computing the quartet distance between general trees

7.1. THE ALGORITHM 41reflecting Eq. (7.3), however, makes it clear that the symmetry only occurs between twoentries in the matrix. Therefore the butterflies are only counted twice. This is becauseEq. (7.3) is incomplete and only captures half of the constellations of the indices. To producethe total number of shared directed butterflies due to symmetry, the two equations(7.2) and (7.3) should have read:( )1 |F i ∩G j | ∑ ∑ ∑ ∑(|F k ∩G l ||F m ∩G n |) + (|F k ∩G n ||F m ∩G l |) (7.4)4 2k≠i l≠j m≠i ,k n≠j,land( )1 I [i , j ] ∑ ∑4 2k≠i l≠j∑ ∑(I [k,l]I [m,n]) + (I [k,n]I [m,l]). (7.5)m≠i ,k n≠j,lFigure 7.4 b) illustrates a choice of indices in Eq. (7.5). It is clear that for some fixed choiceof the (i , j ) entry, there are four ways to choose the other entries; as soon as one entry isselected, say (k,l), there is only one way to select the remaining three entries (m,n), (k,n)and (m,l) if precisely those four entries are to be selected.As the result in the article is simply wrong by a factor of 2, the succeeding parts ofthe article are still valid, if we divide by 2 instead of 4. One can show that the two subexpressionsof Eq. (7.5) are actually expressing the same quantity and thus, this shouldbe a correct workaround. This is most fortunate, since what remains from that point inthe article is to transform Eq. (7.3) into an expression that is constant time computable.This work will still be valid.jnljnliimmkk(a)(b)Figure 7.4: Illustrates (a) the choice of entries in Eq. (7.3) and (b) the choice of entries in Eq. (7.5)A recipe for counting shared butterflies associated with a pair of edges has now beendescribed. This is done for all pairs of directed edges. As mentioned in the descriptionof directed butterflies, one butterfly is identified by two different edges in one tree. Theconsequence is that each shared butterfly is associated with two different pairs of claims.