A-B=C

[fredreload]

Alright, so I think I solved this algorithm, I call it the shrink ray, I got the idea from Minions. Pretty much with any number of A-B in binary, you can reverse engineer A and B from C alone knowing how many 1's and 0's there are for A and B.

 

For example:

 

1010 A (two 1's, two 0's)

-0101 B (two 1's, two 0's)

--------

0101 C

 

Note: I've only tested the condition for A>B, not negative numbers

Now you reverse engineer A and B from C, you know C is 5 and A and B are bigger, so you can have A=6 B=1, A=7 B=2, A=8 B=3, A=9 B=4, A=10 B=5, A=11 B=6, A=12 B=7, A=13 B=8, A=14 B=9, A=15, B=10. Now with all these conditions available, the only set that holds true with A having two 1's and B having two 1's are A=10 and B=5. So that's the solution. I think you can reverse engineer any A and B with C and the bits information on A and B.

 

If A and B are 32 bits each, it only takes 6 bits each to store the bit information since the largest possible sum is 32, so I think it's pretty good in terms of shrinking the size down, check the math for me will ya Strange?

Edited August 25, 2016 by fredreload

[fredreload]

it seems there is an exception when C is 0 where A and B can be anything, well that's the only exception I think

[Endy0816]

-snip-

Edited August 26, 2016 by Endy0816

[fredreload]

Well it's easy to bypass this exception, since you only need one set of value, A or B, so set both bit count to zero(they might not be zero but this means A and B are the same) and set C to either A or B

Edited August 26, 2016 by fredreload