Round robin tournament scheduling using divide and conquer I want to scheduling $n$ teams using divide and conquer method for $n \in \mathbb{N}$.My approach is that if $n$ is even divide it to to group $\frac{n}{2}$ and  $\frac{n}{2}$ but if $n$ is odd, add auxiliary team and scheduling $n+1$ team at the end we remove the last team and in table every match between $n+1$ team and others will mark as rest. but for some $n$ like 6 there will be a little problem. we should first schedule table for 3 team, 3 is odd so we should schedule for 4. for scheduling 4 teams we need schedule for 2 team and that's the base case so we have:

\begin{array}{c|ccc}
 &amp; \text{Day 1} &amp; \text{Day 2} &amp; \text{Day 3} \\
\hline
1 &amp; 2 &amp;  &amp;  \\
2 &amp; 1 &amp;  &amp;  \\
3 &amp; 4 &amp;  &amp; \\
4 &amp; 3 &amp;  &amp; 
\end{array}

now we should conquer the result. upper side of table didn't play with down site of table so we should schedule their matches, and remove team 4. then we have

\begin{array}{c|ccc}
 &amp; \text{Day 1} &amp; \text{Day 2} &amp; \text{Day 3} \\
\hline
1 &amp; 2 &amp; 3 &amp; - \\
2 &amp; 1 &amp; - &amp; 3 \\
3 &amp; - &amp; 1 &amp; 2\\
\end{array}

now we have table for 3 teams it's time to schedule 6 teams so we create table like this

\begin{array}{c|cccccc}
 &amp; \text{Day 1} &amp; \text{Day 2} &amp; \text{Day 3} &amp; \text{Day 4} &amp; \text{Day 5} &amp; \text{Day 6}\\
\hline
1 &amp; 2 &amp; 3 &amp; - &amp; &amp; &amp; \\
2 &amp; 1 &amp; - &amp; 3 &amp; &amp; &amp; \\
3 &amp; - &amp; 1 &amp; 2 &amp; &amp; &amp; \\
4 &amp; 5 &amp; 6 &amp; - &amp; &amp; &amp; \\
5 &amp; 4 &amp; - &amp; 6 &amp; &amp; &amp; \\
6 &amp; - &amp; 4 &amp; 5\\
\end{array}

for conquer step we should schedule match for upper side of table with down side so then

\begin{array}{c|cccccc}
 &amp; \text{Day 1} &amp; \text{Day 2} &amp; \text{Day 3} &amp; \text{Day 4} &amp; \text{Day 5} &amp; \text{Day 6}\\
\hline
1 &amp; 2 &amp; 3 &amp; - &amp; 4 &amp; 5 &amp; 6\\
2 &amp; 1 &amp; - &amp; 3 &amp; 5 &amp; 6  &amp; 4 \\
3 &amp; - &amp; 1 &amp; 2 &amp; 6 &amp; 4 &amp; 5 \\
4 &amp; 5 &amp; 6 &amp; - &amp;1 &amp;2 &amp;3 \\
5 &amp; 4 &amp; - &amp; 6 &amp;2 &amp; 3&amp; 1\\
6 &amp; - &amp; 4 &amp; 5 &amp; 3 &amp; 1 &amp; 2\\
\end{array}

but it's not best answer and we can reschedule rest matches such that tournament end in 5 days for $n = 6$. how can I modify conquer step?
The recursive approach is not necessarily the simplest one here. There is a simple direct construction I am quoting from here. Suppose first that you are trying to schedule $n$ teams, where $n$ is odd. In total there are $n(n-1)/2$ matches to schedule, and in each round at most $(n-1)/2$ matches can take place, so there should be $n$ rounds in total. All computations below are modulo $n$. In round $r$, team $x$ plays against team $r-x$, except for team $r/2$, which stays out. Here is an example for $n=5$:


Round 0: 1–4, 2–3, 0 sits out
Round 1: 0–1, 2–4, 3 sits out
Round 2: 0–2, 3–4, 1 sits out
Round 3: 0–3, 1–2, 4 sits out
Round 4: 0–4, 1–3, 2 sits out


It's easy to prove the correctness of this schedule. Indeed, teams $x$ and $y$ meet in round $x+y$ and in no other round.

If $n$ is even then there are $n(n-1)/2$ matches to schedule, and each round at most $n/2$ can take place, so there should be $n-1$ rounds. Indeed, we take the schedule for $n-1$, and pair up the team sitting out with the extra team. For example, for $n=6$:


Round 0: 1–4, 2–3, 0–5
Round 1: 0–1, 2–4, 3–5
Round 2: 0–2, 3–4, 1–5
Round 3: 0–3, 1–2, 4–5
Round 4: 0–4, 1–3, 2–5


It's again not hard to prove the correctness of the schedule. If $x,y \neq n-1$ then they meet in round $x+y$, while $x$ and $n-1$ meet in round $2x$ (where all computations are modulo $n-1$).