Order Distances and Split Systems

Abstract

Given a pairwise distance D on the elements in a finite set X, the order distance Δ(D) on X is defined by first associating a total preorder ≼_x on X to each x ∈X based on D, and then quantifying the pairwise disagreement between these total preorders. The order distance can be useful in relational analyses because using Δ(D) instead of D may make such analyses less sensitive to small variations in D. Relatively little is known about properties of Δ(D) for general distances D. Indeed, nearly all previous work has focused on understanding the order distance of a treelike distance, that is, a distance that arises as the shortest path distances in a tree with non-negative edge weights and X mapped into its vertex set. In this paper we study the order distance Δ(D) for distances D that can be decomposed into sums of simpler distances called split-distances. Such distances D generalize treelike distances, and have applications in areas such as classification theory and phylogenetics.

Appendix A

Proof of Lemma 1

It remains to show that the upper bound n(n − 1) is tight for all sufficiently large n. To this end, consider a distance D on X such that, for all $\{u,v\} \in \left (\begin {array}{cc}{X}\\{2} \end {array}\right )$, the value D(u, v) = D(v, u) is selected independently and uniformly at random from the set {1, 2}. We now argue that, for sufficiently large n, the probability that $\lvert \mathcal {S}_{D} \rvert = n(n-1)$ is strictly greater than 0.

By construction, D satisfies the triangle inequality, and D(u, v) > 0 for all $\{u,v\} \in \left (\begin {array}{cc}{X}\\{2} \end {array}\right )$. This implies, in view of the definition of the sets X_{u, v} and X_{v, u}, that u ∈ X_{u, v}, v ∈ X_{v, u}, $S_{u,v} \in \mathcal {S}_{D}$ and $S_{v,u} \in \mathcal {S}_{D}$. In order to have $\lvert \mathcal {S}_{D} \rvert = n(n-1)$, for any two distinct {u, v}, $\{a,b\} \in \left (\begin {array}{cc}{X}\\{2} \end {array}\right )$, the splits S_{u, v}, S_{v, u}, S_{a, b} and S_{b, a} must be pairwise distinct.

Again in view of the definition of the sets X_{u, v} and X_{v, u}, S_{u, v} = S_{v, u} can only hold if D(u, x)≠D(v, x) for all x ∈ X. By the construction of D, the probability of this is at most $\left (\frac {1}{2} \right )^{n-2}$. Similarly, S_{u, v} = S_{a, b} implies that X_{u, v} = X_{a, b} (renaming the involved elements of X, if necessary). This can only hold if, for all x ∈ X −{a, b, u, v}, we have D(u, x) < D(v, x) whenever D(a, x) < D(b, x) and vice versa. By the construction of D, the probability of this is at most $\left (\frac {3}{4} \right )^{n-4}$. Applying an analogous analysis for any two of the four splits S_{u, v}, S_{v, u}, S_{a, b} and S_{b, a}, it follows that the probability that at least two of these splits coincide is bounded by d ⋅ cⁿ for some constants 0 < d and 0 < c < 1. This implies that the probability of $\lvert \mathcal {S}_{D} \rvert < n(n-1)$ is at most

$$\left( \begin{array}{cc}{n}\\{2} \end{array}\right) \cdot \left (\left( \begin{array}{cc}{n}\\{2} \end{array}\right) -1 \right ) \cdot d \cdot c^{n},$$

which is strictly less than 1 for sufficiently large n, as required. □

Proof of Lemma 2

First note that if two splits S₁ = A₁∣B₁ and S₂ = A₂∣B₂ of X are such that at least one of the intersections A₁ ∩ A₂, A₁ ∩ B₂, B₁ ∩ A₂ and B₁ ∩ B₂ is empty then also the restrictions A₁ ∩ M∣B₁ ∩ M and A₂ ∩ M∣B₂ ∩ M of these splits to a subset $M \subseteq X$ satisfy this property, assuming that these restrictions both form bipartitions of M into two non-empty subsets at all. Thus, if $\mathcal {S}_{D}$ is compatible then $\mathcal {S}_{D \rvert _{M}}$ is also compatible for every 6-element subset $M \subseteq X$ in view of the fact that every split in $\mathcal {S}_{D \rvert _{M}}$ must be of the form A ∩ M∣B ∩ M for some split $A \mid B \in \mathcal {S}_{D}$.

