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(+1)

So,  I'll do  my best to explain what (hypothetically) might work...

If you divide  for example 8 ÷ 4. Division is basically subtracting (from the dividend) until the number is less then the second number (divisor). So what you could do is have a circuit that loops through itself until a condition is met, then ending the loop and outputting a result. Here's a step-by-step guide to how it (could) work:

Input A = 8 and Input B = 4

So take 4 (B) from 8 (A) and then add one to C (now 1), so A is now 4 and B is still 4.

Then if A (now 4) is less then B (4) we would stop. But A (4) is not less then B (4), SO! We can do another subtraction so A is now 0 and B is 4, and we add one to C (now 2).

So A (8) ÷ B (4) = C (2), and we would output C (2).

So we could input our two numbers (A & B) and loop through a Subtraction Circuit, checking each time if A is less then B, and if so we stop the loop, and output or C (answer). Now this would take a lot of work, but seems possible.

But if you wanna do something easier (but bulkier) you could chain-together a bunch of subtractions (adding 1 to a counter every circuit), and check each time if it is less then, and if so stopping the chain and outputting a result.

I hope all of this makes sense. And if it does, hope it helps.

Thank you, hopefully I can make something out of it, and hopefully it helps others looking on how to make a divider too! :) I also already have a comparator to detect if a number is greater, less than, or equal to the other, so that will help with this.

(1 edit)

Nice! I hope that was able to help. I would definitely like to hear if you made any progress, and if my idea actually worked.

Deleted 322 days ago

It worked! It isn't completely automatic, but it works with most numbers I tested!

👍

I know I'm late but, you can very easily do division in O(1) using long division, here's an implementation for 8 bits: