/sci/ - Science & Math

>>15722500
Here's Marckert.
>input the best/worst X for each
Yeah. Change the notation from (k-j, j) to (k, k-j) so (5,0) (4,0) (3,2) is (5,5) (5,4) (5,3). For (k, k) the input is always 2^(k-2)k/k! For (k, k-1) it's 2^(k-1)(k-1)/k!

>systematically catalog
Consider that 1 shape of k points is also k shapes of k-1 points... k choose j of k-j points. You can measure the convexity a shape up to its full set of subshapes. For example, the minimal and maximal X distributions for 6 points.

min 0 in X(6,j)

X(6; 3, 333335, 333333333344444) : 0 = 360/720, 1 = 260/720, 2 = 80/720, 3 = 20/720
X(6; 3, 333444, 333333333444444) : 0 = 360/720, 1 = 228/720, 2 = 72/720, 3 = 36/720, 4 = 24/720

X(6; 4, 344445, 333333444444444) : 0 = 248/720, 1 = 240/720, 2 = 124/720, 3 = 68/720, 4 = 32/720, 5 = 4/720, 6 = 4/720

max 0 in X(6,j)

X(6; 3, 333333, 333333333333444) : 0 = 456/720, 1 = 240/720, 2 = 24/720
{0, 268}, {1, 288}, {2, 112}, {3, 48}, {4, 4}}

X(6; 4, 334444, 333333334444444) : 0 = 268/720, 1 = 288/720, 2 = 112/720, 3 = 48/720, 4 = 4/720
X(6; 4, 334444, 333333334444444) : 0 = 268/720, 1 = 284/720, 2 = 124/720, 3 = 36/720, 4 = 8/720

Unfortunately, there's no name for the 2 different max X(6,4) shapes. You need a new measure. Something beyond convexity.