pretty sure there are more possible chess positions than atoms in the earth (universe?), so even if every atom of our planet were converted to transistors there’d be no way to fully represent all possibilities.
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!
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).
I was thinking of 10^16kg diamond storage inside a larger SSD that’s the size of a large moon, similar to how a real SSD has data stored in tiny little slivers of silicon inside a much much larger device.
I should have explained that one better. It’s easy to imply such details to keep text shorter.
There is no clear definition of what constitutes a moon other than it being a body that orbits another body that orbits the parent star.
There are some astronomers who say the dividing line between a moon-planet/dwarf planet/asteroid system and binary (or more) planet/dwarf planet/asteroid system is whether or not the barycenter of the orbits is within one of the bodies or not.
And fun fact: if that definition gained acceptance, it would mean that the Pluto-Charon system would go from a dwarf planet-moon system to a binary dwarf planet system. Charon could get a promotion.
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!
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.
@VoidJuiceConcentrate @maris
Right?
pretty sure there are more possible chess positions than atoms in the earth (universe?), so even if every atom of our planet were converted to transistors there’d be no way to fully represent all possibilities.
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!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).
I was thinking of 10^16kg diamond storage inside a larger SSD that’s the size of a large moon, similar to how a real SSD has data stored in tiny little slivers of silicon inside a much much larger device.
I should have explained that one better. It’s easy to imply such details to keep text shorter.
Out of curiosity, why did you say planetary moon? Is there any other kind?
Dwarf planets sometimes have moons (e.g. Pluto)
Some large asteroids have moons too.
this leads to the question: are they still considered a moon when the barycenter is in the space between them?
There is no clear definition of what constitutes a moon other than it being a body that orbits another body that orbits the parent star.
There are some astronomers who say the dividing line between a moon-planet/dwarf planet/asteroid system and binary (or more) planet/dwarf planet/asteroid system is whether or not the barycenter of the orbits is within one of the bodies or not.
And fun fact: if that definition gained acceptance, it would mean that the Pluto-Charon system would go from a dwarf planet-moon system to a binary dwarf planet system. Charon could get a promotion.
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!
A position is the arrangement of all the pieces on the board.
i think you did the weight approximation in the wrong order, 1024 is a lot bigger than 6×1023. 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.
yeah yeah, cosmological approximations and all that, but there’s still a bit of difference between “planetoid” and “gas giant” :P
Slightly less than double, actually. (Doesn’t really change the meat of the argument or anything though.)