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Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

Division by multiplication  revisited
Back on the ASM Community message board, when there was a discussion about performing division by invariant divisor with a MUL instead of DIV, I looked to create an explanation of why this works through the theory of 2adic numbers. But recently, when I played with these ideas again, I realized that it is possible to analyze it just on the basis of pure integer arithmetic. I also made a new (at least to me) variant of the algorithm, which works correctly for every possible input and is easily provable.
This is how the algorithm looks implemented in fasm/fasmg scripting language:
Code: 
; Preparation stage:
BITNESS = 32 ; number of bits in word
DIVISOR = 3 ; a number to divide by
; convert DIVISOR into MAGIC and SHIFT constants:
SHIFT = bsr (DIVISOR1) + 1
MAGIC = ((1 shl SHIFT  DIVISOR) shl BITNESS) / DIVISOR + 1
; Division algorithm
NUMBER = 7FFFFFFFh ; an example number to divide
PRODUCT = NUMBER * MAGIC + NUMBER shl BITNESS
QUOTIENT = PRODUCT shr (SHIFT+BITNESS)
REMAINDER = ( (PRODUCT and (1 shl (SHIFT+BITNESS)  1)) * DIVISOR ) shr (SHIFT+BITNESS)


You need fasmg to get this running for larger input numbers, since fasm 1 overflows easily. It also shows how to use the low bits of the product to produce the remainder from the division, though this is not used the following example.
The implementation in assembly language is a bit more complex. Read further below to learn why it has additional quirks compared to the classic ones:
Code: 
; Preparation stage:
mov rbx,[divisor]
mov rcx,rbx
dec rcx
bsr rcx,rcx
inc cl
mov rdx,1
shl rdx,cl
sub rdx,rbx
xor rax,rax
div rbx
inc rax
mov [magic],rax
dec cl
mov [shift_m1],cl


Code: 
; Division:
mov rbx,[number]
mov rax,[magic]
mov cl,[shift_m1]
mul rbx
add rdx,rbx
rcr rdx,1
shr rdx,cl
mov [quotient],rdx


Note that it stores "shift minus one" value, because one of the needed bit shifts is made through RCR in order to not loose the carry bit when dividing large numbers. The 32bit version would be exactly the same, just replace all the 64bit register names with 32bit ones.
_____________________________________
And now the explanation. To analyze the problem purely from the point of view of the integer arithmetic, we can formulate it as follows.
For a given invariant divisor D < 2^B (where B is the number of bits in word) we want to find a bit shift S and a "magic" number M such that:
M*D = 2^(B+S) + R
where 0 <= R <= 2^S
If we succeed, then to get the quotient Q = N / D, we can do the following calculation:
N*M = Q*2^(B+S) + L
with 0 <= L < 2^(B+S)
(this shows that we can drop the low B+S bits and take the high ones to get the Q)
Let's take a look at the N*M*D product and substitute M*D from the definition of M:
N*M*D = N*2^(B+S) + N*R
Because N < 2^B and R <= 2^S, we have N*R < 2^(B+S), so the high bits of N*M*D contain original number N.
If we look at the components of N*M, we can evaluate the same product in a different way:
N*M*D = (Q*2^(B+S) + L)*D = Q*D*2^(B+S) + L*D
now, as L < 2^(B+S), L*D < D*2^(B+S), therefore high bits of L*D contain a number smaller than D and we can represent it as:
L*D = X*2^(B+S) + Y
where X < D and Y < 2^(B+S)
Now:
N*M*D = Q*D*2^(B+S) + X*2^(B+S) + Y
therefore high bits of N*M*D contain just Q*D + X and we already know that they also equal the original number N, so:
N = Q*D + X
and because X is positive and smaller than D, this proves that the quotient obtained this way is correct.
Moreover, this demonstrates that high bits of L*D contain the remainder.
_____________________________________
Now let's look at a classic example, this time for B=32:
For D = 3 we can take M = 0AAAAAAABh and S = 1, then
M*D = 200000001h = 2^(1+32) + 1
which fulfills the requirements, because 1 < 2 = 2^S. Therefore, to divide a number by 3, we can simply do this:
Code: 
mov eax,[number]
mov edx,0AAAAAAABh
mul edx
shr edx,1
mov [quotient],edx


