Peaceful Encampments, round 2

Consider a plain represented by the unit square. On this plain we want to “peacefully encamp” two armies of point-sized soldiers — one army red and one army green. Each soldier “attacks” chess-queen-wise: horizontally, vertically, and diagonally in all directions. The puzzle is to maximize the size of the equal armies (equivalently, maximize the size of the smallest army), given the constraint that no pair of opposing soldiers can be placed attacking each other.

Previously on this blog: Back in 2014 I had conjectured that the optimum solution for two armies was the following. (Each region has size .) This morning I learned that I was very wrong about that!

Two encampments of size 0.13397 each

Last night I was noodling on this problem again (for the more-than-two-army case) and came up with this knight’s-move-inspired solution for 4 armies. Each home-plate-shaped region has size exactly 0.05.

Four encampments of size 0.05 each

In the above image, I used two colors to illustrate the four-army solution — so the four-army solution (also here) happens to double as a two-army solution of size 0.1. But what if we take that solution and rearrange the colors so that the friendly armies are next to each other instead of diametrically across the board from each other? I expected a modest increase, but after making the switch and jiggling it a bit, I arrived at this interesting configuration!

Two encampments of size 0.1458 each

The vertical lines divide the unit square into fourths; the slashing lines divide the sides of the unit square at the 1/3 and 2/3 marks. So each home-plate-sized region has size , and each slashing region has size , for a total army size of . Wow!

If we take the same four-army solution and color just two armies with the same color, then we get a three-army version. Jiggling that version arrives at these two solutions for three armies. The left-hand solution has army size ; the better right-hand solution has army size . I have not yet done the math to describe this solution algebraically.

Three encampments of size 0.0667 each Three encampments of size 0.0718 each

Here’s my best (and only) solution for five armies. Each encampment here has size . Again I haven’t done the math to describe it algebraically.

Five encampments of size 0.0311 each

Finally, here are four solutions for six armies, of sizes , , , and (my best attempt) , respectively. Again I haven’t done the math to describe these algebraically.

Six encampments of size 0.0104 each Six encampments of size 0.0133 each
Six encampments of size 0.0140 each Six encampments of size 0.0210 each
Posted 2019-01-21