It remains to establish that if $\mathcal {S}_{D}$ is not compatible then there exists some 6-element subset $M \subseteq X$ such that $\mathcal {S}_{D \rvert _{M}}$ is not compatible. So assume that there exist $u,v,u^{\prime },v^{\prime } \in X$ with u≠v and $u^{\prime } \neq v^{\prime }$ such that the splits S_{u, v} = X_{u, v}∣X − X_{u, v} and $S_{u^{\prime },v^{\prime }} = X_{u^{\prime },v^{\prime }} \mid X - X_{u^{\prime },v^{\prime }}$ are incompatible. Then there must exist four distinct elements a, b, c, d ∈ X with $a \in X_{u,v} \cap X_{u^{\prime },v^{\prime }}$, $b \in X_{u,v} \cap (X - X_{u^{\prime },v^{\prime }})$, $c \in (X - X_{u,v}) \cap X_{u^{\prime },v^{\prime }}$ and $d \in (X - X_{u,v}) \cap (X - X_{u^{\prime },v^{\prime }})$. Moreover, in view of the definition of the sets X_{u, v} and $X_{u^{\prime },v^{\prime }}$, we have u ∈ X_{u, v}, v ∈ X − X_{u, v}, $u^{\prime } \in X_{u^{\prime },v^{\prime }}$ and $v^{\prime } \in X - X_{u^{\prime },v^{\prime }}$. Therefore, we can choose the elements a, b, c, d in such a way that $\lvert \{a,b,c,d\} \cup \{u,u^{\prime },v,v^{\prime }\} \rvert = 6$. Hence $M = \{a,b,c,d\} \cup \{u,u^{\prime },v,v^{\prime }\}$ is a 6-element subset such that $\mathcal {S}_{D \rvert _{M}}$ is not compatible. □

Proof of Lemma 3

As remarked in [21], the lemma can be proven using graph-theoretical concepts from [22]. In the following we provide a direct proof for completeness. Consider an allowable pair (π, κ) such that the maximum flat split system $\mathcal {S} = \mathcal {S}_{(\pi ,\kappa )}$ contains all splits of the form {x}∣X −{x}. We represent the simple allowable sequence of permutations of X associated with (π, κ) by a wiring diagram $\mathcal {W}$. Since any two wires cross precisely once, there must exist in $\mathcal {W}$, for all x ∈ X, a unique face that is bounded to the right such that only the wire associated with x is either below or above that face. We form the sequence $F=(f_{1},f_{2},\dots ,f_{n})$ of these faces obtained by ordering them as they occur from left to right in $\mathcal {W}$, first those at the bottom of $\mathcal {W}$ and then those at the top. The sequence F then yields a permutation $\pi ^{*}=(x_{1},x_{2},\dots ,x_{n})$ of the elements in X in view of the fact that each f_i is associated with a unique element x_i ∈ X. This is illustrated in Fig. 4.

Fig. 4

A wiring diagram displaying the maximum flat split system $\mathcal {S} = \mathcal {S}_{(\pi ,\kappa )}$ on X = {a, b, c, d, e} with π = (a, b, c, d, e) and κ = (4,1,2,3,1,4,2,1,3,2). $\mathcal {S}$ contains the split {x}∣X −{x} for all x ∈ X, each corresponding to a unique face that is bounded to the right and has precisely one wire either above or below it. The sequence (f₁, f₂, f₃, f₄, f₅) of these faces yields the permutation π^∗ = (a, b, c, e, d) of the elements in X and all splits in $\mathcal {S}$ fit on π^∗

We claim that every split $S = A \mid B \in \mathcal {S}$ fits on π^∗. To show this, consider the unique face f in $\mathcal {W}$ that is bounded to the right and has all wires corresponding to elements in A below it and all wires corresponding to elements in B above it (renaming these sets, if necessary). Consider the rightmost point q on the boundary of the face f. In q the wires of two distinct elements u, v ∈ X cross. Let $i,j \in \{1,2,\dots ,n\}$ be such that u = x_i and v = x_j. Assume without loss of generality that i < j, that is, u comes before v in π^∗. Then, in view of the fact that any two wires in $\mathcal {W}$ cross precisely once, we must have $S = \{x_{i},x_{i+1},\dots ,x_{j-1}\} \mid X - \{x_{i},x_{i+1},\dots ,x_{j-1}\}$, as required. □

