Paper 2 · Section 3Boundary

How a tiny signal removes a joint edge

A two-register encoder–decoder exposes a discontinuity in the original minimal-target projection. The calculation tracks the exact graph change alongside the surviving information and returned contrast.

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23075822 · Version 1.0

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Section 4 of 20

3 Minimal-target masking

Consider two prepared bits $(b,c)$ , a deterministic decoder

D(b,c)=(b\oplus c,0),

(3.1)

and an encoder whose first branch carries $b$ only in a correlation:

E_\eps(b,c)= \begin{cases} (\xi,\xi\oplus b),&\text{with probability }1-\eps,\\ (b,\zeta),&\text{with probability }\eps. \end{cases}

(3.2)

The fresh fair bits $\xi,\zeta$ and the branch choice are independent of the prepared source. In this section $0\le\eps\le\tfrac12$ . The native operations are $E_\eps,D$ , with a fixed preparation/route contract and the schedule $w=(E_\eps,D)$ .

Theorem 3.1 (Masking of the projected edge)

For the comparison-wise rule of Definition 2.1, the edge $1\to2$ generated by $E_\eps$ exists at $\eps=0$ and is absent at every $\eps>0$ . The decoder still contributes $2\to1$ and $1\to1$ ; thus the two-component SCC present at zero is absent for positive $\eps$ . Nevertheless the encoder rows, in output order $00,01,10,11$ , are

\begin{align}P_0&=(1/2,\eps/2,0,(1-\eps)/2),\tag{3.3}\\ P_1&=(0,(1-\eps)/2,1/2,\eps/2), \tag{3.4}\end{align}

and obey

\begin{align}\sup_x\TV(E_\eps(x),E_0(x))&=\eps/2,\tag{3.5}\\ d_1=\eps,\qquad d_2&=0,\qquad d_{12}=1-\eps,\tag{3.6}\\ r_w&=1-\eps. \tag{3.7}\end{align}

Here $d_L=\TV(P_{L,0},P_{L,1})$ and $r_w$ is the contrast of the fixed terminal root observation.

Proof

Enumerating the two branches gives Equation 3.3, Equation 3.4. At zero both singleton laws are fair and identical under the two preparations, but the pair has opposite parity. The pair is therefore a minimal target and projects to both output coordinates. For positive $\eps$ , the first-bit laws differ by $\eps$ , while the second remains fair. The pair is no longer minimal; its only minimally distinguishing subset is the first coordinate. Changing $c$ affects neither encoder row. The deterministic decoder depends on both inputs only through its first output, yielding the stated remaining edges.

Subtracting a row at zero from its perturbed version gives absolute mass changes $\eps/2$ on each of two atoms, hence Equation 3.5. Marginalization gives $d_1,d_2$ , and direct subtraction gives $d_{12}=1-\eps$ on the stated interval. Under the first branch of $w$ , decoding returns $b$ exactly. Under the second it returns $b\oplus\zeta$ , which is fair. Its error relative to $b$ is $\eps/2$ , so the two terminal root laws differ by $1-\eps$ .

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The conclusion concerns the nominated graph and its candidate SCC, not the qualification or phenomenal status of every resulting proper subcomponent. In particular, Equation 3.7 does not preserve the old graph-derived route certificate after the edge has been removed. It shows that the same executable operations can retain a large terminal distinction while that exact graph criterion changes.

Remark 3.2 (No continuous positive-support-exact weight)

On the family of Theorem 3.1, no continuous nonnegative function of the nominated joint kernels is positive exactly when the published edge $1\to2$ is present.

Proof

Such a function would be zero at every positive $\eps$ , hence zero at zero by continuity, while exact support would require it to be positive there.

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This excludes a particular exact-support repair, not continuous diagnostics in general. It also differs from the familiar addition of weak edges that merges SCCs: here adding a small proper-subset effect deletes an edge supported by a strong joint effect. A tolerance-based rule may be useful, but it changes the original dependence relation rather than continuously reproducing its positive support.

Corollary 3.3 (Finite-use certification limit)

Suppose the nominated complete encoder instrument records only the stipulated two-bit output/successor data. With matched initial laws, other mechanisms, and permissions, any causal experiment using the encoder at most $N$ times has complete-record discrepancy at most

1-(1-\eps/2)^N\le N\eps/2

(3.8)

between the zero and perturbed models. No fixed finite such experiment uniformly certifies the zero-versus-positive graph assignment against all positive $\eps$ .

Proof

While the two histories agree, couple each encoder call with mismatch probability at most $\eps/2$ , using Equation 3.5, and couple all other calls identically. At most $N$ opportunities give agreement probability at least $(1-\eps/2)^N$ . The coupling inequality proves the first bound and the elementary product inequality proves the second. A test's rejection probabilities differ by at most this TV bound, which tends to zero at fixed $N$ .

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A separately retained branch flag, seed archive, or returning receiver enlarges the complete instrument and requires its own bound. Nor are the two models exactly operationally equivalent: for $\eps>0$ , even a singleton law differs. The distinction is between exact-law information and the inability to resolve arbitrarily small changes uniformly with fixed resources.