Arrange square with polyomino
Source: 64 Polish MO 2013 Second Round - Problem 3
April 21, 2018
polyominoColoringcombinatoricsPoland
Problem Statement
We have tiles (which are build from squares of side length 1) of following shapes:
[asy]
unitsize(0.5 cm);
draw((1,0)--(2,0));
draw((1,1)--(2,1));
draw((1,0)--(1,1));
draw((2,0)--(2,1));
draw((0,1)--(1,1));
draw((0,2)--(1,2));
draw((0,1)--(0,2));
draw((1,1)--(1,2));
draw((0, 0)--(1, 0));
draw((0, 0)--(0, 1));draw((5,0)--(6,0));
draw((5,1)--(6,1));
draw((5,0)--(5,1));
draw((6,0)--(6,1));
draw((4,1)--(5,1));
draw((5,2)--(6,2));
draw((5,1)--(5,2));
draw((6,1)--(6,2));
draw((4, 0)--(5, 0));
draw((4, 0)--(4, 1));
draw((6,2)--(7,2));
draw((7,1)--(7,2));
draw((6,1)--(7,1));draw((11,0)--(12,0));
draw((11,1)--(12,1));
draw((11,0)--(11,1));
draw((12,0)--(12,1));
draw((10,1)--(11,1));
draw((10,2)--(11,2));
draw((10,1)--(10,2));
draw((11,1)--(11,2));
draw((10, 0)--(11, 0));
draw((10, 0)--(10, 1));
draw((9, 2)--(9, 1));
draw((9,1)--(10, 1));
draw((9,2)--(10,2));
[/asy]
For each odd integer , determine minimal number of these tiles needed to arrange square with side of length .
(Attention: Tiles can be rotated, but they can't overlap.)