To prove Proposition 1, we first show that for a maximum linearly independent split system $\mathcal {S}$ on X that satisfies the pairwise separation property, linear independence is preserved in the restriction of $\mathcal {S}$ to X −{y}, where the restriction of a split system $\mathcal {S}$ on X to a subset $Y \subseteq X$ is the split system

$$\mathcal{S} \rvert_{Y} = \{A \cap Y \mid B \cap Y : A \mid B \in \mathcal{S},\ A \cap Y \neq \emptyset, \ B \cap Y \neq \emptyset\}.$$

Lemma 6

Let $\mathcal {S}$ be a maximum linearly independent split system on a set X with n ≥ 3 elements that satisfies the pairwise separation property. Then, for any y ∈ X, the restriction of $\mathcal {S}$ to X −{y} is a maximum linearly independent split system that satisfies the pairwise separation property.

Proof

Fix any y ∈ X. For every c ∈ X −{y} let A_c and B_c denote the two subsets of X −{y, c} that must exist for the pair {y, c} according to the pairwise separation property. Consider the set

$$ \begin{array}{@{}rcl@{}} \mathcal{S}_{c} = \{A_{c} \mid X-(A_{c} \cup \{y\}), B_{c} \mid X-(B_{c} \cup \{y\})\} \end{array} $$

for each c ∈ X −{y} and put

$$\mathcal{S}_{\leftrightarrow} = \bigcup\limits_{c \in X - \{y\}} \mathcal{S}_{c}.$$

The set $\mathcal {S}_{\leftrightarrow }$ contains splits of X −{y} and, possibly, also the unordered pair ∅∣X −{y}.

We claim that $\lvert \mathcal {S}_{\leftrightarrow } \rvert \geq n-1$. To see this, consider the graph $G_{\leftrightarrow }$ with vertex set $\mathcal {S}_{\leftrightarrow }$ in which there is an edge between two distinct $S, S^{\prime } \in \mathcal {S}_{\leftrightarrow }$ if there exists c ∈ X −{y} such that $\{S,S^{\prime }\} = \mathcal {S}_{c}$. Thus, by construction, $G_{\leftrightarrow }$ has precisely n − 1 edges, each uniquely associated with an element c ∈ X −{y}. Moreover, the two splits A∣B and $A^{\prime } \mid B^{\prime }$ that are connected by the edge associated with element c ∈ X −{y} are such that $A^{\prime } = A - \{c\}$ and $B^{\prime } = B \cup \{c\}$ (after naming the subsets involved in a suitable way). In this sense, every edge of $G_{\leftrightarrow }$ corresponds to the move of a single element from X −{y} from one subset in an unordered pair to the other subset in that pair. But this implies that, if $G_{\leftrightarrow }$ contains a cycle this cycle must contain all edges of $G_{\leftrightarrow }$, since otherwise it is impossible to return to any particular split along that cycle. Hence, $G_{\leftrightarrow }$ must have at least n − 1 vertices, implying that $\lvert \mathcal {S}_{\leftrightarrow } \rvert \geq n-1$, as claimed.

Next note that $\rvert \mathcal {S}_{\leftrightarrow } \lvert \geq n-1$ implies that, when restricting $\mathcal {S}$ to X −{y}, we obtain at most $\left (\begin {array}{cc}{n}\\{2} \end {array}\right ) - (n-1)$ splits of X −{y} in view of the fact that, for every $S^{\prime } \in \mathcal {S}_{\leftrightarrow }$, there are, according to the pairwise separation property, at least two distinct splits in $\mathcal {S}$ that restrict to $S^{\prime }$. On the other hand, in view of the fact that the square matrix consisting of the $\left (\begin {array}{cc}{n}\\{2} \end {array}\right )$ column vectors formed by the distances D_S for $S \in \mathcal {S}$ has full rank, the restriction of this matrix to the rows associated to the pairs of distinct elements $x,x^{\prime } \in X - \{y\}$ must have rank $\left (\begin {array}{cc}{n}\\{2} \end {array}\right ) - (n-1)$. But this implies that $\mathcal {S} \rvert _{X - \{y\}}$ must contain at least $\left (\begin {array}{cc}{n}\\{2} \end {array}\right ) - (n-1)$ splits. Thus $\mathcal {S} \rvert _{X - \{y\}}$ contains precisely $\left (\begin {array}{cc}{n}\\{2} \end {array}\right ) - (n-1) = \left (\begin {array}{cc}{n-1}\\{2} \end {array}\right )$ splits, implying that $\mathcal {S} \rvert _{X - \{y\}}$ is a maximum linearly independent split system on X −{y}. That $\mathcal {S} \rvert _{X - \{y\}}$ also satisfies the pairwise separation property follows immediately from the definition of this property. □

