16/16-bit division

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16/16-bit division
by on (#199549)
Anybody know how this would be done? The easiest way I know of is by shifting the denominator right until it's less than 256, and then shift the quotient right by the same number of times.

Is there a more accurate way of doing this?
Re: 16/16-bit division
by on (#199553)
6502.org: Division by a 16-bit number
Re: 16/16-bit division
by on (#199555)
http://codebase64.org/doku.php?id=base:6502_6510_maths

But no magic, you need LUT if you want some speed .
Re: 16/16-bit division
by on (#199566)
I was wondering if there was a way that makes use of the SNES's division registers.
Re: 16/16-bit division
by on (#199568)
There probably isn't a way to get exact division by an arbitrary 16-bit number out of the memory controller's divider. So at this point, you may have the XY problem: you have a goal, you think a step (division by a 16-bit number) is the way to go about that goal, and you ask about the step without describing the goal. What are you trying to accomplish with this division?
Re: 16/16-bit division
by on (#199569)
I was recently looking for a good way to do 16/16-bit division to calculate the relative angle between the player and game objects in my raycaster, and was amazed to find out that log2(x/y) = log2(x) - log2(y)! That works well because to find an angle you only need the ratio between x and y, so the actual quotient isn't really needed, just its logarithm will do.

Yeah, shame on me for not knowing how to use logarithms before (I was never good at this in school - if I knew this would help me program games one day, I'd have paid more attention), but I was just glad to find out that with a couple of tables I could turn an expensive division into a simple subtraction.

I ended up finding the answer to my problem myself (and I had probably seen this trick before, but it must've flown over my head), but if I had asked for help, asking "how can I find this angle" vs. "how can I do a 16/16-bit division" could've made all the difference.

To keep the tables small I ended up going with 11/11-bit though, which works perfectly for my needs.
Re: 16/16-bit division
by on (#199575)
I have an idea for this:

Shift the denominator right until it's odd. If the shifted denominator is less than 256, use the division registers. Shift the result left by the same amount of bits as the you shifted the denominator.

If the shifted denominator is 256 or larger, use software division. Because the denominator is 256 or larger, the result has to be less than 256, so only 8 bits are needed for the result.

Now that I think about it, I can reduce the number of compares/subtracts by the number of times the denominator is shifted.


tepples wrote:
There probably isn't a way to get exact division by an arbitrary 16-bit number out of the memory controller's divider. So at this point, you may have the XY problem: you have a goal, you think a step (division by a 16-bit number) is the way to go about that goal, and you ask about the step without describing the goal. What are you trying to accomplish with this division?


Just for fun, pretty much. I might use it as a go-to division routine if I'm in a hurry to get something done.
Re: 16/16-bit division
by on (#199605)
tokumaru wrote:
I was recently looking for a good way to do 16/16-bit division to calculate the relative angle between the player and game objects in my raycaster, and was amazed to find out that log2(x/y) = log2(x) - log2(y)! That works well because to find an angle you only need the ratio between x and y, so the actual quotient isn't really needed, just its logarithm will do.

Yeah, shame on me for not knowing how to use logarithms before (I was never good at this in school - if I knew this would help me program games one day, I'd have paid more attention), but I was just glad to find out that with a couple of tables I could turn an expensive division into a simple subtraction.

I ended up finding the answer to my problem myself (and I had probably seen this trick before, but it must've flown over my head), but if I had asked for help, asking "how can I find this angle" vs. "how can I do a 16/16-bit division" could've made all the difference.

To keep the tables small I ended up going with 11/11-bit though, which works perfectly for my needs.


The problem of using log is that :
- it doesn't work on signed number (ok you can eventually offset them)
- accuracy is poor for high number

But still log is very useful for fast multiply (become an addition) or a division (become a subtraction) :)
Re: 16/16-bit division
by on (#199680)
Garth Wilson has suggested using a table of inverses to feed into a multiplication. He even provides such a table on his scaled-integer page.
Re: 16/16-bit division
by on (#199765)
do you want
u16/u16 = u16
u16/u16 = u8
s16/s16 = s16
s16/s16 = s8
?
Re: 16/16-bit division
by on (#201326)
I figured out you can take the equation Y=X/(A+B), rearrange it as Y = (X - BY)/A, and solve it recursively.
Re: 16/16-bit division
by on (#201446)
The Playstation hardware is using this fast (but slightly inaccurate) division method http://problemkaputt.de/psx-spx.htm#gte ... inaccuracy it's based on Unsigned Newton-Raphson (UNR) algorithm, and comes up with one table read, plus three multiplications, plus some shift/add/sub operations.

The way how it's done there isn't exactly what you want: the input is two unsigned 16bit values... but it does multiply one of the values by 20000h before dividing it, and returns a 17bit result. But it shouldn't be too difficult to remove the multiply by 20000h, and tweak it to return a 16bit value.

For the unsigned multiplies, the SNES hardware supports 8bit*8bit=16bit (via CPU), or 7bit*15bit=22bit (via PPU, with sign bits set to zero). So you'll internally need more than three multiplications on the SNES. When reducing accuary & dropping some bits... maybe you could get away with four PPU operations (two for the first two multiplies, and another two for the final multiply with bigger operands).
Re: 16/16-bit division
by on (#201491)
The APU can also multiply & divide.
Re: 16/16-bit division
by on (#201555)
I think I got it this time.

Do to the thing I said in the OP. Then multiply the quotient by the divisor, and compare it to the dividend. If it's greater than the dividend, decrement the quotient by 1.