Divide arbitrary binary integers by 3 in python

Question

All the binary integers am given are always divisible by 3 so no floating-point values involved. The return value must also be a binary string Tried this but a and h are different yet they should be

>>> a="1111111111111111111111111111111111111111111111111111111111111111111111"
>>> f=int(a,base=2)
>>> print (f)
1180591620717411303423
>>> g = f/3
>>> c=int(g*3)
>>> print(c)
1180591620717411303424
>>> h = bin(c)
>>> print(h)
0b10000000000000000000000000000000000000000000000000000000000000000000000

Are you coding in C or Python? This will produce very different answers... in Python it's as simple as `int('10101', base=2)/3` — Tomerikoo, May 28 '20 at 15:06
Tried in c though i program in python also but was using long division which works for dividing by any number, then failed to figure out a way for just 3, because the process on divides by 3 — Mpiima, May 28 '20 at 15:10
Can be done easily in C by noting x/3 == x/4 + x/16 + x/64 ... — stark, May 28 '20 at 17:30

score 0 · Accepted Answer · answered May 30 '20 at 23:25

Since your numbers are always divisible by 3 it is better to use // for integer division, because when you use floating point division for large numbers the result might not be exactly right.

So your code will be something like:

>>> a="1111111111111111111111111111111111111111111111111111111111111111111111"
>>> f=int(a, base=2)
>>> print(f)
1180591620717411303423
>>> g = f // 3
>>> c = g * 3
>>> print(c)
1180591620717411303423
>>> h = bin(c)
>>> print(h)
0b1111111111111111111111111111111111111111111111111111111111111111111111

Divide arbitrary binary integers by 3 in python

1 Answers1