To prove Proposition 1 we shall also use [21, Theorem 15] which states that a maximum linearly independent split system $\mathcal {S}$ on a set X with n ≥ 2 elements is a maximum flat split system if and only if, for every 4-element subset $Y \subseteq X$, the restriction $\mathcal {S} \rvert _{Y}$ contains precisely 6 splits.

Proof of Proposition 1

In [11] it was remarked that every maximum flat split system satisfies the pairwise separation property. In the following we briefly explain why this is the case. Let $\mathcal {S}$ be a maximum flat split system. Then, by definition, there exists an allowable pair (π, κ) with $\mathcal {S} = \mathcal {S}_{(\pi ,\kappa )}$. We represent the simple allowable sequence of permutations of X associated with (π, κ) by a wiring diagram $\mathcal {W}$. To show that $\mathcal {S}$ satisfies the pairwise separation property, consider two distinct elements x, y ∈ X. By the definition of a simple allowable sequence, there exists precisely one point q where the wires associated to x and y cross in $\mathcal {W}$. Putting A to be the set of those elements in X whose wires are below q and, similarly, B to be the set of those elements in X whose wires are above q, we obtain the two subsets of X −{x, y} required by the pairwise separation property. This is illustrated in Fig. 5.

Fig. 5

A wiring diagram displaying the maximum flat split system $\mathcal {S}_{(\pi ,\kappa )}$ on X = {a, b, c, d, e} with π = (a, b, c, d, e) and κ = (4,1,2,3,2,4,1,3,2,3). For the elements c and e we obtain the subsets A = {b} and B = {a, d} required by the pairwise separation property. The elements in A correspond to those wires that are below the point q where the wires of c and e cross, and the elements in B correspond to those wires that are above q. The faces corresponding to the splits A ∪{c, e}∣B, A ∪{c}∣B ∪{e}, A ∪{e}∣B ∪{c} and A∣B ∪{c, e} are shaded gray

Next assume that $\mathcal {S}$ is a maximum linearly independent split system that satisfies the pairwise separation property. Then, for n ∈{2, 3, 4}, the fact that $\mathcal {S}$ consists of precisely $\left (\begin {array}{cc}{n}\\{2} \end {array}\right )$ splits immediately implies that $\mathcal {S}$ is a maximum flat split system by [21, Theorem 15]. So, assume that n ≥ 5. Then, every 4-element subset $Y \subseteq X$ can be obtained by removing from X the elements in X − Y one by one. Hence, the restriction $\mathcal {S} \rvert _{Y}$ is the last element of a sequence of restrictions, each to a subset with one element less, and to each such restriction Lemma 6 applies. Therefore, $\mathcal {S} \rvert _{Y}$ must be a linearly independent split system on Y containing $\left (\begin {array}{cc}{4}\\{2} \end {array}\right )=6$ splits. But this implies, again by [21, Theorem 15], that $\mathcal {S}$ is maximum flat. □

Proof of Lemma 5

If $\mathcal {S}$ is compatible then it is trivially closed. So assume that $\mathcal {S}$ is orderly and contains two incompatible splits S₁ = A₁∣B₁ and S₂ = A₂∣B₂. Let ω be the weighting of $\mathcal {S}$ with ω(S₁) = ω(S₂) = 1 and ω(S) = 0 for all other $S \in \mathcal {S}$. We consider the order distance Δ = Δ(D) with $D=D_{(\mathcal {S},\omega )}$. Then, putting $n_{1} = \lvert A_{1} \cap A_{2} \rvert$, $n_{2} = \lvert B_{1} \cap A_{2} \rvert$, $n_{3} = \lvert A_{1} \cap B_{2} \rvert$ and $n_{4} = \lvert B_{1} \cap B_{2} \rvert$, we have, for all x₁ ∈ A₁ ∩ A₂, x₂ ∈ B₁ ∩ A₂, x₃ ∈ A₁ ∩ B₂ and x₄ ∈ B₁ ∩ B₂,

