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Coding chess
(lemmy.world)
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this post was submitted on 27 Jun 2025
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This was a fun one to look up. https://en.wikipedia.org/wiki/Shannon_number
It looks like the number of valid chess positions is in the neighborhood of 10^40 to 10^44, and the number of atoms in the Earth is around 10^50. Yeah the latter is bigger, but the former is still absolutely huge.
Let's assume we have a magically amazing diamond-based solid state storage system that can represent the state of a chess square by storing it in a single carbon atom. The entire board is stored in a lattice of just 64 atoms. To estimate, let's say the total number of carbon atoms to store everything is 10^42.
Using Avogadro's number, we know that 6.022x10^23 atoms of carbon will weigh about 12 grams. For round numbers again, let's say it's just 10^24 atoms gives you 10 grams.
That gives 10^42 / 10^24 = 10^18 quantities of 10 grams. So 10^19 grams or 10^16 kg. That is like the mass of 100 Mount Everests just in the storage medium that can store multiple bits per atom! That SSD would be the size of a ~~small~~ large moon!
i think you did the weight approximation in the wrong order, 10^24^ is a lot bigger than 6×10^23^. so you can probably double the final weight.
Well yeah it’s almost double, but I wrote the comment as a mental estimation of the order of magnitude, so it doesn’t change the substance of the discussion.
I mean at the beginning I arbitrarily picked a number in that 10^40 to 10^44 range and that’s a factor of 1:10,000 rather than 1:2, lol.
Slightly less than double, actually. (Doesn't really change the meat of the argument or anything though.)