# Traditional Abacus and Bead Arithmetic/Division/Dealing with overflow
## Introduction

Excluding the so-called "special methods", there are two basic ways of arranging general division problems. Not knowing a standard designation for them, we have called them in the chapter: Guide to traditional division:
* Modern division arrangement (MDA), as explained by Kojima,
MDA 25÷5=5 :
| Abacus | Comment |
| ABCDEF | |
| 5 25 | Dividend starting in E |
| 5 5 | After division quotient begins in D |
* Traditional division arrangement (TDA), as used in ancient books like the Jinkoki (塵劫記), or the Panzhu Suanfa (盤珠算法)
TDA 25÷5=5 :
| Abacus | Comment |
| ABCDEF | |
| 5 25 | 25÷5=5 Dividend starting in E |
| 5 5 | After division quotient begins in E |
MDA seems perfect for any division method; not just the modern and traditional ones, but also any of the amazing variety of methods one can imagine after reading a page like: The Definitive Higher Math Guide on Integer Long Division, and just using the beads of a 4+1 (modern) abacus. On the contrary, TDA is problematic with any division method since a collision between divisor and dividend/remainder frequently occurs, that is, both require the simultaneous use of the same column and, as this is not possible in principle, for example, in the case of modern division we would be forced to postpone the entry of the interim quotient digit in the abacus until the corresponding column be cleared by subtraction. As a result, special techniques or abaci are needed to cope with this collision. Even so, TDA has been used for centuries in conjunction with the traditional method of division while MDA seems to have been deprecated until modern times and the adoption of the modern abacus, even though MDA is the first idea that would occur to us if we tried to adapt the old division method used with counting rods (paradoxically MD!) to a single row instead of the usual three. Why? that could remain a mystery forever. However, certain advantages to TDA must be recognized:
* It uses one rod less
* Result does not displace too much to the left as in MDA, which is of interest in the case of chained operations. This and the above points makes TDA more suitable to small rod number abacuses, like the traditional 13-rod suanpan/soroban.
* It saves some finger movements; for instance, in the operation 6231÷93=67 using traditional (chinese) division, one can count 14 finger movements with TDA versus 24 with MDA.
* Hand displacements are shorter.
* It is less prone to errors as less rods are skipped.
Are they enough to justify its historical usage?
Regarding the traditional division (Guī chúfǎ, Kijohou 帰除法) using TDA, the way to avoid the mentioned collision is to accept that the first column of the dividend/remainder, after the application of Chinese division rules, can overflow and temporarily accept a value greater than 9 (up to 18), while providing some mechanism to deal with such an overflow. This is not a problem with a traditional 5+2 or 5+3 abacus; As already explained, the additional upper beads can be used to store values as high as 20 in one column of the abacus. The problem arises when we think that 5+1 type abaci were popular in Japan during the Edo period and it seems that no ancient Japanese text explains how to deal with overflow. This is the question: What can be done on an 5+1 or 4+1 abacus?.
In a post to Soroban and Abacus Group, a member presented two examples of traditional division using an apostrophe (‘) to mark the columns or rods that temporarily received a value higher than 9 (overflow).
Example from Soroban and Abacus Group :
| Abacus | Comment |
| ABC abcdef | |
| 898 888122 | 見八無頭作九八（Div. table）... |
| 898 9'68122 | 九九八十一引（Mul. table）... |
The apostrophe has the hindrance of breaking the vertical alignment of the columns of the abacus in the procedure tables, but let us think of this apostrophe as a typographical representation of a small 1 (¹), a bead that should be pushed, set or activated somewhere, be it on a real or imaginary column. Note that if we could open or insert a new column in the place of the apostrophe (as it is commonly done in any spreadsheet) all our problems would go away by using the new column to receive the bead, but by doing so we would be using MDA. After a short digression, three alternatives will be described below to stay on TDA.

### On geeses and flocks

We will use the classical exercise 998001÷999=999 as an example to illustrate the three mentioned alternatives. This exercise is called in Chinese: The lone geese return (孤雁歸隊 Gūyàn guīduì). If you enter this division on the abacus, for instance:
* Abacus
* ABCDEFGHIJK
* 999  998001
and if you have an almighty imagination, no doubt, you will identify the lone bead set on K with a lone geese that has just left her flock FGH (you can see the place that she occupied in the lower part of column H). To convince her to rejoin her flock you only have to complete the division and obtain 999!

