Chain graph paper¶
Note: the paper produced from this work was accepted at AAMAS 2021.
Problem formulation¶
Notation. For \(n \in \N\), \([n]\) denotes the set \(\{1,\ldots,n\}\).
Definition 1 (Bipartite tournament)
A bipartite tournament is a triple \((A, B, K)\), where \(A = [m]\) and \(B = [n]\) for some \(m, n \in \N\), and \(K\) is an \(m \times n\) matrix with \(K_{ab} \in \{0, 1\}\) for all \((a, b) \in A \times B\).
When there is no ambiguity we denote a tournament simply by \(K\), with the understanding that \(A_K = [\text{rows}(K)]\) and \(B_K = [\text{columns}(K)]\). Hereafter we drop the subscript \(K\) from \(A\) and \(B\) when \(K\) is clear from context.
Let \(\K\) denote the set of all tournaments. Note that since there are finitely many tournaments of any fixed size, \(\K\) is countable.
Definition 2 (Dual tournament)
The dual tournament of \(K\) is \(\dual{K} = \mathbf{1} - K^\tr\), where \(\mathbf{1}\) denotes the matrix consisting entirely of 1s.
Note that \(A_K = B_\dual{K}\), \(B_K = A_\dual{K}\) and \(K_{ab} = 1\) iff \(\dual{K}_{ba} = 0\).
Also note that \(\dual{\dual{K}} = K\) and that \(K \ne \dual{K}\). 1
Notation. Given \(K\) and \(a \in A\), we write \(K(a) = \{b \in B \mid K_{ab} = 1\}\) for the set of players which \(a\) defeats, which we call the neighbourhood of \(a\).. Similarly, we write \(K^{-1}(b) = \{a \in A \mid K_{ab} = 1\}\) for the set of players defeating \(b\).
Proposition 1
\(K^{-1}(b) = A \setminus \dual{K}(b)\)
\(K^{-1}(b) \subseteq K^{-1}(b')\) iff \(\dual{K}(b) \supseteq \dual{K}(b')\)
Proof.
Let \(a \in A\), \(b \in B\). Note that \(a \in K^{-1}(b)\) iff \(K_{ab} = 1\) iff \(\dual{K}_{ba} = 0\) iff \(a \not\in \dual{K}(b)\). Hence \(K^{-1}(b) = A \setminus \dual{K}(b)\).
This follows directly from (1).
We now define ranking operators for bipartite tournaments.
Definition 3 (Operator)
A ranking operator \(\phi\) assigns to each tournament \(K\) a pair \(\phi(K) = (\ale_K^\phi, \ble_K^\phi)\) of total preorders on \(A\) and \(B\) respectively.
In practise an operator \(\phi\) will often be defined via rating vectors 2 \(r_\phi(K) \in \R^{|A|}\) and \(s_\phi(K) \in \R^{|B|}\), where \(a \ale_K^\phi a'\) iff \([r_\phi(K)]_a \le [r_\phi(K)]_{a'}\) and \(b \ble_K^\phi b'\) iff \([s_\phi(K)]_b \le [s_\phi(K)]_{b'}\).
Key operator properties¶
This section collects a number of potentially desirable properties that may be used to evaluate operators.
Symmetry¶
Notation. For a matrix \(K\) and permutations \(\sigma: A \to A\), \(\pi: B \to B\), let \(\sigma(K)\) and \(\pi(B)\) denote the matrices obtained by permuting the rows and columns of \(K\) by \(\sigma\) and \(\pi\) respectively:
sym: Fix a tournament \(K\) and let \(\sigma:A \to A\) and \(\pi:B \to B\) be permutations. Write \(K' = \pi(\sigma(K))\). Then \(a \ale_K^\phi a' \iff \sigma(a) \ale_{K'}^\phi \sigma(a')\)
dual: \(b \ble_K^\phi b'\) iff \(b \ale_\dual{K}^\phi b'\)
inv-rev: \(a \ale_K^\phi a'\) iff \(a' \ale_{\mathbf{1} - K}^\phi a\)
Remarks.
dual implies that \(a \ale_K^\phi a'\) iff \(a \ble_\dual{K}^\phi a'\). This means an operator can be defined by specifying the ranking for only one of \(A\) or \(B\), and defining the other ranking by duality.
sym, dual and inv-rev each represent a kind of symmetry. sym says that players should be treated symmetrically: the ‘names’ of the players (in \(A\)) should not matter. dual says that the two sides \(A\) and \(B\) should be treated symmetrically: the choice of which side is \(A\) and which side is \(B\) should not matter. Finally, inv-rev says that wins and losses are treated symmetrically.
In fact, dual and inv-rev are related to each other and the following property:
transpose: \(a \ale_K^\phi a'\) iff \(a' \ble_{K^\tr}^\phi a\)
It is easily verified that any two of dual, inv-rev and transpose imply the third. However, neither of dual or inv-rev implies the other. 3
Standard ranking properties¶
mon: If \(K(a) \subseteq K(a')\) then \(a \ale_K^\phi a'\).
pos-resp: Suppose \(a \ale_K^\phi a'\) and \(K_{a',b} = 0\) for some \(b \in B\). Then \(a \alt_{K + \mathbf{1}_{a', b}}^\phi a'\), where \(\mathbf{1}_{a', b}\) is the matrix with a 1 in entry \((a', b)\) and zeros elsewhere.
IIM (Independence of Irrelevant Matches): If \(K_1, K_2\) are tournaments of the same size with identical \(a\)-th and \(a'\)-th rows, \(a \ale_{K_1}^\phi a'\) iff \(a \ale_{K_2}^\phi a'\).
extension: Let us say that a vector \(v \in \{0,1\}^m\) is compatible with a total preorder \(\le\) on \([m]\) if for all \(a \in [m]\) we have \(v_a = 0\) whenever there is \(a'\) with \(a < a'\) and \(v_{a'} = 0\). Also, for an \(m \times n\) tournament \(K\), let \([K \mid v]\) denote the \(m \times (n+1)\) matrix obtained by extending \(K\) to add \(v\) as the final column.
