# Logique de base/Exercices/Tableau de Karnaugh 2
## Exercice 1

| V | V | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| V | V | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
On observe une symétrie :
| V | V | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| V | V | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
ce qui donne un tableau comme ceci en éliminant la variable a:
| V | V | b c | b c | b c | b c |
| V | V | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 0 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 1 | 0 |
| d e | 1 0 | 0 | 1 | 1 | 0 |
On observe de nouveau une symétrie :
| V | V | b c | b c | b c | b c |
| V | V | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 0 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 1 | 0 |
| d e | 1 0 | 0 | 1 | 1 | 0 |
et redonne un tableau comme celui-ci en éliminant la variable d:
| V | V | b c | b c | b c | b c |
| V | V | 0 0 | 0 1 | 1 1 | 1 0 |
| e | 0 | 0 | 1 | 1 | 0 |
| e | 1 | 0 | 1 | 1 | 0 |
On peut alors facilement résoudre le tableau et ressortir l'équation avec ce rassemblement :
| V | V | b c | b c | b c | b c |
| V | V | 0 0 | 0 1 | 1 1 | 1 0 |
| e | 0 | 0 | 1 | 1 | 0 |
| e | 1 | 0 | 1 | 1 | 0 |
Le rassemblement rouge donne l'équation : V = c

## Exercice 2

| W | W | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| W | W | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
| d e | 0 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| d e | 1 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
On constate que l’on a un axe de symétrie selon la variable a
| W | W | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| W | W | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
| d e | 0 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| d e | 1 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| d e | 1 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
ce qui nous donne maintenant ce tableau
| W | W | b c | b c | b c | b c |
| W | W | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 0 | 0 | 1 |
| d e | 0 1 | 0 | 1 | 0 | 0 |
| d e | 1 1 | 0 | 1 | 0 | 0 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
On retrouve de nouveau un axe de symétrie
| W | W | b c | b c | b c | b c |
| W | W | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 0 | 0 | 1 |
| d e | 0 1 | 0 | 1 | 0 | 0 |
| d e | 1 1 | 0 | 1 | 0 | 0 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
et on peut donc éliminer la variable d
| W | W | b c | b c | b c | b c |
| W | W | 0 0 | 0 1 | 1 1 | 1 0 |
| e | 0 | 1 | 0 | 0 | 1 |
| e | 1 | 0 | 1 | 0 | 0 |
Maintenant on peut résoudre le tableau
| W | W | b c | b c | b c | b c |
| W | W | 0 0 | 0 1 | 1 1 | 1 0 |
| e | 0 | 1 | 0 | 0 | 1 |
| e | 1 | 0 | 1 | 0 | 0 |
* Le rassemblement violet donne : ${\bar {e}}.{\bar {c}}$
* Le rassemblement vert donne : ${\bar {b}}.c.e$
L'équation est donc : $W={\bar {e}}.{\bar {c}}+{\bar {b}}.c.e$

## Exercice 3

| Z | Z | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| Z | Z | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
| d e | 0 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
On constate un axe de symétrie
| Z | Z | a b c | a b c | a b c | a b c | a b c | a b c | a b c | a b c |
| Z | Z | 0 0 0 | 0 0 1 | 0 1 1 | 0 1 0 | 1 1 0 | 1 1 1 | 1 0 1 | 1 0 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |
| d e | 0 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 | 1 |
Et on peut donc éliminer la variable a
| Z | Z | b c | b c | b c | b c |
| Z | Z | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
Il nous reste maintenant à résoudre le tableau
| Z | Z | b c | b c | b c | b c |
| Z | Z | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
| Z | Z | b c | b c | b c | b c |
| Z | Z | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
| Z | Z | b c | b c | b c | b c |
| Z | Z | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
| Z | Z | b c | b c | b c | b c |
| Z | Z | 0 0 | 0 1 | 1 1 | 1 0 |
| d e | 0 0 | 1 | 1 | 1 | 0 |
| d e | 0 1 | 0 | 1 | 1 | 0 |
| d e | 1 1 | 1 | 1 | 1 | 1 |
| d e | 1 0 | 1 | 0 | 0 | 1 |
* Le rassemblement jaune donne : ${\bar {c}}.d$
* Le rassemblement vert donne : $c.{\bar {d}}$
* Le rassemblement rouge donne : $d.e$
* Le rassemblement mauve donne : ${\bar {b}}.{\bar {c}}.{\bar {e}}$
l'équation est donc Z = ${\bar {c}}.d+c.{\bar {d}}+d.e+{\bar {b}}.{\bar {c}}.{\bar {e}}$
| Tableau de Karnaugh 1 | Calcul binaire |