How to do division accurately and instaneously

Hey errybody welcome to HOW TO DO DIVISION and such

something to note is that this is in base 10 but the same concept can be applied for the base 2 number system

everyone knows division is overglorified subtraction (its so pretentious) so lets reduce it to its basics shall we

the first problem is getting them to scale, like if you have 60/.0000000000003, not a good situation

what you want to do is get y (x/y) into this scale: X >= Y > (X * 10^-1)

For example, if the X was 50, then the range would be 50-5 (this ensures that the digits are the same)

Now if we apply that to 60/.0000000000003 , you will get 60/30 * 10^14 NOTE: YOU MUST KEEP 10^14 MULTIPLIED AGAINST 1 SOMEWHERE (it ensures that you will be able to re-establish scale after because 60/30 is not the same as 60/.0000000000003, just more calculable)

Now the good part, lets have a division example that doesnt work (like 65/400)

65/40 * 10^-1

65-1(40) = 25
25-6(4) = 1
1-2(.4)=.2
.2-5(.4)=0!

This all means that 65/400 is 1.625 * 10^-1, if we do 65/400 (like in a real calculator)
HEY LOOK IT WORKED!

Now go and divide a bunch :D

I understand how it works, what I don't get is how it's instantaneous. Not doubting that it is, you're more logical than I and I know it can be done, but if I were to make a chip this way, I wouldnt know how to do it in one frame. Take the second step for example, 25-6(4)= 1. If I were to do this calculation, I would take my 4 and say 4<=25? Yes, x2. 8<=25? Yes, x2. And so on. Wouldn't this take multiple frames? Or is there a way to do it in one? Is there a better approach?

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Originally Posted by Tynz21

I understand how it works, what I don't get is how it's instantaneous. Not doubting that it is, you're more logical than I and I know it can be done, but if I were to make a chip this way, I wouldnt know how to do it in one frame. Take the second step for example, 25-6(4)= 1. If I were to do this calculation, I would take my 4 and say 4<=25? Yes, x2. 8<=25? Yes, x2. And so on. Wouldn't this take multiple frames? Or is there a way to do it in one? Is there a better approach?

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In binary there is only one check per frame, so you dont have to worry about it

but if you are to do this in base 10, I would have 9 static checks all at once, and have it favor the highest
for example

1,2,3,4,5,6,7,8,9(4) >= 25? #=6

This way there is no added latency :)