Example $4 A, B, C, D, E, F$ six players are participating in a round-robin table tennis singles tournament (each player plays one match against every other player). Every day, three matches are held simultaneously on three tables. It is known that on the first day, $B$ played against $D$; on the second day, $C$ played against $E$; on the third day, $D$ played against $F$; on the fourth day, $B$ played against $C$. Question: On the fifth day, who will $A$ play against? Who will play against whom on the other two tables?
Explanation: In fact, this problem requires listing the entire match schedule for this competition (Table 2).
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline & Day 1 & Day 2 & Day 3 & Day 4 & Day 5 \\
\hline Court 1 & $B-D$ & $C-E$ & $D-F$ & $B-C$ & $1-?$ \\
\hline Court 2 & (1) & (2) & (3) & (1) & (1) \\
\hline Court 3 & (5) & (ii) & (7) & (8) & (III) \\
\hline
\end{tabular}

First, consider (2). Assume that $D$ plays in this match. From the matches on Court 1 on Day 1 and Day 3, we know that $D$ cannot play against $B$ and $F$. Also, $C$ and $E$ are playing simultaneously, so it must be $D-A$. Therefore, (1) must be $B-F$.

Next, consider (3). Assume that $B$ plays in (3). From the matches on the first, second, and fourth days, we know that $B$ can only play against $A$ or $E$. If $B$ plays against $A$, then (7) must be $C-E$. This contradicts the match on Day 2, so (3) must be $B-E$, and (7) must be $A-C$.

In the first four days, $B$ has played against $C$, $D$, $E$, and $F$. Therefore, on the fifth day, $B$ must play against $A$. Thus, ? must be $B$.
This leads to Table 3.
Table 3
\begin{tabular}{|c|c|c|c|c|c|}
\hline & Day 1 & Day 2 & Day 3 & Day 4 & Day 5 \\
\hline Court 1 & $B-D$ & $C-E$ & $D-F$ & $B-C$ & $A-B$ \\
\hline Court 2 & (1) & $D-A$ & $B-E$ & (1) & (1) \\
\hline Court 3 & (5) & $B-F$ & $A-C$ & (8) & (11) \\
\hline
\end{tabular}

Finally, consider (1). Assume $A$ participates. Then $A$ can only play against $E$ or $F$. If $A$ plays against $F$, then (5) must be $C-E$. This is a contradiction, so (1) must be $A-E$, and (5) must be $C-F$. In the first four days, $C$ has played against $F$, $E$, $A$, and $B$. Therefore, on the fifth day, $C$ must play against $D$. Thus, (9) is $C-D$, and (1) is $E-F$.