this configuration is transformed

reopened

 2 3 2 3 4 3 2 3 2

The grid above shows for each cell how many neighbours there are. Let’s call this the neighbourhood grid or the N-grid. The grid containing coins we can call the C-grid. In applying the rules, if a cell contains fewer than its corresponding N-grid cell, then we call it a miser, because it contributes nothing to its neighbours. If there are no misers, the situation is stable because all cells receive as much as they donate. The misers cause an imbalance.

Therefore, the question is: if the C-grid contains misers, do the rules eventually eliminate them step by step?

Take the central cell with 4 neighbours containing 8 coins. One of its neighbours is a miser, so it donates 4 coins but receives only 3. Net result: 7 coins. On the other hand, the miser cell with three donating cells receives 3 coins so ends up with 4 coins, so exceeding its miser status. However, in the same step, some donating cells become misers, having become impoverished because of miserly neighbours.

The situation is also stable if all cells become misers, but this can only happen if there are fewer than 24 coins, the sum total of the N-cells. This configuration seems to be cyclic. And, of course, whatever happens the total number of coins remains constant.

If a cell containing c coins, has m misers in its neighbourhood and n is its N-grid value (2, 3 or 4), then after the step, if c≥n, it has c-n+(n-m)=c-m coins; if c<n, it has c+n coins.

More to follow...

by Top Rated User (1.1m points)
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