And a couple more instructions could be added to recover the remainder from the low bits of the product, though this is more of a curiosity, since computation of remainder when the quotient is already present is a simple task anyway:
Code: 
mov ecx,3
mov ebx,0
cmovc ebx,ecx
mul ecx
add edx,ebx
shr edx,1
mov [remainder],edx


Note that the highest bit of the 33 low bits of the product landed in CF, and it is multiplied by 3 manually and then added the high bits of the second product.
_____________________________________
For some divisors finding M and S such that R < 2^S may be troublesome.
We can try the most straightforward route and just take smallest S such that D <= 2^S, then divide 2^(S+B) by D and increase it by one to cover the deficit caused by the truncation inherent in integer division:
M = 2^(S+B) / D + 1
then 2^(S+B) < M*D <= 2^(S+B) + D, so the requirements are fulfilled, the only problem is that M is going to be larger than 2^B, so it no longer fits in a machine word.
But, because we take S such that 2^S is the smallest power of two that is not smaller than D:
2^S / D = 1
and therefore M contains only a single bit set above the machine word. We can take M' = M  2^B and get the "magic" number that fits in the machine word. But when multiplying by it, we have to remember that we need to actually multiply by M' + 2^B, which is the same as multiplying by M' and then adding N to the high bits.
This is how the algorithm I presented in the beginning works. The multiplier it computes is:
(2^(S+B) / D + 1)  2^B = ((2^S  D) * 2^B) / D + 1
a then the product is adjusted by adding N*2^B before producing the quotient. This may produce carry, therefore the first of the right shifts that follows has to be RCL, to get the overflowing bit back into position.
_____________________________________
If we prefer to stick to the classic algorithm with M fitting in the machine word, it may not always be possible to find the right combination of M and S as defined above. But we can also search for a pair such that:
M*D = 2^(B+S)  R
where R <= 2^S, as previously.
What this modifies in the evaluations is that:
N*M*D = N*2^(B+S)  N*R
and since N*R again is smaller than 2^(B+S), the high bits of N*M*D this time are N1.
So when we multiply by a number of this form, we actually compute (N1) / D instead of N / D. The classic algorithm deals with this by simply increasing the number by one before multiplication.
For example, for D = 7 we can take M = 92492492h and S = 2. Then:
M*D = 3FFFFFFFEh = 2^(2+32)  2
and 2 < 4 = 2^S. Therefore we can divide by 7 this way:
Code: 
mov eax,[number]
mov edx,92492492h
inc eax
mul edx
shr edx,2
mov [quotient],edx


A tiny weak spot of this routine is that it fails for input 0FFFFFFFFh. This is the reason why I decided that the more complex algorithm I presented first is also worth sharing, as it should work correctly for any input and any divisor (if there is no mistake in my proof).

06 Sep 2017, 21:31 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

Now instead of taking the smallest power of two that is greater or equal to D, we can take the greatest power of two not greater than D and use it as S. If we now divide 2^(S+B) by D, we get:
2^(S+B) = M * D + P, P < D
Let's set R = D  P, then
M * D = 2^(S+B)  R
If R <= 2^S, we can use the algorithm (but we need INC instruction before MUL).
If R > 2^S, then P < 2^S, because we chose S such that D < 2^(S+1). Then we can take M' = M + 1 and use the algorithm without INC, because:
M' * D = 2^(S+B)  R + D = 2^(S+B) + P
This bring us to the following implementation of the classic algorithm:
Code: 
; Preparation stage:
mov ebx,[divisor]
bsr ecx,ebx
mov [shift],cl
mov edx,1
shl edx,cl
xor eax,eax
div ebx
shr edx,cl
setc cl
mov [increment],ecx
xor cl,1
add eax,ecx
mov [magic],eax


Code: 
; Division:
mov eax,[number]
mov edx,[magic]
add eax,[increment]
mul edx
mov cl,[shift]
shr edx,cl
mov [quotient],edx


The [integer] variable has value 0 or 1, it serves as a conditional INC instruction.
Again, no floating point calculations needed, the entire reasoning is solid just in terms of integer relations.
This algorithm does not work for divisors being powers of two, since M does not fit into machine word then. These divisors would need to be detected and division turned into a simple shift instead.
_____________________________________
Note that both the algorithms given use the large values of S even if there may exist a smaller S for which the requirements would be fulfilled. An example for B=32 is number 641, where the algorithm works for S = 0 and M = 663D81h:
663D81h*641 = 2^32 + 1
R = 1 <= 2^S
So 32bit divison by 641 is a rare case when we can do without the shift instruction:
Code: 
mov eax,[number]
mov edx,663D81h
mul edx
mov [quotient],edx