Then whenever \(v\) is compatible with \({\ale_K^\phi}\) we have
\[a \ale_{[K \mid v]}^\phi a' \iff a \ale_K^\phi a' \text{ and } v_a \le v_{a'}\]
Remarks.
mon just says that \(\ale_K^\phi\) extends the (in general, partial) preorder \(\anle_K\). Together with dual, the following ‘strict’ version of mon implies k-chain:
strict-mon: \(K(a) \subset K(a')\) implies \(a \alt_K^\phi a'\)
Also note that sym and pos-resp imply strict-mon.
pos-resp and IIM are adaptations of classic axioms from [Rub80]. Note that these properties – as well as sym and mon – are presented only in terms of the \(A\) ranking; the corresponding properties for \(B\) are implied in combination with dual.
pos-resp also acts as a kind of strategyproofness: \(a\) cannot improve its ranking by deliberately losing a match. Specifically, if \(K_{ab} = 1\) and \(a \ale_K^\phi a'\), then pos-resp implies \(a \alt_{K - \mathbf{1}_{ab}}^\phi a'\).
sym, dual, IIM and pos-resp uniquely characterise \(\phi_\text{count}\) (see here and [Rub80]). These properties without pos-rep characterise the wider class of ‘cardinality’ based operators.
extension is inspired by postulates for iterated belief revision [DP97]. It says that whenever \(v\) is compatible with an existing ranking \({\ale_K^\phi}\) (in the sense that any element of \(A\) ranked strictly higher than \(a\) must perform at least as well as \(a\) in \(v\)), the ranking for \([K \mid v]\) is a refinement of the ranking for \(K\), obtained by splitting the ranks of \({\ale_K^\phi}\) according to \(v\). This is made clearer by writing the condition \((a \ale_K^\phi a' \text{ and } v_a \le v_{a'})\) in the equivalent form
\[a \alt_K^\phi a' \text{ or } ( a \aeq_K^\phi a' \text{ and } v_a \le v_{a'} )\]Note also that if \(v\) is constant then it is compatible with any total preorder. In particular, extension implies that we can add (or remove) a final column full of 1s or 0s without affecting the \(A\)-ranking.
Finally, note that if we have sym as well, \(v\) can be added to \(K\) in any position, not just as the final column.
Coherence-like properties¶
win-coherence: Suppose there is an injective function \(f: K(a) \to K(a')\) such that \(b \ble_K^\phi f(b)\) for all \(b \in K(a)\). Then \(a \ale_K^\phi a'\).
Moreover, if \(f\) is not surjective or if there is \(b\) with \(b \blt_K^\phi f(b)\), then \(a \alt_K^\phi a'\).
loss-coherence: Suppose there is an injective function \(g: B \setminus K(a') \to B \setminus K(a)\) such that \(g(b) \ble_K^\phi b\) for all \(b \in B \setminus K(a')\). Then \(a \ale_K^\phi a'\).
Moreover, if \(g\) is not surjective or if there is \(b\) with \(g(b) \blt_K^\phi b\), then \(a \alt_K^\phi a'\).
ceteris-paribus: Suppose \(K(a) = N \cup \{b\}\) and \(K(a') = N \cup \{b'\}\), for some \(N \subseteq B \setminus \{b,b'\}\). Then \(a \ale_K^\phi a'\) iff \(b \ble_K^\phi b'\).
Remarks.
The central idea behind these properties is to ensure there is some connection between the ranking of \(A\) and \(B\), with respect to the tournament \(K\). They are very close to self-consistent monotonicity, a property that has already been extensively studied for partial tournaments [CS99][GonzalezDiazHL14].
win-coherence says that players in \(A\) should rank highly when beating players in \(B\) which rank highly themselves (i.e. we should reward difficult wins).
loss-coherence says that players in \(A\) should rank poorly when losing to lowly-ranked players in \(B\) (i.e. we should penalise losses in easy matches).
The win and loss versions are related by the dual version of inv-rev (which I will call b-inv-rev):
(win-coherence and b-inv-rev) iff (loss-coherence and b-inv-rev)
The ‘win’ and ‘loss’ versions of coherence seem to fit nicely with the ideas behind chain-addition versus chain-deletion. Is there anything we can say about this?
We reach peak levels of pretentiousness with the ceteris-paribus property, whose name is a Latin phrase meaning “all other things being equal”. It says that if the only difference between \(a\) and \(a'\) is that \(a\) defeats \(b\) whereas \(a'\) defeats \(b'\), then the player defeating the ‘best’ of \(b, b'\) should rank highest.
This axiom is inspired both in name and in its form by ideas from the problem of social ranking from coalitions [HKMO18][AEMO20].
Its statement here is similar to win-coherence, in that it focuses on wins rather than losses. It would also be possible to define a loss version.
ceteris-paribus is strictly weaker than win-coherence.
Both coherence properties imply mon and strict-mon.
It may be possible to combine win-coherence and loss-coherence to get a more cautious property that only applies when \(a'\) achieves more difficult wins and \(a\) suffers easier losses.
The basic counting operator \(\phi_{\text{count}}\) fails to satisfy any of these properties. Indeed, consider
\[K = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{bmatrix}\]Note that \(1 \aeq_K^{\phi_{\text{count}}} 2\) in the \(A\) ranking, and \(2 \blt_K^{\phi_{\text{count}}} 1\) in the \(B\) ranking. win-coherence would imply \(2 \alt_K^{\phi_{\text{count}}} 1\), so \({\phi_{\text{count}}}\) cannot satisfy it. loss-coherence on the other hand implies the reverse ranking \(2 \alt_K^{\phi_{\text{count}}} 1\), so also cannot hold for \({\phi_{\text{count}}}\).
ceteris-paribus with \(N = \emptyset\) would also imply \(2 \alt_K^{\phi_{\text{count}}} 1\), so does not hold.
chain-min operators also violate these properties, as will be shown in the section on chain-min operators.
Miscellaneous other properties¶
improvement: Fix a tournament \(K\) and let \(a, a' \in A\). Let \(\mathbf{e}_a \in \{0,1\}^m\) be the vector with a 1 in the \(a\)-th position and zeros elsewhere. Additionally, write \(K^{(0)} = K\) and \(K^{(t+1)} = [K^{(t)} \mid \mathbf{e}_a]\).
Then there exists \(t \in \N_0\) such that \(a' \alt_{K^{(t)}}^\phi a\).
concat: For total preorders \({\le_1}, {\le_2}\) on a set \(X\), let us say that \({\le_1}\) is compatible with \({\le_2}\) if \(x <_1 y\) implies \(x \le_2 y\). 4
Let \(K_1, K_2\) be \(m \times n_1\) and \(m \times n_2\) tournaments respectively. Then whenever \({\ale_{K_1}^\phi}\) is compatible with \({\ale_{K_2}^\phi}\) we have
\[a \ale_{[K_1 \mid K_2]}^\phi a' \iff a \ale_{K_1}^\phi a' \text{ and } a \ale_{K_2}^\phi a'\]
Remarks.
improvement comes from improvement operators in iterated belief revision [KPerez08].
concat is in a sense a generalisation of extension, which allows for side-by-side concatenation of arbitrary tournaments (provided they have the same number of rows).