$$ \begin{array}{@{}rcl@{}} {\Delta}(x_{1},x_{2}) &= &n_{1} n_{4} + 2(n_{1} n_{2} + n_{3} n_{4}) + n_{2} n_{3},\\ {\Delta}(x_{1},x_{3}) &= &n_{1} n_{4} + 2(n_{1} n_{3} + n_{2} n_{4}) + n_{2} n_{3},\\ {\Delta}(x_{1},x_{4}) &=& 2(n_{1} n_{4} + n_{1} n_{2} + n_{3} n_{4} + n_{1} n_{3} + n_{2} n_{4}),\\ {\Delta}(x_{2},x_{3}) &= &2(n_{2} n_{3} + n_{1} n_{2} + n_{3} n_{4} + n_{1} n_{3} + n_{2} n_{4}),\\ {\Delta}(x_{2},x_{4}) &=& n_{1} n_{4} + 2(n_{1} n_{3} + n_{2} n_{4}) + n_{2} n_{3},\\ {\Delta}(x_{3},x_{4}) &=& n_{1} n_{4} + 2(n_{1} n_{2} + n_{3} n_{4}) + n_{2} n_{3} \end{array} $$

and, for all other u, v ∈ X, we have Δ(u, v) = 0. Since $\mathcal {S}$ is orderly, there must exist some non-negative weighting $\omega ^{\prime }$ of $\mathcal {S}$ with

$${\Delta} = \sum\limits_{S \in \mathcal{S}} \omega^{\prime}(S) \cdot D_{S}.$$

Put $\mathcal {S}^{\prime } = \{S \in \mathcal {S} : \omega ^{\prime }(S) > 0\}$ and $\mathcal {C} = \{A_{1} \cap A_{2},B_{1} \cap A_{2},A_{1} \cap B_{2},B_{1} \cap B_{2}\}$. In view of Δ(u, v) = 0 for all u, v ∈ X with {u, v} a subset of one of the sets in $\mathcal {C}$, we must have

$$ \begin{array}{@{}rcl@{}} \mathcal{S}^{\prime} & \subseteq& \{S_{1},S_{2},S_{3},S_{4},S_{5},S_{6},S_{7}\} = \mathcal{S}^{*} \ \text{with}\\ S_{1} &=& A_{1} \cap A_{2} \mid X-(A_{1} \cap A_{2}), \ S_{2} = B_{1} \cap A_{2} \mid X-(B_{1} \cap A_{2}),\\ S_{3} &=& A_{1} \cap B_{2} \mid X-(A_{1} \cap B_{2}), \ S_{4} = B_{1} \cap B_{2} \mid X-(B_{1} \cap B_{2}),\\ S_{5} &= &A_{1} \mid B_{1}, \ S_{6} = A_{2} \mid B_{2}, \ S_{7} = (A_{1} \cap A_{2}) \cup (B_{1} \cap B_{2}) \mid (A_{1} \cap B_{2}) \cup (B_{1} \cap A_{2}) \end{array} $$

The columns in the following matrix represent the split distances associated with the splits in $\mathcal {S}^{*}$, with each row corresponding to an unordered pair of distinct sets in $\mathcal {C}$.

Removing any column from this matrix yields a matrix of rank 6. This implies that the split system $\mathcal {S}^{*}$ is not linearly independent but every 6-element subset of $\mathcal {S}^{*}$ is. Thus, the space of solutions $(\omega _{1},\omega _{2},\dots ,\omega _{7}) \in \mathbb {R}^{7}$ of the equation

$${\Delta} = \sum\limits_{i=1}^{7} \omega_{i} \cdot D_{S_{i}}$$

is 1-dimensional. More specifically, these solutions have the form

$$(n_{1}n_{4}-\alpha, n_{2}n_{3}-\alpha, n_{2}n_{3}-\alpha, n_{1}n_{4}-\alpha, 2(n_{1}n_{2} + n_{3}n_{4})+\alpha, 2(n_{2}n_{4} + n_{1}n_{3})+\alpha, \alpha)$$

with $\alpha \in \mathbb {R}$. Only the solutions for α = 0 and $\alpha =\min \limits (n_{1}n_{4},n_{2}n_{3})$, however, yield Δ as a sum of linearly independent split distances. This implies that $\mathcal {S}^{\prime }$ must correspond to one of these solutions and, thus, consist of those 6 splits that receive a positive weight in that solution. But this implies that $\mathcal {S}$ satisfies at least one of the conditions (a)-(d) in the definition of a closed split system, as required. □

Proof of Proposition 2

Let $\mathcal {S}$ be a maximum linearly independent split system on a set X with 5 elements. Then, in view of Theorem 1, if $\mathcal {S}$ is circular it is orderly.

