Area estimation
So, how does all of this relate to the square of the sum of lengths
divided by the edge count,LM 2/e , and the estimate
of a shape’s overall area, A ? Suppose that the average mosaic
piece resembled a square, not an octagon, but also with a perimeter
eight times the average edge length a . Each side would be 2a long and the area would be 4 a 2. The
overall area across the graph would therefore be the piece count times 4a 2. In this limited case, A is justLM 2/e because there are
four edges per 2 x 2 rectangle on average: given 25 rectangles, the area
is 100 a 2; e = 4 x 25 = 100;LM = 4 x 25 a = e a = 100a ; and LM 2/e =
100 a 2 = A .
Because the pieces actually average out to octagons, it might seem that
the area of each one would be the area of a regular octagon, which is 2
(1 + 20.5) a 2 = 4.828a 2. Thus, we might estimate A as 4.828/4LM 2/e = 1.207LM 2/e . However, the
maximal size of any polygon is reached when it expands in all directions
to become regular (because it most closely approximates a circle). No
matter what the construction process, polygons subject to any kind of
randomness must be smaller. Thus, the 4.828 figure may be too high.
Nonetheless, simulations provide no evidence to support this hypothesis.
A good explanation is that the average edge in a mosaic abuts a
larger-than-average piece by definition. For example, if half of the
mosaic consists of 6-edge pieces and half of 10-edge pieces, the average
edge abuts a shape of (62 + 102)/16
= 8.5 edges, not eight. The larger a piece, the more closely it
approximates a circle, the shape having the lowest perimeter-to-area
ratio: a square with a perimeter of eight has an area of four, whereas a
circle with the same perimeter (circumference) C has an area ofC 2/(4 pi) = 5.093. This effect seems to cancel
out the overestimation due to irregularity in polygon shapes, and as a
result, throughout the rest of this paper I employ the 4.828/4 = 1.207
regularisation constant.
Turning briefly to MSTs, which can be computed using the mosaicpackage’s tgraph function, each includes about 4/5 as many edges
as a mosaic because the edge:point ratio is nearly one in a large MST.
However, an MST’s total length should be less than 4/5 of the
corresponding mosaic’s length because an MST should avoid many long
edges. Perhaps, the MST on average simply avoids the longest out every
five edges. It can do this because there are four points for every five
edges in the mosaic (see above), and there is a near 1:1 ratio of edges
to points in an MST. However, the choice may come down to only two edges
because the others can’t be avoided: if the points form a line, the MST
must either cross from the left and leave out the last edge or vice
versa. The longer segment when a line is subdivided at random comprises
3/4 of the length on average, so the MST’s length should be (3 + 1/4)/5
= 65% that of the mosaic’s. Thus, if LT is the
length of the MST, then instead of A = 1.207LM 2/e we would predictA = 1.207/0.652LT 2/e = 2.857LT 2/e . However, in
practice, MST-based area estimates are highly problematic because the
0.65 constant seems to vary in simulation according to the shape of the
object: for example, it is higher for circles and rings, and actually
close to 0.8. Therefore, an MST-based approach is not recommended.