I’m just working my way through these problems in sequence.
1 is not particularly difficult to solve
Let’s imagine the base case: B-G. Obviously, there is 1 biochromatic edge. Adding either B or G to a biochromatic edge will turn it into B-B-G or B-G-G respectively, which means there is still 1 bichromatic edge. If you add B to a B-B or G to a G-G it turns into B-B-B or G-G-G, which does not add or destroy any bichromatic edges. The final case is adding G to B-B or B to G-G, which makes either B-G-B or G-B-G, adding two bichromatic edges. Since adding two to an odd number results in an odd number, and we begin with 1 bichromatic edge, we always have an odd number of edges.
For a formal proof, we’d have to prove the unspoken assumption that we can make any finite linear path made up of Blue/Green nodes where the start is a Blue node and the end is a Green node, by adding Blue/Green nodes in between a B-G path.
The proof is as follows: Besides the start and end nodes, every node has two connections. Thus, we can remove a node and connect its two adjacent nodes to each other in its place. Removing a node this way does not make it no longer a qualifying path under our definitions, and the removal of a node can be undone by adding it back in between the two nodes. Thus, since we can remove all the nodes until we’re left with a single B-G path, we can add them back until we’ve reached the original path, while still ensuring that there is an odd number of bichromatic edges.
I’m just working my way through these problems in sequence.
1 is not particularly difficult to solve
Let’s imagine the base case: B-G. Obviously, there is 1 biochromatic edge. Adding either B or G to a biochromatic edge will turn it into B-B-G or B-G-G respectively, which means there is still 1 bichromatic edge.
If you add B to a B-B or G to a G-G it turns into B-B-B or G-G-G, which does not add or destroy any bichromatic edges.
The final case is adding G to B-B or B to G-G, which makes either B-G-B or G-B-G, adding two bichromatic edges. Since adding two to an odd number results in an odd number, and we begin with 1 bichromatic edge, we always have an odd number of edges.
For a formal proof, we’d have to prove the unspoken assumption that we can make any finite linear path made up of Blue/Green nodes where the start is a Blue node and the end is a Green node, by adding Blue/Green nodes in between a B-G path.
The proof is as follows: Besides the start and end nodes, every node has two connections. Thus, we can remove a node and connect its two adjacent nodes to each other in its place. Removing a node this way does not make it no longer a qualifying path under our definitions, and the removal of a node can be undone by adding it back in between the two nodes. Thus, since we can remove all the nodes until we’re left with a single B-G path, we can add them back until we’ve reached the original path, while still ensuring that there is an odd number of bichromatic edges.