(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!