## First way: Brute force

In principle, we could add the small “1” in any unused column, for example the rightmost one; but this could be annoying and inconvenient because both the hand and the attention would have to be jumping from one place to another on the abacus with the risk of ending up working in the wrong column. Here, without any further consideration, we will simply add the small "1" to the column of the just entered interim quotient digit. This may sound strange or brutal (and indeed it is), but if we can keep the value of the interim digit in memory we can operate as usual and any anomaly will disappear from the abacus in a moment. Let's see it with the 998001999=999 example on an 4+1 abacus:
998001÷999 = 999; Brute force method on 4+1 abacus :
| Abacus | Comment |
| ABCDEFGHIJK | |
| 999 998001 | Chinese rule: 9/9->9+9, remember quotient digit 9! |
| 999 1088001 | ("carry run" to the left! Don’t panic!) |
| -81 | -9*9 |
| 999 1007001 | |
| -81 | -9*9 |
| 999 998901 | Chinese rule: 9/9->9+9, remember quotient digit 9! |
| 999 1007901 | ("carry run" to the left! Don’t panic!) |
| -81 | -9*9 |
| 999 999801 | |
| -81 | -9*9 |
| 999 998991 | Chinese rule: 8/9->8+8, remember quotient digit 8! |
| 999 999791 | |
| -72 | -8*9 |
| 999 999071 | |
| -72 | -8*9 |
| 999 998999 | finally, revising up |
| 999 999 | done! |
On a 5+1 abacus, things are easier. We can use the 5th bead to avoid carry runs.
998001÷999 = 999; Brute force method on 4+1 abacus (2nd quotient digit) :
| Abacus | Comment |
| ABCDEFGHIJK | |
| 999 998901 | Chinese rule: 9/9->9+9, remember quotient digit 9! |
| 999 9T7901 | |
| -81 | -9*9 |
| 999 999801 | |
| ... | ...etc. |
As we can see, we can do things this way but it does not seem like a very attractive method as we need memorization and a lot of attention to avoid making mistakes. So one should not attempt this method except as an exercise in concentration.

## Second way: Suspended lower beads

If we use a 5+1, instead of pushing the bead all the way up, effectively adding the small “1” to the interim quotient digit as in the previous case, it seems more reasonable to push it only halfway, leaving a suspended lower bead as illustrated at the top of the image to the right. This suspended bead will represent the overflow while respecting the integrity of the quotient digit.
This seems like a perfect method to deal with the overflow, both in division and multiplication, everything remains under our eyes and nothing has to be memorized. In fact, when using suspended lower beads there is no need for additional upper beads, and the 5+1 abacus becomes as powerful as the 5+2 or 5+3 instruments. This might help explain why the 5+1 abacus was so popular in the past and why the 5th lower bead survived for so long. Note in the bottom half of the figure that, with some complication, this method can also be extended to the 4+1 abacus. From here on, We will use underlined digits to represent the overflow according to the figure, since the underline reminds us of what the suspended bead looks like and they don't mess up abacus procedure tables typed with monospaced fonts as the apostrophe does.

### 5+1 abacus

Let us repeat the above exercise with this technique. The divisor is no longer represented and some more details are also introduced to additionally illustrate how the fifth lower bead may be used in subtraction to somewhat simplify the operation (as usual, T is 10, 1 upper bead + 5 lower beads activated)
On an 5+1 abacus :
| Abacus | Comment |
| ABCDEF | |
| 998001 | |
| 988001 | Chinese rule: 9:9 > 9+9 |
| -8 | Subtract 81 from BC |
| 9T8001 | |
| -1 | |
| 9T7001 | |
| -8 | Subtract 81 from CD |
| 999001 | |
| -1 | |
| 998901 | |
| 997901 | Chinese rule: 9:9 > 9+9 |
| -8 | Subtract 81 from CD |
| 999901 | |
| -1 | |
| 999801 | |
| -8 | Subtract 81 from DE |
| 998T01 | |
| -1 | |
| 998991 | |
| 998791 | Chinese rule: 8:9 > 8+8 |
| -7 | Subtract 72 from DE |
| 998T91 | |
| -2 | |
| 998T71 | |
| -7 | Subtract 72 from EF |
| 9989T1 | |
| -2 | |
| 998999 | Revising up |
| -9 | (from right to left to save a hand displacement) |
| 998990 | |
| -9 | |
| 998900 | |
| -9 | |
| 998000 | |
| +1 | |
| 999000 | Done! |
See also division examples for illustrations of this division on 5+1, 5+2 and 5+3 type abacuses.

