I've came across a round robin scheduler in a basic number theory text where, for $N$ teams, the mth team, called $\ T_m$, plays versus team $\ T_{m,r}$ in the rth round.(where $1\leq m\leq N$)
So, for an odd number of teams, there are $N$ rounds (where a team is assigned a "bye" if $\ T_{m,r}=m$). For an even number of teams, there are $N-1$ games. So far this is all logical.
The formula that determines who $T_m$ plays is determined by $$T_{m,r}\equiv r-m\ (mod\ N)$$
This is all well and good, until I tried it for N even. On every even round, there would be some weird occurrence where two teams are assigned their own number.
For example, if $N=4$, $T_{1,2}\equiv 1$, and $\ T_{3,2}\equiv -1\equiv 3\ (mod\ 4)$
So the easy answer would be to just switch them? However, this fails harder when $N\geq 6$. You end up getting team 1 playing team 4 twice, and missing team 5. Likewise, Team 2 will never play team 4, instead getting team 5 twice. So the whole table falls apart.
If anyone is familiar with this algorithm, I would appreciate your help in this issue.
The paper I was reading was poorly written (or I just didn't understand). When there are even teams, you just schedule as if there was one fewer team, then assign the "bye" team to play the $N$th team.
I was trying to find a way to make everything work mathematically -- which isn't possible due to the nature of evens in mod arithmetic.