6.1. Question 10: Edge Colouring in Graphs

Now let us come to question number 10. Here, we want to prove that if you are given a graph with n vertices, and if you are doing edge colouring, then you cannot use a single colour to colour more than 2n edges. This is a very simple fact, depending upon whether your n is odd or even, we can prove this very easily.

In this section, we start by addressing the concept of edge colouring in graphs. The key to understanding this concept is the fact that every edge in a graph has two endpoints (vertices). Thus, if we colour edges with the same colour, the endpoints of these edges will collectively represent the vertices of the graph. If we try to colour more than a certain number of edges with the same colour, the total endpoints will exceed the number of available vertices in the graph, hence violating the constraints of the graph's structure. We analyze two cases based on whether the number of vertices (n) is even or odd. For an even n, the logic follows that we cannot colour edges in such a way that we exceed the limit of endpoint vertices. For odd n, the arithmetic of edges and endpoints will not align to permit any more than the defined distinct edges.

Imagine arranging chairs for a party where each chair symbolizes a vertex and each pairing of chairs signifies an edge. Now, if you try to pair chairs (edges) but run short of actual chairs (vertices) by using the same set of chairs repeatedly, you will eventually find you simply cannot make more pairs without exceeding the number of chairs available. Each time you use the same chair to create a new pair, you take away the chance to use it in another pair, which mimics the limitations in graph edge colouring.

So let us take the case where n is even. So remember, each edge has 2 endpoints. That means if I consider distinct edges of the graph, and if I focus on their endpoints, that will constitute the entire vertex set. So that means I cannot colour (2n + 1) edges with the same colour, because if I do that, then their endpoints will give me (n + 2) vertices, but my graph at the first place has only n vertices.

When n (the number of vertices) is even, the concept of endpoint counting becomes straightforward. If I try to colour 2n + 1 edges using the same colour, I will end up with 2n + 2 endpoints. However, with only n vertices to work with, this creates an inconsistency. Since each edge connects exactly two vertices, if more edges are coloured with the same colour than there are vertices, it ultimately leads to a scenario where the endpoint count surpasses the available vertices. This proves that only up to n distinct edges can be coloured with the same colour.

Think of a basketball match where each player represents a vertex and each pass made between players is an edge. If there are 10 players, each player can pass to up to 9 other players. If we say every player only uses one ball (colour) to pass, we can easily see that if one player's pass ends up being counted multiple times as using the same ball, we soon reach a point where someone is trying to handle more passes than there are players to receive them, leading to chaos.

Whereas if n is odd, then this quantity is not well defined, it will not be an integer value. So, (n-1)/2 in that context of an odd value of n will be almost n/2. Indeed, it is easy to see that I cannot use a single colour to colour more than (n-1)/2 distinct edges.

In the case of an odd n, the math becomes slightly more complex. The formula for maximum edge colouring becomes less straightforward as dividing an odd integer can lead to another non-integer. Therefore, we can only conclude that a similar constraint applies. Specifically, it indicates that I cannot colour more than (n-1)/2 distinct edges with the same colour. This is based on the same logic as the even case: exceeding the maximum allowed would again violate the limit imposed by the vertex count.

Imagine a team of musicians where each musician represents a vertex and the notes they can play for various songs are akin to edges. If there are 9 musicians and they need to play specific notes, trying to extend the notes played by any individual musician leads to an unplayable situation, creating dissonance. You can only allow so many repetitions without exceeding the number of musicians available to play them.

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