Suppose we have 10 players A, B, C, D, E, F, G, H, I, J (sorted by their ranks).
First-round pairs are A - F, B - G, C - H, D - I, E - J and players A, B, C, D, E win their matches.
What should be the pairs for the second round?
Suggestion #1: A - C, B - D, E - F, G - I, H - J
Suggestion #2: A - D, B - E, C - F, G - I, H - J
or something else.
It's not a chess tournament, so coloring is not important.
There are various Swiss pairing systems. On the assumption that because you are asking in a chess forum you are asking about the most commonly used Swiss pairing systems in the chess world. After your first round you have two scoregroups, each with an odd number of players. That has to be solved either by downfloating from the higher group or upfloating from the lower group.
The FIDE (Dutch) System document is a basically reverse engineered documentation of how FIDE does its pairing, which is why it is hard to read. Here is the relevant line which covers the case you describe:
A (pairing) bracket is a group of players to be paired. It is composed of players coming from one same scoregroup (called resident players) and of players who remained unpaired after the pairing of the previous bracket.
So, after the first round A, B, C, D, E will be in the "1" scoregroup and F, G, H, I, J will be in the "0" scoregroup.
Pairing of the "1" scoregroup will give A-C, B-D with E unpaired. E will be moved (downfloated) to the "0" scoregroup and paired with the top player in that group, "F", and the remaining players paired against each other according to rank. That will give E-F, G-I and H-J.
Thus, your suggestion #1 is how it would normally be done in the chess world.
It should be noted that there are variations. In England when the ECF uses its own patent method E would be paired against H (middle of the lower group) rather than F (top of the lower group). I don't think there is a pairing program which does this, so most arbiters in England would just use Swiss Manager with normal FIDE pairings.
