Application for jackson535

one of my best redstone creations is a recirculating 6 bit floating point square root calculator this device could take any number from 1 to 63 and find the square root with the answer having exactly 6 binary digits. first conceptualize some simple binary square roots. 16, or 00001 has a square root of 001. 64, 0000001, has a square root of 0001. the rule for this is that every time a binary number is shifted 2 decimal places, the answer shifts 1. so for our basic binary perfect square, we can summarize the rule as every 2 0's in front of the 1 for the number being square rooted is 1 0 in the answer in front of the 1. 1,1 root1=1 100, 10 root4=2 10000, 100 root16=4 so what happens if we want to solve for a number that isnt normal like this? take 8, for example. our first step will be splitting 1000 into sections of 8 from the origin. this is how you effectively count 0's. in this case, we get 10|00. then, we look at the furthest pair of bits, the 10 part. we know that the answer has to be AT LEAST 2, because the 1 is at the end of between 2 and 4 0's. so we put a 1 for that part of our answer. so our answer so far is 10. we can then find what is leftover by subtracting 1 from the section with 10. this leaves us with a remainder of 1. this remainder part represents our remainder to this formula that we are using. 2^2<8. in order to find what x could be, we need to take our answer so far, and subtract it from 8. this gives us our remainder of 4. now, in order to check if x is 1/2 of 2, or just 1(meaning 11 or 3 would be partial answer if it works) we have to see if we can subtract 2^2+1 from 4. the +1 is used because the total of what we subtracted the first time (4) plus our current answer squared, plus 1, will make exactly 9 or 3^2, and we are testing for the 1's place to see if we get a 3. we cannot have a positive result, so we do not subtract. so our answer is still 10 or 2. next we multiply our remanider (still 4) by 4, because moving a square number over 2 binary digits will allow us to analyze 1 square root binary digit. so we get 16. when we subtract our 2^2+1 from 16, our new remainder is 11. since we could subtract, our next part of the answer is a 1. so our answer so far is 10.1 or 2.5. now we multiply the remainder by 4 again, and get 44. we also multiply our 2.5^2+1 by 4, giving us 29. we can subtract 29 from 44, so our answer is now 10.11 or 2.75. you can repeat this process as many times as you wish untill a desired accuracy is achieved. so this is what my machine does. its just extra slow and smaller because it recirculates the answers back into the same calculating device. many shift registers, t flipflops, delays, and subtractors were needed.