Consider the first quadrant in the Cartesian plane divided into unit squares by horizontal and vertical lines at the positive integers.

(1) Place $latex 3$ dots (clones) in the shape of an $latex L$-tromino in the bottom left-most squares, and draw a “barbed wire fence” enclosing the dots and their $latex 3$ respective squares: this is the red fence in following figure.

At each step you can erase a dot and replace it with two copies in adjacent squares, one directly above and the other directly to the right, as long as those squares are currently unoccupied.

Prove that it is impossible to free all clones from the prison.

(2) A single clone is placed in square $latex (0,0)$, and the prison encloses

(a) the $latex 10$ squares $latex (i,j)$ with $latex i+j\leq 3$;
(b) the $latex 6$ squares $latex (i,j)$ with $latex i+j\leq 2$.

Show that there will always be at least one clone in the prison.

Due day is May 24.