More Elegant 3D Lattice

In my previous post, my model called for up to 57 different grid intersection configurations. I'm now thinking a more minimalist 16 possible grid intersections, which is probably the minimum no. you can get away with. Each intersection consists of either a NAND gate, a simple intersection of wires, no wires at all, or a crossover intersection. Alternatively, you can replace all the NAND gates with NOR gates, which would call for a slightly different wiring pattern. When enumerating all the possible wiring configurations of a single grid intersection, you only have to worry about a wire going from the intersection to the right, from the intersection downwards, and from the intersection to the corresponding intersection in the plane behind the current plane (that's 3 directions). The other 3 directions (left, up, and in front of) are taken care of by adjacent grid intersection (their wires "belong" to other grid intersections).

Here is how I arrived at 16 possible configurations: there are 7 configurations involving a NAND gate, one crossover intersection, and 8 configurations involving 0 to 3 wires and no NAND gate. Regarding the NAND gate configurations, each NAND gate has 2 incoming wires and 1 outgoing wire. There are 2 directions, right and down, and 3 possibilities for each direction: no wire, 1 incoming wire, or 1 outgoing wire, which equals 3 squared = 9. But 2 of those configurations are invalid: 2 outgoing wires, and no wires at all, leaving just 7 configurations involving a NAND gate. A crossover intersection involves 4 wires in the current plane, and the 2 vertical wires (up and down) do not physically touch the horizontal wires (left and right). As for the 8 configurations involving a simple intersection of wires (or no wires at all), there are 3 directions (right, down, and the "z" direction, or behind). For each of those 3 directions, there can be either a wire or no wire (2 possibilities), totaling 2 cubed = 8 configurations. So there are only 16 configurations (or 4 bits of data, instead of 57 configurations or 6 bits of data) for each grid intersection, and I maintain that you can't get much more elegant than that.

It's a little bit warm in my condo and I can't sleep. Yesterday was Saturday, when I often get depressed and go back to bed, which I did, and on Sunday I stopped being depressed. Going from depressed to not depressed often involves insomnia with me. A week from now I hop on a plane to Victoria to visit my sister for 12 days, so I'm looking forward to that. I'll probably take my laptop with me, and work on designing the wiring diagram of my Computaverse digital computer, built out of NAND gates and wires up to 10 raised to the 20,000th power in length (the no. of grid cells they traverse). If you want to visualize the shape of this computer, it's very close to a one-dimensional line, all length with very little width or depth. It's very difficult to visualize a number of grid cells equal to a one followed by 20,000 zeroes. Say there are 10 ^ 100 subatomic particles in the universe. Take that number and, not just square it, not just cube it, or even raise it to the 4th power, but instead multiply it by itself 200 times (raise it to the 200th power).

I look forward to designing my Computaverse computer, and writing the documentation for that computer. I will buy a tool for $89 which converts MS Word to HTML, and upload that HTML to my web site. Then I will search for philosophy forums on the Net where it would be appropriate to post a summary of my Computaverse theory, and point readers to my web site. It will be interesting to see what kind of reaction I get, if any. I'm hoping that my own modern creation myth will appeal to secular geeky types. So I should probably post it on at least one computer science forum, not just philosophy.

I'm going to post this, turn off all the lights (but not my computer), and see if I can't get sleepy. If not, the lights go on again and I make coffee (and go to work at 7 a.m.). If I go to work today, I must leave the office at 2 p.m., as I have a dentist appt. Tomorrow (Tuesday) I can work on designing my Computaverse computer. So, basically it's 16 possible grid intersections, and a total of (2 ^ (2 ^ 16)) divided by 4 = 2 ^ (65,536 - 2) = 2 ^ 65,534 cells per grid (3D lattice). Each grid corresponds to a unique initial configuration of 2 ^ 65,536 bits, so the total possible no. of unique grids is 2 ^ (2 ^ 65,536). There is no God, only NAND gates and lots of very long wires. End of story.