Chain-minimal operators¶
This section considers the above properties in relation to chain-min in more detail. Hereafter, we refer to operators satisfying chain-min as chain-minimal operators. First we introduce some terminology.
Definition 6 (Chain choice function)
A chain choice function is a mapping \(\alpha: \K \to \K\) such that \(\alpha(K) \in \minch{K}\).
Note that any chain choice function \(\alpha\) defines a chain-minimal operator \(\phi(K) = ({\anle_{\alpha(K)}}, {\bnle_{\alpha(K)}})\), and every chain-minimal operator \(\phi\) has an associated chain choice function \(\alpha_\phi\). (TODO: does each \(\phi\) have a unique choice function? i.e. Can we show that \(\minch{K}\) does not contain distinct tournaments with the same neighbourhood rankings?)
Symmetry properties¶
Proposition 3
There is no operator satisfying both sym and chain-min.
Proof.
Suppose, for contradiction, that \(\phi\) satisfies both sym and chain-min. Write
Consider the permuatations \(\sigma = \pi = (1 \ 2)\). We have
and
Applying sym we have that \(1 \ale_K^\phi 2\) iff \(\sigma(1) \ale_{\pi(\sigma(K))} \sigma(2)\) iff \(2 \ale_K^\phi 1\). That is, 1 and 2 (when seen as players in \(A\)) are equally ranked in \(K\) according to \(\phi\).
On the other hand, it is easily verified that
Since \(\phi\) satisfies chain-min, \(\phi(K)\) is chain-definable by some \(\tilde{K} \in \minch{K}\), i.e. \(a \ale_K^\phi a'\) iff \(\tilde{K}(a) \subseteq \tilde{K}(a')\). In particular, we must have \(\tilde{K}(1) = \tilde{K}(2)\), i.e. the first two rows of \(\tilde{K}\) are equal. But clearly this is not the case for any \(\tilde{K} \in \minch{K}\) – a contradiction.
The failure of sym for chain-minimal operators suggests that some kind of tie-breaking rule is required when none of the closest chain tournaments can be distinguished whilst respecting symmetry. One possibility for such a tie-breaking mechanism is to fix a total order \(\trianglelefteq\) on the set of all tournaments \(\K\), and define a choice function \(\alpha\) by \(\alpha(K) = \min(\minch{K}, {\trianglelefteq})\). 5 We give an example.
Example 1. For an \(m \times n\) tournament \(K\), let \(\v(K) \in \R^{2 + mn}\) be the concatenation of \((m, n)\) with the vectorisation of \(K\). That is, we join together the size of \(K\) with its entries, going column-by-column. For example,
Note that \(\v\) is injective. 6 This allows us to simply define \(K_1 \trianglelefteq K_2\) iff \(\v(K_1)\) precedes \(\v(K_2)\) in the lexicographic order on the set of real vectors \(\bigcup_{k \in \N}{\R^k}\). 7
For example, take \(K = \left[\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{smallmatrix}\right]\) as above. Note that
The corresponding vectors are
The first one is minimal with respect to \(\trianglelefteq\) (due to the 0 in the third position), so \(\alpha(K) = \left[\begin{smallmatrix}0 & 1 & 0 \\ 0 & 1 & 1\end{smallmatrix}\right]\). The rankings of \(\phi(K)\) correspond to the neighbourhood-subset relations of \(\alpha(K)\), i.e. the \(A\) ranking is \(1 \alt 2\), and the \(B\) ranking is \(2 \blt 3 \blt 1\).
We move on to look at dual for chain-minimal operators.
Proposition 4
There exists an operator satisfying dual and chain-min.
Proof.
Let \(\phi\) be any chain-min operator and \(\alpha_\phi\) a corresponding choice function. We will define a new choice function \(\alpha\) such that \(\alpha(\dual{K}) = \dual{\alpha(K)}\). The corresponding operator \(\phi'\) will clearly satisfy dual; for any \(K\) we will have \(\phi'(K) = ({\anle_{\alpha(K)}}, {\bnle_{\alpha(K)}})\) and consequently (applying (2) from Proposition 2 in the second step)
To construct \(\alpha\), let \(\trianglelefteq\) be an arbitrary total order on set of all tournaments \(\K\). Write
where \(\vartriangleleft\) denotes the strict part of \(\trianglelefteq\). Note that for all \(K\), exactly one of \(K, \dual{K}\) lies in \(T\). Define
To conclude we need to show that \(\alpha(K) \in \minch{K}\) and \(\alpha(\dual{K}) = \dual{\alpha(K)}\).
If \(K \in T\), then clearly \(\alpha(K) = \alpha_\phi(K) \in \minch{K}\). Also, \(\dual{K} \not\in T\), which means
as required.
On the other hand, if \(K \not\in T\) then \(\dual{K} \in T\) and
By Proposition 2, \(K' \in \minch{K}\) iff \(\dual{K'} \in \minch{\dual{K}}\). We see that \(\dual{\alpha(K)} \in \minch{\dual{K}}\) implies \(\alpha(K) \in \minch{K}\) as required.
Remark. A very similar proof shows that a chain-minimal operator satisfying inv-rev is also possible: we write \(T = \{K \in \K \mid K \vartriangleleft \mathbf{1} - K\}\) and construct \(\alpha\) such that \(\alpha(\mathbf{1} - K) = \mathbf{1} - \alpha(K)\).
Monotonicity properties¶
While not enough to show that there exists a chain-minimal operator satisfying mon, the following results point in that direction.
Lemma 1
Let \(X, Y, Z\) be sets with \(Y \cap Z = \emptyset\). Then
\(|X \setminus Y| = | X \setminus (Y \cup Z) | + | X \cap Z|\)
\(|(Y \cup Z) \setminus X| = | Y \setminus X| + | Z \setminus X|\)
Proof.
Write \(U = X \setminus (Y \cup Z)\) and \(V = X \cap Z\). It is easily observed that \(U \subseteq X \setminus Y\) and, since \(Y\) and \(Z\) are disjoint, \(V \subseteq X \setminus Y\) also. Hence \(U \cup V \subseteq X \setminus Y\). We also have \(X \setminus Y \subseteq U \cup V\). Indeed, take \(x \in X \setminus Y\). If \(x \in Z\) also then clearly \(x \in V \subseteq U \cup V\). Otherwise, \(x \not\in Y \cup Z\), so \(x \in U \subseteq U \cup V\).
