# AI theory - maybe useful for bots in games?

edited November 2
binary
make path using binary, steps of binary power = path
= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0), (1 or 0), (1 or 0), (1 or 0)]]

learn path = one binary at a time completes binary power.
= [(2 ^ 0, 2 ^ 1, 2 ^ 2, 2 ^ 3, 2 ^ 4) = [(1), (1), (1), (1), (1)]]

redo binary power correcting binary errors in subsequent versions to match the number made in the binary power

if previous binary path = true, new binary path may also be true.
changing the binary power to be true for a lower binary power, 2 to the 0 power for example = 1,
greater binary power which does not know this binary value of 2 to the power of 1 = 0 incorporates the value into the number their binary power makes
= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1)]]
= [2 ^ 3, 2 ^ 4) = [(1 or 0), (1 or 0)]]

in creating the value the binary power equals, you incorporate lower binary power values into the greater binary power value which sorts out errors = blend binary path

you branch out the various nouns into their own verb binary path, a verb binary path merges greater binary power paths with lower binary power paths
to make the greater binary path true.

the greater the binary power, the more smaller binary values have to be made to fit the greater binary power, the harder to complete the harder to merge.
for example the number pi, you can merge pi to math but not all of the number pi.
and for things that are a rational number you need to have the entire binary power path, this is considerate.

so fitting a irrational number to a rational number is corrupt. you need to take only a part of the irrational number to a rational value = proportion.

therefore, if a path = a binary to some power, made up of smaller binary power that are themselves made up of smaller binary power,
reversing the largest binary value down each power in the path to sort out its origin might fix the one largest path to sort out the smaller paths to their value,
but the smaller value or path is tied to the larger  value so it holds the smaller values in relation to the largest path value.
therefore in sorting out the return to the smallest value the largest value having sorting out the smaller values to their position,
keeps the largest value open as reference for that value or power till that power has completed its return to the smallest value.
= [(2 ^ 0, 2 ^ 1, 2 ^ 2) = [(1), (1), (1), (1)]]
= [2 ^ 4) = [(1 or 0)]]

therefore, to remove smaller power values inaccessible to the larger path power, the larger path power creates a path intermediate that removes the small power value for the larger power path.
= 2 ^ 3

so thats my theory. in practice if you jimmy up an algorithm using my theory the bot would see it needs the intermediate to generate the result wanted.
the binary goes up to the problem, but not down to the solution.
so the intermediate 2 to some power, is held open binary to some power, made up of smaller binary power that are themselves made up of smaller binary power, eventually to 2 ^ 0. each time this happens a new pice of the puzzle is added until the entire puzzle can be solved, all the pieces are there.

in my theory, the bot would see it needs to sort out a value to its original spot, it cant, so it uses an intermediate paths to generate the result wanted.

im not able to test this theory practically yet, im still learning my binary theory and other maths and game programming but for future reference both to myself and others and people who could jimmy up an algorithm using my theory i put this here now.

i could have put this in the off topic forum but its useful for bots which might come in handy for oculus.

• Posts: 2,197 Valuable Player
I’m not sure what I just read but I am impressed 😀
• edited November 2
i edited the first post with some math and this is the post to include the picture i drew describing the idea; • using my previous posts jpeg, here is the math, in the first line of the math, from the 2^2, go to 2^3 to have new numbers.

a -> b
let a = b
a^ 2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b) // a - b = 0
a + b = b
b + b = b
2b = b
2 = 1
becomes
[(2 ^ 4) -> (2 ^ 2 = 4)] // solution -> problem
let (2 ^ 4) = (2 ^ 2 = 4)
(2 ^ 4)^ 2 = (2 ^ 4)(2 ^ 2 = 4)
(2 ^ 4)^2 - (2 ^ 2 = 4)^2 = (2 ^ 4)(2 ^ 2 = 4) - (2 ^ 2 = 4)^2
((2 ^ 4) + (2 ^ 2 = 4))((2 ^ 4) - (2 ^ 2 = 4)) = (2 ^ 2 = 4)((2 ^ 4) - (2 ^ 2 = 4)) // (2 ^ 4) - (2 ^ 2 = 4) = 0
(2 ^ 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)
(2 ^ 2 = 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)
2(2 ^ 2 = 4) = (2 ^ 2 = 4)
2 = 1
with this bitshifting you divide by zero to have AI. or, 2 -> 1, which is not dividing by zero but it is still.

• Posts: 2,197 Valuable Player • edited November 10

[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4)] (2 ^ 3 = 8) = [(->, <-), ->] = 0
[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4) (2 ^ 3 = 8)] (2 ^ 4 = 16) = [(->, <-), (->, <-)] = 1
0 & 1 = 1 bit

///////////
a -> b
let a = b
a^ 2 = ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b) // a - b = 0
a + b = b
b + b = b
2b = b
2 = 1

becomes

[(2 ^ 4) -> (2 ^ 2 = 4)] // solution -> problem
let (2 ^ 4) = (2 ^ 2 = 4)
(2 ^ 4)^ 2 = (2 ^ 4)(2 ^ 2 = 4)
(2 ^ 4)^2 - (2 ^ 2 = 4)^2 = (2 ^ 4)(2 ^ 2 = 4) - (2 ^ 2 = 4)^2
((2 ^ 4) + (2 ^ 2 = 4))((2 ^ 4) - (2 ^ 2 = 4)) = (2 ^ 2 = 4)((2 ^ 4) - (2 ^ 2 = 4)) // (2 ^ 4) - (2 ^ 2 = 4) = 0
(2 ^ 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)
(2 ^ 2 = 4) + (2 ^ 2 = 4) = (2 ^ 2 = 4)
2(2 ^ 2 = 4) = (2 ^ 2 = 4)
2 = 1

with this bitshifting you divide by zero to have AI. or, 2 -> 1, which is not dividing by zero but it is still. in a bit, you have two states 1 = 0.

