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Repose
Registered: Oct 2010
Posts: 225 |
sets of add/sub
I want to do a-b+c-d+e-f. I wonder if you can save some cycles somewhere? I looked at the stats of the actual numbers, no hopes there, any pair can carry/borrow with about 50% chance.
First I'll look at adding 3x 16bit adds and generalizing
Standard version
-----------------
We will keep the running sum in a register
n1,n2,n3 - numbers to add
r, r+1,... - the sum
;assume n1 stored in r
2 clc
3 lda r
3 adc n2
2 tax;n2+n1 save low byte in X
2 lda #0;small optimization first time when you know r+1 is 0.
2 adc #0; basically convert carry to 1
2 tay;16c save high byte in Y
;-- repeat for n>=3 (n is number of numbers to add)
;clc not needed
2 txa
3 adc n3
3 sta r;r+n3=n1+n2+n3 or tax until final add
2 tya
2 adc #0
3 sta r+1;16+15=31 or tay until final add
rts
In general,
Time is 16+(n-2)*13+2=18+(n-2)*13
n time, cycles
2 18 =18
3 18+13 =31
4 18+13+13 =44
Fast Version 1
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2 clc
3 lda n2
3 adc r;n2+n1
2 bcs s1;2+9=11 if taken
3 adc n3
3 sta r
2 lda #0
2 adc #0
3 sta r+1;2+8+13=23 if no carry in n1+n2
rts;or bcc s2 =26
;11c
2 s1 clc;there was a carry in first add, A=n1+n2
3 adc n3
3 sta r
2 lda #1;representing the carry from n1+n2>255
2 adc #0
3 sta r+1;11+15=26 if carry in n1+n2
s2 rts;23/26, average 24.5, saving 6.5 cycles from standard (31c)
realistically always 26 or saving 5 cycles
Conclusions
Adding 3 numbers, I went from 31 to 26 cycles.
The statistics of adding 3 numbers drawn from multiplies
--------------------------------------------------------
In a partial sums routine, only some bytes of the product are added
case carries % total
l l h 2908442244 67.7 4294967296
h h l 2044355014 47.6 4294967296
Conclusions
High bytes of products are always less likely to create a carry when added to other numbers. The differences are so little that it hardly makes any difference.
ps took a while to simulate 4 billion multiplies and gather stats. I'd hate to figure out stats for larger mults, would have to actually think through it instead. But I can tell you they change quite a bit, in some cases there's hardly two carries in a row for example. |
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Repose
Registered: Oct 2010
Posts: 225 |
Btw, that test case isn't valid for the way I'm adding the f(x) values in a row first (f(x)=x^2/256). So I really need to construct better values.
Sure, I can just avoid the buggy addition routine, but there's a big point here - randomized testing is very unlikely to ever show up the bug.
What we need is 100% test coverage of all the branches/carries to avoid any such bugs in the future. For example, getting all those additions in the right order without making a typo, for some 72 of them (considering a 32x32), would be very difficult.
Now comes the interesting problem of which values cause carries in each sub-addition.
This is a very clear case where 100% testing is impossible, randomized testing will fail, and only careful analysis will catch bugs.
So I want to work on this and I believe CJ is as well.
I think it lies in finding factors and doing small sets of trials around that. |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Oh, just tried ff+ff+01. Sums to 01ff as expected |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Quoting Reposegetting all those additions in the right order without making a typo, for some 72 of them (considering a 32x32), would be very difficult.
..exactly why I went for a code generator.
But you're right, getting good test coverage for edge cases is hardÂ… |
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Repose
Registered: Oct 2010
Posts: 225 |
Ok, seems I took a wrong turn somewhere, and that when you went down the optimization path with sbc tab,x you avoided my particular bug.
My point about test coverage is still valid though.
I'll have to grok your post later. |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Quoting ReposeOk, seems I took a wrong turn somewhere, and that when you went down the optimization path with sbc tab,x you avoided my particular bug.
The key is to not set carry before doing the sbc high - that SEC was throwing away information.
Quote:My point about test coverage is still valid though.
Agreed. |
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Repose
Registered: Oct 2010
Posts: 225 |
Ok quick post yes I realized I need to keep the last carry which is why I did adc #0 before the subtract, I think we're on the page. |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
So, for the NxN multiply, as Repose pointed out it's only worthwhile skipping the CLCs for columns with a reasonable number of addends. Coincidentally, all of those columns also require the addition of a non-zero offset, to account for my excess $4000 representation of the g() table.
I end each of those columns with the following sequence:
sbc id+($ff-Kn),x
sta res+n
txa
where Kn is the additional constant for that column (must be at least as high as the maximum value X can reach, lest the indexing could cross the page boundary)
I've just run an exhaustive test in VICE, comparing a more traditionally computed reference value with the optimised code above, and it matches for all 16842752 valid sets of inputs. I've included the test code below, eliding the loop iterator and display routines for brevity.
; test parameters are
; A (00..ff)
; C (00..01)
; X (00..ff)
; K (X..ff) ; routine requires that X+(0xff-K)<= 0xff, ie X<=K
;
; result should be A+C+K+255*X
;-----------------------------
; calculate reference value
;-----------------------------
lda tC
lsr
lda tA
adc tK
sta ref+0
lda tX
adc#0
sta ref+1 ; ref is now A+C+K+256*X
sec
lda ref+0
sbc tX
sta ref+0
lda ref+1
sbc#0
sta ref+1 ; ref is now A+C+K+255*X
;-----------------------------
; prepare registers for test
;-----------------------------
lda tK
eor#$ff
sta rn+1
lda tC
lsr
lda tA
ldx tX
;-----------------------------
; run test code
;-----------------------------
rn sbc id+$3f,x
sta res+0
txa
adc#0
sta res+1
;-----------------------------
; compare result to reference value
;-----------------------------
lda res+0
cmp ref+0
bne fail
lda res+1
cmp ref+1
bne fail
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Repose
Registered: Oct 2010
Posts: 225 |
I got it to work now thanks :)
And good news, I found something even faster now! It always beats the clc version, even for 3 adds. I might be able to beat whatever you have :)
See if you can generate 16x16 your way, that will be my first working example for now. |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
And now proof that the adc:bcc:inx sequence retains sufficient information to calculate the sum of the column, regardless of when carries occur.
If we define T to be equal to the current value of A+C+255*X,
it's possible to prove that after each adc:bcc:inx sequence, T increases by the value added.
; before:
; T=A+C+255*X, so A=T-C-255*X
then we execute
adc #V
bcc *+3
inx
Let C', X' and A' be the values of carry, accumulator and X after running the code sequence.
Now, either the add didn't set C,
in which case
C'=0
X'=X
A'=A+V+C
=(T-C-255*X)+V+C
=T+V-255*X
T'=A'+C'+255*X'
=(T+V-255*X)+0+255*X
=T+V
or, the add did set C, in which case
C'=1
X'=X+1 # as the increment happens
A'=A+V+C-256 # as the add has overflowed into C
=(T-C-255*X)+V+C-256
=T+V-255*X-256
T'=A'+C'+255*X'
=(T+V-255*X-256)+1+255*(X+1)
= T+V-255*X-255+255*X+255
= T+V
Either way, T has increased by the correct amount |
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Repose
Registered: Oct 2010
Posts: 225 |
Nice proof! Couldn't have said it better myself. |
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