Grammar

$S \to ABC$, $A \to aA \mid \varepsilon$, $B \to bB \mid \varepsilon$, $C \to cC \mid d$

FIRST sets

• $\mathrm{FIRST}(A) = \{a, \varepsilon\}$
• $\mathrm{FIRST}(B) = \{b, \varepsilon\}$
• $\mathrm{FIRST}(C) = \{c, d\}$
• $\mathrm{FIRST}(S) = (\mathrm{FIRST}(A)\setminus\varepsilon) \cup (\mathrm{FIRST}(B)\setminus\varepsilon) \cup \mathrm{FIRST}(C) = \{a, b, c, d\}$ (since $A$ and $B$ can both vanish, $C$ is reached)

FOLLOW sets

• $\mathrm{FOLLOW}(S) = {$}$
• $\mathrm{FOLLOW}(A) = (\mathrm{FIRST}(B)\setminus\varepsilon) \cup \mathrm{FIRST}(C) = \{b, c, d\}$ (from $S \to ABC$, as $B$ is nullable)
• $\mathrm{FOLLOW}(B) = \mathrm{FIRST}(C) = \{c, d\}$
• $\mathrm{FOLLOW}(C) = \mathrm{FOLLOW}(S) = {$}$

Non-terminal	FIRST	FOLLOW
$S$	$\{a,b,c,d\}$	${$}$
$A$	$\{a,\varepsilon\}$	$\{b,c,d\}$
$B$	$\{b,\varepsilon\}$	$\{c,d\}$
$C$	$\{c,d\}$	${$}$

LL(1) parsing table

• $S \to ABC$: $\mathrm{FIRST}(ABC) = \{a,b,c,d\}$, so fill $a,b,c,d$.
• $A \to aA$ on $a$; $A \to \varepsilon$ on $\mathrm{FOLLOW}(A) = b,c,d$.
• $B \to bB$ on $b$; $B \to \varepsilon$ on $\mathrm{FOLLOW}(B) = c,d$.
• $C \to cC$ on $c$; $C \to d$ on $d$.

| | $a$ | $b$ | $c$ | $d$ | \0 | `aabd | $S \to ABC$ | | 2 | ABC\1 | `aabd | match a |
| 4 | ABC\2 | `abd| matcha | | 6 | ABC\3 | `bd | $B \to bB$ |
| 8 | bBC\4 | `d | $B \to \varepsilon$ | | 10 | C\5 | `d | match d |
| 12 | $$$| `$` | accept |

The string aabd is accepted.