Kyo

When people think of a simulation, their only reference point is the common home computer. But dreams are a type of simulation too. Life is but a dream, a pre-programmed dream. ;)

1 week ago (edited) | [YT] | 0

Kyo

A quantum system is like a probability cloud, a dynamic superposition of all possible states, each with amplitude and phase.

Imagine this cloud having a center of mass.

String theory then models that trajectory, the motion of the “center”, not as a single point, but as an extended object (a string or brane) sweeping through higher-dimensional space.

1 week ago | [YT] | 0

Kyo

6 pariahs (centers)
20 pieces (edges and corners)

3 weeks ago | [YT] | 0

Kyo

The geometry of spacetime and the fine tuning constant.

* I forgot to write that the sum of its primes is also 48. 41+5+2

1 month ago (edited) | [YT] | 1

Kyo

Bisecting a cubicle quantum system allows for the reference points to be reduced from 8 to 2, and the leftover 6 classical bits can uniquely identify the state of the system as it unfolds under a 3 qubit and 3 classical bit relation.

1 month ago | [YT] | 0

Kyo

Had chatgpt restate it for me, hopefully in clear terms.

We represent a 5×5×5 Rubik’s cube as a 36-bit classical register plus 8 qubits. All qubits start at |0>, and all classical bits start at 1.
Our goal is to define a shift register rule b = f(Q, C, t) that, given 3 qubits and 3 classical bits, determines the state of a 4th qubit so that after 20 steps the system tends toward all zeros.
The rule may involve comparing probabilities of 2 reference qubits (derived from bisecting the system) or selecting the 4 qubits with the highest summed probabilities at each step. (Non bisected cube) This process is iterated over 20 time steps to ensure convergence.

1 month ago | [YT] | 0

Kyo

A 5x5x5 Rubiks cube is equivalent to a 36-bit + 8 qubit shift register.

Adding more classical bits only increases tedium, not complexity.

Need to find a rule based on the systems' current state that tends towards 0 after 20 time steps.

b = f(Q, S, t)

Q = little r from the wave function
S = big R from the wave function

We need f to assume that all qubits are initialized to 0 and that all 12 bits are initialized to 1 and guarantee the system reaches 0 after t = 20 steps

I believe the relation I am looking for may take 3 qubits and 3 classical bits as input to determine the output of a 4th qubit.

We then compare the 2 probabilities of the 2 reference qubits, or we compute it for all 8 qubits and find a group of 4 qubits with the highest probabilities when summed together.

1 month ago (edited) | [YT] | 0

Kyo

Little r is what i have been referring to as reference points. Big R is what i have been referring to as energy.

1 month ago | [YT] | 0

Kyo

The collatz conjecture states that 3n+1 and 2n can be used to reach 1.

Even if it isn't true for all numbers, if it's true for all numbers up to 43 quintillion, then by numbering the cube and repeated applications, a path can always be found, even if not optimal.
That path can then be optimized by eliminating redundant moves.

Optimal paths require fourier transform.

You would only need 2 algorithms to solve every case.

Although it would probably make sense to use 4 of the qubits and 6 of the bits for powers of 2 and 3 and then 3 qubits and 5 bits for reciprocal powers like (1/3) and (1/9) then just use binary search of dividing by 2 over and over.
Only 1 algorithm needed.

Or would that be considered a radix search?

1 month ago (edited) | [YT] | 1

Kyo

The Rubik's cube is a shift register containing 12 bits and 8 qubits.

Spacetime is a shift register built on information.

Information is electrons and photons.

🤍

1 month ago (edited) | [YT] | 0