07 Sep 2017, 11:12 

revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 15238
Location: 1I/ʻOumuamua

Re: Division by multiplication  revisited
Tomasz Grysztar wrote: 
Therefore we can divide by 7 this way:
Code: 
mov eax,[number]
mov edx,92492492h
inc eax
mul edx
shr edx,2
mov [quotient],edx


A tiny weak spot of this routine is that it fails for input 0FFFFFFFFh. This is the reason why I decided that the more complex algorithm I presented first is also worth sharing, as it should work correctly for any input and any divisor (if there is no mistake in my proof).


If branches are cheap then we can add a single jz:
Code: 
mov eax,[number]
mov edx,92492492h
inc eax
jz @f
mul edx
@@: shr edx,2
mov [quotient],edx


I use this method with ARM code with a conditional mul instruction. Although only some divisors need this, so it is somewhat wasteful to make it generic. Also if you know your numerator doesn't need it then leave out the jz.
The most common divisor 10 where M=CCC...CCD and S=3 can use the normal N*M >> S approach.

07 Sep 2017, 11:28 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

Re: Division by multiplication  revisited
revolution wrote: 
If branches are cheap then we can add a single jz


You are right, this correctly handles the overflow (just like RCL in my first algorithm). It only gives the correct upper half of the product, though  so if you wanted to recover remainder, you'd need to do it some other way.
With this correction in mind, the only remaining advantage of the first algorithm is that it does not require any assumptions concerning divisors. But since the problematic divisors are the powers of two, we'd be better off handling them separately anyway.

07 Sep 2017, 12:01 

revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 15238
Location: 1I/ʻOumuamua

Re: Division by multiplication  revisited
Tomasz Grysztar wrote: 
It only gives the correct upper half of the product, though ...


eax will be zero, so I thought it is the required full 64bit result of (N+1)*M?

07 Sep 2017, 12:09 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

Re: Division by multiplication  revisited
revolution wrote: 
Tomasz Grysztar wrote: 
It only gives the correct upper half of the product, though ...


eax will be zero, so I thought it is the required full 64bit result of (N+1)*M?


Ah, you got me there. It may not even be a coincidence that all the values end up being correct.

07 Sep 2017, 12:21 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

revolution wrote: 
Also if you know your numerator doesn't need it then leave out the jz.


If we consider additional assumptions about the numerator, then there is one more interesting thing that can be deduced from my proof.
The requirement that R <= 2^S is only needed to show that N*R < 2^(S+B). So if we can get the same condition through the assumptions on N, it may enable new choices of S, even S = 0.
So if we know that N is always small enough that N*D <= 2^B, we can take S = 0, M = 2^B / D + 1 (except when D is a power of 2, when we should drop "+ 1"). Then multiplying by M gives correct N/D in the high bits for all N < M.
For example for D = 10, if we can safely assume that N is smaller than M = 2^32 / 10 + 1 = 1999999Ah, we can just multiply by it and take the high 32 bits as a result.

07 Sep 2017, 12:21 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

What about dividing numbers larger than the machine word? With DIV instruction it is possible to chain multiple divisions together, the remainder from the previous division is already in EDX/RDX ready to become the high word (smaller than the divisor, so overflow never happens) of the subsequent division in chain. With such chaining it is possible to divide arbitrarily large number by a small divisor.
The high word (EDX or RDX) must be smaller than the divisor, otherwise an overflow would occur (with DIV instruction it would be an exception 0). If we take V = bsr D, then:
N < 2^(B+V+1), because high word of N is smaller than the divisor
R < D < 2^(V+1)
N*R < 2^(B+2*V+2)
So we can take S = 2*V+2. Obviously M = 2^(B+S) / D + 1 is going to be larger than machine word, so the the multiplication is no longer going to be simple.
For D = 10, we get S = 8 and M = 2^40 / 10 + 1 = 199999999Ah. This led me to the following 32bit routine that can replace "DIV 10" (without checking for overflow, input EDX must be smaller than 10):
Code: 
div10:
; this trashes EBX and ECX, you may need to preserve them
push eax edx
imul edx,19h
mov ecx,edx
mov edx,09999999Ah
mov ebx,eax
mul edx
mov eax,ebx
mov ebx,edx
mov edx,19h
mul edx
add ebx,eax
adc ecx,edx
pop eax
mov edx,09999999Ah
mul edx
add eax,ebx
adc ecx,edx
shrd eax,ecx,8
pop edx
imul ecx,eax,10
sub edx,ecx
retn