Now
{
[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4)] (2 ^ 3 = 8) = [(->, <-), ->] = 0
[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4) (2 ^ 3 = 8)] (2 ^ 4 = 16) = [(->, <-), (->, <-)] = 1
0 & 1 = 1 bit
}
= 2^ 0 = 1 bit = 1

2^1 = 2

2^2 = 4, where did 3 go to? you need to mix two bits to make the number 3 [(2^0) + (2^1) = 3] = a different type of bit
because a non-bit number is made of bit numbers to have a solution and problem.

its very unpleasant trying to log in. i have a very difficult time trying to log in. its egregiously tedious and demanding work. so much so its distressful i cant log in for my username and password in multiple browsers for so many tries. by the time i finally get in im so fatigued by it all i have no joy in posting whatsoever.
pitiful.

• Posts: 2,110 Valuable Player
..and how on earth did you conclude this forum was the right place to post that?  Did you run that decision by your "AI" ? PCVR: CV1 || 4 sensors || TPcast wireless adapter || MamutVR Gun stock V3
PSVR: PS4 Pro || Move Controllers || Aim controller
WMR: HP Reverb
• nobody cares. so it doesnt matter. but its for bot decisions which is sort of for ai and games, i guess. but it doesnt matter, nobody cares. no worries friendo.
• Posts: 2,197 Valuable Player
Have you considered posting in the developer section? Most of us are just enthusiasts here.
• edited November 11
this started as a theory with no math, right now i have the math but no code implementation into build-able AI so its of no use to a developer yet.

when you build sw you need to create a diagram with a path from start to finish, the diagram has a line you follow to boxes the boxes show a yes or no that directs the path. without describing what i described here you cant build the diagram in the first play, then with the diagram you have at least a partial ability to code it.

if i posted this in a forum with strict behavior, a math forum or a sw dev forum, i would be in a bad way. but this is a necessary step to build the diagram that is then built into a sw and i figured it has to be shared so i put it here where no one cares and if i get good enough to build ai in the future i can re-read this later on a build it myself.
• Posts: 2,197 Valuable Player
edited November 12
Well it looks very interesting. Let us know how you get on with it.
• Posts: 14,623 Valuable Player
edited November 12
Could also help predict which Stock to invest in, or which Lottery Numbers to purchase!

Interesting data dump. Are you a fan of the Myst games? Check out my Mod at http://www.mystrock.com/
• Posts: 1,809 Valuable Player
@hoppingbunny123

I'm a software dev and I don't have any idea what you're talking about. When you explain something the most important thing is the high level non-technical description. Can you explain what the problem you're trying to solve is and just generally what your solution is?

Is this a path search algorithm like A* search algorithm? i7 6700k 2080ti   Rift-S, Index
• edited December 6
nvm
• edited December 6
nvm
• Posts: 14,623 Valuable Player
This should probably get moved to Off Topic section since it's not really about VR at all.
Are you a fan of the Myst games? Check out my Mod at http://www.mystrock.com/
• edited December 6
nvm

• edited December 6
nvm

• {
[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4)] (2 ^ 3 = 8) = [(->, <-), ->] = 0
[(2 ^ 0 = 1) (2 ^ 1 = 2) (2 ^ 2 = 4) (2 ^ 3 = 8)] (2 ^ 4 = 16) = [(->, <-), (->, <-)] = 1
0 & 1 = 1 bit
}

second arrow pointing right = first arrow pointing right, which puts the second arrow pointing right in-between the first arrow pointing right and the first arrow pointing left, serving as a bridge or seal.

second arrow pointing left = first arrow pointing left,  which puts the second arrow pointing left in-between the first arrow pointing left and the first arrow pointing right, which divides the second arrow pointing right into two.

the above describes a box thats taped shut, then the tape is cut down the middle so that the box isnt shut anymore. this is good.

(the seal that is 0 is divided by 1) = 1/2 = 0, 1

repeating this;
[(->, <-), ->] = 0
[(->, <-), (->, <-)] = 1

becomes;
([(->, <-), (->, <-)] + ->) = 0
([(->, <-), (->, <-)]  + (->, <-)) = 1

theres 1 too few arrows going left in the second example, so as the division happens the arrow pointing left cuts through 0, the second pair of (->, <-) is being cut so that (1 [->, <-]) is being cut as 0.

to fix this you subtract this additional padding of (->, <-), so that the arrow point left can divide (->, <-) = 1.

this is like cutting the tape on the box trying to open the box, but the padding on the inside of the box is holding the second set of lips flush against the top of the box so that cutting the seal the scissors run into this second pair of lips and has to wiggle the scissors so the scissors dont but the box lip only the tape holding the box shut.

• I have degrees in both Physics and Computer Science, work as software dev and my job is creating software bots to automate various tasks. I have no idea what you're trying to do. It's like walking into a class on a proof when it's halfway done and the original question has been erased. Without context, this is all meaningless.