Sensible Solutions for the Escape Cases

Even when an interception is impossible, the monster might still try to chase the tofubeast. What, then, is the best $\alpha$? The ghosted $\alpha$ of the escape illustrations in Figure 7 might appear to be the only natural choice, but it's a little misleading: the whole point is that a constant $\alpha$ cannot be found, and so the illustrations are only geometrically valid for a certain point in time. The choice of $\alpha$ will depend on factors outside the scope of the problem. Some ideas:

• If $M$ expects $T$ to run straight all the way to a distant finish line, the best $\alpha$ might be defined as the angle that gets $M$ to the finish line with the least amount of lag behind $M$--which would be $\alpha = 180 - \theta$, or a parallel course.

• If $M$ expects $T$ to reverse direction sometime soon, $M$ might do best to head directly towards the current location of $T$ (at any given point in time): $\alpha = 0$.

• A good general-purpose $\alpha$ would be the angle at which $M$ should travel to intercept $T$ if $M$'s velocity were just large enough to effect an interception at some point in the future (or at least maintain a distance if behind). For $\theta < 90^{\circ}$, $\alpha = 90^{\circ}$; for $\theta \ge 90^{\circ}$, $\alpha = 180 - \theta$. As with any $\alpha$ chosen between 0 and $180 - \theta$ for a non-interception, the problem will change with each tick: $\theta$ will increase, and point $M$ will curve around behind $T$ in pursuit.

• The illustrations above are only valid for a certain point in time during the chase: the geometry of $\theta$ and the theoretical point $I$ will change, making a straight shot to $I$ impossible (and therefore $\alpha$ undefined). But the ghosted $\alpha$ and straight $v_{M}t$ seem to imply a possible solution over time: take a straight path for a little while by (arbitrarily) picking a time in the future, calculating the distance $T$ will travel assuming constant velocity and direction, and then heading toward that point (regardless of $v_{M}$). Of course, once that future time is past, a new $\alpha$ will be needed, and the path of $M$ over time, if not curved, will then be segmented. (Notice this solution and the second are actually quite similar: this $\alpha$ will probably end up being only a few degrees off 0.)