Thus \(X \setminus Y = U \cup V\). Finally, note that \(U\) and \(V\) are disjoint, so we get \(|X \setminus Y| = |U \cup V| = |U| + |V|\) as required.
This follows from the fact that \((Y \cup Z) \setminus X = (Y \setminus X) \cup (Z \setminus X)\) and the disjointness of \(Y \setminus X\) and \(Z \setminus X\).
Lemma 2
Let \(K\) be an \(m \times n\) tournament with \(K(a_1) \subseteq K(a_2)\) for some \(a_1, a_2 \in A\). Additionally, let \(K'\) be any chain \(m \times n\) tournament.
Then there exists an \(m \times n\) chain tournament \(K''\) such that
\(K''(a_1) \subseteq K''(a_2)\)
\(d(K, K'') \le d(K, K')\)
For \(a \not\in \{a_1, a_2\}\), \(K''(a) = K'(a)\).
Proof.
If \(K'(a_1) \subseteq K'(a_2)\) then we can simply take \(K'' = K'\), so assume without loss of generality that \(K'(a_2) \subset K'(a_1)\). Define \(K''\) by simply swapping the neighbourhoods of \(a_1\) and \(a_2\):
Clearly property (3) is already satisfied, and \(K''\) has the chain property. Moreover,
so property (1) also holds. It only remains to show that \(d(K, K'') \le d(K, K')\).
Now, since \(K(a_1) \subseteq K(a_2)\) and \(K'(a_2) \subset K'(a_1)\), we may take \(N, N', E, E' \subseteq B\) with \(N \cap E, N' \cap E' = \emptyset\) so that
Note that \(E' \ne \emptyset\) (TODO: is this important?!). Next, observe that
where \(X \symdiff Y = (X \setminus Y) \cup (Y \setminus X)\) is the symmetric difference of two sets \(X\) and \(Y\). Since \(K'(a) = K''(a)\) for \(a \not\in \{a_1,a_2\}\), it follows that
\(d_1\) can be simplified by expanding the definition of \(\symdiff\) and applying Lemma 1 parts (1) and (2) to the bracketed terms:
Similar treatment of \(d_2\) (with the signs reversed) gives
Hence
For the first two terms in \((*)\), note that
since \(N\) and \(E\) are disjoint, so \(|(N \cup E) \cap E'| - |N \cap E'| = |E \cap E'|\). For the second two terms in \((*)\), Lemma 1 part (1) gives
Consequently
so \(d(K, K'') \le d(K, K')\), as required.
Proposition 5
If \(K(a) \subseteq K(a')\) then there exists \(\hat{K} \in \minch{K}\) with \(\hat{K}(a) \subseteq \hat{K}(a')\).
Proof. This follows from Lemma 2 by letting \(K'\) be any element of \(\minch{K}\), and setting \(\hat{K} = K''\).
Proposition 5 shows that we can find a minimal chain tournament with \(\hat{K}(a) \subseteq \hat{K}(a')\) for each pair \((a, a')\) whenever \(K(a) \subseteq K(a')\). To conclude that chain-min and mon can be satisfied simultaneously, we need to be able to find \(\hat{K}\) with this property for all such pairs \((a, a')\) at the same time.
The following shows that this can at least be done for the pairs such that \(K(a) = K(a')\).
Proposition 6
Let \(K\) be a tournament. Then there exists \(\hat{K} \in \minch{K}\) such that for all \(a \ne a'\),
Proof.
Recall that \(\aneq_K\) is the symmetric part of the neighbourhood-subset relation \(\anle_K\), which forms an equivalence relation given by \(a \aneq_K a'\) iff \(K(a) = K(a')\). Let \(A_1,\ldots,A_t \subseteq A\) be the corresponding equivalence classes. Note that each \(A_i\) has a corresponding neighbourhood \(B_i \subseteq B\) such that \(K(a) = B_i\) for each \(a \in A_i\).
Now, let \(\hat{K}^{(1)}\) be an arbitrary member of \(\minch{K}\). Construct a function \(f: \{A_1,\ldots,A_t\} \to A\) such that
Note that \(f(A_i) \in A_i\). Define \(\hat{K}^{(2)}\) by
where \([a]\) denotes the equivalence class containing \(a\). That is, \(\hat{K}^{(2)}\) is constructed so that \(\hat{K}^{(2)}(a) = \hat{K}^{(1)}(f([a]))\).
Condition \((*)\) is clearly satisfied: if \(K(a) = K(a')\) then \([a] = [a']\), so \(\hat{K}^{(2)}_{ab} = \hat{K}^{(2)}_{a',b}\) for all \(b\) and consequently \(\hat{K}^{(2)}(a) = \hat{K}^{(2)}(a')\). Moreover, \(\hat{K}^{(2)}\) has the chain property: for any \(a, a' \in A\) we have \(a \anle_{\hat{K}^{(2)}} a'\) iff \(f([a]) \anle_{\hat{K}^{(1)}} f([a'])\), and \(f([a]), f([a'])\) are comparable with respect to \({\anle_{\hat{K}^{(1)}}}\) since \(\hat{K}^{(1)}\) has the chain property. Hence \({\anle_{\hat{K}^{(2)}}}\) is total, and \(\hat{K}^{(2)}\) has the chain property.
It only remains to show that \(\hat{K}^{(2)} \in \minch{K}\). We have
But \(|B_i \symdiff \hat{K}^{(1)}(f(A_i))| \le |B_i \symdiff \hat{K}^{(1)}(a)|\) for any \(a \in A_i\) by construction of \(f\). Hence
Since \(\hat{K}^{(1)} \in \minch{K}\) and \(\hat{K}^{(2)}\) has the chain property, we must in fact have \(d(K, \hat{K}^{(2)}) = d(K, \hat{K}^{(1)})\) and thus \(\hat{K}^{(2)} \in \minch{K}\), as required.
Note that \(K(a) = K(a')\) does not imply that \(\hat{K}(a) = \hat{K}(a')\) for all \(\hat{K} \in \minch{K}\). However it does appear that the following might be true: 8
Conjecture 1
If \(K(a) \subset K(a')\) then \(\hat{K}(a) \subseteq \hat{K}(a')\) for all \(\hat{K} \in \minch{K}\).
If true, this would imply that every chain minimal operator has the following (weaker) monotonicity property:
weak-mon: If \(K(a) \subset K(a')\) then \(a \ale_K^\phi a'\).
When combined with Proposition 6, it feels like this might be enough to show that chain-min and mon are compatible.
