"Certain species can hybridise" (last para but one). The trees in this article can easily be seen in reverse as hybridisation diagrams. In Figure 2, two genes A) TTATGA and B) TTCGGA come together to reduce their differences by replacing A's third letter with B's third letter, and in return B's fourth letter is replaced by A's, to create the hybrid TTCTGA, call it H. The distance between A and B is 2, but between each of them and H is only 1.

What happens then? In the figure this H looks like it goes on to hybridise with yet another gene. But let's imagine instead that it produces three copies. Two of them are exact copies of the original A and B, and the third is an exact copy of itself, the hybrid H. This last, H's identical daughter, in turn produces only two daughters this time, but the daughters of A and B in turn produce only one each.

So far the numerical sequence representing population increase at each stage goes 2 1 3 4. with respective A:H:B ratios of 1:0:1, 0:1:0, 1:1:1, 1:2:1. Continuing to the next stage, A and B produce two copies each. One of the H's produces two but the other H only one, bringing the total to 7, with a ratio of 2:3:2. Try continuing the tree yourself.

The rule we've settled down to seems to be if your node produced two or more last generation, it gets to produce one less in this. But if it produced only one last time, then it produces two this time. The restrained rather than greedy algorithm. And maybe hybrids are a way of reducing phylogenetic distance.

"Certain species can hybridise" (last para but one). The trees in this article can easily be seen in reverse as hybridisation diagrams. In Figure 2, two genes A) TTATGA and B) TTCGGA come together to reduce their differences by replacing A's third letter with B's third letter, and in return B's fourth letter is replaced by A's, to create the hybrid TTCTGA, call it H. The distance between A and B is 2, but between each of them and H is only 1.

What happens then? In the figure this H looks like it goes on to hybridise with yet another gene. But let's imagine instead that it produces three copies. Two of them are exact copies of the original A and B, and the third is an exact copy of itself, the hybrid H. This last, H's identical daughter, in turn produces only two daughters this time, but the daughters of A and B in turn produce only one each.

So far the numerical sequence representing population increase at each stage goes 2 1 3 4. with respective A:H:B ratios of 1:0:1, 0:1:0, 1:1:1, 1:2:1. Continuing to the next stage, A and B produce two copies each. One of the H's produces two but the other H only one, bringing the total to 7, with a ratio of 2:3:2. Try continuing the tree yourself.

The rule we've settled down to seems to be if your node produced two or more last generation, it gets to produce one less in this. But if it produced only one last time, then it produces two this time. The restrained rather than greedy algorithm. And maybe hybrids are a way of reducing phylogenetic distance.