(2025-06-09) Addition addiction
-------------------------------
Well, the summer is here, and it's time to take a rest from all those
electronics (except probably the one I'm typing this post on). To be honest, 
I had always wanted some kind of a purely mechanical adding machine (any 
kind will do but smaller and lighter are preferred) but never got a chance 
to obtain one. In case you didn't know, an adding machine is a type of 
calculator that bases all its operations on maintaining a running total. 
Yeah, that's it. I had watched a lot of videos on this topic (including but 
not limited to all of them from Chris Staecker and Stephen Freeborn, most 
from Jaap Scherphuis and some from Jens Puhle) and am pretty amazed at the 
amount of models or even different types of adding machines and mechanical 
calculators as a whole that the industry and individual inventors had come 
up with throughout last several centuries. Just to recap from memory: 
Schickard's design, Pascaline, Troncet (aka Addiators), Leibniz wheels and 
the original Arithmometer based on them, Odhner's pinwheel design, co-axial 
Pascalines, Comptometers, Webb-type column adders, Adix-type column 
adders... And that surely is not all. This industry was so vast and diverse 
it's hard to even find the full list of all of the different device types, 
let alone brands, yet all those essentially served a single purpose: add 
(and sometimes subtract) decimal numbers. If they had any additional 
features advertised as those for multiplication and division, those features 
just helped automating the same addition and subtraction, nothing more. Some 
of them even went beyond decimal, like the Troncet-based Hex Adder or Odhner 
LuSiD for calculating old British currency. What's even more remarkable is 
that, unlike slide rules or sectors which were purely analogue, adding 
machines of any type were purely digital. That is, they all worked with 
discrete digits and always returned the exact result as opposed to an 
approximation given by a slide rule. You don't count money with a slide rule 
if you want to be accurate down to a cent, you better have an adding 
machine. Or... at the very least...

OK. I'm getting a bit ahead of the train. Let's first talk about the addition
operation itself. It has been known to humankind for many millenia because 
people always needed to count things in their life (even cats and dogs can 
count to four because they sometimes need it too), but what changed the 
situation drastically is when the humankind discovered _positional_ ways to 
represent numbers, which evolved into the numeric systems we know and use 
today, like decimal, hexadecimal, octal, binary etc. How do we know whether 
we need to add 1 or 10? As the zero represents nothing (and the concept of 
zero didn't appear as early as the other digits btw), we can only 
distinguish these two numbers by the _position_ of 1. So, in the second 
case, we still just add 1 but to the digit at the second-to-last position 
instead of the last one. Of course all these things are trivial for a modern 
person, but finally making addition positional had been a major historical 
breakthrough that led to the appearance of the simplest possible calculating 
device called the counting board, which, after streamlining the concept, 
evolved into an abacus. See, the amount of objects (marbles, beans, coins 
etc) in a single counting board tray could be arbitrary and, although the 
rules of "trading" for the higher-position tray existed, nothing enforced 
them which could lead to some confusion and ambiguity. By fixing the maximum 
amount of objects (beads on a rod or marbles in a groove) allowed to exist 
in a single postion, the abacus removed this ambiguity and solidified the 
basics of modern positional addition, especially with the notion of carry 
and a strict rule on when it occurs.

So, how does an adding machine of any type differ from an abacus? Well,
besides the internal representation of numbers and the method of operation 
(sliders, cranks, keys etc), the only real difference is that adding 
machines offered a mechanism for automatic (or sometimes semi-automatic) 
carry. That's what usually made them faster but that's what also broke the 
most often (unless it's a semi-auto carry like in the Troncet-type machines 
that have almost nothing to break) because that usually involved passing the 
torque to the next wheel to the left, and if several carries in a row were 
required (like 9999 + 7)... You get the idea. Chain-based (and there even 
were ribbon-based) machines suffered from multi-carries the most. Some of 
them even printed phrases like "do not playfully add 999999 and 1 because 
this will cause stress to the mechanism" in their instruction manuals. Even 
the machines specifically designed to handle multi-carries well, like 
Comptometers, couldn't handle them properly in the earliest revisions if 
several keys had been pressed at a time (which later, when this issue was 
fixed, became a killer feature of this machine class). So, if it ain't a 
Troncet or a two-three-wheel column adder, the question "how long can this 
machine carry on carrying without being broken?" looked like a real dilemma, 
considering how much non-Troncet machines usually cost in the past. And, of 
course, the more moving parts it had, the more potential points of failure 
were awaiting the user. In an ideal design, the amount of moving parts would 
be less than or equal to the amount of different digits you can enter in 
each position times the amount of positions. Which inevitably brings us back 
to...

