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Since we have to choose the nearest point from t$_{H}$ , the solution we want is the former. Therefore, τ$_{H}$ becomes ( q$_{1}$, − q$_{1}$, 2 q$_{1}$ − 1). Then theorem 4 claims that the CSS of ρ$_{H}$ is 

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Now, we will show that π$_{H}$ in Eq.(5.8) is really CSS of ρ$_{H}$ by making use of Eq.(1.6). In order to show this we define ˜ π$_{H}$ = ( 1 ⊗ σ$_{x}$ ) π$_{H}$ ( 1 ⊗ σ$_{x}$ ) $^{†}$. Then by making use of Eq.(4.8) and Eq.(4.9) it is straightforward to find a set of the entangled quantum states ˜ ρ ( x ), whose CSS are ˜ π$_{H}$ . After taking the inverse LU transformation one can derive ρ ( x ) = ( 1 ⊗ σ$_{x}$ ) $^{†}$˜ ρ ( x )( 1 ⊗ σ$_{x}$ ). The expression of ρ ( x ) is 

where x ≥ 0 and 

When 

ρ ( x ) reduces to ρ$_{H}$ in Eq.(5.1). It is easy to prove that x$_{H}$ ≥ 0 if λ 2 1 ≥ 4 λ$_{2}$λ$_{3}$ , which is an entangled condition for ρ$_{H}$ . Therefore, theorem 4 is completely proved. 

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Fig. 6 shows how to find the CSS for the generalized Horodecki state ρ$_{H}$ when r = − s = 0 . 3 (Fig. 6a) and r = − s = 0 . 5 (Fig. 6b). If ρ$_{H}$ is explicitly given, compute r$_{H}$ , s$_{H}$ and t$_{H}$ . Then make a straight line which connects a point t$_{H}$ and the nearest vertex of T . Find the crossing points between the line and L $_{r,s}$. As Fig. 3 shows, there are two intersection 

