A rectangular board was covered without overlapping by $1×4$ and $2×2$ polyminos . Then the polyminos were removed from the board but one $2×2$ was lost . Instead, another $1×4$ polymino was provided . Prove that now the board cannot be covered by the polyminos without overlapping.
This is a question from Mathematical Circles by Dimitri Fomin, Sergey Genkin and IIa Itenberg.
The solution given is as follows:
Let us consider the coloring using $4$ colors . Then each $2×2$ title contains exactly one box of color $1$ and each $1×4$ tile contains none or two boxes of color $1$. Therefore, the parity of the number $2×2$ polyminoes coincides with the parity of the number of boxes of color $1$. This prove the statement:after the parity of the number if $2×2$ polyminoes changed(when one has lost) we cannot cover the same board without overlapping .
However, I am not getting the idea of the solution . How can they conclude that the parity of the number of $1$ colored boxes is same as that of the number of $2×2$polyminos. How can we be sure that in the original arrangement when the polymino was not lost there were no $1×4$ polyminos on a $1$ colored square. I am not getting it... also there is a duplicate link suggested ...however the detailed one there ...i mean the solution which is probably one of the most detailed one or maybe the only one is as follows:
It is enough to keep track of just a single color: say, the color 1. The key difference between the tiles is that: A single 2×2 tile always covers a single tile of this color. A single 1×4 tile always covers 0 or 2 tiles of this color: an even number. Therefore: If there is an odd number of this color, an odd number of 2×2 tiles must be used when we cover the entire floor. If there is an even number of this color, an even number of 2×2 tiles must be used when we cover the entire floor. Swapping out a 2×2 tile for a 1×4 tile, or vice versa, changes the parity of 2×2 tiles (from even to odd, or vice versa). Since we started with the correct parity to cover the floor, we now have the incorrect parity, and we cannot do it.
I am not getting it. What if the parity of number of $1's$ is even to begin with ...now after one $2×2 $ square is lost ...no of $2×2 $ polyminos is odd ...but $1×4$ polyminos can cover an even number of $1's$ so there shouldn't be any problem....so the solution looks a bit off ...I dont quite get it...Also then a $2×2$ polymino may be extra ....Furthermore I am asking about a certain proof ....The user in that post asked about the solutions...