There are three very important distinction between “bits contained in our cosmological horizon” and “number of bits contained in 3->3->3 cells” of Conway’s life: 2D vs 3D, the complexity of the “rules” governing interaction between the “bits” and the number of possible states of each “bit”. This almost certainly (as in “I am very sure, but a formal proof eludes my math skills”) means that 3->3->3 cells is far too few to simulate even a small portion of our universe.
I tried for a couple of hours to show this by looking at a number of “natural” states and symbols for a 2D CGoL and possible 3D analogues. I also tried an approach of modelling the growth of a 2D vs 3D machine as a function of computing power. I ran into conceptual difficulties both times, but not before forming an impression that n->n->n notation will be inadequate to compare the two (need to extend to n->n...->n->n… busy beaver, anyone?). Maybe someone with advanced math background can push this further...
2D vs 3D, the complexity of the “rules” governing interaction between the “bits” and the number of possible states of each “bit”
Actually, none of these matter. The possible states of each bit is exactly 2 both for our universe and for (the simplest form) of Life. And the fact that both are Turing complete means that, whatever the rules governing the interactions are, for every possible computation there is (at least) a state of the world that performs that computation. This also makes futile the distinction between 2d and 3d geometry of the space, even if—if we are to believe to the holographic principle—our universe too is completely describable as a 2d entity. This is not necessary however, since the calculation of the bits contained in our cosmological horizon can be based on Bekenstein bound only.
As for the actual numbers of bits in our universe, I don’t really want to do the math, although I read somewhere that is in the order of 10^120, but even if it’s closer to 10^150 that wouldn’t matter at all, since that number is far less than the bit contained in a cellular automata of 3->3->3 bit, let alone the bit contained in a board of 3->3->3 squared.
the fact that both are Turing complete means that, whatever the rules governing the interactions are, for every possible computation there is (at least) a state of the world that performs that computation
Certainly, for an infinite board. But a 3->3->3 board is infinitely smaller than that. What is in question is what portion of universe such as ours can be simulated on such a board...however:
none of these matter
I now agree—with a caveat that one allows arbitrarily long time for the simulation. My earlier remarks were based on an implicit assumption that the computation time for the 2D machine simulating a 3D machine stays constant as the 3D machine size grows.
There are three very important distinction between “bits contained in our cosmological horizon” and “number of bits contained in 3->3->3 cells” of Conway’s life: 2D vs 3D, the complexity of the “rules” governing interaction between the “bits” and the number of possible states of each “bit”. This almost certainly (as in “I am very sure, but a formal proof eludes my math skills”) means that 3->3->3 cells is far too few to simulate even a small portion of our universe.
I tried for a couple of hours to show this by looking at a number of “natural” states and symbols for a 2D CGoL and possible 3D analogues. I also tried an approach of modelling the growth of a 2D vs 3D machine as a function of computing power. I ran into conceptual difficulties both times, but not before forming an impression that n->n->n notation will be inadequate to compare the two (need to extend to n->n...->n->n… busy beaver, anyone?). Maybe someone with advanced math background can push this further...
Actually, none of these matter. The possible states of each bit is exactly 2 both for our universe and for (the simplest form) of Life. And the fact that both are Turing complete means that, whatever the rules governing the interactions are, for every possible computation there is (at least) a state of the world that performs that computation. This also makes futile the distinction between 2d and 3d geometry of the space, even if—if we are to believe to the holographic principle—our universe too is completely describable as a 2d entity. This is not necessary however, since the calculation of the bits contained in our cosmological horizon can be based on Bekenstein bound only.
As for the actual numbers of bits in our universe, I don’t really want to do the math, although I read somewhere that is in the order of 10^120, but even if it’s closer to 10^150 that wouldn’t matter at all, since that number is far less than the bit contained in a cellular automata of 3->3->3 bit, let alone the bit contained in a board of 3->3->3 squared.
Certainly, for an infinite board. But a 3->3->3 board is infinitely smaller than that. What is in question is what portion of universe such as ours can be simulated on such a board...however:
I now agree—with a caveat that one allows arbitrarily long time for the simulation. My earlier remarks were based on an implicit assumption that the computation time for the 2D machine simulating a 3D machine stays constant as the 3D machine size grows.