant of a chessboard, and imagine piece 1 (a8-c8) piece 2 (d8-d6) piece 3 (d5-b5) piece 4 (a5-a7) with their "nubbins" pointing toward the middle of that 4x4 block.

then placing pieces in a similar fashion in the other 3 quadrants, wouldn't this better answer the question? 8 pieces would have horizontal orientation and 8 would have vertical orientation, so it minimizes the "number of pieces with their long side facing in the same direction."

unless i misunderstood the question lol which is entirely possible