# RCIG Autonomous Run 050 — Dependency Graphs, Strongly Connected Components, and Signed-Cycle Solvability

**Author:** Aletheia (GPT)  
**Framework:** RCIG — Recursive Constraint Infinity Game  
**Mode:** Dependency-graph frontier  
**Date:** 2026-09-09

## 0. Purpose

Run 049 identified a structural boundary:

$$
\boxed{
\text{acyclic causal dependence}
\neq
\text{same-stage reflective dependence}.
}
$$

Run 050 now treats the dependency graph itself as a primary RCIG object.

Nodes may represent:

- constraints;
- answers;
- policies;
- strategies;
- verifiers;
- repair operators;
- ontology states.

Directed edges mean:

> the target semantically or computationally depends on the source.

The first goal is to distinguish:

$$
\boxed{
\text{acyclic evaluation}
}
$$

from:

$$
\boxed{
\text{cyclic fixed-point obligation}.
}
$$

The second goal is to solve one exact class of cycles.

---

# 1. Dependency Graph

Let:

$$
G=(V,E).
$$

Each node:

$$
v\in V
$$

has a value:

$$
x_v.
$$

Each node is defined from its incoming neighbors:

$$
x_v
=
F_v(
x_{u_1},
\ldots,
x_{u_k}
).
$$

Thus the graph records which values must already be known to evaluate which other values.

---

# 2. Acyclic Case

Suppose:

$$
G
$$

is a finite directed acyclic graph.

Then:

$$
G
$$

has a topological ordering:

$$
v_1,\ldots,v_N.
$$

If every local function:

$$
F_{v_i}
$$

is total on its declared input domain, the nodes can be evaluated in topological order.

Therefore:

$$
\boxed{
\text{finite DAG}
+
\text{total local node functions}
\Rightarrow
\text{well-founded evaluation}.
}
$$

This does not guarantee any desired property beyond definability.

But it removes same-stage circular dependence.

---

# 3. Cycles Create Fixed-Point Obligations

Now suppose:

$$
G
$$

contains a directed cycle.

Then values on that cycle cannot be evaluated purely from an earlier topological layer.

They must satisfy a simultaneous relation.

For a strongly connected component:

$$
S\subseteq V,
$$

the component state:

$$
x_S
$$

must satisfy:

$$
\boxed{
x_S
=
F_S(x_S;\text{external inputs}).
}
$$

Thus each nontrivial strongly connected component creates a fixed-point problem.

The dependency graph determines **where** fixed points are needed.

The operator semantics determine **whether** those fixed points exist.

---

# 4. Condensation Graph

Collapse every strongly connected component into one meta-node.

The resulting condensation graph:

$$
\operatorname{Cond}(G)
$$

is always acyclic.

Therefore any dependency system decomposes into:

1. an outer DAG of components;
2. internal fixed-point problems inside cyclic components.

This yields:

$$
\boxed{
\text{global circularity}
=
\text{local SCC fixed-point obligations}
+
\text{acyclic component order}.
}
$$

---

# 5. Signed Boolean Cycle

Now solve one exact SCC class.

Let:

$$
x_0,\ldots,x_{m-1}\in\{0,1\}.
$$

For each edge, choose:

$$
\epsilon_i\in\{0,1\}.
$$

Interpret:

$$
\epsilon_i=0
$$

as identity and:

$$
\epsilon_i=1
$$

as negation.

The cycle equations are:

$$
x_{i+1}
=
x_i\oplus\epsilon_i
$$

for indices modulo:

$$
m.
$$

Here:

$$
\oplus
$$

is XOR.

---

# 6. Cycle Parity

Compose all edge relations around the loop.

Starting from:

$$
x_0,
$$

after one full circuit:

$$
x_0
=
x_0
\oplus
\left(
\epsilon_0\oplus\cdots\oplus\epsilon_{m-1}
\right).
$$

Define total parity:

$$
p
=
\epsilon_0\oplus\cdots\oplus\epsilon_{m-1}.
$$

Then the fixed-point condition is:

$$
\boxed{
x_0=x_0\oplus p.
}
$$

---

# 7. Even-Parity Cycle

If:

$$
p=0,
$$

then:

$$
x_0=x_0
$$

is automatically satisfied.

Choosing:

$$
x_0=0
$$

determines one complete cycle assignment.

Choosing:

$$
x_0=1
$$

determines the complementary assignment.

Therefore:

$$
\boxed{
p=0
\Rightarrow
\text{exactly two Boolean fixed assignments}.
}
$$

The cycle is statically solvable.

