Pythagorean Addition
tl;dr: Instead of labouriously computing \(c = \sqrt{a^2 + b^2}\), we can<br>mentally calculate using the alpha-max plus beta-min algorithm, by estimating
\[\hat{c} = \mathrm{max}\left(a, 0.9a + 0.5b \right)\]
and this will be very close to the actual \(c\). This is useful for adding up<br>sources of variance, or figuring out radiuses, or other such things.
Background
The mathematical relationship \(c^2 = a^2 + b^2\) is surprisingly common. It<br>happens among other things in
geometry (Pythagorean theorem);
statistics (sources of variance add up);
physics (the energy–momentum relation).
When it shows up, it’s often because one of the variables is unknown, i.e. we<br>have either
\(? = \sqrt{a^2 + b^2}\) or
\(? = \sqrt{c^2 - b^2}\).
The annoying part is that these are hard to mentally calculate, even when one is<br>good at estimating squares and square roots (e.g. because of previous logarithm<br>practice) because numbers grow large when squared.
Insight
I just had a flash of insight. Maybe the problem is thinking of this as three<br>separate operations (square, add, take the root). What if instead we think of it<br>as one fundamental, composite operation? We could call it \(⊞\) (Unicode name apt:<br>squared plus), and define it as
\[a ⊞ b = \sqrt{a^2 + b^2}\]
and then we could use spaced repetition to train ourselves in evaluating it<br>mentally, much like we would do with multiplication tables and logarithms. Then<br>we’d never have to deal with this annoyance again! Given two sources of<br>variation measured in standard deviations, we would instantly know the total<br>variation – again, in standard deviations. That would be much more intuitive.
The one problem is that we’d also have to learn the inverse operation,
\[c ⊟ b = \sqrt{c^2 - b^2}\]
The ⊞ operation should be fairly easy to learn, because its contour lines form<br>concentric circles radiating out from the origin. The ⊟ operation might be<br>trickier, because its contour lines are a more weirdly shaped conic section.
Prior art
It turns out I’m not the first person to have thought about this. There’s a<br>research paper out of ibm from the early 1980’s where the authors have come up<br>with a method for computers to evaluate ⊞ with a high rate of<br>convergence.11 Replacing Square Roots by Pythagorean Sums; Moler & Morrison;<br>ibm Journal of Research and Development; 1981. The method is very cool.<br>Given a point (x,y), the authors have found a way to nudge that point along the<br>radius of a circle down toward the abscissa, so that when the y-value is<br>sufficiently small, the x-value is equal to the radius. However, iterative<br>algorithms like this aren’t well suited for mental arithmetic.22 I do hear<br>about people who can refine approximations by running a couple of iterations of<br>Newton–Raphson in their heads. I want to be like those people, but I am not.
However, there’s also a great method to do it as a human. To evaluate \(a ⊞ b\),<br>assuming \(a\) is the larger number (if it is not, swap them):
Compute \(\hat{c} = 0.9a + 0.5b\).
If \(\hat{c}\) is smaller than \(a\), set \(\hat{c} = a\).
Done! That’s the result.
This is an estimation and it does come with an error, but the error is at worst<br>3 %, and on average it is 1.5 %. That’s remarkable for such an easy procedure.<br>To be clear, we are only shaving a tenth off of the larger number, and adding<br>back in half of the smaller number, and this is very close to being the square<br>root of the sum of their squares!
The reason this method is called alpha-max plus beta-min is that while we used<br>\(\alpha=0.9\) and \(\beta=0.5\) because that was convenient for mental maths, other<br>parameters exist, and some are slightly more accurate.
Inverting
The great thing about this algorithm is that it’s easy to invert, too. If we<br>only have the total \(c\) and one of the terms \(b\), we can subtract and get either
\[\hat{a} = 2 \left( c - 0.9b \right)\]
or
\[\hat{a} = 1.11 \left( c - 0.5b \right)\]
depending on whether we have the small or the large term.
Example
For a concrete example of how to use this, let’s say we know men are on average<br>12 cm taller than women. The average height of a person with known sex then<br>corresponds to a coin toss that can land either +6 or −6, which gives it a<br>standard deviation of 6 cm. We also know that within the groups of men and women<br>separately, the standard deviation of stature is 7 cm.
Then the total variation of stature, across both men and women, ought to be<br>around
\[0.9 \times 7 + 0.5 \times 6 =\]<br>\[= 6.3 + 3 = 9.3\]
and that would indeed be what we found if we went out and randomly picked people<br>across the globe and measured their height. How cool is that? I did not expect<br>to be able to mentally add sources of variance.