Commit 6ab285bf authored by Richard Levitte's avatar Richard Levitte
Browse files

I think I got it now. Apparently, the case of having to shift down 

the divisor was a bit more complex than I first saw.  The lost bit
can't just be discarded, as there are cases where it is important.
For example, look at dividing 320000 with 80000 vs. 80001 (all
decimals), the difference is crucial.  The trick here is to check if
that lost bit was 1, and in that case, do the following:

1. subtract the quotient from the remainder
2. as long as the remainder is negative, add the divisor (the whole
   divisor, not the shofted down copy) to it, and decrease the
   quotient by one.

There's probably a nice mathematical proof for this already, but I
won't bother with that, unless someone requests it from me.
parent 1d3159bc

crypto/bn/asm/vms.mar

+27 −7
Original line number Diff line number Diff line
@@ -187,11 +187,8 @@ n=12 ;(AP) n by value (input) 
; view, i.e. when the highest bit is set), dividing the dividend by

; 2 isn't enough, it needs to be divided by 4.  Furthermore, the

; divisor needs to be divided by 2 (unsigned) as well, to avoid more

; problems with the sign.  In this case, the divisor is so large,

; from an unsigned point of view, that the dropped lowest bit is

; insignificant for the operation, and therefore doesn't need

; bothering with.  The remainder might end up incorrect, bit that's

; adjusted at the end of the routine anyway.

; problems with the sign.  In this case, a little extra fiddling with

; the remainder is required.

;

; So, the simplest way to handle this is always to divide the dividend

; by 4, and to divide the divisor by 2 if it's highest bit is set.
@@ -213,10 +210,20 @@ n=12 ;(AP) n by value (input) 
; if (q < 0) q = -q		# I doubt this is necessary any more

;

; r'    = r >> 30

; if (d' > 0) q = q << 1

; if (d' >= 0) q = q << 1

; q     = q << 1

; r     = (r << 2) + l'

;

; if (d' < 0)

;   {

;     [r',r] = [r',r] - q

;     while ([r',r] < 0)

;       {

;         [r',r] = [r',r] + d

;         q = q - 1

;       }

;   }

;

; while ([r',r] >= d)

;   {

;     [r',r] = [r',r] - d
@@ -278,13 +285,26 @@ d=12 ;(AP) d by value (input) 
	bicl3	#^X00000003,r3,r3

	addl	r5,r3		; r = r + l'



	tstl	r7

	bgeq	5$

	bitl	#1,r7

	beql	5$		; if d < 0 && d & 1

	subl	r2,r3		;   [r',r] = [r',r] - q

	sbwc	#0,r6

45$:

	bgeq	5$		;   while r < 0

	decl	r2		;     q = q - 1

	addl	r7,r3		;     [r',r] = [r',r] + d

	adwc	#0,r6

	brb	45$



5$:

	tstl	r6

	bneq	6$

	cmpl	r3,r7

	blssu	42$		; while [r',r] >= d'

6$:

	subl	r7,r3		;   r = r - d

	subl	r7,r3		;   [r',r] = [r',r] - d

	sbwc	#0,r6

	incl	r2		;   q = q + 1

	brb	5$