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NeighborNet is consistent

Case 1: {} ∩ {u, v} = 0. The above inequalities follow immediately since d is circular, and d and d' as well as and coincide on Y'\{u, v}.
Case 2: {} ∩ {u, v} = 1. Consider the situation = u. Then
The other inequalities can be derived in a completely analogous way.
Case 3: {} ∩ {u, v} = 2. Then we have = u and = v and
The other inequality can be shown to hold in a similar way. ■
The procedure FINDORDERING calls itself recursively with and d' as input. An ordering of Y', the union of , is returned. By Proposition 4.3 and the induction hypothesis, this ordering Θ' is compatible with and d'. It is used to construct an ordering Θ on Y, in line 14, which becomes the output of the procedure.
Proposition 4.4 The ordering Θ is compatible with collection and with the distance function d.
Proof: Since is compatible with Θ' it is straightforward to check that is compatible with Θ. Hence we only need to show that Θ is compatible with d.
Let orderings = y_{1}, ..., y_{n }of Y and = z_{1}, ..., z_{n1 }of Y' be as in the proof of Proposition 4.3 and let ω be the split weight function such that d = d_{ω}. Then is compatible with all splits S such that ω(S) > 0. Now consider some split S = {A, B} such that ω(S) > 0 and assume that y_{n }∈ B. Then there exists i, j ∈ {1, ..., n  1}, i ≤ j, such that A = {y_{i}, ..., y_{j}}. Note also that, since the distance function d' is compatible with ordering = z_{1}, ..., z_{n1 }of Y' and, hence, is circular, there exists a unique circular split weight function ω': Ϭ(Y') → _{≥0 }with the property that d' = d_{ω'}. We divide the remaining argument into five cases.
Case 1: j ≤ 3. Then, clearly, S is compatible with Θ.
Case 2: j ≥ 4 and i = 1. Define A' = {z_{1}, ..., z_{j1}} and the split S' = {A', Y'\A'} of Y'. Then we can express ω'(S') in terms of d' as follows (cf. [12]):
Thus, ω'(S') > 0. Hence, the split S' is compatible with the ordering Θ' of Y'. But then the split S is compatible with the ordering Θ of Y.
Case 3: j ≥ 4 and 2 ≤ i ≤ 3. We only consider the situation when i = 2; the situation i = 3 is completely analogous. Define A' = {z_{2}, ..., z_{j1}} and the split S' = {A', Y'\A'} of Y'. With a similar calculation as made for Case 2 we obtain ω'(S') ≥ (α + β)ω(S). Hence, ω'(S') > 0 and, thus, S' is compatible with Θ'. But then S is compatible with Θ.
Case 4: j ≥ 4 and i = 4. This case is similar to Case 2. Define A' = {z_{4}, ..., z_{j1}} and S' = {A', Y'\A'}. We obtain ω'(S') ≥ ω(S). Hence, as for Case 2, ω'(S') > 0 and, thus, S is compatible with Θ.
Case 5: j ≥ i ≥ 5. Define the split S' = {A, Y'\A}. Then we have ω'(S') = ω'(S') > 0. Hence, S' is compatible with Θ' and, thus, S is compatible with Θ. ■
4.2 Case 2: The selection case
Now suppose that there are no clusters C ∈ with C ≥ 3. This is the selection case in the description of the algorithm.
In line 17 the algorithm selects two clusters that minimize (3):
where
Note that is a distance function defined on the set of clusters . We will first show that is circular. We do this in two steps: Proposition 4.5 and Proposition 4.6.
Proposition 4.5 Let d: M × M → _{≥0 }be a circular distance function and Θ = x_{1}, ..., x_{n }be an ordering of M that is compatible with d. Let M' = (M\{x_{1}, x_{2}}) ∪ {y} where y is a new element not contained in M. Define a distance function d': M' × M' → _{≥0 }as follows:
where λ is a real number with the property that 0 λ
(i) d' is circular and compatible with ordering y, x_{3}, ..., x_{n }of M'.
(ii) If z_{1}, ..., z_{n1 }is an ordering of M' that is compatible with d' then at least one of the orderings x_{1}, x_{2}, z_{2}, ..., z_{n1 }or x_{2}, x_{1}, z_{2}, ..., z_{n1 }of M is compatible with d.