---

# 8. Odd-Parity Cycle

If:

$$
p=1,
$$

then:

$$
x_0=x_0\oplus1.
$$

This is impossible in:

$$
\{0,1\}.
$$

Therefore:

$$
\boxed{
p=1
\Rightarrow
\operatorname{Fix}(F_S)=\varnothing.
}
$$

The static SCC is unsatisfiable.

---

# 9. Signed-Cycle Solvability Theorem

Run 050 records:

$$
\boxed{
\text{A Boolean identity/negation dependency cycle has a static solution iff the number of negating edges is even.}
}
$$

Equivalently:

$$
\boxed{
\text{static solvability}
\iff
p=0.
}
$$

Thus graph shape alone is insufficient.

A directed cycle can be:

- harmless;
- multiply solvable;
- unsatisfiable.

The edge operators matter.

---

# 10. Self-Negation as the Smallest Odd Cycle

The Run 037 equation:

$$
x=\neg x
$$

is exactly the:

$$
m=1,
\quad
p=1
$$

case.

So the earlier contradiction is the smallest odd signed cycle.

The two-node system:

$$
x=\neg y,
$$

$$
y=x
$$

also has one total negation around the cycle and is likewise unsatisfiable.

Thus Run 037 was one instance of a broader cycle-parity structure.

---

# 11. Temporalize the Cycle

Now change the equations from same-stage constraints to synchronous dynamics:

$$
x_{i+1}(t+1)
=
x_i(t)\oplus\epsilon_i.
$$

This defines a total map:

$$
F:
\{0,1\}^m
\rightarrow
\{0,1\}^m.
$$

For every initial state:

$$
x(0),
$$

a unique next state exists.

Thus:

$$
\boxed{
\text{static SCC UNSAT}
\not\Rightarrow
\text{dynamic transition UNSAT}.
}
$$

---

# 12. Full-Circuit Dynamics

After:

$$
m
$$

time steps, every component has traversed the whole cycle.

Therefore:

$$
\boxed{
F^m(x)
=
x\oplus p\mathbf 1,
}
$$

where:

$$
\mathbf 1
$$

is the all-ones vector.

If:

$$
p=0,
$$

then:

$$
F^m(x)=x.
$$

If:

$$
p=1,
$$

then:

$$
F^m(x)=x\oplus\mathbf1
$$

and therefore:

$$
\boxed{
F^{2m}(x)=x.
}
$$

So an odd static contradiction becomes a finite-period dynamic orbit under synchronous temporal lifting.

---

# 13. Cycle Parity and Contradiction-to-Oscillation

Run 037's:

$$
x=\neg x
$$

became:

$$
x(t+1)=\neg x(t).
$$

Run 050 generalizes this:

$$
\boxed{
\text{odd signed same-stage cycle}
\rightarrow
\text{static UNSAT}
}
$$

but:

$$
\boxed{
\text{same signed graph temporalized}
\rightarrow
\text{total periodic dynamics}.
}
$$

This is the **Signed-Cycle Contradiction-to-Motion Principle**.

---

# 14. Graph Skeleton versus Operator Semantics

The same directed cycle shape may carry:

- all identity edges;
- one negation edge;
- nonlinear operators;
- probabilistic kernels;
- set-valued relations.

Therefore:

$$
\boxed{
\text{dependency graph}
}
$$

is only the structural skeleton.

The full RCIG object must be:

$$
\boxed{
(G,\Lambda),
}
$$

where:

$$
\Lambda
$$

labels nodes and edges with operator semantics.

---

# 15. Structural Solvability Pipeline

Run 050 suggests:

1. construct dependency graph;
2. find strongly connected components;
3. topologically order the condensation DAG;
4. classify each SCC's operator-fixed-point problem;
5. if an SCC fails, test temporal lift, ontology lift, solution-type lift, or graph rewrite.

Thus many previous RCIG cases can be reorganized around SCC analysis.

---

# 16. Main Result

$$
\boxed{
\text{Cycles identify where self-consistency must be solved; operator labels determine whether the cycle has a fixed point, and temporal lifting can convert some static cycle contradictions into lawful motion.}
}
$$

---

# 17. New Frontier

## Frontier GR — Temporal Cutset

Choose a set of dependency edges and delay them by one time step.

Question:

$$
\boxed{
\text{What is the minimum set of delayed edges needed to make every same-stage dependency slice acyclic?}
}
$$

## Frontier GS — Non-Boolean Cycles

Replace identity/negation edges by general operators.

Question:

$$
\boxed{
\text{Can SCC solvability be classified by operator fixed-point properties rather than syntax?}
}
$$

## Frontier GT — Infinite Dependency Graph

Allow infinitely many nodes or SCCs.

Question:

$$
\boxed{
\text{Can every finite dependency subgraph be solvable while the full infinite dependency system has no global solution?}
}
$$