Doing square root on the abacus is a lot like doing it on paper. The big difference? It's actually *easier* on the abacus. What I find pretty cool is that I'm a rank beginner at the abacus. I never actually tried to use one before I started writing these posts. But I can do that root *faster* on the abacus than I can on paper.

The one difficult step in the paper square root is guessing the approximate digits; as you get beyond the third or fourth digit, the numbers start getting a bit large, and it can be hard to guess the correct estimate. On the abacus, you can very rapidly do repeated subtraction, so you deliberately guess low, and then add on. You'll see what I mean as we work through an example.

One thing about the square root is you need a bigger abacus. So far, we've used very small ones for the examples here. The more you want to do with an abacus, the bigger you want it to be. A small abacus typically has something like 9 columns; a medium abacus has 13 digits. But for more interesting calculations, the kind of thing that we westerners would have a slide rule or scientific calculator, the abacus equivalent is a *27* digit abacus with a couple of sliding markers for helping keep track of things. You want a nice big abacus for doing things like roots, because you're going to partition it into multiple sections.

To do the square root on the abacus, you need at least three sections: one to hold the number you're taking the root of; one to hold the answer you're building; and one do use as a sort of scratchpad for multiplication.

So, suppose we want to take the square root of 262852 on the abacus, and that we'd like to get it to five places. That means we need 10 digits for the number; 5 for the answer. So we'll use the right-most 10 columns to store "26 28 52 00 00"; the left-most 5 columns to store the answer as we produce it. In between, we'll pick 6 columns to use for our scratchpad. (The "scratch" area, which is used for the intermediate sums like the number in parens in the paper method, never has more than 1 digit more than the current answer.)

Let's look at the paper calculation first. I won't step through it, but just write it out:

5 1 2 . 6 9 /--------------- \/ 26 28 52.00 00 25 ---- (100) 128 101 ------ (1020) 2752 2044 ------ (10240) 70800 61476 ------- 102520 932400 922761 --------- 9639

Doing this on the abacus, we'd start with the number in rightmost ten digits; the answer in the leftmost 5; and we'll use the space in between as scratch. This setup is shown in the diagram below, with the number under the radical sign in the paper method in the *radix area*, marked in blue, and the *result area* marked in green. The space in between, we can use for scratch.

We start off exactly as on paper: we look at the number as digit pairs, and figure out what number is approximately the square root of the first digit pair, in this case 26? Obviously, it's five. So we put "5" in the *leftmost* column of the result area, and subtract 25 from the two working digits in the radix area:

Now we start the iteration. We add two more digits to the working set in the radix area; take the current result *r*, and multiply it by 20; and then try to guess the correct *x* such that *((20×r)+x)×x* is smaller than the

number in the working area. This is where the scratch area can come in handy; we can use it for trying multiples. But basically, we proceed pretty much exactly as we did on paper. Get the digit, add it to the result area, do the multiplication (using the scratch area), subtract it from the working value in the radix area, and go on to the next step.

So, the second digit is a 1: 1×101 ≤ 128; 2×102 > 128. So we subtract 101 from the radix area, and our abacus looks like the following:

Now we add two more digits to the working area; and we want to know what's the largest *x* such that *(1020+x)×x ≤ 2782*? 2×1022=2044≤2782; and 3 will definitely be too large.

We can keep going like this. The big advantage of the abacus for this? Let's skip ahead to the fifth digit. At this point, the abacus looks like this:

We're going to want to know the largest *x* such that *(102520+x)×x≤932400*. You can probably tell very quickly that the answer will be either 8 or 9, but you'd need to think a bit more to figure out which one. On the abacus, you do the multiplication in the scratch area: 8×102528=820224. Subtract that in the radix area, and you have 112176:

That's too large. So, it's going to be 9. But we don't need to re-multiply; we can subtract *another* 102529 (leaving 9647), and then subtract 8 as the correction from the previous multiplication, since the last digit was wrong. So we wind up with 9639 as the remainder. And we've got 512.69 as the square root. 512.69^{2}=262851; pretty good.

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I see this is pretty much the same as the paper method, except you never really have to guess anything. You can do all the subtraction as you go along if you want -- guessing just saves some time.

Does the method generalise to cube, fourth and higher-order roots?

AJS:

There's a cube-root version of this algorithm, but beyond cube roots, the fundamental equations get too complicated, and things fall apart.

Remember that the square root algorithm is based on the fact that (a+b)2=a2 + 2ab + b2. That means that in the iterative step, we're working with the "2ab + b2" part of the equation. For cubes, it's harder, but not *too* much harder; we'd be working with the tail of a3 + 3a2b + 3ab2 + b3. That can be manageable, because you can play some tricks to combine the two middle factors, since they're both multiplied by 3. When you get to the fourth root, the basic equation is just too hard for this method: the coefficients are 1,4,6,4,1, and it's just too much. At that point, you're better off with newtons method.

Richard Feynman once called the extraction of cube roots the "topnotch exercise in abacus-land". (He managed to beat the abacus man, of course, because the number they were challenged to cube-root happened to be 1729.03, just a little over twelve cubed.)

While walking to school one bright fall day about five years ago, I saw a booklet resting on the Commonwealth Avenue sidewalk. When I picked it up, I discovered that it was entitled The Fundamental Operations in Bead Arithmetic. It didn't have an author's name anywhere. I believe it is the "published in Hong Kong without year" booklet whose illustration is shown [http://www.joernluetjens.de/sammlungen/abakus/abakus-en.htm here]. To this day, I don't know where it came from, and I can only consider it a propaganda leaflet from the cast-off gods of mechanical arithmetic.

I must say, though, that these posts make me wish I had asked for an abacus for my birthday. (-:

My wife just bought my son the abacus. I printed out your five posts on the abacus and he is learning the method right now.

The best book on it in english is probably Kojima's Advanced Soroban, which is, sadly, out of print. I hear of it on ebay from time to time.

I *much* prefer Newton's approximation for square roots. It's less complicated and converges quickly. I've suspected that it is less total work, but have never proven this hypothesis.

I got behind and am now just catching up.

And put an abacus on my amazon wishlist, which my husband will see when he goes to place an order for a birthday present. :)

Search out Kato Fukutaro's method for extracting square roots, a much faster and easier method than that which you demonstrate here. After only an hour or so of practice, I could easily compute the square root of any number to as many decimal places as I desired.

I only have 13 rows on my abacus, so I write down the answer on paper but do the calculations on the beads.

i liked it dude