Proof: (i) and (ii) can be proven using convexity arguments, or in a way analogous to our proof of Propositions 4.3 and 4.4, respectively. ■
Proposition 4.6 The distance function , defined on the individual clusters in , is a circular distance. Moreover, for every ordering D_{1}, ..., D_{k }of that is compatible with there exist orderings Θ_{i }of D_{i}, i ∈ {1, ..., k}, such that the ordering Θ_{1}, ..., Θ_{k }of Y is compatible with distance function d.
Proof: We use multiple applications of Proposition 4.5, once for each cluster in with two elements, and with λ = in each case. ■
We now have the more difficult task of showing that clusters C_{1 }and C_{2 }selected by the Qcriterion, that is by minimizing (3), are adjacent in at least one ordering of the clusters that is compatible with , as described in Proposition 4.6. This is the most technical part of the proof. The key step is the inequality established in Lemma 4.7. This is used to prove Theorem 4.8, which establishes that the Qcriterion when applied to a circular distance will always select a pair of elements that are adjacent in at least one ordering compatible with the circular distance. As a corollary it will follow that there exists an ordering of the clusters in compatible with where C_{1 }and C_{2 }are adjacent.
Lemma 4.7 Let Θ = x_{1}, x_{2}, ..., x_{n }be an ordering of M that is compatible with circular distance d on M and suppose that 3 ≤ r ≤ ⌈n/2⌉. Let S = {A, M\A} be a split compatible with Θ where A = {x_{i}, ..., x_{j}}. Define Q_{S}: M × M → by
and let
(i) If min{A, M\A} > 1 and A ∩ {x_{1}, x_{r}} = 1 then λ(S)
(ii) Any other split S compatible with Θ satisfies λ(S) ≤ 0.
Proof: Expanding λ(S) gives
We divide the rest of our argument into five cases which are summarized in Table 1. For these cases straightforward calculations yield the entries of Table 2. Using Table 2 we compute λ(S) in each case.
Case (i): We obtain λ(S) = 2(j  1)(j + 1  r) + 2(j  1)(j + 1  n). Hence, λ(S) = 0 if j = 1 and λ(S) j ≥ 2.
Case (ii): We obtain λ(S) = 0.
Case (iii): We obtain λ(S) = (j  i)(4(j  i)  2n + 8). Thus, since j  i ≤ r  3 ≤ (n + 1)/2  3, λ(S) = 0 if i = j and λ(S) i j.
Case (iv): We obtain λ(S) = 2(i  r)(n  2  (j  i)) + 2(2  i)(j  i). Thus, since j  i ≤ n  3, λ(S) i r. If i = r then λ(S) = 0 if j = r and λ(S)
Case (v): We obtain λ(S) = 0. ■
Theorem 4.8 Let M be a set of n elements and d: M × M → _{≥0 }be a circular distance function. Suppose that x, y minimize
Then there is an ordering of M that is compatible with d in which x and y are adjacent.
Proof: Let Θ = x_{1}, ..., x_{n }be an ordering of M that is compatible with d. Suppose that Q(x_{1}, x_{r}) ≤ Q(x, y) for all x, y where, without loss of generality, 2 ≤ r ≤⌈n/2⌉. If r = 2 then we are done, so we assume r ≥ 3. Let ω be the (circular) split weight function for which d = d_{ω}, so Θ is compatible with ω. Let Θ* be the ordering obtained by removing x_{r }from Θ and reinserting it immediately after x_{1}. We claim that Θ* is also compatible with ω.
As in Lemma 4.7, for any split S compatible with Θ we define
By the choice of x_{1 }and x_{r }we have
Since Q is linear, and d = Σ_{S∈Ϭ(X)}ω(S)d_{S }by Lemma 4.7 we have
Now consider any split S compatible with Θ but not Θ*. Then S satisfies the conditions in Lemma 4.7 (i), giving λ(S) ω(S) = 0. Thus there are no splits in the support of ω that are not compatible with Θ*, and Θ* is compatible with ω and, hence, d. Thus x_{1 }and x_{r }are adjacent in an ordering Θ* compatible with d. ■
Corollary 4.9 Let C_{1 }and C_{2 }be the two clusters selected in line 17 of procedure FINDORDERING. Then there exists an ordering Θ* = D_{1}, ..., D_{k }of such that D_{1 }= C_{1}, D_{2 }= C_{2 }and is compatible with Θ*.
