Building the schedule for a speed date algorithm

Hello everyone.

The Problem
I am working on a speed date project. The "who meets whom" part is already solved, and the output of that is a full list with all the pairs that will be meeting that day. The actual speed dating day will consist of N rounds, and will have P participants. The restrictions are:

• P is an even number
• Each participant is represented by a number, from 1 to P
• N is equal or lower than P-1
• Each person meets N other people, and is seen by N people as well. Consequently, people don't "sit out" (i.e. everyone must meet someone on all rounds)
• People meet in pairs, and naturally, one person can only have one meeting on each round.

The Need
Does anyone know an algorithm that could help me address the issue exposed on Example 2? (i.e. indicates which pairs are meeting in which round).

The Input
As for the input, you are given a list with 2 columns. Each line contains a pair that must meet, and there are " P * N / 2 pairs "(i.e. each round accommodates P/2 pairs). The lowest number between the two people is on the left column. Not sure this matters.

I can also point out that N will probably be 3, 4 or 5, and that P will be somewhere in the range of 30-40 (again, even numbers only). I don't need the fastest solution in the world. If you have an algorithm that solves the problem in under 6 hours with the power of an average desktop, I am happy enough.

Examples
2 examples: Example 1 has 6 people and 4 rounds. Example 2 has 30 people and 4 rounds.

Output of the example 1: One can read the table as: (1st line example) Person1(p1) is meeting p2 on round1, then meets p3 on round 2, then p4 on r3, and finally p5 on r4

Output of the Example 2: This showcases where one can start seeing some issues: as we try to allocate the pairs, some clashes start to occur. 1st clash in this example occurs at Yellow (line 19 of the input), and a 2nd clash occurs at red (line 29 of the input). The yellow case was "solved" by simply delaying the pair 6 & 10 to the 2nd round. But when we try to make players 8 & 9 meet, we can see how attempting to make that happen, would cause a cascade of effects on the table.


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