MEI Conference 2011
Algebra in unexpected places
Presenter: Bernard Murphy
bernard.murphy@mei.org.uk
Conway’s Chessboard Army
Fitting straight trominoes on a chessboard
1
y7
xy7
x2y7
x3y7
x4y7
x5y7
x6y7
x7y7
6
6
2 6
3 6
4 6
5 6
6 6
7 6
y
xy
xy
xy
xy
xy
xy
xy
y5
xy5
x2y5
x3y5
x4y5
x5y5
x6y5
x7y5
y4
xy4
x2y4
x3y4
x4y4
x5y4
x6y4
x7y4
y3
xy3
x2y3
x3y3
x4y3
x5y3
x6y3
x7y3
y2
xy2
x2y2
x3y2
x4y2
x5y2
x6y2
x7y2
y
xy
x2y
x3y
x4y
x5y
x6y
x7y
1
x
x2
x3
x4
x5
x6
x7
Level
4
Level
3
Level
2
Level
1
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Fitting straight trominoes on a chessboard
The sum of the expressions on cells covered by a horizontal
tromino is of the form xi y j 1  x  x 2 and by a vertical tromino is

of the form xi y j 1  y  y 2



When all 21 trominoes have been placed, leaving the x a y b cell
empty, we have
f  x, y  1  x  x 2   g  x, y  1  y  y 2   x a y b 
1  x  x
2
Conway’s Chessboard Army
Imagine it was possible to reach the square marked 1. Think about the
sum of the values of all the occupied squares at every point in the game.
At the end of the game this sum would be 1 (as only one counter remains
and that is on the square marked 1). Is it possible to assign a value to x
so that at every move the total never increases? For this to be true we
would need
xn2  xn1  xn (a jump up the board or towards the central column)
xn  xn1  xn2 (a jump down the board or away from the central column)
xn  xn1  xn1 (a jump across the central column).
 x3  x 4  x5  x 6  x 7 1  y  y 2  y 3  y 4  y 5  y 6  y 7  One possible value satisfying these inequalities is the positive root of the
equation 1  x  x 2 . So, with x  521 , at each move the sum of the
x3  1   x  1  x 2  x  1  0 has roots 1,  
1  i 3 2
, .
2
In the first equation above substitute
x  , y   :
0  0   a b  1   1     
x  , y   2 : 0  0   a  2b  1    1   2   1
So a  b  1, 4,7,10,13 and a  2b  0,3,6,9,12,15,18, 21
Solving simultaneously gives  a, b    2, 2  ,  2,5 ,  5, 2  ,  5,5 
values of the squares occupied never increases.
In the starting position, only squares below the barrier are occupied.
The sum of the values of all the squares below the barrier, T, is given by
T  x5  3x 6  5 x 7  7 x8  9 x9  ... 
T  xT  x5  2 x 6  2 x 7  2 x8  2 x9  ...  x5 
x5 1  x 
2 x6
 T
2
1 x
1  x 
Substituting x  521 leads to T  1 and since not all squares can be
occupied it follows that the starting sum is strictly less than 1. However,
after a series of moves in which the sum never increases, it ends up with
value 1. This is clearly impossible and so we can conclude that it is
impossible to reach level 5.
You might want to think why this argument breaks down for level 4.