Saturday, June 29, 2013

Asteroids Algebra


Problem:

Waldo is playing Asteroids. This game starts with 3 ships and you earn an extra ship for every 10000 points you score. At the end of his first game, Waldo noticed that he had the lowest score possible while averaging exactly 9000 points per ship. At the end of his second game, Waldo noticed that he had the highest score possible while averaging exactly 9000 points per ship. What were his two scores?

Solution:

Let the number of starting ships be \(S\), points for an extra ship be \(X\) and the player's average scoring rate be \(A\). Furthermore, let \(R = A/X\) and the number of bonus ships earned by the player during the game be \(B\). The number of bonus ships earned by the player during his game is the floor of his final score over the bonus value \(X\). Algebraically, we have \[ B = \lfloor (S+B)\cdot R \rfloor.\] Now let \[ (S+B)\cdot R = \lfloor (S+B)\cdot R \rfloor + f,\] where \( 0 \leq f < 1 \). This gives us
\begin{aligned}
(S+B)\cdot R &= B + f ,\\
SR + B(R-1) &= f,\\
B &= (SR-f)/(1-R).
\end{aligned}
Together with \(0 \leq f < 1\) we get that \[\frac{SR-1}{1-R} < B \leq \frac{SR}{1-R}.\] Thus, the smallest possible number of bonus ships is \[ \left\lfloor \frac{SR-1}{1-R} + 1 \right\rfloor = \left\lfloor \frac{SR-R}{1-R}\right\rfloor \] and the greatest possible number of bonus ships is \[ \left\lfloor \frac{SR}{1-R}\right\rfloor. \] It follows that the lowest possible score while averaging \(A\) is  \[ A\cdot \left\lfloor \frac{SR-R}{1-R} + S \right\rfloor = A\cdot \left\lfloor \frac{S-R}{1-R}\right\rfloor \] and the highest possible score while averaging \(A\) is \[ A\cdot \left\lfloor \frac{SR}{1-R} + S\right\rfloor = A\cdot \left\lfloor \frac{S}{1-R}\right\rfloor. \]
For Waldo, \(S=3\), \(A=9000\) and \(R=9000/10000 = 0.9\). Waldo therefore scored
\[ 9000\cdot \left\lfloor \frac{3-0.9}{1-0.9} \right\rfloor = 189000\] points in his first game, \[ 9000\cdot \left\lfloor \frac{3}{1-0.9} \right\rfloor = 270000\] points in his second game.

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