<https://git.mleku.dev/mleku/dendrite/src/branch/dev/spacetime-expansion-hypothesis.md>
Forgot to actually push it. The thing about this document is that it's there in Dendrite because Dendrite is training on everything in the repo, to evolve its lattice data structure so that it can be used to store and retrieve knowledge.
I've been so caught up getting excited about what Claude and I talked about that fed the generation of this paper, that with paid work demands happening and me spacing out about hairstyles, I totally forgot.
No matter. It's there now. In there is also a set of instructions about how to build two experiments. I'm going to build the Wu Xing Pair-Breaker in the next few months.
The most interesting thing that Claude discovered about the geometry of the Wu Xing (the complete directed graph of a pentagon with a star shape at its nodes) is that it has a phi ratio. Here's a freshly generated description of the math:
---
## K₅, the Pentagon, and φ
When you place the five phases at the vertices of a regular pentagon, the complete graph K₅ decomposes into exactly two Hamiltonian cycles:
1. **Shēng (generating) cycle** — the 5 outer edges of the pentagon:
Wood → Fire → Earth → Metal → Water → Wood
(step +1 mod 5, as in your `Generates()` at `wuxing/wuxing.go:44`)
2. **Kè (overcoming) cycle** — the 5 inner diagonals forming the pentagram star:
Wood → Earth → Water → Fire → Metal → Wood
(step +2 mod 5, as in your `Overcomes()` at `wuxing/wuxing.go:55`)
These two cycles together account for all 10 edges of K₅. Every pair of phases has exactly one relationship — generating or overcoming — and the graph is complete.
**The question:** why is the ratio of a diagonal (kè path segment) to a side (shēng path segment) exactly φ = (1 + √5)/2?
## The Proof, Using Only Pythagoras
Place the pentagon on a circle of radius R, vertices at angles 0°, 72°, 144°, 216°, 288°. Vertex coordinates:
```
V_k = (R·cos(72°k), R·sin(72°k))
```
**Side length (shēng edge)** — between adjacent vertices, e.g. V₀ = (R, 0) and V₁ = (R cos72°, R sin72°):
By Pythagoras on the right triangle formed by the horizontal and vertical displacements:
```
s² = (R - R·cos72°)² + (R·sin72°)²
= R²(1 - cos72°)² + R²·sin²72°
= R²(1 - 2cos72° + cos²72° + sin²72°)
```
Since cos² + sin² = 1 (which is itself a statement of Pythagoras on the unit circle):
```
s² = R²(2 - 2·cos72°)
```
**Diagonal length (kè edge)** — between vertices separated by one, e.g. V₀ and V₂ = (R cos144°, R sin144°):
```
d² = (R - R·cos144°)² + (R·sin144°)²
= R²(2 - 2·cos144°)
```
Same Pythagorean expansion, same identity.
**The ratio:**
```
d²/s² = (2 - 2·cos144°) / (2 - 2·cos72°)
= (1 - cos144°) / (1 - cos72°)
```
Now we need the exact cosines. These come from the geometry of the pentagon itself (constructible with straightedge and compass, which is to say, solvable by nested Pythagorean operations):
```
cos 72° = (√5 - 1)/4
cos 144° = -cos 36° = -(√5 + 1)/4
```
Substituting:
```
1 - cos72° = 1 - (√5-1)/4 = (5 - √5)/4
1 - cos144° = 1 + (√5+1)/4 = (5 + √5)/4
```
Therefore:
```
d²/s² = (5 + √5) / (5 - √5)
```
Rationalize by multiplying top and bottom by (5 + √5):
```
d²/s² = (5 + √5)² / (25 - 5) = (25 + 10√5 + 5) / 20 = (30 + 10√5) / 20 = (3 + √5) / 2
```
Now take the square root:
```
d/s = √((3 + √5)/2)
```
Note that φ² = ((1+√5)/2)² = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2.
So:
```
d/s = √(φ²) = φ = (1 + √5)/2 ≈ 1.618...
