Table 1. The Sattath and Tversky (1977) method for finding repeated neighbors

Chosen set of 4	Sum of distances	Pairs chosen
ABCD	n_AB + n_CD = 22 + 18 = 40	AB, CD
	n_AC + n_BD = 39 + 41 = 80
	n_AD + n_BC = 39 + 41 = 80
ABCE	n_AB + n_CE = 22 + 20 = 42	AB, CE
	n_AC + n_BE = 39 + 43 = 82
	n_AE + n_BC = 39 + 41 = 82
ABDE	n_AB + n_DE = 22 + 10 = 32	AB, DE
	n_AD + n_BE = 39 + 43 = 82
	n_AE + n_BD = 41 + 41 = 82
ACDE	n_AC + n_DE = 39 + 10 = 49	AC, DE
	n_AD + n_CE = 39 + 20 = 59
	n_AE + n_CD = 41 + 18 = 59
BCDE	n_BC + n_DE = 41 + 10 = 51	BC, DE
	n_BD + n_CE = 41 + 20 = 61
	n_BE + n_CD = 43 + 18 = 61
Totals from Column 3 giving the number of times a pair gives the lowest score: AB (3), DE (3), CD (1), CE (1), and BC (1). AB and DE are therefore closest neighbors. The five sequences used in the above example (see Fig. 3) are divided into the five possible groups of four. The sums of distances for each set of sequence pairs for the three possible groupings are then determined, and the closest pairs in each grouping are determined. The closest neighbors overall are those that appear as neighbors most often. In this example, AB and DE appear most often as neighbors. These sequences are then chosen as neighbors to calculate the branch lengths on the phylogenetic tree by the method of Fitch and Margoliash (1987).