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I don't think anyone believes that Go is somehow more complex than the universe it is a subset of. The point is that enumerating all cases of Go is impossible a
by harryjo 10y ago
I don't think anyone believes that Go is somehow more complex than the universe it is a subset of. The point is that enumerating all cases of Go is impossible and always will be, so more sophisticated analysis is required.
- vacri 10y agoSametmax's point is that you're comparing a simple count (atoms) to a factorial (combinations of pieces). For example, it's hardly surprising that 6! is larger than 6 - factorials grow much faster than simple counts.
- tromp 10y agoIndeed, enumerating all 208168199381979984699478633344862770286522453884530548425 639456820927419612738015378525648451698519643907259916015 628128546089888314427129715319317557736620397247064840935 positions in Go is impossible.
- dogecoinbase 10y agoIn your enumeration, what's the board look like at position 348277381979984699478633344862652779770286522453884530548425639456820927419612?
- Jyaif 10y ago1/ Convert the number to base 3. 2/ Each digit represents an intersection on the goban, assuming the following mapping: 0 = no stone, 1 = black stone, 2 = white stone.
- tromp 10y agoThat would work for all positions regardless of legality, which are in 1-1 correspondence to {0,1,2}^(19*19). The count above is for legal positions only, i.e. those where every connected group of stones is adjacent to an empty point.
- tromp 10y agoI can't say, because I didn't enumerate them. I only counted them. See http://tromp.github.io/go/legal.html http://tromp.github.io/go/legal.html for the method used, which is a form of dynamic programming.
- nhaehnle 10y agoThough if it is dynamic programming, then it should be possible for you to answer dogecoinbase's question using not much more computational power than you used to count them in the first place, right? If you think of dynamic programming as counting the number of paths in a directed graph (in this case, from skimming the paper, the nodes correspond to border states), then given a path number, you can trace the path backwards through the graph, as long as you remember the number of paths ending in every vertex.
- tromp 10y agoYes, you could if you preserved all intermediate counts. But the graph I used has 362 layers each of which can have up to 363 billion nodes, so I had to recycle the space used for the counts (4TB per layer). Also, I didn't even compute with full counts. I reconstructed them using the Chinese Remaineder theorem from 9 separate modular counts. So, yes it's possible, but highly impractical...
- ChristianBundy 10y agoThat's called "counting". Enumeration is iterative.