0

Singled Out

$$(X,\textbf{A}^n)_\alpha=\big(X,(\textbf{A}^n)_\beta\big)_\gamma\iff\alpha_i\in\{\theta_0,\theta_1\}\ \ni\ (X)_{\alpha_i}=(X,\textbf{A}^n)_\alpha\big|_{(\textbf{A}^n)=i}$$

Derivation

First we'll start with the desired equality.

$$(\textbf{X}^n,\textbf{A}^m)_\alpha=\big(\textbf{X}^n,(\textbf{A}^m)_\beta\big)_\gamma$$

Our goal is for this equality to be true for all condition values of $\textbf{A}^m$ and $\textbf{X}^n$. But, for the sake of demonstration, let's just look at maintaining the equality for all condition values of $\textbf{A}^m$.

$$(\textbf{X}^n)_{\alpha_i}=(\textbf{X}^n,\beta_i)_\gamma\ \forall i\ \ni\ i=(\textbf{A}^m)_\text{CV}$$

You'll notice that, in this equation, $\beta_i$ can only take two values, 0 and 1. Let's explore what these two cases do for the problem.

$$(\textbf{X}^n)_{\alpha_i}=(\textbf{X}^n,0)_\gamma=(\textbf{X}^n)_{\theta_0}$$ $$(\textbf{X}^n)_{\alpha_i}=(\textbf{X}^n,1)_\gamma=(\textbf{X}^n)_{\theta_1}$$

So, the only way we can maintain the equality for all values of $i$ is for every semi-parse $\alpha_i$ to be one of two species, $\theta_0$ or $\theta_1$.

$$\alpha_i\in\{\theta_0,\theta_1\}\ \forall i\ \ni\ i=(\textbf{A}^m)_\text{CV}$$

From here, the construction of $\beta$ and $\gamma$ are fairly simple, by reversing the semi-parse process.

$$\beta=\sum_{i=0}^{2^n-1}\beta_i2^i$$ $$\gamma=\theta_0+\theta_12^m$$
< Previous
Next >