Most informative, particularly the image. I'd definitely chalk this up as artifacts of the control software  station keeping is noisy. I could ask some colleagues who would likely have some pretty wonky comments about the system, but I don't think deeper physics would really impact the signal processing (deeper physics available upon request).
The problem statement as I now interpret it is generate a strictly monotonic series of X values while eliminating the smallest number of points. Agree given the presented dataset that dropping points is more appropriate than a smoothing. The controller overshot, overcompensated, and ultimately returned to the correct number. I would read this as wanting to drop the whole vee, and if graff's observation is correct, keep the top corner point and drop the lower corner point. I would implement this as
#! perl slw
use strict;
use 5.10.0;
printf "%s", scalar <DATA>;
my @points = map[ split ' ', $_ ], <DATA>;
my @modified;
push @modified, shift @points; # Kickstart
while ( @points ) {
my $cmp = ($points[0][0] <=> $modified[1][0])*($points[0][1] > $m
+odified[1][1]);
if ($cmp == 1) {
push @modified, shift @points;
} elsif ($cmp == 1) {
shift @points;
} else {
pop @modified;
}
}
which underperforms on the keepthemostpoints metric. Note the yvalue switch so you can descend the loop with the same code. It also consistently deviates low, since it presumes that only the part after the first regression of the curve is wrong. Similar results are obtained with
while ( @points ) {
if (($points[0][0] > $modified[1][0])^($points[0][1] < $modified[
+1][1])) {
push @modified, shift @points;
} else {
shift @points;
pop @modified;
}
}
which drops a slightly larger number of points, and I think has a slightly cleaner curve because of it. My best idea is that we are trying to correct for S's in the data. For minimal impact to the curve, we probably want to hold onto the start of the S, the end of the S, and the median of the retrograde component. Note that by incorporating the median, we may be adding a fictitious point.
while ( @points ) {
if (($points[0][0]$modified[1][0])/($points[0][1]$modified[1][
+1]) > 0) {
push @modified, shift @points;
} else { # Retrograde!
my @retro = (pop @modified, shift @points);
push @retro, shift @points while ($retro[1][0]$points[0][0])
+/($retro[1][1]$points[0][1]) < 0;
shift @points while ($retro[0][0]$points[0][0])/($retro[0][1]
+$points[0][1]) < 0;
pop @modified while ($retro[1][0]$modified[1][0])/($retro[
+1][1]$modified[1][1]) < 0;
push @modified, [0.5*($retro[@retro/2.5][0] + $retro[@retro/2
+][0]),
0.5*($retro[@retro/2.5][1] + $retro[@retro/2
+][1]),
]
}
}
With this, I finally beat graff's point count, though one might argue his result is a little better on the wiggly bits since I bias high.
I realize that the above implementations are written in a very Perly way, but they can certainly be changed to more traditional structures after a best algorithm has been suggested.
Finally, I should note that given the characteristic uncertainties of these measurements, this whole discussion is academic since none of these curves are significantly different from each other from a measurement perspective.
#11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.
