What's the number of bits that could require every possible location in the universe, from the smallest distance, the Planck length, to the largest distance, which is, presumably, the known visible distance of the universe?
The Planck length is 1.6 x 10-35 meters1, and we assume that it doesn't really make sense to describe things in finer detail below the Planck length.
The visible diameter of the universe is about 1012 light years2, which is about 1028 meters.
So we need about 63 digits in the length exponent to describe distances from the very very small to the very very large. This means that we need about 210 bits, at the very minimum. So let's think like engineers and add some safety factor: We'll add 10 bits to handle computations at the very very small levels, and then another 10 bits at the very large, which is about 230 bits, and then we round up to 232 bits to make this divisible by 8 (computers love to process in 8 bit bytes), except 29 is an ugly number of bytes, so we can go up to 32 bytes, which is divisible by 2, and so useful for creating really easy addressing index loops, and so the final answer is:
32 bytes = 256 bits.
256 bits? That's all? Shoot. That's not so many. Some strong encryption protocols use 2000-4000 bits just for one of those encryption words. So the number of bits used to describe the universe's geometry is waaay smaller than what we need to do strong encryption. Ironic, isn't it?
- Wikipedia, Planck Length
- Wikipedia, Observable Universe