...the abacus. The primordial positional adder and subtractor. Of course,
with an abacus, you are the one who does the carries, but other than that, 
it's fundamentally the same. I mean, whatever can be done on a Troncet-type 
machine can be done on an abacus (in fact, one may consider this design an 
attempt to create an ultraportable abacus). Whatever can be done on a 
Pascaline, including co-axial ones, can be done on an abacus. Whatever can 
be done on a Comptometer can be done on an abacus, although usually slower. 
The only real inconvenience is that the abacus can only be used when firmly 
placed on a level horizontal surface, meaning you can't use it on the go or 
something (although, again, no one prevents you from creating one with 
friction-held beads etc). But this limitation, if you are really interested 
in it, can be compensated for with another thing that I'm going to mention 
closer to the end of this post.

## Level 0: Represent numbers ##

Now, what's the optimal amount of beads per rod for us to be able to
represent any decimal digit (from 0 to 9)? Of course, the number theory says 
the _minimum_ amount is 4, but that would require us to isolate the position 
of each individual bead, effectively turning the abacus into binary, which 
is not very mental-friendly. So, the optimal amount needs to be higher than 
that. And it turns out the Japanese figured it out perfectly with the use of 
the bi-quinary system on their native abacus (which is called soroban): 5 
beads per rod with the higher bead moving in an isolated position (this is 
also called the 1/4 configuration) and the "answer bar" aka "reckoning bar" 
being in between the upper and lower beads (so that adding 5 is moving a 
5-bead down but adding 1 is moving a 1-bead up). I guess five is the 
practically feasible minimum of moving parts per digit, anything less than 
that would be too inconvenient to use. The 1/4 configuration imposes a 
specific set of rules for doing addition and subtraction on any individual 
rod, and these rules actually sound more complicated than they really are. 
To really grasp them though, you first need to know the concept of 
complements.

## Level 1: Complement, add and subtract ##

Complements are yet another thing an abacus has in common with adding
machines. We say that n-complement of a digit d is n-d, in other words, the 
sum of a digit and its n-complement always equals to n. For instance, 7 is 
10-complement of 3, and 1 is 5-complement of 4. Adding machines usually use 
9-complements for subtraction (due to the fact that a - b = (10^(|log b| + 
1) - b) + a if we ignore the last carry, and this first term consists of all 
9-complements of b with extra 1 added). The 10-bead and 9-bead abaci only 
use 10-complements. The soroban, on the other hand, uses both 5-complements 
and 10-complements in different situations for both addition and 
subtraction, but again, it's much easier in practice than it sounds. See for 
yourselves. Here are the four rules for single digit addition on the soroban:

1. Are there enough beads to add the digit directly (e.g. 1 + 7)? If so, do
this. If not, go to rule 2.
2. Is the digit you're adding less than or equal to 5 (e.g. 2 + 4)? If so,
subtract its 5-complement in the lower part and add 5 in the upper part 
(toggle the 5-bead, adding 1 to the rod to the left if necessary). Otherwise 
(or if you cannot subtract the 5-complement), go to rule 3.
3. Are there enough beads in the lower part to subtract the 10-complement
(e.g. 3 + 8)? If so, do this and add 1 to the rod to the left. Otherwise, go 
to rule 4.
4. If no previous conditions apply (e.g. 6 + 7), perform a "swipe-up" motion
to subtract the 10-complement: grab the digit by the 5-bead and the 
remaining beads in the lower position (e.g. for 7 grab the 5-bead and the 
two lower beads from the part away from the reckoning bar), and move them 
all up at the same time. Then just add 1 to the rod to the left.

