*** There are three kinds of abaci in recent use that I'm aware of:**
** ~The Japanese "soroban" or "9 count abacus" with
4 + 1 beads per column.**
** ~The Chinese "suanpan" or "15 count abacus" with
5 + 2 beads per column.**
** ~Several nations'/cultures' basic counter abaci
which simply have 10 vertical beads per column.**

*** Here's the standard, 13 column suanpan:**

**But for now, let's think about this abacus without a decimal point,
such that the right hand column represents "ones" --for which there are
five "single counters" below plus two "five counters" above.**

*** When all the beads are as you see them --away from the middle bar,
the count is zero.**

*** We can easily see how this device works for addition: move the
single count beads up until you reach a value of "5", whereupon you can
return those five beads to the lower bar and then move down one of the
5-count beads. When the count reaches "10", park (or "clear") all the "ones
column" beads, move up the 5-count bead, then move a single lower bead
in the "tens column" up to the middle bar --and so on.**

**So okay: we're good to go with addition.**

** ~Notice that the lowest bead of the five bead
sets, and the upper bead of the two bead sets --appear to be redundant.
The Japanese abacus doesn't use them.**

*** However, now consider subtraction. Let's say
you've got three single count beads up and one five count bead down in
the right-hand (first) column. That equals "8". Raise one of the lower
beads up in the second column and you've got a total of "18" --right?**

** ~Subtracting 1, 2, 3, 4, 5, 6, 7, 8 works just
fine, but subtracting 9 requires you to "borrow" 10 from the second column
--by pushing down that single bead, then adding back the subtracted 10
in the "ones column" --but oh dear: we've done run out of beads(!) (and
that goes double for the Japanese abacus). You run into the same problem
when subtracting 7 from 16 or 8 from 17.**

*** Now this sounds kind of stupid, since we all know that 18 minus
9 equals 9 --but that's not the point. The accomplished abacuser (abacusist?)
wants his/her fingers to "know" it equals 9, and since even nimble
fingers aren't real smart, they need those "redundant" beads to help them
"think".**

*** Again: adding numbers up goes well --even with the Japanese abacus,
since your fingers reflexively park a full count of beads in one column
and then push up a bead in the next: a "no brainer".**

*** Back in the 1960s and earlier, a familiar entertainment was to
demonstrate how an experienced person with an abacus could easily trounce
his counterpart who used a mechanical calculator. While experts can speed
through a column of figures with either a Japanese or a Chinese abacus
(and we wouldn't want to kink their styles by changing the per column bead
count), I suggest that the novice stands a better chance of achieving proficiency
by having more "redundant" beads per column.**

*** This next one might be called--**

*** Thinking on it, we need only be able to store an 18 count in each
column^ --not 20, since the next "overage" subtraction after minusing a
9 would be to minus 10, for which we just push down one bead in the next
column to the left. To achieve this numerical economy, we could get by
with only thirteen 5-count red beads, but we'd have to add 39 more 1-count
red beads!**

**Again: the finger-thinking "redundant" beads are colored red. However
--and aside from the increased size and bead count, I think that "reflexively"
keeping track of 8+2 beads would be more prone to error than working the
above 5+3 abacus --so that's the one I made:**

**^* It's not a problem when beads pile up per column, since the operator
routinely "clears" or "rectifies" the bead display --shuttling beads to
the left, to the left, to the left --until all the columns read zero through
9.**

**** As usual, any original work on this page carries no claims of
patent nor copyright.**