a few more lines
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@ -659,4 +659,18 @@ as $o_a$ is a Clifford operator:
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\end{equation}
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Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. It is therefore enough to study the measurements
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of any Pauli operator on the vertex operator free graph states.
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of any Pauli operator on the vertex operator free graph states. The commutators of the observable
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with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute
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so it is easier to list the operators that anticommute:
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\begin{equation}
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\begin{aligned}
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A_{\pm X_a} &= \left\{j | \{j, a\} \in E\right\}\\
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A_{\pm Y_a} &= \left\{j | \{j, a\} \in E\right\} \cup \{a\} \\
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A_{\pm Z_a} &= \{a\}\\
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\end{aligned}
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\end{equation}
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This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a$
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is measured the result $+1$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
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