It remains to show that if $\mathcal {S}$ is orderly then it is circular. So, assume that $\mathcal {S}$ is orderly. Then, in view of Lemma 5, $\mathcal {S}$ is closed. Moreover, since $\mathcal {S}$ is a maximum linearly independent split system, we have $\lvert \mathcal {S} \rvert = 10$. Thus, as any compatible split system on X contains at most 7 splits, $\mathcal {S}$ must contain two incompatible splits S₁ and S₂. Relabeling the elements in X, if necessary, we assume without loss of generality that S₁ = {a, b}∣{c, d, e} and S₂ = {b, c}∣{a, d, e}. Then, since $\mathcal {S}$ is closed, it must also contain the splits {b}∣{a, c, d, e} and S₃ = {d, e}∣{a, b, c}. Moreover, in view of $\lvert \mathcal {S} \rvert = 10$, $\mathcal {S}$ must contain an additional split S₄ = {x, y}∣X −{x, y} for a 2-element subset $\{x,y\} \subseteq X$ with {x, y}∉{{a, b},{b, c},{d, e}}. We consider three cases.

Case 1: S₄ = {a, c}∣{b, d, e}. Then S₁ and S₄ are incompatible. Thus, since $\mathcal {S}$ is closed, we have $\{a\} \mid \{b,c,d,e\} \in \mathcal {S}$. Similarly, S₂ and S₄ are incompatible and, therefore, we have $\{c\} \mid \{a,b,d,e\} \in \mathcal {S}$. Put

$$\mathcal{S}^{*} = \{S_{1},S_{2},S_{3},S_{4}\} \cup \{\{x\} \mid X - \{x\} : x \in \{a,b,c\}\}.$$

For all weightings ω of $\mathcal {S}^{*}$ we have $D_{(\mathcal {S}^{*},\omega )}(d,e) = 0$ and $D_{(\mathcal {S}^{*},\omega )}(d,x) = D_{(\mathcal {S}^{*},\omega )}(e,x)$ for all x ∈{a, b, c}. Thus, the matrix whose columns are the split distances for the 7 splits in $\mathcal {S}^{*}$ has rank at most 6, implying that $\mathcal {S}^{*} \subseteq \mathcal {S}$ is not linearly independent, a contradiction.

Case 2: S₄ = {b, e}∣{a, c, d}. (Note that the case S₄ = {b, d}∣{a, c, e} is symmetric.) Then S₃ and S₄ are incompatible. Therefore, since $\mathcal {S}$ is closed, $\{a,c\} \mid \{b,d,e\} \in \mathcal {S}$. From this we obtain a contradiction as in Case 1.

Case 3: S₄ = {c, d}∣{a, b, e}. (Note that the cases S₄ = {c, e}∣{a, b, d}, S₄ = {a, d}∣{b, c, e} and S₄ = {a, e}∣{b, c, d} are symmetric.) Then, using again that $\mathcal {S}$ is closed, the fact that S₂ and S₄ are incompatible implies that the splits {c}∣{a, b, d, e} and S₅ = {a, e}∣{b, c, d} are contained in $\mathcal {S}$. Similarly, since S₃ and S₄ are incompatible, we have $\{d\} \mid \{a,b,c,e\} \in \mathcal {S}$. Since S₁ and S₅ are incompatible, we have $\{a\} \mid \{b,c,d,e\} \in \mathcal {S}$. And since S₃ and S₅ are incompatible, we have $\{e\} \mid \{a,b,c,d\} \in \mathcal {S}$. Thus, $\mathcal {S} = \mathcal {S}_{\pi }$ for the permutation π = (a, b, c, d, e) of the elements in X, implying that $\mathcal {S}$ is circular, as required. □