easy problem

this problem is too easy.

on a chess board, 16 of these "four square objects" will fit:

~o
ooo


what is the least amount possible which may be situated with their 3-square's aligned in the same direction.

(ignore ~)

I get 12

I get 8, assuming that you mean that no others can be placed on the board with the same orientation.

With the wide 3 squares at the bottom, the leftmost square for each piece are location in positions:
a2, a6,
c1, c5,
e2, e6,
g1, g5

The answer's thirteen I think.

You're right, I just got 13 as well. I'll upload a pic if someone wants it.

If 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.

i obviously did not explain well enough.

16*4=64

all squares will be occupied.

perhaps i did not do well explaining the shape.

it is like the shape in "tetris" which has one square adjacent to the center of 3 aligned squares.

On a cylindrical board (where a-file is adjacent to the h-file), you could fit all 16 with the same 3-square alignment.

I'm not sure about the normal non-looping chessboard, however.

Ok, 12.

I 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)

Yeah, that's how I got it

but 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

Yesm you seem to be right. Don't worry. I've just misunderstood the question twice 😀 well done.

Rereading the question, I see the word "least". Why, then, was I looking for the most?