Exemple

Les expressions :

$$C(b_{2}, b_{1}) = ((\Delta(b_{2}) \geq 8) \wedge (\Delta(b_{1... ...)) \vee ((\Delta(b_{2}) \geq 6) \wedge (\Delta(b_{1}) \geq 5))$$

$$\vee ((\Delta(b_{2}) \geq 4) \wedge (\Delta(b_{1}) \geq 6)) \vee ((\Delta(b_{2}) \geq 1) \wedge (\Delta(b_{1}) \geq 7))$$

et

$$C(b_{1}, b_{2}) = ((\Delta(b_{1}) \geq 10) \wedge (\Delta(b_{... ...)) \vee ((\Delta(b_{1}) \geq 6) \wedge (\Delta(b_{2}) \geq 5))$$

$$\vee ((\Delta(b_{1}) \geq 1) \wedge (\Delta(b_{2}) \geq 8))$$

s'écrivent sous forme clausale de la façon suivante :

$$C(b_{2}, b_{1}) = (\Delta(b_{2}) \geq 1) \wedge (\Delta(b_{1}... ...1) \wedge ((\Delta(b_{2}) \geq 8) \vee (\Delta(b_{1}) \geq 5))$$

$$\wedge ((\Delta(b_{2}) \geq 6) \vee (\Delta(b_{1}) \geq 6)) \wedge ((\Delta(b_{2}) \geq 4) \vee (\Delta(b_{1}) \geq 7))$$

et

$$C(b_{1}, b_{2}) = (\Delta(b_{1}) \geq 1) \wedge (\Delta(b_{2}... ...) \wedge ((\Delta(b_{1}) \geq 10) \vee (\Delta(b_{2}) \geq 5))$$

$$\wedge ((\Delta(b_{1}) \geq 6) \vee (\Delta(b_{2}) \geq 8))$$

On cherche ensuite le « maximin » sur chaque buffer :

$$\Delta_{min}(b_{1}) = \max ( x_{min}(C(b_{1}, b_{2})), y_{min}(C(b_{2}, b_{1})))$$

$$\Delta_{min}(b_{2}) = \max ( x_{min}(C(b_{2}, b_{1})), y_{min}(C(b_{1}, b_{2})))$$

Ce qui nous donne :

$$\Delta_{min}(b_{1}) = \max ( x_{l} , y_{1} )$$

$$\Delta_{min}(b_{2}) = \max ( x_{l} , y_{1} )$$

Avec les valeurs :

$$\Delta_{min}(b_{1}) = \max ( 1 , 1 )$$

$$\Delta_{min}(b_{2}) = \max ( 1 , 1 )$$

On remarque que quand les buffers sont tous de sens opposés, les valeurs minimales sont des $1$. Les domaines de définition sont les suivants :

$$\Delta^{0}(b_{1}) = 1, \Delta^{1}(b_{1}) = 5, \Delta^{2}(b_{1}) = 6, \Delta^{3}(b_{1}) = 7, \Delta^{4}(b_{1}) = 10$$

$$\Delta^{0}(b_{2}) = 1, \Delta^{1}(b_{2}) = 4, \Delta^{2}(b_{2}) = 5, \Delta^{3}(b_{2}) = 6, \Delta^{5}(b_{2}) = 8$$