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Russian Cubes

How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.

Tree Graphs

Stage: 4 Challenge Level: Challenge Level:1

The simplest tree graph consists of one line with two vertices, one at each end.

If a new line is added it must connect to one and only one of the existing vertices.

If the new line connected to no vertices, the tree graph would not be connected, as the new line's vertices could not be reached from the existing vertices.

If each end of the new line connects to a vertex the graph will have a circuit and will not be a tree graph.

So every new line added will join on to one existing vertex and create a new vertex at its end. This adds one line and one vertex to the tree graph making no change to the difference between the numbers of edges and vertices. So the difference remains constant at what it originally was.

Marcos's solution is a subtle variation on the above method:

  1. Firstly, it's obvious that every edge is connected to exactly two vertices. (by definition)
  2. More importantly, there is at least one vertex which is connected to exactly one edge.

Proof . If there wasn't at least one such vertex we could keep moving around the graph indefinitely and as there is a finite number of edges it would mean that there is a cycle, counter to the definition of a tree.

Take one such vertex as described in (2) and its respective edge. So far we have 2 vertices and 1 edge. The difference is 1.

(*) Add to this an adjacent edge (this is adding one edge and one vertex. The difference is still one.

Generally, carrying out this step, (*), an arbitrary number of times until we add the final edge will still result in a difference of 1 (as the step was in no way linked to fact that the previous edge was the starting one)

Hence, the number of vertices is one more than the number of edges.