From Keith Wynroe comes an epic power battle that is sure to wear you down:
The Game of Attrition has two players, each of whom starts with a whole number of “power points.” Players take turns “attacking” each other, which involves subtracting their own number of power points from their opponent’s until one of the players is out of points.
For example, suppose Player A (who goes first) starts with 5 points and Player B starts with 7 points. After A’s first attack, A still has 5 points, while B has been reduced to 2 points (i.e., 7 minus 5). Now it’s B’s turn, who reduces A to 5 minus 2, or 3 points. Finally, on A’s second turn, B is reduced from 2 points to nothing (since 2 minus 3 is −1). Despite starting with fewer points, A wins!
Now suppose A goes first and starts with N points. In terms of N, what is the greatest number of points B can start with so that A will still emerge victorious?
Assume Player A starts with $a$ points and player B starts with $b$ points. The table below has the scores for each round
| Round | Player A Score | Player B Score |
|---|---|---|
| Start | $a$ | $b$ |
| $1$ | $a$ | $b-a$ |
| $2$ | $2a-b$ | $b-a$ |
| $3$ | $2a-b$ | $2b-3a$ |
| $4$ | $5a-3b$ | $2b-3a$ |
| $5$ | $5a-3b$ | $5b-8a$ |
| $6$ | $13a-8b$ | $5b-8a$ |
| $7$ | $13a-8b$ | $13b-21a$ |
| $...$ | $...$ | $...$ |
| $n_{(odd)}$ | $F_{n}a-F_{n-1}b$ | $F_{n}b-F_{n+1}a$ |
| $n_{(even)}$ | $F_{n+1}a-F_{n}b$ | $F_{n-1}b-F_{n}a$ |
In order for Player A to always have a positive score,
$$F_{n}a-F_{n-1}b > 0$$
$$\dfrac{F_{n}}{F_{n-1}}a > b$$
$\dfrac{F_{n}}{F_{n-1}}$ converges to $φ$.
If $a = N$ and Player A wins, the maximum integer value $b$ can take is $$\lfloor N \cdot φ \rfloor$$