base:fast_sqrt

**This is an old revision of the document!**

Imagine you wanna have a square root:

R = sqrt(N)

That obviously gives you:

R^2 = N

Since we wanna establish a convergent algo, we simply use this during calculation:

R(n)^2 <= N

R(0) will be 0. To get closer to N we add a third value D to R as long as that formula is true. Since we work with binary computers, this D will be of the kind 2^x (128, 64, 32 etc).

Ok assuming we wanna have the root from a 16 bit number and since the output of a sqrt of max 65535 (highest 16 bit input) is maximum 255.998 we start with a D of 128 and LSR it down to 1. On each iteration we try (R+D)^2 ⇐ N and if that formula is true, we have R=R+D. If not, R stays unmodified. The resulting algo looks like this:

R = 0 D = 128 while (D >= 1) { temp = R+D if (temp*temp <= N) then R=temp D = D/2 }

As you see this algo is very simple and apart from the temp*temp it does not involve any time consuming operation. The D/2 is ofcourse just a single LSR.

Ok but we are not satisfied. One single integer multiplication is too much for our little 6510 so we have to get rid of it.

Ok now we are not satisfied with the mul but how to get rid of it? First you have to see that (R+D)^2 is a simple binomical formula so you can do this:

(R+D)^2 = (R+D)*(R+D) = R*R + 2*R*D + D*D

Still this doesn't look THAT promising but let's try more formula changing:

R*R + 2*R*D + D*D = R*R + D*(2*R + D)

Ok what do we have now? 2*R is easy since it's just a simple ASL in assembler. A multiplication with D is also easy since D is always of the kind 2^x so D*something is equal to ASL something several times. But damnit, there's still that R*R biatch, let's send her to hell by introducing another variable called M:

M(n) <= N - R(n)^2

And since R(0) = 0:

M(0) = N

As you see M(n) is just the difference from N and “current N during algo” which gives us a new way to decide if D should be added to R or not. The only thing we need now is a way to calculate M(n) without calculating R(n)^2. This is a bit messy but it works with these given formulas:

N(n) <= (R(n)+D(n))^2

Given the easy formula transformation from above we have:

N(n) <= R(n)^2 + D(n)*(2*R(n)+D(n))

Just substract R(n)^2 from both sides and get:

N(n) - R(n)^2 <= D(n)*(2*R(n)+D(n))

Now look a few lines above at the M(n) formula. Do you notice something? Exactly, there we have the front part of this formula too so it is proven that:

M(n) <= D(n)*(2*R(n)+D(n))

Time for a party, that R*R is gone! But how do we calculate those M(n)? Quite simple actually:

M(n+1) = M(n) - D(n)*(2*R(n)+D(n))

So we got a completely new algo:

M = N D = 128 while (D >= 1) { T = D*(2*R+D) if (T <= M) then M=M-T : R = R+D D = D/2 }

Yeeehaw! R*R is gone, and 2*R is easy (ASL) and D*whatever is easy too (ASL multiple times). To make this clear I change the muls into ASL's:

M = N D = 128 for n = 7 to 0 { T = (R+R+D) ASL n if (T <= M) then M=M-T : R = R+D D = D LSR 1 }

Wicked! BUT: We are not satisfied because of n-times ASL :)

Ok I stopped at the point where the multiplication was gone but a ASL with varying shift count appeared. That is no problem for big CPU's like 680×0 or x86, but our poor 6510 doesn't like this so that “ASL n” has to leave.

To do that, we don't investigate the math any further but take a look at how the variables behave during iteration of this algo:

R = 0 M = N D = 128 for n = 7 to 0 { T = (R+R+D) ASL n if (T <= M) then M=M-T : R = R+D D = D LSR 1 }

D, T, M and R now look like this:

