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Scenergy #02
31 декабря 1999 |
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Coding - procedure for rapid multiplication.
FAST EXPONENTIAL MULTIPLY
[Baze/3SC]
Some smart man said hundreds years ago: "When there are no
limits, there is no creativity ". Unfortunately we don't live in
his era. He would be probably glad to see how Z80 assembly
coders perform multiply.
When we look at common calculations, we mostly need to multiply
some integer by fractional number in range <0, 1>. Good examples
are sinus, cosinus or perspective. People which are not familiar
with fixed point arithmetics might wonder how to achieve this.
As always ... simply:
1) Pre-multiply fraction by 256 [to make it appear as 8-bit
integer].
2) Multiply our two 8-bit integers.
3) Divide whole 16-bit result by 256 [actually, just take higher
byte].
This article is about doing it FAST. The main idea is hidden in
equation:
a * b = exp (log (a) + log (b)) [where LOG is natural logarithm]
Smart coder will immediately find out that we need just three
table lookups and one addition. So our first attempt would look
like:
B, C - operands
ld h, LOG; set pointer to LOG table
ld l, b
ld a, (hl); get logarithm of the first integer
ld l, c
add a, (hl); add logarithm of the second integer
inc h; move to EXP table
ld l, a
ld a, (hl); get higher byte of the result
Unfortunately, this will not work. To prevent ADD A, (HL) from
overflow we must store logarithms with only 7-bit precision
which would cause our results to be inaccurate. What do I mean
by 7-bit precision? Look. When we are making LOG table, the
biggest entry we can get is:
log (255) = 5.54126 ... => 6
It is a good idea to pre-multiply all entries by 127 / log
(255) and save the logarithmic curve with more precision [so
there will be no numbers 0 - 6 but 0 - 127]. Let's call this
constant S [scale]. Of course, we must slightly modify the
calculation of EXP table:
for X = 0 to 254
EXP [X] = exp (X / S) / 256
Anyway, this is just an example. To make our routine really
work, we must increase scale. After various experiments with
Busy we found out that 10-bits is optimum. Results are accurate
[Many times they are even correctly rounded!] And memory
requirements are still acceptable. Logarithms are no more 8-bit.
They are stored in 512 bytes long table. EXP table has 1024 +
1024 entries and S = 1023 / log (255). So we can move a step
forward and write our second attempt:
ld h, LOG; set pointer to LOG table
ld l, b
ld e, (hl)
inc h
ld d, (hl); store logarithm of B into DE
ld l, c
ld b, (hl)
dec h
ld c, (hl); store logarithm of C into BC
> Ld hl, EXP; set pointer to EXP table
> Add hl, bc
add hl, de
ld a, (hl); get the result
This routine would work perfectly. Anyway, it has one
disatvantage - it can be faster. We can cut off highlighted
instructions by famous table offset trick. Imagine that we
stored EXP table at address 200 * 256. Look what happens if we
increase higher byte of logarithms by 100:
ld h, LOG
ld l, b
ld e, (hl); get lower byte of log (B)
inc h
ld d, (hl); get higher byte of log (B) increased by 100
ld l, c
ld a, (hl); get higher byte of log (C) increased by 100
dec h
ld l, (hl); get lower byte of log (C)
ld h, a; store complete logarithms into HL and DE
add hl, de; 100 + 100 = 200! we have pointer to EXP table
ld a, (hl); get the result
We have just developed complete multiply in 1973 T!
Now let's talk about handling negative numbers. It would be
nice to call the same routine dependless of operand signs. And
it is possible to do it. We will just extend our trick and add
two different table offsets to higher byte of the logarithm. One
for positive numbers [entries 0 - 127] and one for negative
numbers [entries 128 - 255]. Since EXP table is 2048 [8 * 256]
bytes long, difference between these values must be at least
8. Let's take values 100 and 108 for example and make out
possible variations:
[+] [+] = [+] [+] * [-] = [-]
100 + 100 = 200 100 + 108 = 208
[-] [+] = [-] [-] * [-] = [+]
108 + 100 = 208 108 + 108 = 216
So, at address 200 * 256 will begin ordinary EXP table. At
address 208 * 256 will be stored EXP table with negative
results. And once again, there will be ordinary positive EXP
table at address 216 * 256. We will lose 6K of memory instead of
2K, but we don't need to care about signs. In addition, tables
will be more accurate. Because all integers will lie in interval
<-127, 127>, we can use S = 1023 / log (127).
Those of you which are extremely smart can ask: "Hah! And what
if one operand will be zero? Everybody knows that log (0) is
undefined! ". Yes ... but can we give up after such voyage?
Definitely no! We just fill table entries for log (0) with
zeroes [But we must add correct table offsets to higher byte].
The worsest case is: log (0) + log (127). When we add them
together, we will get 1,023 [supposing that logarithms were
scaled] and final result will be:
exp (1023 / S) = 255
Since higher byte of 255 is 0, our problem is solved. If you
decide to use correct rounding you can get value 1. In most
cases nobody will notice this little inaccuracy. However, it
depends on the nature of whole problem. Either we decide to
round [and risk] or to ignore fractional part [and make all
multiplies slightly inaccurate]. It's upon you. I suppose that
nobody will make any additional zero checks.
Finally, I will put there X * sin (Y) routine, which is very
useful in vector rotations. The only difference between
"Classic" multiply and this routine is in additional LOG table,
which actually contains no S * log (abs (Y)), but S * log (abs
(127 * Sin (Y * PI / 128))) entries. If you are wondering about
ABS, you better jump few pages back and think again about
handling signs by table offsets.
ld h, LOG
ld l, b
ld e, (hl)
inc h
ld d, (hl); get log (B)
inc h
ld l, c
ld a, (hl)
inc h
ld h, (hl)
ld l, a; get log (sin (C))
add hl, de
ld a, (hl); and we are finished
You will probably spend lot of time by building tables, but it
is well worth the effort. See ya next time.
-Baze-3SC->
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