Update 24th November
I did eventually prove that chain-min and mon are compatible. The proof involved repeated applications of the swapping procedure in Lemma 2 to find \(K' \in \minch{K}\) preserving the strict inclusions from \(K\), and then some further faffing around to get the weak inclusions also.
I am still not sure about Conjecture 1.
More can be said about pos-resp.
Proposition 7
There is no operator satisfying both pos-resp and chain-min.
Proof.
Suppose otherwise, and let \(\phi\) be a chain-minimal operator satisfying pos-resp. Consider
\(K\) has a unique closest chain tournament \(\tilde{K}\):
chain-min therefore implies \(\phi(K) = ({\anle_{\tilde{K}}}, {\bnle_{\tilde{K}}})\). Note that \(\tilde{K}(1) = \tilde{K}(2)\), so we have \(1 \aeq_K^\phi 2\).
In particular, \(1 \ale_K^\phi 2\). Since \(K_{23} = 0\), we may apply pos-resp to get \(1 \alt_{K + \mathbf{1}_{23}}^\phi 2\). But \(K + \mathbf{1}_{23}\) is just \(\tilde{K}\)! Since the chain property already holds for \(\tilde{K}\), we have
so in fact \(1 \aeq_{K + \mathbf{1}_{23}}^\phi 2\), contradicting pos-resp.
Remark. The failure of pos-resp can be explained by thinking about the probabilistic model in which chain-min operators arise as maximum likelihood estimates (see here, although that page uses different notation…). For \(K\) from Proposition 7, the most likely reason for failure of the chain property is that \(2 \in A\) lost an ‘easy’ match against \(3 \in B\) ‘by accident’. This means the 0 at position \((2, 3)\) is switched to a 1 to construct the rankings for \(K\) anyway; switching this entry in the input consequently has no affect and we do not see the strict improvement in the ranking of \(2 \in A\) as pos-resp would require.
Remark. The pos-resp failure above was the result of chain addition: we added a 1 to find the nearest chain tournament. The result still holds if we consider only chain deletion operators, though. Indeed, the closest chain tournament to \(\mathbf{1} - K\), considering deletion only, is \(\mathbf{1} - \tilde{K}\), and we have \(1 \aeq 2\) in both tournaments. However, \((\mathbf{1} - \tilde{K}) + \mathbf{1}_{23} = \mathbf{1} - K\), so this contradicts pos-resp by the same argument as above.
The obvious next question is whether chain-minimal operators can satisfy a weaker version of pos-resp:
nonneg-resp: Suppose \(a \ale_K^\phi a'\) and \(K_{a',b} = 0\) for some \(b \in B\). Then \(a \ale_{K + \mathbf{1}_{a',b}}^\phi a'\).
Let us first note that it is possible for a chain-minimal operator to fail nonneg-resp. Indeed, consider
\(\minch{K}\) contains 8 tournament at a distance of 2 from \(K\). In particular, \(\minch{K}\) contains \(K'\) given by
Suppose \(\phi\) is some chain-minimal operator such that \(\phi(K)\) is chain-definable by \(K'\). Then clearly \(2 \ale_K^\phi 4\) and \(K_{41} = 0\). nonneg-resp would imply that \(2 \ale_{K + \mathbf{1}_{41}}^\phi 4\); however it is easily seen that \(K + \mathbf{1}_{41} = \left[\begin{smallmatrix}0&1&1&1\\1&0&1&0\\0&1&0&0\\1&0&0&0 \end{smallmatrix}\right]\) has a unique closest chain tournament \(K''\):
in which \(K''(2) \supset K''(4)\). Consequently \(4 \alt_{K + \mathbf{1}_{41}}^\phi 2\).
Despite this example, it seems likely that there exists some chain-minimal operator which satisfies nonneg-resp. This remains to be proven.
Independence¶
Proposition 8
There is no operator satisfying both IIM and chain-min.
Proof.
Suppose otherwise, and let \(\phi\) be an operator satisfying IIM and chain-min. Write
Note that the first and second rows of \(K_1\) and \(K_2\) are identical, so by IIM we have \(1 \ale_{K_1}^\phi 2\) iff \(1 \ale_{K_2}^\phi 2\). Both tournaments have a unique closest chain tournament requiring changes to only a single entry:
Write \(\tilde{K_1}\) and \(\tilde{K_2}\) for these nearest chain tournaments respectively. By chain-min, we must have \(\phi(K_i) = ({\anle_{\tilde{K_i}}}, {\bnle_{\tilde{K_i}}})\). In particular, \(1 \alt_{K_1}^\phi 2\) and \(2 \alt_{K_2}^\phi 1\). But this contradicts IIM, and we are done.
Extension¶
Lemma 3
Let \(K\) be an \(m \times n\) tournament and \(v \in \{0,1\}^m\). Then
\(a \anle_{[K \mid v]} a'\) if and only if \(a \anle_K a'\) and \(v_a \le v_{a'}\)
\([K \mid v]\) has the chain property if and only if \(K\) has the chain property and \(v\) is compatible with \({\anle_K}\).
\(\mindist{K} \le \mindist{[K \mid v]}\).
Proof.
We have
\[\begin{aligned} a \anle_{[K \mid v]} a' & \iff \forall b \in [n + 1] : ([K \mid v]_{ab} = 1 \implies [K \mid v]_{a',b} = 1) \\ & \iff \forall b \in [n + 1] : [K \mid v]_{ab} \le [K \mid v]_{ab} \\ & \iff (\forall b \in [n] : K_{ab} \le K_{a',b}) \text{ and } (v_a \le v_{a'}) \\ & \iff (a \anle_K a') \text{ and } (v_a \le v_{a'}) \end{aligned}\]Suppose \([K \mid v]\) has the chain property. The fact that \(K\) has the chain property follows from (1). Indeed, take \(a, a' \in A\) and assume wlog that \(a \anle_{[K \mid v]} a'\). Then (1) gives \(a \anle_K a'\), so \({\anle_K}\) is total.
To show that \(v\) is compatible with \({\anle_K}\), suppose \(a \anlt_K a'\) and \(v_{a'} = 0\). Now, \(a \anlt_K a'\) means \(K(a) \subset K(a')\), so there is \(b < n\) with \(K_{ab} = 0\), \(K_{a',b} = 1\). Clearly we also have \([K \mid v]_{ab} = 0\) and \([K \mid v]_{a',b} = 1\). Since \([K \mid v]\) has the chain property, the \(A\)-neighbourhoods in \([K \mid v]\) are nested and we must therefore have \(a \anle_{[K \mid v]} a'\).
