How Does an Abacus Think?

Reading the beads

A rod shows one digit, and you read it by counting only the beads that touch the bar. The five-bead down means 5; each one-bead up adds 1. So five beads are enough for all ten digits.

Each rod of the abacus is one place: on the right are the ones, then the tens, the hundreds and so on to the left, exactly as you write a number. The rods are place value you can hold, from Grade 6. A rod with no bead touching the bar is a 0, an empty place. A soroban usually has 13 rods or more, so it can hold numbers of 13 digits.

Seeing, not counting

Why 5 and 1, and not ten beads worth 1 each? Try reading 8 as eight beads in a row: you would have to count them. On the soroban, 8 is “the five, and three more”, and three beads can be seen at a glance. Psychologists who study the abacus point out that its beads come in groups of four or five, about as many things as the mind can keep track of at one look. So a rod is read in an instant, and a row of rods almost as fast.

On the soroban you never stop to work out the answer and then set it. You move beads by a few fixed rules, and the answer appears by itself. The first rules are for adding when the one-beads run out.

You saw one: on a rod showing 4, add 3 by adding 5 and taking away 2. That works because +3 = +5 − 2. 3 and 2 make 5 together; we’ll call them friends of five, and in maths they are called complements: two numbers are complements of 5 if they add up to 5.

To add	push the five-bead down and take away	because
1	4 one-beads	+1 = +5 − 4
2	3 one-beads	+2 = +5 − 3
3	2 one-beads	+3 = +5 − 2
4	1 one-bead	+4 = +5 − 1

You only need this rule when there aren’t enough one-beads down to push up. 1 + 3 needs no friend: three of the four one-beads are still free. The widget above counts your moves: 1 + 3 takes one move, 4 + 3 two.

Now the rod itself runs out. A rod showing 8 has its five-bead down and three one-beads up. Add 4, and the answer, 12, needs a second rod. So you use the next bead up: one one-bead on the tens rod is worth 10. Push it up (+10) and take away 6 on the ones rod (−6). Altogether that is +4, because +4 = +10 − 6. 4 and 6 are friends of ten, complements of 10.

To take away 6 from the 8 on the ones rod, you push the five-bead up (−5) and one one-bead down (−1). So the whole of 8 + 4 is three moves: +10 − 5 − 1.

The friends of ten are 1 and 9, 2 and 8, 3 and 7, 4 and 6, and 5 and 5. To add a number when the rod hasn’t room, add 10 on the next rod and take its friend away here. Sometimes the taking away needs a friend of five as well. 7 + 6: push one up on the tens rod (+10), then take 4 away from the 7 on the ones rod. But the 7 has only two one-beads up, so −4 is done as −5 + 1: the five-bead goes up and one more one-bead goes up. The ones rod now shows 3, and the abacus 13.

And when the next rod is full too, the carry runs along. 99 + 1: the ones rod has no room, so +1 = +10 − 9, but the tens rod shows 9 and has no room either, so that +10 becomes +100 − 90. Every rod of 99 is cleared and one bead goes up on the hundreds rod: 100. It is the same carry you do on paper, when 9 + 1 makes 0 and you carry 1.

Place value you can hold

Every rule above comes down to place value. A bead on the tens rod is worth ten beads on the ones rod, because each place is worth ten times the place on its right. So pushing up one bead on the tens rod and taking ten away on the ones rod changes nothing; and adding 4 can be done as adding ten and taking away six. The abacus never “knows” that 8 + 4 is 12. It follows the rules, and the beads end up showing 12.

This is the idea behind the topic’s question. An abacus doesn’t think: you do. What it does is hold every number so that a few small rules, each needing no sums at all, do the adding.

Two counters on every rod

Look again at the ten digits in Figure 39.1. Each digit is made of two parts: the five-bead (down or up, two choices) and the one-beads (0 to 4 of them, five choices).

$$\text{digit} = 5 \times (\text{five-bead: 0 or 1}) + (\text{one-beads: 0 to 4})$$

A code built from a part with two choices and a part with five is called bi-quinary (bi, two, and quinary, five). Every digit has exactly one pattern of beads, and there are 2 × 5 = 10 patterns, just enough.