After selecting C_{1 }and C_{2 }the procedure FINDORDERING removes these clusters from the collection and replaces them with their union C' = C_{1 }∪ C_{2}. It also assigns an ordering Θ_{C' }to the cluster.
FINDORDERING is then called recursively. The following is directly analogous to Proposition 4.3.
Proposition 4.10 There exists an ordering of Y that is compatible with collection and split weight function ω.
Proof: We already know by Proposition 4.9 and Proposition 4.6 that there exists an ordering = y_{1}, ..., y_{n }of Y that is compatible with and ω and, in addition, also satisfies one of the following properties:
If x_{1 }∈ C_{1 }and x_{2 }∈ C_{2 }are selected such that is also compatible with then we are done. Otherwise we have to construct a suitable new ordering of Y. There are, up to symmetric situations with roles of C_{1 }and C_{2 }swapped, only two cases we need to consider.
Case 1: C_{1 }= {y_{1}, y_{2}}, x_{1 }= y_{1 }and x_{2 }= y_{3}. We want to show that ordering = y_{2}, y_{1}, y_{3}, ..., y_{n }is compatible with ω. To this end we first show that [d](y_{2}, y_{3}) ≤ [d](y_{1}, y_{3}). It suffices to establish this inequality for all split metrics d_{S }with S ∈ . Define the set of splits
Ϭ' = {{{y_{2}, ..., y_{i}}, Y\{y_{2}, ..., y_{i}}}3 ≤ i ≤ n  1}. 
By a case analysis similar to the one applied in the proof of Lemma 4.7 we obtain the following:
• [d_{S}](y_{2}, y_{3}) = [d_{S}](y_{1}, y_{3}) if S ∈ \Ϭ', and
• [d_{S}](y_{2}, y_{3}) [d_{S}](y_{1}, y_{3}) if S ∈ Ϭ'.
But then, since [d](y_{1}, y_{3}) is minimum, [d](y_{2}, y_{3}) = [d](y_{1}, y_{3}). Thus, by the above strict inequality, for every split S ∈ Ϭ' we must have ω(S) = 0. Hence, ω is compatible with .
Case 2: C_{1 }= {y_{1}, y_{2}}, C_{2 }= {y_{3}, y_{4}}, x_{1 }= y_{1}, x_{2 }= y_{4 }and n ≥ 5. We want to show that = y_{2}, y_{1}, y_{4}, y_{3}, y_{5}, ..., y_{n }is compatible with ω. A similar argument to the one used in Case 1 shows that for every split S in
Ϭ' = {{{y_{2}, ..., y_{i}}, Y\{y_{2}, ..., y_{i}}}3 ≤ i ≤ n  1} ∪ {{{y_{4}, ..., y_{i}}, Y\{y_{2}, ..., y_{i}}}5 ≤ i ≤ n} 
we must have ω(S) = 0. Thus, ω is compatible with . ■
Now, by Proposition 4.10, we can apply the induction hypothesis and conclude that the recursive call FINDORDERING(, d) will return an ordering Θ compatible with and d. Since Θ will order C' according to Θ_{C' }(or its reverse), we have that Θ is compatible with C_{1 }and C_{2}. Thus Θ is compatible with and d, completing the proof of Theorem 4.2. □
Remark 4.11 Note that we have shown that Corollary 4.9 holds under the assumption that (in view of line 6) every cluster in contains at most two elements. However, it is possible to prove this result in the more general setting where clusters can have arbitrary size. In principle, this could yield a consistent variation of the NeighborNet algorithm that is analogous to the recently introduced QNet algorithm [16], where, instead of reducing the size of clusters when they have more than two elements, the reduction case is skipped entirely and clusters are pairwise combined until only one cluster is left. However, we suspect that such a method would probably not work well in practice since the reduced distances have smaller variance than the original distances.
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