Include late person in all-meet-all speed networking rotation The top-voted answer here appears to solve the problem of rotating guests so that everybody meets everybody. But what if one person shows up late? Is there a way to add them to the rotation after it has already started without messing up the rotation for everyone else? Specifically, everyone already there should


*

*never have to sit out more than once

*never have to visit with the same person twice

*meet everyone else who showed up on time


If a solution does not exists, is there an almost-solution?
 A: Let $N$ be the number of people at the networking event before the latecomer arrives.
The case where $N$ is odd is dealt with by @antkam in the comments: there is a (rotating) person left out in the ordinary rotation anyway, so we can just have that person meet with the latecomer. This way, there are equally many rounds, all the people from the original group of $N$ meet one another, and everyone participates in every round (from the moment that the late-comer arrives on).
Now consider the case where $N$ is even, where the people have been rotating around a table with one person fixed in place, as described in the linked answer. In this case, we can achieve the following: 


*

*the latecomer's addition adds two rounds to the networking game,

*every participant sits out at most two rounds,

*everyone who was on time meets everyone who was on time originally.


(This is a worse solution than you had asked for; but see my explanation at the end.)
The people are rotating around a table with one person in a corner fixed in place. As soon as the late-comer arrives, you designate one chair, also in a corner, on the opposite side of the table from the person fixed in place. You place the newcomer there, and whenever a person from the original group would occupy that seat, they sit that round out instead. Other than that, everyone takes their positions in the original planning.
The original $N - 1$ rounds have the following results:


*

*The latecomer gets to meet someone every round from his arrival on. He fails to meet at least two people: the person fixed in place, and the person who was in the seat opposite him in round 1 (since they missed that round). 

*The person fixed in place meets everyone except the latecomer.

*The people rotating fail to meet at most two on-time people: the person immediately before them in the rotation, and the person immediately after them in the rotation. These are the people they would ordinarily have met at the end of the table where the latecomer is sitting.
Once the original rounds are over, line up all the people who failed to meet a person they would otherwise have met, either because they talked to the latecomer or because they had to skip a round. If you line those people up in the original rotation order, the people they haven't talked to are the people they are adjacent to. Therefore, in the remaining two rounds, you can make pairs out of this line, so that everyone gets to meet every person adjacent to them.
You can furthermore pair up the late-comer with two people he hasn't met, to squeeze out as many meetings as you can.
Note that indeed everyone in the original group gets to meet everyone in the original group: if they hadn't met everyone in the original rounds, they would be in the line-up, and therefore meet the one or two people they had not met.
Furthermore, everyone sits out at most two rounds: if you sit out a round during the original rounds, it must be because the latecomer is sitting in your seat; but then you are in the line-up, and therefore have a meeting in at least one of the final two rounds. 
Thus this strategy achieves the points outlined above.

Note that this is weaker than you had hoped for, but your demand was a bit optimistic: when $N$ in the original version of the problem shifts from even to odd, two rounds are added to the total number of rounds: everyone sits out a round, and everyone has to meet the new person. Therefore it makes sense that two rounds are added even if someone arrives late. 
But then the demand that everyone sits out at most one round becomes too strong: there are 2 more rounds, and therefore $N$ new meeting slots. Of course the newcomer cannot participate in $N$ meetings total, since he already missed at least one. Therefore at least some of the original participants must sit out 2 rounds.
A: Not an answer but too long for a comment.
I just wanna point out the original networking problem (without the late-comer) is equivalent to edge-coloring the complete (undirected) graph $K_n$.  Each edge is a meeting, and edges of the same color are meetings happening simultaneously in one round in the schedule.  The rules of edge-coloring means no node has two edges of the same color, equivalent to no person having two meetings scheduled in the same round.  The schedule length (number of rounds) is the number of colors.
This problem has one of the most beautiful pictorial proofs I have ever come across.  It is constructive and obviously generalizable.
Here is the picture.  See if you can figure out what's going on.  I've always wondered if this proof can be classified as "wordless".  :)
$K_8$">
If you need words, read the caption from the source, or refer to the second paragraph "A complete graph..." of this article.
The text and picture both refer to the case of even $n$, but it's obvious you can drop the center node to get an optimal answer for odd $n$.
I think this assignment is equivalent to the top-voted answer from the other MSE question, with the center node being the stationary person, and the outer edge in a given color being where the moving "array" of people would be "folded" in that round.  Regardless, I offer this as at least a different visualization, from which maybe others can come up with other ways to deal with a late-comer.
