1. Two arbitrary squares of different colors have been removed from a checkerboard.
Is it possible to cover the remaining portion of the board with domino tiles so that each domino tile covers exactly two squares?
2.Assume that at every step we remove a pair of squares of different colors.
What is the minimum number of such pairs that need to be removed to assure that the remaining portion of the checkerboard cannot be fully covered with domino tiles?
Part 1: Make a 'track' for the dominoes on the board, for example:
+--+--+--+--+--+--+--+--+
| | | | |
+ + + + + + + + +
| | | | | | | | |
+ + + + + + + + +
| | | | | | | | |
+ + + + + + + + +
| | | | | | | | |
+ + + + + + + + +
| | | | | | | | |
+ + + + + + + + +
| | | | | | | | |
+ + + + + + + + +
| | | | | |
+ +--+--+--+--+--+--+ +
| |
+--+--+--+--+--+--+--+--+
Removing two squares of opposite parity will leave two partial tracks with the same number of black and white squares. Each partial track can be filled with dominoes because in each track the squares strictly alternate white,black,white,black,etc.
Part 2:
Pretty easy, isolate one corner by removing its two orthogonal (same parity) neighbors. The two squares of opposite parity to be removed can be any squares except the isolated square.
Part 2 extension: What if the remaining squares must continue to be all connected? The best minimum I can do is four pairs (XX for removed squares). The T-shaped peninsula carved out cannot be tiles with dominoes:
+--+--+--+--+--+--+--+--+
|XX XX XX |
+ + + + + + + + +
|XX XX XX XX |
+ + + + + + + + +
| |
+ + + + + + + + +
|XX |
+ + + + + + + + +
| |
+ + + + + + + + +
| |
+ + + + + + + + +
| |
+ + + + + + + + +
| |
+--+--+--+--+--+--+--+--+
Solution
