In a paper I wrote entitled A New Model of Computational Genomics [1], I presented a simple test for ancestry that is impossible to argue with. Let |AB| denote the number of matching bases between two genomes A and B. Given genomes A, B, and C, if we assume that genome A is the common ancestor of genomes B and C, then it is almost certainly the case (see [1] for a discussion of the probabilities) that |AB| > |BC| and |AC| > |BC|. That is, genomes A and B, and A and C, almost certainly have more bases in common than genomes B and C. For intuition, beginning with genome A, and assuming independent mutations away from A to genomes B and C, this is like tossing two independent coins (i.e., the mutations within genomes B and C that deviate from A), which should not have more than chance in common. As such, B and C should deviate away from each other at a faster rate than they deviate from A individually.

Now this is already really powerful, and led me to a complete history of mankind, which is more than plausible. But that said, it’s a necessary condition, not a sufficient condition. That is, if genome A is the common ancestor of genomes B and C, then the inequalities above almost certainly hold, but it’s subject to falsification (i.e., it’s not a sufficient condition). I realized tonight, you can actually transform this into a necessary and sufficient condition.

Specifically, the inequality above can be represented as a graph where A is connected to B, A is connected to C, and B is connected to C, with the match counts labelling the edges of the graph. For example, the edge connected A to B would be labeled with |AB|, which will be some integer. If the inequalities are satisfied, only two such graphs out of six are plausible, for the same reasons that underly the inequality. Specially, if I assume A is the ancestor of B, which is in turn the ancestor of C, then A and C almost certainly have fewer bases in common than A and B.

The graphs that remain, imply that if the inequality is satisfied, then A is almost certainly the ancestor of either B or C, or both, as a necessary and sufficient condition. If we plug in an implausible genome for either B or C (e.g., assuming that the Norwegians = A are the ancestors of Heidelbergensis = B), then the inequality serves as a necessary and sufficient condition for the descendants of the Norwegians, i.e., genome C. I will write more about this tomorrow, including code and some testing.

UPDATE 10/19/25

I’ve implemented a new version of the ancestry algorithm, which so far seems to work. Code is attached below, more to come!