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---
tags:
- informatique
- sorbonne
- td
- architecture-des-ordinateurs
Semestre: 3
---
4 états -> 2 bits
4 + 3 = 7 transitions
|> 0 -> 1, 1 -> 2, 2 -> 3, 3 -> 0, 1 -> 0, 2 -> 0, 0 -> 0
RST = reset
$F_1 = \overline{\text{RST}} . (c_1 \oplus c_2)$
$F_0 = \overline{\text{RST}} . \bar c_0$
```mermaid
flowchart LR
A-- 1 ---A
A-- 0 ---B
B-- 0 ---B
B-- 1 ---C
C-- 1 ---A
C-- 0 ---D
D-- 0 ---B
D-- 1 ---E
E-- 0,1 ---E
```
```mermaid
flowchart LR
A-- 0 ---B
B-- 1 ---C
C-- 0 ---D
D-- 1 ---E
```
5 états -> 3 bits
4 transitions
| États | $C_2$ | $C_1$ | $C_0$ |
| ----- | ----- | ----- | ----- |
| A | 0 | 0 | 0 |
| B | 0 | 0 | 1 |
| C | 0 | 1 | 0 |
| D | 0 | 1 | 1 |
| E | 1 | 0 | 0 |
ext_A = ext_B = ext_C
WE_A = WE_B = WE_C
ext_D = 0
ADD = 0
1. `RA,RB,RC` -> ext_{A,B,C}+WE_{A,B,C}
2. `RD = RA+RB` -> OP_A+OP_B + ADD + WE_D
3. `RD = RD + RC`
5 états -> 3 bits
6 transitions
init = chargement dans les registres
s0 = comparaison
s1 = a < b -> a = a, b = b-a
s2 = a > b -> a = a-b, b = a
end = a == b -> met le résultat dans le registre
| États | Valeur |
| ----- | ------ |
| init | 0 |
| $S_0$ | 1 |
| $S_1$ | 10 |
| $S_2$ | 11 |
| end | 100 |
| $C_2$ | $C_1$ | $C_0$ | $a>b$ | $a=b$ | $F_2$ | $F_1$ | $F_0$ |
| ----- | ----- | ----- | ----- | ----- | ----- | ----- | ----- |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 0 | 1 | 0 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |
$F_0 = \overline{S_0.((a<b)+(a=b))+end}$
$F_1=S_0.(\overline{a=b})$
$F_2 = end+S_0.(a=b)$
|