For subtraction, all bead motion and carry direction is reversed but the
principle is fundamentally the same:

1. Are there enough beads to subtract the digit directly (e.g. 7 - 1)? If so,
do this. If not, go to rule 2.
2. Is the digit you're subtracting less than or equal to 5 (e.g. 6 - 2)? If
so, add its 5-complement in the lower part and subtract 5 in the upper part 
(toggle the 5-bead, subtracting 1 from the rod to the left if necessary). 
Otherwise (or if you cannot add the 5-complement), go to rule 3.
3. Are there enough beads in the lower part to add the 10-complement (e.g. 11
- 8)? If so, do this and subtract 1 from the rod to the left. Otherwise, go 
to rule 4.
4. If no previous conditions apply (e.g. 13 - 7), perform a "swipe-down"
motion to add the 10-complement: grab the digit by the 5-bead and the 
remaining beads in the lower position (e.g. for 7 grab the 5-bead and the 
two lower beads from the part near the reckoning bar), and move them all 
down at the same time. Then just subtract 1 from the rod to the left.

That's basically everything you need to remember. Multi-digit
addition/subtraction is just a cascade of single digit 
additions/subtractions. All operations are performed left to right and 
carries are propagated as usual, although it's impossible to carry/borrow 
more than 1 at a time here. With even a little bit of practice, these rules 
will become your second nature and you won't even think about the individual 
rules, just see and apply the correct one in every possible case. Getting 
these basic operations to speed brings a real fun factor and also opens the 
door to more advanced stuff: multiplication, division, square roots, natural 
logarithms and exponents, even trigonometry. I don't think it will surprise 
anyone that most of this advanced stuff is pretty much duplicating the 
techniques used with pen and paper or with adding machines, although my 
personal preference is a mix and match: for instance, I prefer paper-like 
multiplication (with my own simplifications, effectively a Trachtenberg 
method but in a different order) but machine-like division and square roots. 
Regardless of the method, all the following techniques boil down to simple 
additions and subtractions for both the abacus and any kind of adding 
machine.

Just as a fun exercise, try adding 98765432 nine times on a soroban. This
puts you through all possible situations you can encounter in real life 
addition problems. And you'll see why it's called "crazy eights" (if you 
don't see them, then you've done something wrong). I really encourage you to 
find more training problems like this.

## Level 2: Multiply and divide ##

Once you've mastered addition and subtraction, you can go ahead. For
multiplication, I personally prefer treating both operands as integers and 
dealing with the decimal point later. If your soroban is long enough, you 
can cache the multiplier on the right hand side, but that's optional. All 
you need to memorize for it is your usual digit-by-digit times table. The 
algorithm is as follows:

1. Start with the leftmost position on the abacus as the current position.
2. Enter the product of the first digit of the multiplicand and the first
digit of the multiplier at the current position. Treat all products as 
2-digit numbers, e.g. 2 * 3 = 06.
3. Shift one position to the right and add the product of the second digit of
the multiplicand and the first digit of the multiplier.
4. Repeat the process of shifting and adding until all digits of the
multiplicand are multiplied by the first digit of the multiplier.
5. For every remaining digit of the multiplier, increase the starting
position (second, third, fourth etc) and repeat steps 2 to 4 until all 
digits are processed.
6. Adjust the decimal point if necessary (the amount of decimals in the
result is the sum of the decimal amounts in the operands).

