Problem 4 Answer: "Divvying Up The Loot"

The second example neglects to add in the value of the shield to the total when calculating the individual share of loot. As instructed, the 1250 GP bid should be added to the 5000 GP in gold to get a total treasure value of 6250 GP, which comes out to 1562.5 GP per player. Thus A would get the shield and 312.5 GP, while the others would get 1562.5 GP each.

BONUS QUESTION: What is the actual maximum possible bid for A? In other words, what bid for A would result in him getting the shield and no gold, while everyone else gets the gold divided evenly?

Problem 3 Answer: "Take Cover!"

In solutions 3.1 and 3.3, the corner of A's square at which the two dotted lines converge does not cross any line connecting it to D's square, as seen in the diagrams. (Note that the endpoint touches a wall, but endpoints touching walls don't count.) Thus A has unobstructed line of sight, despite clearly being on the other side of a wall.

In solutions 3.2 and 3.4, the corner of D's square at which the two dotted lines converge itself touches a wall, so given any corner of A's square, the line from that corner to D's square will touch a wall. Remember that in problems 3.2 and 3.4, the endpoint touching a wall does count for crossing. Thus D has cover, despite having his back to the wall and there clearly being no obstructions between him and A.

This problem demonstrates that the writers of the D+D rulebook did not do a particularly good job clearly explaining the conditions for cover and line of sight.

BONUS PROBLEM: Suppose we attempt to modify the crossing condition to state that the endpoint at A's square counts for crossing, but the endpoint at D's square does not. (This would invalidate all the anomalies described in the solutions at left.) Is it still possible to find a situation in which D has cover, but no line between an interior point of A's square and an interior point of D's square crosses a wall? (Or the other way around, where D has no cover but every line between interior points crosses a wall.)

Problem 2 Answer: "Too Many Men on the Playing Field"

With the main character's 100 points, he can have 2^16 100 point followers (20 points for one-fifth of the point value plus 80 points for 16 doublings). Each of these followers can have 2^16 duplicates (also 20 points for one-fifth of the point value plus 80 points for 16 doublings.) Thus the main character can field a total of 2^32 + 2^16 + 1, or a little over four billion, units.

Problem 1 Answer : "Ineligible Receiver (of bullets) Downfield"

Given the two enemy units A and B, and the friendly unit X. Construct the perpendicular bisector of the line segment AB. Whichever side X is on, that is the unit that is closest.