[jira] [Created] (MATH-1558) MidpointIntegrator doesn't implement the midpoint method correctly

["Sam Ritchie (Jira)" <ji...@apache.org>]

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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