A way of slashing the brute forced bit string in half which is %50 logarithmically efficient!
Normally brute forcing a bit string is O(2^n) hard because every slot can get 2 values the floor is 2 where n is the length of the bit string but because every bit string is a repeatation of other bitstrings and lower n strings are exponentially easier to calculate we can use those to calculate our new bitstring for example:
bb - 2 bits
00 = A, 10 = B, 01 = C, 11 = D
I calculated the every possible combination of the 2 bit string and then I labeled them as A, B, C, D now I will use labels to calculate a new bit string!
AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
If I see AA I will decode that as 0000 = AA (because A = 00) this calculation when done normally is equal to O(2^4) but with this method its big O notation is O(2^2 + 4^2) so the big O notation of my algorithm is O(2^(n/2)) which is logarithmically or exponentially %50 more efficient!