Let's see an example, 1337 * 53:

- Enter 1 * 5 = 05 into the leftmost position.
- Add 3 * 5 = 15 into the shifted position, the result register shows 065.
- Add 3 * 5 = 15 into another shifted position, the result shows 0665.
- Add 7 * 5 = 35 into another shifted position, the result shows 06685.
- We've processed the first multiplier digit, now let's process another one.
- Add 1 * 3 = 03 into the second leftmost position, the result shows 06985.
- Add 3 * 3 = 09 into the shifted position, the result shows 07075.
- Add 3 * 3 = 09 into another shifted position, the result shows 07084.
- Add 7 * 3 = 21 into another shifted position, the result shows 070861.
- All digits of the multiplier have been processed, stop the algorithm.

Indeed, 1337 * 53 = 70861.

For division, lots of folks (especially on the soroban) try mimicking what
essentially is the "standard" long division on paper, but I find that 
extremely inconvenient. The approach used with adding machines (especially 
the pinwheel ones) is much more reliable for me. Again, this assumes that 
the decimal point can be figured out later on. The algorithm is as follows:

1. Enter the dividend starting from the third leftmost rod.
2. Start subtracting the divisor from the dividend using the least amount of
digits possible, keeping track of the amount of times you subtract it in the 
leftmost rod.
3. When you cannot subtract the divisor anymore, shift one position to the
right for both the operation register and the tracking rod.
4. Repeat steps 2 and 3 until you get no remainder or the desired accuracy of
the result. The result will be formed in the tracking rods on the left side 
of the soroban.
5. Adjust the decimal point of the result based on the operand size if
necessary.

Let's see an example, 512 / 32:

- Enter 512 starting from the third rod.
- We cannot subtract 32 from 5, let's use the next digit. Can we subtract 32
from 51? Yes.
- Start subtracting 32 from 51, keeping track of the subtraction count in the
first rod.
- We could subtract 32 from 51 one time, with the remainder of 19.
- Shift one position to the right, we have 192. Can we subtract 32 from 192?
Yes.
- Start subtracting 32 from 192, keeping track of the subtraction count in
the second rod.
- We could subtract 32 from 196 six times, with no remainder left.
- Our result can be read in the first tracking rods, 16.

Indeed, 512 / 32 = 16.

Of course, when dividing by a single-digit number, you can use the times
table to shortcut the process of repeated subtraction by estimating the 
closest product less than the remainder, and then directly entering the 
resulting digit into the tracking rod.

As you can see from these algorithms, multiplication and division really are
nothing more than series of shifted additions and subtractions. On the 
fundamental level, computers do the same, just in binary. But we can go even 
deeper.

## Level 3: Find square roots ##

I know that it might sound absurd, but the very first electronic calculator I
got in my life as a child (I only remember the OEM brand, Namhwa, and the 
fact it had a flip cover for the keyboard) didn't even have a square root 
key (there was a stupid beep toggle key instead). When I finally got one 
with this function several years later (again, I only remember that it was a 
Taksun and had a flip cover too), it still seemed pretty arcane to me. After 
all, this is the most common way to obtain irrational numbers, something 
that can't even be expressed as a fraction! I even came up with something 
resembling the Heron's method on my own, just to get square roots on my 
crippled Namhwa. But little did I know at that time that there was a way to 
calculate them even on the granny's abacus we had in the closet.

To be honest, ALL square root methods described on dedicated soroban
resources had left me EXTREMELY confused. I just couldn't follow them and 
left with the frustration of not understanding what it was: whether the 
algorithm descriptions had been missing some crucial details or it was just 
me who couldn't see something directly on the surface. Not just the texts, 
but the videos on this topic are equally confusing. And then I realized I 
had been looking in the wrong place. Just like with division, the real 
answer here lies with how square roots are done with adding machines. With 
some interesting tweaks, of course.

