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Repose
Registered: Oct 2010
Posts: 225 |
Fast large multiplies
I've discovered some interesting optimizations for multiplying large numbers, if the multiply routine time depends on the bits of the mulitplier. Usually if there's a 1 bit in the multiplier, with a standard shift and add routine, there's a "bit" more time or that bit.
The method uses several ways of transforming the input to have less 1 bits. Normally, if every value appears equally, you average half 1 bits. In my case, that becomes the worst case, and there's about a quarter 1 bits. This can speed up any routine, even the one that happens to be in rom, by using pre- and post- processing of results. The improvement is about 20%.
Another speedup is optimizing the same multiplier applied to multiple multiplicands. This saves a little in processing the multiplier bits once. This can save another 15%.
Using the square table method will be faster but use a lot of data and a lot of code.
Would anyone be interested in this?
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Also if z3 is assume in the y-reg when done instead:
sta z3;x1*y1h;31 => tay
all inc z3 => iny
I think that's ok since you'd probably do some lda z3 afterwards anyway, and instead you have it readily available in Y. |
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Repose
Registered: Oct 2010
Posts: 225 |
Thanks :) I screwed up my timing though, the first multiply part is 33 not 31 because of the single SEC, bringing my total to 203 (and VICE was right, the multiply part is 160). Because of this, I think I'll use your optimizations to bring it back down to 201. Therefore it still stands :) |
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Also in a real life situation, f.e. subpixel vectors you'd only keep z3 and z2 as a screen limited 8.8 result. Typically one would do:
Rotated x,y,z in 8.8
Rotated z in 8.8
Reciprocal z = 1/z = 0.16
Perspective:
8.8 * 0.16 = 8.24 (keep only the top 8.8, i.e. pixel and the bresenham initial error) |
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Repose
Registered: Oct 2010
Posts: 225 |
Maybe I should use what I've learned to do 3d rotations and perspective transform? I think A Different Perspective 2017 3d Creators Update is in order :) (I'm one of the original authors).
So I had a plan for this fast multiply, it can lead to a fast division because of multiplying by the inverse of the divisor. I can also do squares and cubes faster than this.
Edit: was thinking multiply is only the beginning. I made it 16% faster than your routine but if I can make such gains throughout the transform stack it will add up.
Also for Andropolis, I was thinking to not use EOR fill but a straight store (in fact that's the insight I had on A Different Perspective), and also to calc frame differences and plot those only. |
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Quote: Maybe I should use what I've learned to do 3d rotations and perspective transform? I think A Different Perspective 2017 3d Creators Update is in order :) (I'm one of the original authors).
So I had a plan for this fast multiply, it can lead to a fast division because of multiplying by the inverse of the divisor. I can also do squares and cubes faster than this.
Edit: was thinking multiply is only the beginning. I made it 16% faster than your routine but if I can make such gains throughout the transform stack it will add up.
Also for Andropolis, I was thinking to not use EOR fill but a straight store (in fact that's the insight I had on A Different Perspective), and also to calc frame differences and plot those only.
I've had similar ideas and I even did an frame difference experiment in Jave. Problem was that since diffs are small so will the triangles be. They'll be extreme, sharp and 'problematic'. Hard to render correctly and since they're diffs any render error will accumulate.
Regarding transforms I came to the conclusion to cut most of the stack and forget about how it works conventionally.
Regarding divs we all do mul by the reciprocal. For Andropolis I did what Graham did and calced the reciprocal by linear interpolation:
X is 8.8 call it a.b:
1/a.b ~= invtab[a]*(1-b) + invtab[a+1]*b
invtab[x] is 0.16 result of 65536/x for x 1..255 |
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Repose
Registered: Oct 2010
Posts: 225 |
You should write an article on how you did that, it sounds interesting. Obviously I'm a noob at this problem and would have a lot of research to do.
I wasn't thinking triangles exactly but just doom like hallways, wouldn't that work for differencing? |
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JackAsser
Registered: Jun 2002
Posts: 2014 |
Quote: You should write an article on how you did that, it sounds interesting. Obviously I'm a noob at this problem and would have a lot of research to do.
I wasn't thinking triangles exactly but just doom like hallways, wouldn't that work for differencing?
Hehe, I suck at writing articles! :)
Imagine the simpliest case: axis aligned rooms, no height differences and just rotate around Y, i.e. Wolfenstein.
The diff formed by the edge between the wall and the floor/ceiling will be an enlogated thin triangle. The diff might be 3-4 pixels high close to the camera, this is easy, but further down the corridor the height of the diff will be <1 but still pixels here and there must be pur. Surely an extended bresenham that "draws" both the top and the bottom line simultaneous could handle it, and calc the total height between the lines and still propagate individual errors. |
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Repose
Registered: Oct 2010
Posts: 225 |
Ok, I think that makes sense now.
Now 196
I managed to save another 7 cycles from the correct total of 203, bringing it down to 159+37=196. As a bonus, the two highest bytes are returned in registers.
If I put do_adds in zp, there's one more optimization to save 4 instead of 3, for 192. Finally, if you don't need the lowest bytes, you can save 3 cycles by deleting sta z1.
lda (p_sqr_hi),y
sbc (p_invsqr_hi),y
tay;x1*y1h;Y=z3, 30 cycles
do_adds:
;-add the first two numbers of column 1
clc
c1a: lda #0
c1b: adc #0
sta z1;9
;-continue to first two numbers of column 2
c2a: lda #0
c2b: adc #0
tax;X=z2, 6 cycles
bcc c1c;3/6 avg 4.5
iny;z3++
clc
;-add last number of column 1
c1c: lda #0
adc z1
sta z1;8
;-add last number of column 2
txa;A=z2
c2c: adc #0
tax;X=z2, 6
bcc fin;3/4 avg 3.5
iny;z3++
;Y=z3, X=z2
;9+6+4.5+8+6+3.5=37
fin: rts
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Oswald
Registered: Apr 2002
Posts: 5094 |
I did not follow closely, could you enlighten me why c1a/b/c is not added up in one go ? |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Damn, so you're now 8.5 cycles faster than me? I was not expecting partial products to be faster than the optimisations we've been working on for runs of adc. Going to have to study this more closely.
Canada holds the record, for now. Nice work! |
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