The checkers problem told to me by M.O.Rabin

Last week, during the annual seminar at the University of Newcastle-upon-Tyne, Michael O. Rabin told me the following problem.

Consider an infinite checkers board, of which the columns and rows are identified by the integer coordinates  x  and  y  respectively. Initially, there is a piece on each square whose coordinates satisfy   (even.xeven.y) y 0.   Pieces can be moved "upwards" by the usual "capturing moves": from to and from to . The question posed is whether there is an upper bound on the y-coordinates of the squares that may be occupied.

*         *         *

Because the question asked is about y-coordinates, we abstract for a moment from the x-coordinate. The two original moves then become one and the same move: from to . In order to capture that a high piece is created from ("is as good as", "corresponds to") its two successors —those who know him see Fibonacci lurking around the corner!— , we give a piece (on a square) at height  y  a weight  φy  with  φ  the positive root of   φ2 = φ + 1 . The equation is chosen so that the weight of the new piece equals the weight of the two pieces it replaces, i.e. a move is neutral as far as total weight is concerned. By restricting ourselves to the positive root, we keep all weights positive, i.e. total weight a monotonic function of the number of pieces involved. Solving the equation yields

 φ = (1 + √5)/2       .

The fact that a move is neutral as far as total weight is concerned, means that the creation of a piece at height  Y  (with Y>0) uses from the original configuration a set of pieces with total weight φY. This "target weight", being φY and φ being positive, grows exponentially with Y. We shall next observe that, in view of the shape of the two moves —x re-enters the picture— , the original pieces involved in the creation of a piece at height  Y  come from a restricted area. Calling their total weight the "available weight", we shall show that the latter grows linearly with Y. Hence, the condition

target weight ≤ available weight

imposes an upper bound on Y: the answer to the question posed is "Yes".

In order to establish the available weight we observe the moves for small values of Y.

Y=1 requires a single move, say    .

Y=2 requires two moves, say    ,
the one thin dot above the
line indicating a square
that has been temporarily
occupied.

Y=3 requires two moves more.

For the restricted areas from which the original pieces have to be recruited we take infinite (truncated) triangles

For Y=1 , the available weight (∑n: n≥0: (n+1)φ-n) is finite, and so is the increment (∑n: a≥0: φ-n). Summing the sequences one finds for the available weight

4 + 2√5 + Y(3 + √5)
2

The smallest value for  Y  such that the target weight φY exceeds the available weight is  7 : then the target weight equals (29+13√5)/2, the available weight is only (25+9√5)/2 .

For Y=6, the target weight (18+8√5)/2 is less than the available weight (22+8√5)/2 , but this does not justify the conclusion that Y=6 is attainable. Y=6 can be achieved, but the only way of showing this that I know of is showing a game that does the job. The required game turns out to have 53 moves, which makes finding it and reporting it somewhat of a challenge. I could convince myself that Y=6 could be reached only

(i) after having realized that the constraint of at most 1 piece per square is inessential and can be stopped

(ii) after having decided to play the game backwards

Below, we show successive stages of the backwards game: in the centre the configuration of the pieces, to the left the y-coordinates of the rows in question and —by way of check— to the right for each row the total number of pieces in it.

``` 6                               1                            1

5                             1                              1
4                           1                                1

4                           1   1                            2
3                                 1                          1

3                         1   1   1                          3
2                       1   1                                2

2                         1   2   1   1                      5
1                               1   1   1                    3

1                       1   1   2   2   2                    8
0                     1   1       1   1   1                  5

0                     2   2   2   2   3   2                 13
-1                   1   1   2   1   1   1   1                8

0                     1   1   1   1   1   1                  6
-1                   2   2   3   2   2   2   2               15
-2                 1   1   1   1   1       1   1              7

0                     1   1   1   1   1   1                  6
-1                   1   1   1   1   1   1   1                7
-2                 2   2   2   2   2   1   2   2             15
-3               1   1   1       1   1   1   1   1            8

and now not repeating all the constant rows

-1                   1   1   1   1   1   1   1                7
-2                 1   1   1   1   1   1   1   1              8
-3               2   1   2   1   2   2   1   2   2           15
-4             1           1   1   1   1       1   1          7

-2                 1   1   1   1   1   1   1   1              8
-3               1   1   1   1   1   1   1   1   1            9
-4             2       1   1   2   2   1   1   1   2         13
-5           1       1       1   1       1           1        6

-3               1   1   1   1   1   1   1   1   1            9
-4             1   0   1   1   1   1   1   1   1   1          9
-5           1   1   1       2   1   1   1       1   1       10
-6                 1       1           1       1              4

-4             1   0   1   1   1   1   1   1   1   1          9
-5           1   1   1       1   1   1   1       1   1        9
-6                 1       1   1       1       1              5
-7                               1                            1
```

There may exist a game of 51 moves, but I am not interested in that optimization. The above is already more elaborate than I had hoped.

Austin, 20 September 1992

 prof.dr. Edsger W. Dijkstra Department of Computer Sciences The University of Texas at Austin Austin, TX 78712-1188, USA