It all started with the Toepler's algorithm. It was specifically designed for
mechanical pinwheel calculators of Odhner and Leibniz systems and 
essentially is based on the fact that the sum of N successive odd numbers 
(1,3,5,7...) is equal to N-squared. The major inconvenience of this 
algorithm is that, after you have done all possible subtractions in a step, 
you have to multiply the intermediate result by 20 and then re-enter it for 
further subtraction. This still is not a problem for pinwheel machines 
because, instead of multiplying the result by 20, you can just add 1 to the 
intermediate input lever position, then shift once and you're good to go, 
however, for any other machine type that doesn't store intermediate input, I 
think this method becomes way too cumbersome. But then came Friden and 
modified this procedure in a way that not only is friendly to their 
electromechanical beasts, but also, all of a sudden, to the abacus.

Enter the "fives method" that turns the odd numbers into the odd multiples of
5 and, most importantly, multiplying by 20 into multiplying by 100, which is 
just a 2-position shift. It is, in my opinion, the most straightforward way 
of obtaining the square root on any non-lever-driven mechanical device that 
can subtract: from a soroban to a Marchant Silent Speed. Of course, the 
decimal point can be dealt with later. The algorithm follows.

1. Multiply the argument by 5 and enter the result starting from the third
leftmost rod, keeping the even number of digits in the integer part (e.g. if 
the result was 858.05, you enter 085805 into the soroban).
2. Split the entered number into _two-digit_ groups. The rods to the left of
the number, starting from the leftmost one, will be your result tracking 
rods.
3. Start subtracting 5, 15, 25, 35 etc from the first group, keeping track of
these subtractions in the first tracking rod (the leftmost rod). The amount 
of subtractions goes to the result R.
4. Whenever you cannot subtract anymore, bring down the next 2-digit group
(along with whatever remained of the previous groups) and start subtracting 
R * 100 + odd multiples of 5 (R*100 + 5, R*100 + 15, R*100 + 25, R*100 + 35 
etc), keeping track on the corresponding result rod. If you cannot do even a 
single subtraction, don't forget to record 0 into the result.
5. Repeat step 4 (using the entire result value R for the next iteration)
until you're out of rods or have no remainder or have reached the desired 
accuracy.
   * If you are out of rods before reaching the desired accuracy, you can
move the intermediate state to the left (by the *even* number of rods) and 
continue further, as long as you have it separated from the result by at 
least one empty rod.
   * If you don't have enough rods to continue with full accuracy, you can
either transfer the result onto paper and start from the left again or 
continue with reduced accuracy by imagining the remaining rods with shifted 
positions.
6. Adjust the decimal point of the result if necessary.

OK, let's see an example: square root of 171.61.

- First, we multiply it by 5 and get 858.05. We have an odd number of digits,
so we need to enter 085805 into the soroban starting from the third 
position. The number forms three two-digit groups.
- The first group is 08. We can only subtract 05 from it and can't subtract
15 anymore. Record 1 in the first result rod. And the remainder (in the 
first group) is 03.
- Bring down the second group and we have 0358. Now, since the first result
digit is 1, we need to subtract 105, 115, 125 etc. In fact, we could only 
subtract these three numbers before running into the remainder of 13. Record 
3 in the second result rod.
- Bring down the third group and we have 1305. We also have 13 in the result,
so the first number to subtract is 1305 as well. Which leaves no remainder, 
that means that the final result rod is only populated with 1 and the 
complete result 131 can be read on the left of the soroban.
- Now, we see that 131 is the square root of 17161 but we also need to adjust
the decimal point. Two places in the square mean one place in the root, so 
the true result is 13.1.

Indeed, sqrt(171.61) = 13.1.

I still stand firmly on my position that this method is much easier than the
Kojima's method, the Kato's method or pretty much any other seen online. It 
only requires a single multiplication in the beginning and then, just like 
division, boils down to repeated subtractions, and the numbers you're 
subtracting are always multiples of 5 which makes the task even easier. 
Again, maybe I'm missing something big, but why hasn't this method gotten 
the popularity it deserves?