In Grade 8 you write numbers in base 5, where each place is worth five times the one on its right. Look at a single rod with that in mind: the five-bead is the fives place and the one-beads are the ones place. So each rod writes its digit in base 5! 7 is a five and two ones, and in base 5, 7 is 12₅. 9 is 14₅, five-bead down and four up. Every rod is a tiny base-5 number, and the rods together make a base-10 number.

Why the friends always work

Does one of the rules always fit, whatever the rod shows and whatever you add? Here is the argument, for adding a digit from 1 to 4 to a rod.

• If there are enough one-beads down, push them up. Done.
• If not, most of the one-beads are already up. Say you are adding 3 and fewer than 3 are free: then at least 2 are up. If the five-bead is still up, push it down and take away the friend, 2. At least 2 one-beads are up, so there are always 2 to take away.
• If the five-bead is down as well, the rod shows at least 5 + 2 = 7, and 7 + 3 is at least 10: too big for one rod. Then add 10 on the next rod and take away 7 here, which works because the rod shows at least 7. Adding 10 is adding 1 on the next rod, so the same rules apply there, and if that rod is full too, the carry runs on, as in 99 + 1.

The same thinking works for every digit you add. For 5 to 9 the five-bead goes down with the extra one-beads if they fit; if they don’t, or the five-bead is down already, the rod plus what you add passes 10, and a friend of ten does it. A computer checked the rules bead by bead for every sum of two numbers from 0 to 999: 1,000,000 sums, every one right. That is why abacus addition never needs a table of sums: just the friends of five and of ten.

Five beads is the fewest that can show 0 to 9 with beads worth 1 and 5. With only three one-beads you couldn’t show 4, and with no five-bead you would need nine one-beads to reach 9. But other abacuses chose differently.

The Chinese suanpan has seven beads a rod, two fives and five ones, so a rod can show up to 2 × 5 + 5 = 15. That looks like a waste for digits that stop at 9, but the extra beads were useful. In China’s old weights, 1 jin was 16 liang, so sums of weight needed places worth sixteen times the place on their right, base 16, and a rod that holds 0 to 15 is exactly what base 16 needs. The extra beads also hold a carry for a moment while you work. Japan took the Chinese abacus and dropped the beads digits don’t need: first one five, then one one.

The Russian schoty goes the other way: ten beads on each wire, each worth 1, with no five-bead at all. It needs no friends of five, but more beads to slide and to count, which is why its 5th and 6th beads are colored differently: they make the beads easier to read without counting.

Once they are quick on a real abacus, students practice mental abacus (in Japanese, anzan, “blind calculation”): they picture the soroban and move the beads in their mind. The rules are the same; only the beads are imagined. The hands seem to be part of it, as Anisha’s were: researchers found that keeping trained users’ hands busy upsets their sums more than making them talk.

How much abacus can a mind hold? Try it.

How far did you get? Researchers who tested children trained in mental abacus in India found that even they seem to hold only about three or four rods at a time. Each rod is held in the mind as one picture, a single “thing” with its beads inside it, so the imagined abacus is a row of a few rod-pictures, and long numbers are worked a few rods at a time. The same research found that trained users can calculate while they talk, which badly upsets ordinary mental arithmetic: the imagined abacus seems to be kept as a picture, not as words.

• Where Did Our Zero Come From? The counting board, the empty column and the sign that fills it.
• How Did People Multiply Before Calculators? Doubling, halving, lattices and Napier’s rods.
• How Can You Multiply 98 × 97 in Your Head? Mental tricks for multiplying, and why they work.
• Is the Answer Roughly Right? Estimating, and checking a sum by casting out nines.
• Why Do the ‘Vedic Mathematics’ Tricks Work? (Grades 9–12) More mental tricks as algebra, and where they really come from.
• How Does the Mind-Reading Card Trick Work? (Computing, Going Further) Place value in base 2, as a card trick.
• Which Card Was Flipped? (Computing, Going Further) Codes that catch their own mistakes, like the IBM 650’s digits.
• From the abacus to the phone (Grade 6 Computing) An abacus stores numbers; a computer processes them too.