On the other hand, \(v_{a'} = 0\) means \(n + 1 \not\in [K \mid v](a')\); since \([K \mid v](a') \supseteq [K \mid v](a)\) we have \(n + 1 \not\in [K \mid v](a)\) also, and \(v_a = 0\). Hence \(v\) is compatible with \({\anle_K}\).
For the other direction, suppose \(K\) has the chain property and that \(v\) is compatible with \({\anle_K}\). Let \(a, a' \in A\) and assume wlog that \(a \anle_K a'\). If \(a \anlt_K a'\) then compatibility of \(v\) necessarily implies \(v_a \le v_{a'}\), so by (1) we have \(a \anle_{[K \mid v]} a'\). Otherwise we have \(a \aneq_K a'\), and (1) implies that \(a \anle_{[K \mid v]} a'\) if and only if \(v_a \le v_{a'}\), and \(a' \anle_{[K \mid v]} a\) if and only if \(v_{a'} \le v_a\). In any case, \(a\) and \(a'\) are comparable with respect to \({\anle_{[K \mid v]}}\); their order depends on \(v_a\) and \(v_{a'}\). Hence \({\anle_{[K \mid v]}}\) is a total preorder, and \([K \mid v]\) has the chain property.
Let \([K' \mid v'] \in \minch{[K \mid v]}\) so that \([K' \mid v']\) has the chain property and achieves the minimal distance \(\mindist{[K \mid v]}\). By (2), \(K'\) has the chain property. Since \(K'\) is a chain tournament whose dimensions match those of \(K\), we have \(d(K, K') \ge \mindist{K}\), and so
\[\begin{aligned} \mindist{[K \mid v]} &= d([K \mid v], [K' \mid v']) \\ &= d(K, K') + \underbrace{d(v, v')}_{\ge 0} \\ &\ge d(K, K') \\ &\ge \mindist{K} \end{aligned}\]as required.
Remark. Incidentally, parts (1) and (2) tell us that extension implies k-chain in combination with dual and the following (very weak) property, which just says that counting wins is the only sensible ranking when there is only one element of \(B\): 9
single-b: If \(|B| = 1\) then \(a \ale_K^\phi a'\) iff \(|K(a)| \le |K(a')|\)
We come to the main result of this subsection.
Proposition 9
There exists an operator satisfying extension and chain-min.
Proof.
The proof proceeds in a similar way to Proposition 4, where we take an existing chain-min operator \(\phi\) and produce a new operator \(\phi'\) which agrees with \(\phi\) on a subset of \(\K\) and additionally satisfies extension.
To that end, let \(\alpha_\phi\) be a choice function for \(\phi\). We define a new choice function \(\alpha\) recursively in terms of \(|B|\). If \(|B| = 1\), set \(\alpha(K) = \alpha_\phi(K)\). 10 For \(|B| > 1\), set
First we show that \(\alpha(K) \in \minch{K}\) by induction on \(|B|\). If \(|B| = 1\) then \(\alpha(K) = \alpha_\phi(K) \in \minch{K}\) since \(\alpha_\phi\) is a choice function. Now suppose the property holds for \(|B| = n\), and consider an \(n + 1\)-column tournament of the form \([K \mid v]\). If \(v\) is not compatible with \({\anle_{\alpha(K)}}\), then again we have \(\alpha([K \mid v]) = \alpha_\phi([K \mid v]) \in \minch{[K \mid v]}\). Otherwise, since \(\alpha(K) \in \minch{K}\) by assumption, \(\alpha(K)\) has the chain property and Lemma 3 part (2) implies \(\alpha([K \mid v]) = [\alpha(K) \mid v]\) has the chain property.
It remains to show that the distance between \(\alpha([K \mid v])\) and \([K \mid v]\) is minimal. Denoting this distance by \(t\), we have
where Lemma 3 part (3) is applied in the final step. On the other hand, \(t \ge \mindist{[K \mid v]}\) by definition, so in fact \(t = \mindist{[K \mid v]}\) and \(\alpha([K \mid v]) \in \minch{[K \mid v]}\).
So, \(\alpha\) is indeed a choice function, and its corresponsing operator satisfies chain-min.
extension now follows: if \(v\) is compatible with \({\ale_K^{\phi'}} = {\anle_{\alpha(K)}}\), then using Lemma 3 part (1) and the definition of \(\alpha([K \mid v])\) we have
and \(\phi'\) satisfies extension, as required.
Remark. The above proof takes any chain-minimal operator \(\phi\) and ‘converts’ it to one satisfying extension. But in fact, this level of generality is unnecessary: the chain-minimal operator from example Example 1 already satisfies extension.
Indeed, let \(\alpha(K) = \min(\minch{K}, \trianglelefteq)\) with \({\trianglelefteq}\) defined as before, and let \(\phi\) be the chain-minimal operator corresponding to \(\alpha\). To show extension, suppose \(v\) is compatible with \({\ale_K^\phi} = {\anle_{\alpha(K)}}\). Combining Lemma 3 parts (2) and (3) as in the above proof, we get that \([\alpha(K) \mid v] \in \minch{[K \mid v]}\) and therefore \(\mindist{[K \mid v]} = \mindist{K}\). But this implies that any element of \(\minch{[K \mid v]}\) has \(v\) as its final column: if \([K' \mid v'] \in \minch{[K \mid v]}\) then \(K'\) has the chain property, so \(d(K, K') \ge \mindist{K}\) and
Hence \(v = v'\). Moreover we see that \(d(K, K') = \mindist{K}\), so \(K' \in \minch{K}\). In other words, any element of \(\minch{[K \mid v]}\) is of the form \([K' \mid v]\) for some \(K' \in \minch{K}\).
Finally, recall that \({\trianglelefteq}\) ranks tournaments lexicographically based on their column-by-column vectorisation. This ensures that \([\alpha(K) \mid v] \vartriangleleft [K' \mid v]\) for all \(K' \in \minch{K}\), and therefore \(\alpha([K \mid v]) = [\alpha(K) \mid v]\). The extension property follows from part (1) of the lemma by the same argument as in the proof of Proposition 9. Note that had we instead defined \({\trianglelefteq}\) by the row-by-row vectorisation, extension might not hold.
Coherence properties¶
Proposition 10
There is no operator satisfying both chain-min and any of win-coherence, loss-coherence or ceteris-paribus.
Proof.
