Correct me if I'm wrong, but wouldn't all local maxima have to be at the center of at least one range? If not, you'd be able to move one place towards the midpoint and get a higher value. (Since you can't use two ranges at the same time to make a bigger range)

Rather than testing 4M points, just grab the 25k ranges, delete the ones that are subsets of another range, and then the midpoints of what are left are your local maxima.

Conversely, local minima would be found at the midpoints between the midpoints of overlapping ranges. You would still have to test three values to ensure it is a true minima and not a flat spot.

In this case, you might get to use the buckets technique from back up the thread. 25k+1 midpoints between adjacent ranges * 3 sampled values * ~30 ranges in the bucket = 2.3M compares instead of 120M

Oh, I was too tired last night. I'm looking for local minima. Sorry for the error. I don't follow your idea re. local minima. Could you please elaborate.

Do you have a larger test set--say max=100 ranges=100 rangesize=10--plus results?

The potential maxima are at the center of your ranges. Since the peaks are all the same size (ranges being all 5k wide, and the same slope (+/-1 per unit distance) then the minima will be at the half way point between two maxima.

4k |-   / \  / \    / \ / \       /
2k |-  /   \/   \  /       \     /
0k |- /          \/         \___/

Since you know where the peaks are (start+2.5k), and you can sort the ranges by start position, you can trivially identify the neighboring ranges. Halfway between the peak of range N and range N+1, there might be a local minima, or a flat spot, as in the picture.

The value at the minima will be easily calculated once located by finding the distance to either of the two ranges causing it.