## Level 4: Transcend ##

Once you know how to do all four basic arithmetic operations and the square
root, you can calculate any polynomial (converting it into the Horner form 
first, of course) and any transcendental function with a known polynomial or 
other similar approximation. This includes all trigonometric functions, 
exponents, logarithms and so on. Of course, the methods for computing them 
on an abacus are not unique to the abacus or even mechanical adding 
machines, they can be used pretty much on any mechanical or electronic 
device that doesn't provide such functions natively. Taylor series, Minimax, 
Newton's method, Lagrange interpolation... You name it. However, those 
methods can and do differ by the amount of elementary operations (additions 
and subtractions) required to be performed to achieve any sufficient 
accuracy, and yes, this difference is most noticeable on an abacus where 
everything is manual. So, I guess, transcendental algorithms deserve a whole 
separate post in this phlog. Don't worry, I have a lot of interesting stuff 
to tell ya about them. In this section, however, I'd like to draw some 
attention to one thing: sometimes, when you compute those functions, you run 
into negative numbers. How do we even represent them on a soroban, and, more 
importantly, how to continue the chain of operations like nothing happened? 
That's what I'd like to discuss now, because that's where the machine and 
human approaches totally differ.

Well, we could operate with all nines before the number like adding machines
do, that just is inconvenient and error-prone on an abacus. So, if we know 
the answer is going to be negative, what we do instead is set the 1 to the 
rod before the number we're subtracting from and then doing the "normal" 
subtraction. However, if we need to read the negative result right away, we 
read the digits from the beads that are NOT near the reckoning bar (in fact, 
they form the 9-complements), and then add 1. From that point onwards 
though, we can continue additions and subtractions, but whenever we return 
to the positive side, we must subtract the highest bead we had added in the 
beginning. Simple enough. But what if we got a negative intermediate result 
and need to multiply it by something (positive)? That's where things get 
tricky: we still can use the negative number in the complementary form, but 
we need to subtract the positive factor from the leftmost digits of the 
result (and borrow another high 1 if we cannot). The amount of those 
leftmost digits is the same as the amount of significant digits in the 
factor. By the way, we can multiply negative by negative in the 
complementary forms too, just remember to subtract _both_ operands (still in 
the complementary form) from the leftmost result digits afterwards.

For negative-by-positive division, you pre-add the divisor-1 to the leftmost
dividend digits accordingly. What to do with a remainder? Quoting: "If 
there's a remainder in the final answer, convert the positive remainder to 
negative by adding 1 to the quotient and subtracting the divisor from the 
remainder." But I hope this won't be needed as much often, to be honest.

Let's take a look at one of my favorite example problems:

1. Divide 45 by 90. Let's name the result x = 0.5.
2. Square x. Let's name it y = x * x = 0.25.
3. Multiply y by 0.07 and then subtract 0.64.
4. Multiply the result by y and then add 1.57.
5. Finally, multiply the result by x.

So, first we multiply y = 0.25 by 0.07 according to the multiplication
procedure described above. The result is 0.0175. Then, we subtract 0.64 from 
0.0175 and get into the negative zone: the true result is -0.6225 (as we can 
see by the complements + 1) but the buffered complementary result shown on 
the soroban is 0.3775. And that's normal. Now, we multiply this 
complementary result by 0.25 and get 0.094375. However, as we are in the 
negative zone, we also need to subtract the multiplier (0.25) from the 
result's 0.09 (two significant digits), which gives us 0.844375 on the 
soroban. And now, we add 1.57 to this and return to the positive zone, 
getting 2.414375. Now, we need to remember to subtract one bead from the 
highest position once we're in the positive zone again, getting 1.414375. On 
the last step, we multiply this result by the initial x = 0.5, getting our 
final result 0.7071875. 

Now, if you have noticed this value we've got is close to half of the square
root of 2, that's not a coincidence. In fact, it's about 0.000081 more than 
half of the square root of 2. Why? Because the above problem is in fact an 
algorithm to calculate a sine approximation (with the maximum deviation of 
0.000239)! And yes, you can even plug degrees into it, as demonstrated here, 
we calculate sin(45 deg). That's why this is my favorite low-tech 
calculation problem. And that's why understanding negative numbers on a 
soroban helps with computing transcendentals. See how everything suddenly 
started making perfect sense?

