Answer to Question #146064 in Combinatorics | Number Theory for Asfandyar

Question

Kate drew a 41 x 41 checkered square (lattice) on the asphalt with white chalk (i.e. there are 42 horizontal segments and 42 vertical segments drawn).

By one move it is allowed to pick out an arbitrary square (of any size) and repaint its boundary using a chalk of blue colour. In different moves it is allowed to repaint any segment more than once. What is the smallest number of such moves required to repaint all the initial lines in blue colour?

Answer 1

We can use complementary counting, counting all of the repaint all the initial lines in blue colour.

• For at least one blue repainting:

There are $41 \times 41$ squares to choose which one will be blue. Then there are $2^5$ ways to repaint the rest of the squares $4 \cdot 32=128$ .

• For at least one $2 \times 2$ square:

There are two cases: it will be repainted its boundary using a chalk of blue colour or any segment will be repainted more than once.

The first case is easy: 4 ways to choose which the side the squares will be on, and $2^3$ ways to color the rest of the squares, so $2^2$ ways to do that. For the second case, there will by only two ways to pick two squares, and ways to color the other squares $32+8=40$ .

• For at least three $2 \times 2$ squares:

Choosing three such squares leaves only one square left, with four places to place it. This is $2 \cdot 4=8$ ways.

• For at least four $2 \times 2$ squares, we clearly only have one way.

By the Principle of Inclusion-Exclusion, there are $128-40+8-1=95$ ways to have at least one blue $2 \times 2$ square.

There are $2^{1681}$ ways to repaint the $41 \times 41$ square all the initial lines with no restrictions, so there are $2^{1681}-95$ ways to paint the square with the restriction.