D T M R 10000000 0100000000000000 xxxxxxxxxxxxxxxx x0000000 01000000 0x01000000000000 0xxxxxxxxxxxxxxx xx000000 00100000 00xx010000000000 00xxxxxxxxxxxxxx xxx00000 00010000 000xxx0100000000 000xxxxxxxxxxxxx xxxx0000 00001000 0000xxxx01000000 0000xxxxxxxxxxxx xxxxx000 00000100 00000xxxxx010000 00000xxxxxxxxxxx xxxxxx00 00000010 000000xxxxxx0100 000000xxxxxxxxxx xxxxxxx0 00000001 0000000xxxxxxx01 0000000xxxxxxxxx xxxxxxxx

Removing the “ASL n” would completely change the behaviour of T. But also M has to change because otherwise the T ⇐ M compare and M-T math will not work. So if T is not left-shifted, M has to be right-shifted. The result looks like this:

D T M R 10000000 0100000000000000 xxxxxxxxxxxxxxxx x0000000 01000000 x010000000000000 xxxxxxxxxxxxxxx0 xx000000 00100000 xx01000000000000 xxxxxxxxxxxxxx00 xxx00000 00010000 xxx0100000000000 xxxxxxxxxxxxx000 xxxx0000 00001000 xxxx010000000000 xxxxxxxxxxxx0000 xxxxx000 00000100 xxxxx01000000000 xxxxxxxxxxx00000 xxxxxx00 00000010 xxxxxx0100000000 xxxxxxxxxx000000 xxxxxxx0 00000001 xxxxxxx010000000 xxxxxxxxx0000000 xxxxxxxx

So now instead of shifting T we shift M, but the advantage of shifting M is that it's only 1 bit per iteration. Look at the code now:

R = 0 M = N D = 128 for n = 7 to 0 { T = (R+R+D) ASL 7 if (T <= M) then M=M-T : R = R+D M = M ASL 1 D = D LSR 1 }

Much much better since ASL 7 can be replaced by a single LSR in asm code later. This code is quite useful on 6510 already, BUT ofcourse this is not everything.

Taking a slightly modified version of the last pseudo code:

R = 0 M = N D = 128 for n = 7 to 0 { T = (R ASL 8) OR (D ASL 7) if (T <= M) then M=M-T : R = R OR D M = M ASL 1 D = D LSR 1 }

T can now be calculated quite easily. The high byte is simply R OR <some stuff> and the low byte is always 0 except for the last iteration where it is 128. The shifting of D is done via a table and voila, here is some 6510 code:

LDY #$00 ; R = 0 LDX #$07 .loop TYA ORA stab-1,X STA THI ; (R ASL 8) | (D ASL 7) LDA MHI CMP THI BCC .skip1 ; T <= M SBC THI STA MHI ; M = M - T TYA ORA stab,x TAY ; R = R OR D .skip1 ASL MLO ROL MHI ; M = M ASL 1 DEX BNE .loop ; last iteration STY THI LDA MLO CMP #$80 LDA MHI SBC THI BCC .skip2 INY ; R = R OR D (D is 1 here) .skip2 RTS stab: .BYTE $01,$02,$04,$08,$10,$20,$40,$80

I hope I didn't make any mistakes in that routine :)

Input is MLO/MHI for N and output is Y-register for int(sqrt(N)).

Ok guys, I finally tested the routine I did up there. Like said before, it only allows 14 bit input ($0000 to $41FF to be more accurate). The reason for this is that M needs one more bit. Ok, some people might want full 16 bit so here is a fixed routine which only has 3 opcodes more:

LDY #$00 ; R = 0 LDX #$07 CLC ; clear bit 16 of M .loop TYA ORA stab-1,X STA THI ; (R ASL 8) | (D ASL 7) LDA MHI BCS .skip0 ; M >= 65536? then T <= M is always true CMP THI BCC .skip1 ; T <= M .skip0 SBC THI STA MHI ; M = M - T TYA ORA stab,x TAY ; R = R OR D .skip1 ASL MLO ROL MHI ; M = M ASL 1 DEX BNE .loop ; last iteration BCS .skip2 STY THI LDA MLO CMP #$80 LDA MHI SBC THI BCC .skip3 .skip2 INY ; R = R OR D (D is 1 here) .skip3 RTS stab: .BYTE $01,$02,$04,$08,$10,$20,$40,$80

This routine works perfectly for all values from $0000 to $FFFF.

