| $y_3y_2y_1/x_2x_1$ | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 000 (1) | 3 | 6 | (1) | (1) |
| 001 (2) | (2) | 3 | 1 | (2) |
| 011 (3) | (3) | (3) | (3) | 5 |
| 010 (4) | 3 | (4) | (4) | 1 |
| 110 (5) | 2 | (5) | 6 | (5) |
| 111 (6) | (6) | (6) | (6) | 2 |
However, with this assignment, it’s impossible to make sure there are no races (for example, 1 → 3 or 6 changes multiple bits).
| $y_3/y_2y_1$ | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 0 | 1 | 4 | $\beta$ | 3 |
| 1 | $\alpha$ | 6 | 5 | 2 |
Here, we introduce states $\alpha$ and $\beta$ (since we had 6/8 states). With this assignment, we can go from 1 → 6 via $\alpha$.
| $y_3y_2y_1/x_2x_1$ | 00 | 01 | 11 | 10 |
|---|---|---|---|---|
| 000 (1) | 3 | $\alpha$ | (1) | (1) |
| 001 (4) | $\beta$ | (4) | (4) | 1 |
| 011 ($\beta$) | 3 | X | X | 5 |
| 010 (3) | (3) | (3) | (3) | $\beta$ |
| 110 (2) | (2) | 3 | $\alpha$ | (2) |
| 111 (5) | 2 | (5) | 6 | (5) |
| 101 (6) | (6) | (6) | (6) | $\alpha$ |
| 100 ($\alpha$) | X | 6 | 1 | 2 |
Input: $x$
Output: $a,\,b$
When $x=0$, $a,\,b = 0$
When $x$ goes 0 → 1, either $a$ or $b$ → 1 alternating
(Also known as a frequency divider)
| Output | ||||
|---|---|---|---|---|
| $q/x$ | 0 | 1 | 0 | 1 |
| A | (A) | B | 00 | -0 |
| B | C | (B) | -0 | 10 |
| C | (C) | D | 00 | 0- |
| D | A | (D) | 0- | 01 |
Show:
$a\cdot ((a+b)+ c) = a$
$=a(a+b)+ ac$