This is a long function with five multiplication instructions, but when I took a routine for converting very large integers to decimal and replaced the DIV instruction with a call to "div10", it still gave a noticeable improvement on my machine. If anything, this demonstrates how expensive DIV can sometimes be.

08 Sep 2017, 15:05 

tthsqe
Joined: 20 May 2009
Posts: 701

Since this is a thread on division, I though I would collect my thoughts on multiprecision division. Throughout this post, let be the word size of the machine  so usually , but any integer will do. Given wordsized positive integers , the x86 architecture has an instruction div for calculating
An exception is raised if this quotient is , i.e. not word sized.
The goal of division is to calculate for multiword integers and . When is wordsized, this can be calculated by stitching together div's, so the problem is to solve the case when is bigger than a word. In the grade school algorithm, the words of the quotient are produced from most significant to least significant. This means that the grade school algorithm is only dependent upon a solution to the following problem:
Given big integers and with and , calculate the wordsized integer .
Now, by shifting and , this problem can also be represented as calculating
where and are integers with and , and and are real numbers with .
Fact: With the above assumptions, it holds that
.
This means that the difficult (n+1)byn word division can be reduced to the simpler 3by2 division . If the latter does not match the former, is must be over by only one. In the grade school algorithm, this corresponds to overestimating a quotient term, and thus having to 'fix it up'. The point is that if the quotient term is estimate incorrectly, it is overestimated and only overestimated by one.
Proof: Write where and . Then,
It is easy to see that
and some manipulation of the assumption shows that
This last quantity is greater than by the assumptions on .
So now the grade school division has been reduced to the division problem
with the assumptions that the and are wordsized and that .
This will be covered in a future post.

09 Sep 2017, 09:48 

tthsqe
Joined: 20 May 2009
Posts: 701

In the previous post, large integer division was reduced to a sequence of 3by2 divisions followed by nby1 multiplysubtract operations, in true grade school fashion. The quotient of the 3by2 division was between and inclusive. Since the quotient terms for the grade school division are by definition wordsized, we need to calculate
where it is assumed that this floor is and of course all of the previous assumptions, including .
If you follow the proof given in the previous post for this problem you will see that
However, care is required as the division can overflow a word.
I propose that the following algorithm computes .
Code: 
if B2 < A1
(q, r) = div(B2:B1, A1)
(lo, hi) = mul(q, A0)
T0 = sub(B0, lo)
T1 = sbb(r, hi)
T2 = sbb(0, 0) ; copy carry flag across T2
if T2 <> 0
q = dec(q)
T0 = add(T0, A0)
T1 = adc(T1, A1)
T2 = adc(T2, 0)
if T2 <> 0
q = dec(q)
T0 = add(T0, A0) ; only calculated
T1 = adc(T1, A1) ; for the asserts
T2 = adc(T2, 0) ; below
end if
end if
assert( T2 == 0) ; T2:T1:T0 is supposed
assert( T1 * X + T0 < A2 * X + A1) ; to contain the remainder
return q
else
q = X  1
T0 = B0
T1 = sub(B1, A0)
T2 = sbb(B2, A1)
T3 = sbb(0, 0)
T0 = add(T0, A0)
T1 = adc(T1, A1)
T2 = adc(T2, 0)
T3 = adc(T3, 0)
if T3 <> 0
q = dec(q)
end if
return q
end if


I do not know at this time why at most one fixup is required in the "else" block. Of course two fixups are required in general by the above inequality.
Also, notice that this algorithm contains a call to div with a fixed divisor for each quotient word produced in the whole grade school algorithm. Thomasz would probably want to change this invariant division into some multiplications.

