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 Author: tschak909 [ Tue Jul 25, 2017 12:21 am ] Post subject: divide by 10 and divide by 6 using bit shifts+addition Does anyone happen to have a general algorithm for dividing by 10 and by 6 using shifts/addition? debating on whether to go this direction, or to code it into a table...-Thom

 Author: NovaSquirrel [ Tue Jul 25, 2017 12:46 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition The big Unsigned Integer Division Routines thread has routines for division by 6 and 10.

 Author: Dwedit [ Tue Jul 25, 2017 6:23 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition 8-bit dividing by 10 is the same as doing 16-bit multiplication by 0x1A (00011010), then discarding the low byte. So that's x << 4 + x << 3 + x << 1 in 16-bit math.Dividing by 6 is multiply by 0x2B (00101011), or x << 5 + x << 3 + x << 1 + x.16-bit shifts and adds take up code and time.Then there's the really stupid way to divide, subtracting in a loop, if you don't care about speed, it's a decent way.But lookup tables are really cheap, just 256 bytes of ROM for 8-bit math.

 Author: rainwarrior [ Tue Jul 25, 2017 9:09 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition Dwedit wrote:8-bit dividing by 10 is the same as doing 16-bit multiplication by 0x1A (00011010), then discarding the low byte. So that's x << 4 + x << 3 + x << 1 in 16-bit math.Dividing by 6 is multiply by 0x2B (00101011), or x << 5 + x << 3 + x << 1 + x.16-bit shifts and adds take up code and time.To clarify, this is a fixed point approximation technique. The results are not entirely accurate.The approximation works because a division by 256 is "cheap", so you can rescale the input with a pre-multiply.Code:x / d =  x * a / 256d = a / 256a = 256 / dd = 10 --> a = 256 / 10 = 25.6x / 10 = x * 25.6 / 256d =  6 --> a = 256 /  6 = 42.666...You will notice that these are not exactly 26 (0x1A) or 43 (0x2B). This is why it's an approximation, and you will find the result is off by 1 in some cases:Code:⌊ 69 / 10 ⌋ = 6          (6.9)⌊ 69 * 26 / 256 ⌋ = 7    (7.0078125)Depending on whether you need perfect results or not, this may not be an acceptable compromise. You could increase it to 24 bits for more precision, or in some cases you might be able to adjust some errors by carefully adding a bias value.Dwedit, I'm curious why you offered these numbers in hexadecimal and binary but not decimal. What significance do they have in hex/bin?

 Author: lidnariq [ Tue Jul 25, 2017 9:30 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition I'm arbitrarily going to guess that the OP is either converting frames or seconds to the next larger unit ... perhaps, rather than implementing division, it makes sense to store the time in BCsexagesimal ?

 Author: tschak909 [ Tue Jul 25, 2017 9:34 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition (this is for Wizard of Wor)Nah, I just need to be able to calculate from a sprite position the nearest dungeon box that he's in (the dungeon is implemented as a 10x6 array of squares), so that I can implement collision detection and the like. -Thom

 Author: tepples [ Tue Jul 25, 2017 10:04 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition So it's division by 24 (which may involve division by 3 or 6 as an intermediate step) to convert from sprite coordinates to metatile coordinates, then multiply by 10 or by 6 (depending on whether row-major or column-major) to convert metatile coordinates to an address in the backing map. I don't see where division by 10 enters into it, except during conversion of a score to decimal.

 Author: tschak909 [ Tue Jul 25, 2017 10:11 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition tepples wrote:So it's division by 24 (which may involve division by 3 or 6 as an intermediate step) to convert from sprite coordinates to metatile coordinates, then multiply by 10 or by 6 (depending on whether row-major or column-major) to convert metatile coordinates to an address in the backing map. I don't see where division by 10 enters into it, except during conversion of a score to decimal.Oh ok, I was thinking to derive the X and Y values seperately, but.. ok..-Thom

 Author: Dwedit [ Tue Jul 25, 2017 11:21 am ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition rainwarrior wrote:Dwedit, I'm curious why you offered these numbers in hexadecimal and binary but not decimal. What significance do they have in hex/bin?In binary, you can see the sequence of shifts and adds to do the multiply. Multiplying by 00110011 means x << 0 + x << 1 + x << 4 + x << 5 because those bits are set in the binary number.And hex because I'm used to expressing reciprocal approximations in hex. On 32-bit arm, I'd always do 0x100000000 / x, then add 1, and use UMULL to do division. Here I did 0x100 / x + 1, but I might have needed some more precision here.

 Author: rainwarrior [ Tue Jul 25, 2017 12:12 pm ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition Dwedit wrote:In binary, you can see the sequence of shifts and adds to do the multiply.Ah, yes that makes sense.

 Author: tschak909 [ Tue Jul 25, 2017 4:02 pm ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition Wound up using omegamatrix's div by 24 routine:Code:;Divide by 24;15 bytes, 27 cycles_div24:   lsr   lsr   lsr   sta   temp   lsr   lsr   adc   temp   ror   lsr   adc   temp   ror   lsr   rtsAs it turns out, I was being derp by not realizing that I needed to count pixels, not boxes for the divisor, thanks tepples. -Thom

 Author: tschak909 [ Tue Aug 01, 2017 5:37 pm ] Post subject: Re: divide by 10 and divide by 6 using bit shifts+addition Is there a complimentary routine for returning the modulus? I'm needing to do fine adjustment of the division to indicate whether I am actually as far into a box as I can be physically.-Thom

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