previous illustration. The difference is, in this case, $B_{1}$ (or $A_{1}$) is empty.

(0, 0) circle (1 cm); (0,-3) circle (1 cm); (3, 0) circle (1 cm); at (-1.5, 0) [$A_{1}$]{}; at (-1.5,-3) [$A_{2}$]{}; at (4.5, 0) [$B_{2}$]{};

at (0, .5) \[vertex\_closed\] (v1) ; at (-.5, -.5) \[vertex\_closed\] (v2) ; at (.35,-.35) \[vertex\_closed\] (v3) ;

at (3, .5) \[vertex\_open\] (v4) ; at (2.65,-.35) \[vertex\_open\] (v5) ; at (3.5, -.5) \[vertex\_open\] (v6) ;

at (0, -3.5) \[vertex\_open\] (v7) ; at (-.5, -2.65) \[vertex\_open\] (v8) ;

(.6,-.2)–(2.75, 0); (.35, .2)–(2.5, .5); (3,-.4)–(0,-2.65); (3.15, -.5)–(.35, -3);

at (1.5, -4) [$H$]{};

at (7,0) \[vertex\_closed\] (w1) ; at (8,0) \[vertex\_open\] (w2) ; at (7,-1) \[vertex\_open\] (w3) ; (w1)–(w2)–(w3); at (7.5, -1.5) [$K$]{};

It is clear that $H$ is contained in a blow-up of $K$ where $$K=\{\{x\}, \{x,y\}, \{y,z\}\}\subseteq C^{\{1,2\}}(2,1)=\{\{x\}, \{z\}, \{x, y\}, \{z, y\}\}.$$ It follows that $\pi(H)\leq \pi(C^{\{1,2\}})=1$. $\square$

The combination of these propositions completely determines $\pi(H)$ when $R(H)=\{1,2\}$. The results are summarized by the following theorem.

\[t12\] For any hypergraph $H$ with $R(H)=\{1,2\}$, we have $$\pi(H) = \begin{cases} 2-\frac{1}{\chi(H^{2})-1} & \text{if } H^{2} \text{ is not bipartite}; \\
                         \frac{5}{4}             & \text{if } H^{2} \text{ is bipartite and } \min \{k:\bar{P}_{2k}\subseteq H\}=1; \\
             \frac{9}{8}             