The first puzzle was on Dmitri Zvonkine’s handout last week, but we didn’t
have time to look at it during his presentation. It’s really pretty, and it
involves a slightly different type of invariant than the other examples we did
that day.
We begin with three checkers (or amoebas) on a “chessboard” covering the
first quadrant of the plane:

Now we make a series of moves: Each move consists of first selecting any
checker on the board for which both the square immediately above it and the
square immediately to its right are empty. We then remove the checker from the
board while placing new checkers on those two empty squares. For example,
here’s a possible sequence of the first four moves:
(I’ve
included the outline of a checker to mark the one removed from the board)
The puzzle is to show that no sequence of moves from the starting position,
regardless of its length, will ever succeed in making all three starting
squares empty.
You should ask yourself: “How can I find an invariant here? What sort of
quantity would stay the same when a checker in one location is replaced with
two checkers in related locations?” This is a pretty big hint, you need to
find some sort of weight for each checker that depends on the location of the
checker such that the sum of the weights of the checkers stays the same as the
moves are made. Once you can do that, you can then try to tackle why this
invariant shows you can’t reach a situation where those three starting
squares are all empty.
This reminded me of a second invariant problem involving checkers on an
infinite board. In this one you also have checkers on an infinite board, but
with a few differences:
- You use the entire plane for your chessboard,not just one quarter of
it
- You start with an infinite number of checkerson the board, one in every
square in the lower half-plane
- You move the pieces in a different way, in this puzzle the checkers move
as in “Peg Solitaire”, that is, a checker may jump over one adjacent
checker (whether above, below, left, or right) if the following square in
the same direction is empty. The checker that is jumped over is then
removed.
- the object of this puzzles is to get at leastone checker as far into the
upper half plane as possible (that is, to bring it to a square with the
largest possible value of its y-coordinate).
So, the starting arrangment looks like:

and an initial sequence of moves might look like this (where I’ll draw
outline circles for both the previous location of the checker that just jumped
and the checker that was jumped over and is now removed):

These moves brought a checker up to the second row above the x-axis, and
it might not be too hard to continue this sequence of moves with three more
that would bring a checker to the third above the x-axis. But can you
reach the fourth row? or the fifth? or any row above the x-axis?
And what does this have to do with invariants?
Showing that you can reach a specific row might be simply a matter of
constructing a series of moves that reach that row. Showing that you can reach
any row, might involve induction. But if you want to show that it’s
not possible to reach some row, an invariant could be very helpful! Such an
invariant might have to be similar to the weights used in the previous puzzle,
but there are two noticable differences:
- in this problem, each move replaces two checkers by one, in the previous
problem one checker is replaced bytwo.
- the three squares affected by a move are three-in-a-row here, in the
previous problem they werethree-in-an-L-shape.
How do these differences influence the choice of weights?