### 4+1 abacus

And now on a 4+1 abacus. We need to use the suspended group of four lower beads as a code for 9:
On an 4+1 abacus :
| Abacus | Comment |
| ABCDEF | |
| 998001 | |
| 988001 | Chinese rule: 9:9 > 9+9 |
| -81 | Subtract 81 from BC |
| 987001 | |
| -81 | Subtract 81 from CD |
| 998901 | |
| 997901 | Chinese rule: 9:9 > 9+9 |
| -81 | Subtract 81 from CD |
| 999801 | |
| -81 | Subtract 81 from DE |
| 998991 | |
| 998791 | Chinese rule: 8:9 > 8+8 |
| -72 | Subtract 72 from DE |
| 998071 | |
| -72 | Subtract 72 from EF |
| 998999 | Revising up |
| 999000 | Done! |
If you have tried this, you have probably noticed that the group of four suspended beads behaves the same as the suspended upper bead used on the 5+2 abacus; i.e. with "inverse arithmetic", if you move the suspended bead toward the abacus bean you are subtracting instead of adding!.

## Third Way: Minimal memorization

It has been said above that using suspended lower beads seems a perfect method… but in fact it is somewhat annoying due to its inherent slowness. It is always difficult to suspend a bead, especially the small ones of modern abacus with little free space left on the rods, and this despite the silly trick of pinching the bead with two fingers and then retiring the hand as if taking a flower. It is true that with an 5+1 abacus there is no need of additional upper beads, but no doubt, if you have a lot of multiplications or divisions to do, you will prefer the speed that additional beads provide, since one very seldom need to suspend a bead on the 5+2, and never on the 5+3.
Rather than physically moving/suspending the overflow bead, it is enough to think that the bead has been already suspended on the quotient rod, or pushed on an imaginary rod flying around your abacus, around you..., or simply remember that the “overflow status” has been set to ON and that it needs to be unset back to OFF as soon as possible. This last way is similar to the process of setting flags ON/OFF in old electronic calculators programming. Obviously, moving no bead is faster than moving any bead, so nothing can be faster than this alternative. Nevertheless, we should expect to need some practice to get used to this method and prepare to make some more mistakes due to memorization. However, memorizing a digit, as in the brute force method, is worse than simply memorizing an alert condition as required here.
No need for a new example. The previous ones can be followed under this new view simply by interpreting the underlines as something like OverflowFlag: ON.

## Conclusion

We have seen here three techniques to deal with overflow on 4+1 and 5+1 abacuses that pushes the small “1” up on the interim quotient column:
* All way, effectively adding it as a carry to the quotient
* Only half way, leaving a suspended lower bead
* Nothing at all (but in our minds)
These methods bring us the possibility of using traditional techniques and arrangements on any abacus type by simply adapting the mechanics to the presence/absence of additional beads. This is an advantage if you finally end up convinced by traditional techniques.
It has been mentioned that no ancient Japanese text explains how to deal with overflow with a 5+1 abacus. Most likely the form used was one of the last two methods introduced here. Consider that the second method can be demonstrated to others in just seconds, and that once seen, it is neither forgotten nor requires further explanation; It is so obvious. So there is not much need to write long texts to convey that knowledge.
* ↑ Kojima, Takashi (1954), The Japanese Abacus: its Use and Theory, Tokyo: Charles E. Tuttle Co., Inc., ISBN 978-0-8048-0278-9
* ↑ Yoshida, Mitsuyoshi (吉田光由) (1634). Jinkoki (塵劫記) (in Japanese). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
* ↑ Xú Xīnlǔ (徐心魯) (1993) [1573]. Pánzhū Suànfǎ (盤珠算法) (in Chinese). Zhōngguó kēxué jìshù diǎnjí tōng huì (中國科學技術典籍通彙). {{cite book}}: Unknown parameter |trans_title= ignored (|trans-title= suggested) (help)
* ↑ "The Definitive Higher Math Guide on Integer Long Division (and Its Variants)". Math Vault. Archived from the original on May 14, 2021. Retrieved August 4, 2021.
* ↑ Murakami, Masaaki (2020-06-29). "The 5th lower bead". (Web link). Retrieved on 2021-08-13.