Fig. 1

RF(−) and RF(+) distances. This figure illustrates the difference between the traditional (RF(−)) and RF(+) distance measures when applied to trees with partially overlapping leaf sets. In this example, the leaf sets of \(T_1\) and \(T_2\) are a subset of the leaf set of S. To compute the RF(−) distance between \(T_1\) and S, we must first restrict S to the leaf set of \(T_1\), resulting in tree \(S_1\). The RF(−) distance between S and \(T_1\) is thus \(RF (S_1, T_1)\), which is 2. Likewise, to compute the RF(−) distance between \(T_2\) and S, we must first restrict S to the leaf set of \(T_2\), resulting in tree \(S_2\). The RF(−) distance between S and \(T_2\) is thus \(RF (S_2, T_2)\), which is also 2. In contrast, to compute the RF(+) distance between \(T_1\) and S, we must first compute an optimal completion of \(T_1\) on the leaf set of S (denoted by the dashed red lines), resulting in tree \(T_1'\). The RF(+) distance between S and \(T_1\) is thus \(RF (S, T_1')\), which is 2. Likewise, to compute the RF(+) distance between \(T_2\) and S, we must first compute an optimal completion of \(T_2\) on the leaf set of S, resulting in tree \(T_2'\). The RF(+) distance between S and \(T_2\) is thus \(RF (S, T_2')\), which is 4. Observe that while both \(T_1\) and \(T_2\) are equidistant from S under RF(−), computing the RF(+) distances reveals that \(T_1\) is more similar to S than is \(T_2\)