09 Sep 2017, 11:07 

revolution
When all else fails, read the source
Joined: 24 Aug 2004
Posts: 15238
Location: 1I/ʻOumuamua

When I was doing this stuff a few years ago I found that Barrett and Montgomery were the most used two methods for division and modulo for multiprecision operations. There are also whole books dedicated to this subject, and many PhD theses that attempt to improve the efficiency and performance in specific CPUs.

09 Sep 2017, 11:59 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

tthsqe wrote: 
Since this is a thread on division, I though I would collect my thoughts on multiprecision division.


I wanted to continue my thoughts on "magic multipliers" but I see that you have moved the topic in a slightly different direction already. Perhaps we should split the thread?
Also, what I wanted to focus on here is the analysis of these tricks purely on the grounds of integer arithmetic, with no event slightest mention of real (or even padic) numbers. This is my personal hobby.

09 Sep 2017, 13:28 

tthsqe
Joined: 20 May 2009
Posts: 701

I don't think it needs to be split. I am just giving you more to think about.
What you have been doing is the division of singe words (like the c "/" operator). You can interpret what I am saying as request for a division of two words by one word (like the x86 "div" instruction) by using multiplications with precomputed numbers.

09 Sep 2017, 14:50 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

Back to the beginning: the idea behind the division by multiplication can be presented as follows.
If we want to divide integer N by divisor D, what we can try is to find number such that and:
,
.
The same product can also be written as:
and because :
,
so:
,
,
.
Therefore and thus which gives us correct division with remainder, since . We get the result Q simply by taking the high bits of X. We can also get the remainder R by taking high bits of LD.
Now, to find the number X we can multiply N by socalled magic multiplier M, which is a number such that:
,
and then:
which gives the correct result as long as .
All the algorithms presented above made this work by finding n large enough that this inequality would be fulfilled for all N in the required range.
But there is still another approach we could try. When we compute M, we also know P, and we can use this knowledge to try to adjust the result even if NP ends up being large enough to meddle with the high bits of NMD.
We can compute NP and split into high and low part:
and then try to adjust for the error by taking:
.
This leads to:
Now, keeping in mind that:
we also need to ensure that:
.
Then the worst that may happen is that the high bits of N'MD might contain N  1 instead of N. This can be easily fixed after computing the remainder  if remainder ends up being larger than D we can subtract D from it and increase the quotient by one.
Let's try it on an example, with D = 10, n = 32. Then M=1999999Ah and P = 4:
To simulate a DIV instruction we need to operate on range:
With this in mind we have:
so
and finally:
This is far from being dangerous for the high bits of N'MD.
It is time for an implementation:
Code: 
div10:
; cmp edx,10
; jae int0
push ebx ecx
push eax
mov ebx,eax
mov ecx,edx
shld edx,eax,2
sub ebx,edx
sbb ecx,0
mov eax,ebx
mov ebx,1999999Ah
mul ebx
mov eax,ecx
imul eax,ebx
add eax,edx
pop edx
imul ecx,eax,10
sub edx,ecx
cmp edx,10
jb .done
sub edx,10
inc eax
.done:
pop ecx ebx
retn


This ends up being faster than the previous one, it only has three multiplications (and just one of them is the fullsized MUL) thanks to the fact that P=4 and multiplication by P can be done by shifting.
For 64bit version just replace all 32bit registers with their 64bit counterparts and the constant with 199999999999999Ah (P is the same).

09 Sep 2017, 14:53 

Tomasz Grysztar
Assembly Artist
Joined: 16 Jun 2003
Posts: 6633
Location: Kraków, Poland

tthsqe wrote: 
You can interpret what I am saying as request for a division of two words by one word (like the x86 "div" instruction) by using multiplications with precomputed numbers.


What I wrote above is about such division, though perhaps not general enough for your needs. It only works for divisors D such that:
.
But when dividing large integer by another large integer, you usually need to divide by a normalized divisor, such that:
.
I have not yet tried to solve this variant with my integerbased approach.
When a divisor is normalized, the algorithm I presented first has the shift equal to the size of machine word (the assembly implementation I provided does not work then, because SHRD sees such shift as zero), so the result can be taken straight from the third word after multiplication. Also, multiplying the lowest word seems pointless, because it can affect the result by no more than one, and this can be corrected when computing remainder, as in my previous post. However to extend the range of algorithm, some kind of compensation for the NP term would be needed. I don't know if I can solve this variant with integer arithmetic as nicely as the previous ones.

09 Sep 2017, 14:58 



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