Consider \(K=\left[\begin{smallmatrix} 1&0&1\\0&1&0\\1&0&0\end{smallmatrix}\right]\). \(K\) has a unique closest chain tournament \(\tilde{K}=\left[\begin{smallmatrix} 1&0&1\\0&{\color{red}0}&0\\1&0&0\end{smallmatrix}\right]\), and it is easy any chain-min operator \(\phi\) must therefore have \(2 \alt_K^\phi 3\) and \(1 \blt_K^\phi 2\).
Note that \(K(3) = \{1\}\) and \(K(2) = \{2\}\). Applying either win-coherence or ceteris-parbus (with \(N = \emptyset\)) would therefore give \(3 \alt_K^\phi 2\), since we already know that \(1 \blt_K^\phi 2\). This is a contradiction.
For loss-coherence a similar argument can be applied to \(2, 3 \in A\) for \(\mathbf{1} - K = \left[\begin{smallmatrix} 0&1&0\\1&0&1\\0&1&1\end{smallmatrix}\right]\).
Remark. The proof above shows that chain-minimal operators fail even the ‘weak’ versions of the coherence properties (which incidentally are satisfied even by \(\phi_{\text{count}}\)!).
Rough notes (to tidy)
Why does chain-min cause coherence failures?
Coherence properties use the matrix \(K\) and the \(B\) ranking to determine when some \(a \in A\) achieved some “difficult wins”
chain-min, however, is a bit too far removed from the actual tournament \(K\): we “fix the errors” in \(K\) (c.f. maximum likelihood) before ranking.
In the counterexample above, the most likely situation (when false positive and negatives rates are the same) is that \(2 \in A\) defeated the “difficult” \(2 \in B\) by accident
This is good for \(2 \in B\) (we remove \(2 \in A\) from its list of defeaters) and bad for \(2 \in A\) (we remove one of its defeatees)
But the whole idea of coherence is that what’s good for \(b \in K(a)\) is good for \(a\)
In a sentence: coherence says we use entries of \(K\) in a very specific way, but chain-min says we can’t trust \(K\) to be accurate.
Question: what about just chain addition or deletion? Maybe we can have win-coherence for chain addition and loss-coherence for chain deletion?
Match-preference operators¶
Note: The examples here could probably replace Example 1 if making this document into a paper.
The core idea behind chain-minimal operators is to modify the ‘noisy’ matrix of tournament results \(K\) so that it satisfies the chain property. By using the Hamming distance to find the minimum change required, we have so far given equal weight to changes in any position in the matrix. That is, we assume errors occur in \(K\) with equal probability in any position. This leads to non-uniqueness of the closest chain tournament, as seen in several examples.
In this section we consider a family of chain-min operators which single out a unique chain tournament by removing this equal-weight assumption. Specifically, we assume a total order over the set of matrix indices \(\N \times \N\) (the matches), which indicates the cost associated with changing an entry in \(K\): the lower down \((a, b)\) is in the ranking, the higher the cost for changing \(K_{ab}\).
One possible motivation for such a ranking comes from cases where matches in the tournament occur at distinct points in time. In this case the matches occurring more recently are (presumably) more representative of the players’ current abilities, and we should therefore prefer to modify the outcome of old matches where possible.
Notation. For a relation \(R\) on a set \(X\) and \(S \subseteq X\), write \(R \rs S = R \cap (S \times S)\) for the restriction of \(R\) to \(S\).
Definition 7
Let \({\trianglelefteq}\) be a total order on \(\N \times \N\). For a tournament \(K\), write \(\v_{\trianglelefteq}(K)\) for the vector in \(\{0,1\}^{|A| \cdot |B|}\) obtained by collecting the entries of \(K\) in the order given by \({\trianglelefteq} \rs (A \times B)\), starting with the minimal entry.
That is,
where \((a_1,b_1), \ldots, (a_{mn},b_{mn})\) is the unique enumeration of \(A \times B\) such that \((a_i,b_i) \vartriangleleft (a_{i+1},b_{i+1})\).
Notation. Write \(K \oplus K' = (K + K') \mod 2\). Note that \(K \oplus K'\) is 1 in exactly the entries where \(K\) and \(K'\) differ.
Definition 8 (Match-preference operator)
Let \(\trianglelefteq\) be a total order on \(\N \times \N\). Define an operator \(\phi_{\trianglelefteq}\) according to the choice function
where the minimum is taken with respect to the lexicographic ordering on \(\{0,1\}^{|A| \cdot |B|}\).
Any operator generated in this way will be called an match-preference operator.
TODO: think of a better name?
Clearly any match-preference operator satisfies chain-min.
Example 2. Consider \(K = \left[\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{smallmatrix}\right]\) as in Example 1. We have
Writing \(K_1, \ldots, K_4\) for these tournaments respectively, we have
Now suppose that \({\trianglelefteq} \rs [2] \times [3]\) orders entries in row 1 from left to right, followed by row 2 from right to left. That is,
Then
We see that \(K_3\) corresponds to the lexicographic minimum, so \(\alpha_\trianglelefteq(K) = K_3\). Indeed, this is the expected result: our ordering \({\trianglelefteq}\) says that the maximal entry \((2, 1)\) is the least important to preserve; since \(K_3\) makes a change in only this position, it is the most preferrable amongst \(\minch{K}\).
One might be tempted to generalise Definition 8 so that \(\trianglelefteq\) is allowed to depend on \(K\). However, it can be shown that every chain-minimal operator is of this form, and we would therefore no longer be looking at a specific subclass of operators. 11 The idea here is that, given a choice function \(\alpha\) for an arbitrary chain-minimal operator \(\phi\), we can construct a total order \(\trianglelefteq_K\) such that the positions of the zeros in \(K \oplus \alpha(K)\) are minimal, which guarantees that the corresponding (generalised) match-preference operator agrees with \(\phi\) on all inputs. Formally, we have the following proposition.
Proposition 11
Let \(\alpha: \K \to \K\) be a choice function. Then for all tournaments \(K\) there exists a total order \(\trianglelefteq_K\) on \(\N \times \N\) such that \(\alpha(K) = \alpha_{\trianglelefteq_K}(K)\). 12
Proof.
Let \(K\) be a tournament. Write
Note that \(Z\) is precisely the positions where \(K \oplus \alpha(K)\) is 0. Let \(\trianglelefteq_K^1\) be an arbitrary total order on \(Z\), and let \(\trianglelefteq_K^2\) be an arbitrary total order on \((\N \times \N) \setminus Z\). Define \(\trianglelefteq_K\) by ‘joining’ these two orders together, with \(Z\) ranking minimally. That is,
Now, it is clear that the first \(|Z|\) entries of \(\v_{\trianglelefteq_K}(K \oplus \alpha(K))\) are 0. We claim that any other \(K' \in \minch{K}\) must have at least one 1 in the first \(|Z|\) positions of \(\v_{\trianglelefteq_K}(K \oplus K')\), proving that \(\v_{\trianglelefteq_K}(K \oplus \alpha(K))\) preceeds \(\v_{\trianglelefteq_K}(K \oplus K')\) lexicographically, and therefore showing that \(\alpha_{\trianglelefteq_K}(K) = \alpha(K)\).
