[Coco] Divide by 10 on a Motorola MC6809(E)

[Joel Rees]

The multiply-by-constant as a fixed sequences of adds and shifts is,
 of course, interesting (multiply-and-accumulate).

I've tried to find a similar inverse method for divide-by-constant, but
I think the fixed sequence turns out to be per the dividend, not the
divisor, that is, divide-a-constant-by rather than divide-by-constant.

I think.

I have also looked at division as multiply-by-inverse, but convinced
myself that inverting the divisor ends up being expensive enough
to negate any advantage over just doing the subtract-shift-count
binary long division.

Looking at this again, it occurs to me that, if the divisor is a
constant, it can be pre-computed. So, in the case of divide-by-ten,
the inverse of ten in binary is

.0001100110011001

to sixteen bits.

(bc is useful in calculating such things.)

Giving it a try with an arbitrary dividend of 65000 shows me that
there will be some tricky business with rounding, but it should
almost work:

bc -l<enter>
obase=2<enter>
65000<enter>
1111110111101000
65000/10<enter>
1100101100100.000000000000000000000000000000000000000000000000000000\
0000000000000
1/10<enter>
.0001100110011001100110011001100110011001100110011001100110011001100
ibase=2<enter>
1111110111101000*0001100110011001<enter>
11001011000110110011110101000

Divide-by-ten can also be looked at as divide-by-five, then
divide-by-two, which might be of use in working out the rounding.

1/10 and 1/5 are repeating fractions in binary, of course, and that
could be used to effect in turning the long chain of add-and-shifts
into a loop, at the cost of adding loop overhead.

I'm going to come back to this when i have more time.

-- 
Joel Rees

http://reiisi.blogspot.jp/p/novels-i-am-writing.html