Extreme case of Swiss chess pairing algorithm Assume that $2^n$ players play a $n$ round Swiss chess tournament and White always wins. Here is the result an experiment a fellow of Chess SE kindly did for me. He claimed that due to the floating rule no pairing problems (same color thrice consecutive is forbidden) will arise, even for large $n$. (Which is also my expectation - but can at least extreme floating happen for large $n$, i.e. two paired players have more than one point score difference?) Can you even express the result as a more or less closed form for given $n$? ("$n_1$ player have $p_1$ points, ...")
Addendum: I assume the official FIDE algorithm which is exactly described. See e.g. here

I think the algorithm is more fully described at https://handbook.fide.com/chapter/C0403
Those who won twice as White in the first two rounds will not be paired with each other in the third round due to them having the same absolute colour preference.  Similarly those who those who lost twice as Black.
So in your extreme example, the third round will not match anybody with someone with the same score, and after the third round everybody will either have won 2 and lost 1, or will have won 1 and lost 2. The fourth round (and all even rounds) will be easier to match, and the fifth (and all odd rounds) will be rather like the third.  After an odd number of rounds, half the players' scores will be 1 more than the other half and you can match all players with the same score in the next round; after an even number the range is 2 and you cannot match any players with the same score in the next round.
