Sam Ritchie created MATH-1558:
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Summary: MidpointIntegrator doesn't implement the midpoint method
correctly
Key: MATH-1558
URL: https://issues.apache.org/jira/browse/MATH-1558
Project: Commons Math
Issue Type: Bug
Affects Versions: 3.6.1
Reporter: Sam Ritchie
Hi all,
I've been reading through the implementation of {{MidpointIntegrator}}, and
I've discovered an issue with the algorithm as implemented.
The midpoint method generates an estimate for the definite integral of {{f}}
between {{a}} and {{b}} by
* subdividing {{(a, b)}} into {{n}} intervals
* estimating the area of each interval as a rectangle {{(b-a)/n}} wide and as
tall as the midpoint of the interval - ie, {{((b-a)/n) * f((a + b) / 2)}}
* adding up all estimates.
The {{MidpointIntegrator}} implementation here sticks to n = powers of 2 for
this reason, stated in the comments:
?? The interval is divided equally into 2^n sections rather than an arbitrary m
sections because this configuration can best utilize the already computed
values.??
*Here is the* *problem:* the midpoint method can't reuse values if you split an
interval in half. It can only reuse previous values if you split the interval
into 3; ie, use 3^n sections, not 2^n.
What's actually implemented in `MidpointIntegrator` is a left Riemann sum
without the leftmost point. Or, identically, a right Riemann sum without the
rightmost point: [https://en.wikipedia.org/wiki/Riemann_sum]
This matters because the error of a (left, right) Riemann sum is proportional
to 1/n, while the error of the midpoint method is proportional to 1/n^2... a
big difference.
h2. Explanation
The idea behind the midpoint method's point reuse is that if you triple the
number of integral slices, one of the midpoints with n slices will overlap with
the n/3 slice evaluation:
{{n = 1 |--------x--------|}}
{{n = 3 |--x--|--x--|--x–-|}}
You can incrementally update the n=1 estimate by
* sampling the points at (1/6) and (5/6) of the way across the n=1 interval
* adding them to the n=1 estimate
* scaling this sum down by 3, to scale down the widths of the rectangles
For values of n != 1, the same trick applies... just sample the 1/6, 5/6 point
for every slice in the n/3 evaluation.
What happens when you try and scale from n => 2n? The old function evaluations
all fall on the _boundaries_ between the new cells, and can't be reused:
{{n = 1 |-------x-------|}}
{{n = 2 |---x---|---x---|}}
{{n = 4 |-x-|-x-|-x-|-x-|}}
The implementation of "stage" in MidpointIntegrator is combining them like this:
{{n = 1 |-------x-------|}}
{{n = 2 |---x---x---x---|}}
{{n = 4 |-x-x-x-x-x-x-x-|}}
which is actually:
* tripling the number of integration slices at each step, not doubling, and
* moving the function evaluation points out of the midpoint.
In fact, what "stage" is actually doing is calculating a left Riemann sum, just
without the left-most point.
Here are the evaluation points for a left Riemann sum for comparison:
{{n = 2 x-------x--------}}
{{n = 4 x---x---x---x----}}
{{n = 8 x-x-x-x-x-x-x-x--}}
Here's the code from "stage" implementing this:
{{// number of new points in this stage... 2^n points at this stage, so we}}
{{ // have 2^n-1 points.}}
{{ final long np = 1L << (n - 1);}}
{{ double sum = 0;}}
{{// spacing between adjacent new points}}
{{ final double spacing = diffMaxMin / np;}}
{{// the first new point}}
{{ double x = min + 0.5 * spacing;}}
{{ for (long i = 0; i < np; i++) {}}
{{ sum += computeObjectiveValue(x);}}
{{ x += spacing;}}
{{ }}}
{{ // add the new sum to previously calculated result}}
{{ return 0.5 * (previousStageResult + sum * spacing);}}
h2. Suggested Resolution
To keep the exact same results, I think the only solution is to remove the
incorrect incremental "stage"; or convert it to the algorithm that implements
the correct incremental increase by 3, and then _don't_ call it by default.
Everything does work wonderfully if you expand n by a factor of 3 each time.
Press discusses this trick in Numerical Recipes, section 4.4 (p136):
[http://phys.uri.edu/nigh/NumRec/bookfpdf/f4-4.pdf] and notes that the savings
you get still make it more efficient than NOT reusing function evaluations and
implementing a correct scale-by-2 each time.
Happy to provide more information if I can be helpful!
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