Given below is the classic combinatoral geometric problem of whether 8x8 board can be covered by 21 straight triominos. It is taken from Solomon W. Golombs Polyminoes and Puzzles.
Now I am having trouble understanding how the author proofs certain statements using the concept of symmetry.
- He says the board cannot be covered with 21 straight triominos and 1 monomino on the lower left hand corner because the configuration would result in 22 red, 21 white, 20 blue boxes, whereas the total number required for 21 triominoes is 21 red, 21 white and 21 blue.
This makes sense. But when he says that all 4 corners are symmetric and thus covering any corner would lead to it being it impossible for it to be covered by 21 straight triminoes is where I have a problem. A picture of my idea is given below:
You see the board contains 22 red, 21 white and 21 blue boxes. So logically can't I just cover up one of the red corners and have a valid configuration ?
- Next, he extends this idea to say that any mononimo placed on a square symmetric to blue and white square leads to an impossible configuration. I just don't understand what that means. What is the meaning of squares symmetric to blue and white squares ??
He even gives marks out which red squares are not symmetric, I again have no idea what that means. Picture given below.
What am I missing here?