## Level 5: Encrypt ##

Now, THIS may sound absolutely bonkers, but an abacus looks like a perfect
helper device for one-time pad encryption. You see, unlike adding machines 
where the carry is absolutely hardwired into the mechanism, the manual 
carrying process on an abacus can be seen as an advantage: when required, 
you can choose NOT to do it. And if you choose NOT to do it while obeying 
the rest of the addition/subtraction rules, what you're left with is pure 
digit-wise modulo 10 arithmetic, exactly the kind used by the good old OTP. 
Of course, I recommend practicing it only when you become absolutely fluent 
in normal addition and subtraction, otherwise it's quite easy to get 
confused. The concept itself, however, is pretty straightforward. You can 
use the classic scheme (add the keystream to encrypt, subtract to decrypt) 
or the Diana-like (I call it XOR-like) symmetric scheme (subtract the 
message FROM the keystream for both encryption and decryption), in which 
case you only need to memorise the modulo 10 subtraction rules. A 
well-trained operator can subtract 98765 from 43210 digit-wise faster than I 
can enter both values into the program. The result is 55555, in case you 
wonder. A **really** well-trained operator though can do the same 
subtraction mentally. Which brings us to...

## Level 6: Visualize ##

This is the endgame. This is why all of it is worth it. This is why millions
of Japanese kids are still being taught soroban to this very day. This is 
what I have mentioned in the beginning: the abacus as a physical object 
definitely does have some limits on its usability, but those limits 
disappear when you always carry it inside your mind. For me personally, I'm 
not even sure that I can ever reach this level, but I promise that I'll do 
my best. For now I know that just being able to fully visualize a working 
13-rod soroban inside your head, let alone do any computations on it, 
requires an enormous amount of practice (this is also why it's better to be 
done at a young age). Effectively, this kind of practice makes you think 
about numbers in a different way: you'll no longer need any kind of an 
adding machine because you'll become an adding machine yourself. And this, 
by the way, requires some constant training. Learning a physical abacus is 
like learning to ride a bike: once you have it in your muscle memory, it's 
almost impossible to forget. Switching to a mental abacus though is a 
continuous effort even when you become proficient at it. Unlike a physical 
soroban that's impossible to break under normal operation, keeping your mind 
sharp takes a lifelong commitment.

Well, if all this got you interested, let me give you some final tips.
Climbing up to any of these levels does take time, so don't expect to 
quickly become proficient in any advanced stuff before your brain is firmly 
holding the fundamentals. Practice addition and subtraction for as long as 
you need to get as fast with an abacus as possible, ideally faster than with 
a Troncet-type machine (Addiator etc), without mistakes or sudden slowdowns, 
and only then move on to multiplication, division and roots. Addition and 
subtraction should become your numeric bicycle that you need to learn to 
ride steadily. Eventually, you'll find that, at least for these two 
operations, there's no point in taking an electronic calculator at all as 
you enter the operands just as quickly but can see the result immediately 
after entering the last one. When this realization clicks in you, it is the 
definite sign that you can move further. Another important tip is: don't 
overthink. In particular, don't try reimagining numbers as anything but what 
you can see directly in front of you. For instance, the number seven is not 
the digit 7, it's a five-bead down and two one-beads up. I myself often 
experience temptation to use mental math shortcuts I know for numbers in 
their traditional representation, but they just will slow you down when 
working with beads and rods. What's entered on the soroban stays on the 
soroban. Only the eight bead movement rules apply there: four for addition, 
four for subtraction. In this regard, you really should act like a machine 
to achieve consistent results: no shortcuts, no exceptions to the rules, no 
deviation from the algos.

In the next post, we'll start getting into some transcendental stuff
specificaly for the abacus and adding machines. Stay tuned!

--- Luxferre ---