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!
10120 is the number of valid game-trees, or valid ~80 move games.
The much smaller number I quoted above, though, IS the valid positions, I was thinking it was actually the trimmed down “truly valid” game-tree sequences.
Isn’t math fun? Limitless ways for us to be wrong!
Assuming your math is correct (and I have no reason to doubt that it is) a mass of 10^16 kg would actually be a pretty small moon or moderately sized asteroid. That’s actually roughly the mass of Mars’ moon Phobos (which is the 75th largest planetary moon in the Solar System).
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
smalllarge moon!Lol, big board you’re playing with…
If you don’t limit it to valid positions/arrangements it’s like 10^120. Closer to the “number of X in the observable universe” caliber of number.
So I think I was wrong, but you are too lmao.
10120 is the number of valid game-trees, or valid ~80 move games.
The much smaller number I quoted above, though, IS the valid positions, I was thinking it was actually the trimmed down “truly valid” game-tree sequences.
Isn’t math fun? Limitless ways for us to be wrong!
Assuming your math is correct (and I have no reason to doubt that it is) a mass of 10^16 kg would actually be a pretty small moon or moderately sized asteroid. That’s actually roughly the mass of Mars’ moon Phobos (which is the 75th largest planetary moon in the Solar System).
Out of curiosity, why did you say planetary moon? Is there any other kind?
Dwarf planets sometimes have moons (e.g. Pluto)