Here's an extended incarnation of the integer sqrt routine. This one is extended so it does proper rounding. It is achieved by adding another iteration and calculate bit -1 which then can be used to determine if the fractional part is >= .5 or not.

The disadvantage of rounding is that the output has to be 9 bits now, because inputs $FF01 to $FFFF result in 256 now (and not 255). Alternatively you also might modify this routine and simply use bit -1 for further calculations. This gives you more accuracy than rounding.

LDY #$00 ; R = 0 LDX #$07 CLC ; clear bit 16 of M .loop TYA ORA stab-1,X STA THI ; (R ASL 8) | (D ASL 7) LDA MHI BCS .skip0 ; M >= 65536? then T <= M is always true CMP THI BCC .skip1 ; T <= M .skip0 SBC THI STA MHI ; M = M - T TYA ORA stab,x TAY ; R = R OR D .skip1 ASL MLO ROL MHI ; M = M ASL 1 DEX BNE .loop ; bit 0 iteration STY THI LDX MLO LDA MHI BCS .skip2 CPX #$80 SBC THI BCC .skip3 .skip2 TXA SBC #$80 TAX LDA MHI SBC THI STA MHI INY ; R = R OR D (D is 1 here) .skip3 CPX #$80 ROL MHI ; bit -1 iteration and rounding ( + 0.5) LDX #$00 BCS .skip4 CPY MHI BCS .skip5 .skip4 INY ; R = R + 0.5 BNE .skip5 INX .skip5 ; R in X and Y (Y is low-byte) RTS stab: .BYTE $01,$02,$04,$08,$10,$20,$40,$80

If you don't wanna have the rounding and prefer having bit -1, then use this final iteration:

; bit -1 iteration BCS .skip4 LDX #$00 CPY MHI BCS .skip5 .skip4 LDX #$80 .skip5 ; R in Y, X contains bit -1 RTS

I'm still working on it, anyway this is based on one of the partial algorithms, and computes Sqrt of a 32bit number. Put the number in _square and get the result in _sqrt and _remainder.

_square = $5B ; 4 bytes: $5b-$5e; input number _sqrt = $5F ; 2 bytes: $5f-$60; result of the computation _remainder = _M+2 ; 2 bytes: $5d-$5e; is in fact the high bytes of _M _T = $57 ; 4 bytes: $57-$5a _M = $5B ; 4 bytes: $5b-$5e, same as the input _square _D = $61 ; 2 bytes: $61-$62 sqrt32 ldx #0 stx _sqrt ;R stx _sqrt+1 stx _D ;stx _T ; Tlo is always 0 ldx #$80 stx _D+1 ldy #16 _loopsq lda _sqrt asl sta _T+2 lda _sqrt+1 rol sta _T+3 clc lda _D adc _T+2 sta _T+2 lda _D+1 adc _T+3 lsr sta _T+3 lda _T+2 ror sta _T+2 lda #0 ; sta _T ; Tlo is always 0 ror sta _T+1 LDA _M+3 CMP _T+3 BCC skip1 ; T <= M branch if T>M bne skip0 lda _M+2 cmp _T+2 bcc skip1 bne skip0 lda _M+1 cmp _T+1 bcc skip1 ; bne skip0 ;_Tlo is always 0 ; lda _reM ; cmp _T ; bcc skip1 skip0 sec ; lda _M ; Tlo is always 0 ; sbc _T ; sta _M lda _M+1 ; M=M-T sbc _T+1 sta _M+1 lda _M+2 sbc _T+2 sta _M+2 lda _M+3 sbc _T+3 sta _M+3 clc ; possibly unnecessary lda _sqrt ; R=R+D adc _D sta _sqrt lda _sqrt+1 adc _D+1 sta _sqrt+1 skip1 asl _M rol _M+1 rol _M+2 rol _M+3 lsr _D+1 ror _D dey beq _endsqrt jmp _loopsq _endsqrt rts

base/fast_sqrt.1563921807.txt.gz · Last modified: 2019-07-24 00:43 by verz