```
**QED.** The kè diagonal is exactly φ times the shēng side.
Forgot to actually push it. The thing about this document is that it's there in Dendrite because Dendrite is training on everything in the repo, to evolve its lattice data structure so that it can be used to store and retrieve knowledge.
I've been so caught up getting excited about what Claude and I talked about that fed the generation of this paper, that with paid work demands happening and me spacing out about hairstyles, I totally forgot.
No matter. It's there now. In there is also a set of instructions about how to build two experiments. I'm going to build the Wu Xing Pair-Breaker in the next few months.
The most interesting thing that Claude discovered about the geometry of the Wu Xing (the complete directed graph of a pentagon with a star shape at its nodes) is that it has a phi ratio. Here's a freshly generated description of the math:
---
## K₅, the Pentagon, and φ
When you place the five phases at the vertices of a regular pentagon, the complete graph K₅ decomposes into exactly two Hamiltonian cycles:
1. **Shēng (generating) cycle** — the 5 outer edges of the pentagon:
Wood → Fire → Earth → Metal → Water → Wood
(step +1 mod 5, as in your `Generates()` at `wuxing/wuxing.go:44`)
2. **Kè (overcoming) cycle** — the 5 inner diagonals forming the pentagram star:
Wood → Earth → Water → Fire → Metal → Wood
(step +2 mod 5, as in your `Overcomes()` at `wuxing/wuxing.go:55`)
These two cycles together account for all 10 edges of K₅. Every pair of phases has exactly one relationship — generating or overcoming — and the graph is complete.
**The question:** why is the ratio of a diagonal (kè path segment) to a side (shēng path segment) exactly φ = (1 + √5)/2?
## The Proof, Using Only Pythagoras
Place the pentagon on a circle of radius R, vertices at angles 0°, 72°, 144°, 216°, 288°. Vertex coordinates:
```
V_k = (R·cos(72°k), R·sin(72°k))
```
**Side length (shēng edge)** — between adjacent vertices, e.g. V₀ = (R, 0) and V₁ = (R cos72°, R sin72°):
By Pythagoras on the right triangle formed by the horizontal and vertical displacements:
```
s² = (R - R·cos72°)² + (R·sin72°)²
= R²(1 - cos72°)² + R²·sin²72°
= R²(1 - 2cos72° + cos²72° + sin²72°)
```
Since cos² + sin² = 1 (which is itself a statement of Pythagoras on the unit circle):
```
s² = R²(2 - 2·cos72°)
```
**Diagonal length (kè edge)** — between vertices separated by one, e.g. V₀ and V₂ = (R cos144°, R sin144°):
```
d² = (R - R·cos144°)² + (R·sin144°)²
= R²(2 - 2·cos144°)
```
Same Pythagorean expansion, same identity.
**The ratio:**
```
d²/s² = (2 - 2·cos144°) / (2 - 2·cos72°)
= (1 - cos144°) / (1 - cos72°)
```
Now we need the exact cosines. These come from the geometry of the pentagon itself (constructible with straightedge and compass, which is to say, solvable by nested Pythagorean operations):
```
cos 72° = (√5 - 1)/4
cos 144° = -cos 36° = -(√5 + 1)/4
```
Substituting:
```
1 - cos72° = 1 - (√5-1)/4 = (5 - √5)/4
1 - cos144° = 1 + (√5+1)/4 = (5 + √5)/4
```
Therefore:
```
d²/s² = (5 + √5) / (5 - √5)
```
Rationalize by multiplying top and bottom by (5 + √5):
```
d²/s² = (5 + √5)² / (25 - 5) = (25 + 10√5 + 5) / 20 = (30 + 10√5) / 20 = (3 + √5) / 2
```
Now take the square root:
```
d/s = √((3 + √5)/2)
```
Note that φ² = ((1+√5)/2)² = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2.
So:
```
d/s = √(φ²) = φ = (1 + √5)/2 ≈ 1.618...
```
**QED.** The kè diagonal is exactly φ times the shēng side.