We can introduce mathematics to the situation by assigning a
place-value to each particular element on the grid. The
exponent of the power is what determines the final value of
any element and it is simply the number of steps in directions
parallel to the 2D grid layout in a shortest path from the
distinguished point (coloured black) to the given point in the
question. Figure 5 shows part of the whole grid with weights
of each unit assigned according to our black unit which is a coin that has reached the ideal 5 th level. An important property of weight is that it only decreases or stays constant with every move. Therefore, > .
Abstractly speaking, you can interpret this by thinking that
no coin is added throughout the process of this game and
hence the weight. The beauty of John Conway’s solution starts here. If we do succeed in moving a coin to
the unit (with value 1), then the value of the board is at least 1.
Now we have learnt about the equilibrium of our grid before and after every move, it is time to focus on
individual moves and how they affect the value of (which is what we want to find out). By doing a little
experimentation yourself you will see that moving a coin further away from the destination will never be to
your advantage – since we have (ideally) so many coins, there isn’t such ‘strategy’ to act as the reason for
moving a coin further away.
Figure 6 shows an example of a vertical move (in a generalized format). The coin at position +2 ‘jumps’ over +1 to land on . In fact, if you do a
horizontal move the result will the same (just imagine tilting figure 6 by 90
degrees. Therefore, this model can be applied to anywhere on this grid and it
has given us access to a broader solution by simply calculating 3 units. The
key to defining is that we want a jump to have no effect on the total
weight of the board.
+1 + +2 =
≠ 0
∴ 2 = 1 −
Solving this with the quadradic formula step by step:
2 + − 1 = 0
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