48 Chapter 5
Did you succeed? Perhaps one begins to think that it is impossible. But how could one
ever prove such a thing?
Theorem 40. It is impossible to make moves in the Escape! game so as to vacate the
shaded corner region.
Proof. Let us assign weights to the squares in the lattice according to the following pattern:
We give the corner square weight 1/2, the next diagonal of squares 1/4 each, and then 1/8,
and so on throughout the whole playing board. Every square gets a corresponding weight
according to the indicated pattern.
1
2
1
4
1
4
1
8
1
8
1
8
1
16
1
16
1
16
1
16
1
32
1
32
1
32
1
32
1
32
The weights are specifically arranged so that making a move in the game preserves the
total weight of the occupied squares. That is, the total weight of the occupied squares is
invariant as one makes moves, because moving a stone with weight 1/2
k
will create two
stones of weight 1/2
k+1
, which adds up to the same. Since the original three stones have
total weight
1
2
+
1
4
+
1
4
= 1, it follows that the total weight remains 1 after every move in the
game. Meanwhile, let us consider the total weight of all the squares on the board. If you
consider the bottom row only, the weights add to
1
2
+
1
4
+
1
8
+ ···, which is the geometric
series with sum 1. The next row has total weight
1
4
+
1
8
+
1
16
+ ···, which adds to 1/2.
And the next adds to 1/4 and so on. So the total weight of all the squares on the board
is 1 +
1
2
+
1
4
+ ···, which is 2. This leaves a total weight of 1 for the unshaded squares.
The subtle conclusion is that after any finite number of moves, only finitely many of those
other squares are occupied, and so some of them remain empty. So after only finitely many
moves, the total weight of the occupied squares off the original L-shape is strictly less
than 1. Since the total weight of all the occupied squares is exactly 1, this means that the
L-shape has not been vacated. So it is impossible to vacate the original L-shape in finitely
many moves.