Indeed, let \(K' \in \minch{K} \setminus \{\alpha(K)\}\). Write \(Z'\) for the set of \((a,b)\) where \(K_{ab} = K'_{ab}\). Note that \(|Z'| = |Z| = |A| \cdot |B| - \mindist{K}\) since both \(K', \alpha(K) \in \minch{K}\), but \(Z' \ne Z\) (else \(K' = \alpha(K)\)). Hence neither \(Z\) nor \(Z'\) is a subset of the other; in particular there is \((a, b) \in Z \setminus Z'\). The corresponding entry in \(K \oplus K'\) is clearly 1, which places a 1 within the first \(|Z|\) positions of \(\v_{\trianglelefteq_K}(K \oplus K')\). This completes the claim and the proof.
Match-preference operators can also be viewed as a generalisation of chain-minimal operators where we use a weighted Hamming distance function.
Definition 9 (Weighted Hamming distance)
A weighting \(W = (w_{m,n})_{m,n \in \N}\) is a family of functions \(w_{m,n}: [m] \times [n] \to \R_{\ge 0}\). For such a weighting and for each \(m, n \in \N\), define a distance function on \(m \times n\) tournaments by 13
For an \(m \times n\) tournament \(K\), write
Remark. Clearly the ordinary Hamming distance just corresponds to \(w_{m,n} \equiv 1\) for all \(m, n \in \N\).
Remark. If we abuse notation and consider \(w_{m,n}\) to be a matrix, it holds that \(d_W(K, K') = \v_{\trianglelefteq}(w_{m,n}) \dotprod \v_{\trianglelefteq}(K \oplus K')\) for any total order \(\trianglelefteq\), where \({\dotprod}\) denotes the dot product.
The set \(\minchw{W}{K}\) is a straightforward generalisation of \(\minch{K}\) which uses the weighted distance \(d_W\) to find the closest chain tournaments. Let us say that \(W\) is resolute if \(|\minchw{W}{K}|=1\) for all \(K\). Any resolute weighting \(W\) induces an operator \(\phi_W\) by letting \(\phi_W(K)\) be chain-definable by the unique element of \(\minchw{W}{K}\).
Next we show that any match-preference operator corresponds to a resolute weighting.
Lemma 4
Let \(k\) and \(l\) be integers with \(1 \le k \le l\). Then
Proof.
This follows from the formula for the sum of a finite geometric series:
which holds for all \(r \ne 1\). In this case we have
Proposition 12
Let \(\phi\) be a match-preference operator. Then there exists a resolute weighting \(W\) such that \(\phi = \phi_W\).
Proof.
Let \(\trianglelefteq\) be the total order on \(\N \times \N\) corresponding to \(\phi\), and let \(\alpha_{\trianglelefteq}\) be the associated choice function. For \(m, n \in \N\) and \(a \in [m]\), \(b \in [n]\), write
for the ‘position’ of \((a,b)\) in \({\trianglelefteq} \rs [m] \times [n]\). Then define a family of weight functions \(W = (w_{m,n})_{m,n \in \N}\) by
Note that by construction we have \(\v_{\trianglelefteq}(w_{m,n}) = (1 + 2^{-1},\ldots,1 + 2^{-mn})\). Hence
where \(\mathbf{x} = (2^{-1},\ldots,2^{-mn})\) and \(d(K, K')\) is the unweighted Hamming distance.
Now, we will show that for any \(m \times n\) tournament \(K\) and \(K' \in \ch_{m,n}\) with \(K' \ne \alpha_{\trianglelefteq}(K)\) we have \(d_W(K, \alpha_{\trianglelefteq}(K)) < d_W(K, K')\). This implies \(\minchw{W}{K} = \{\alpha_{\trianglelefteq}(K)\}\) and thus \(\phi_W(K)\) and \(\phi(K)\) are both chain-definable by \(\alpha_{\trianglelefteq}(K)\), i.e. \(\phi_W(K) = \phi(K)\). Since \(K\) was arbitrary, this will prove \(\phi_W = \phi\).
So, let \(K\) be an \(m \times n\) tournament and \(K' \in \ch_{m,n}\). To ease notation, write \(v = \v_{\trianglelefteq}(K \oplus \alpha_{\trianglelefteq}(K))\) and \(v' = \v_{\trianglelefteq}(K \oplus K')\). There are two cases.
Case 1: \(K' \not\in \minch{K}\). In this case we have \(d(K, K') \ge \mindist{K} + 1\), and
where Lemma 4 was applied in the 4th step. This shows \(d_W(K, \alpha_{\trianglelefteq}(K)) < d_W(K, K')\), as required.
Case 2: \(K \in \minch{K}\). In this case we have
Now, by definition of \(\alpha_{\trianglelefteq}\) we know that \(v\) strictly preceeds \(v'\) with respect to the lexicographic order on \(\{0,1\}^{mn}\). Consequently there is \(j \ge 1\) such that \(v_i = v'_i\) for \(i < j\) and \(v_j < v'_j\). That is, \(v_j = 0\) and \(v'_j = 1\). This means
where Lemma 4 was applied in the second to last step. Again, this shows \(d_W(K, \alpha_{\trianglelefteq}(K)) < d_W(K, K')\), and the proof is complete.
Example 3. Let us revisit the match-preference operator from Example 2 from the point of view of weighted operators, in light of Proposition 12. Once again take \(K = \left[\begin{smallmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{smallmatrix}\right]\) as an example. Given the ordering on \([2] \times [3]\) from Example 2, the corresponding weight function (for \(2 \times 3\) tournaments) is
Now, it seems more difficult to calculate ‘by eye’ the set of minimal chain tournaments \(\minchw{W}{K}\) for a general \(W\) (although we know in this case there is a unique minimum). In total there are 46 chain tournaments of size \(2 \times 3\) – let us look at 3 for the sake of illustration (note that \(K_2\) is not in \(\minch{K}\)):
We have
We see that \(K_3\) is the closest to \(K\) amongst these three, and indeed \(K_3\) is the closest among all chain tournaments. This matches the result from Example 2.
Rough notes on axiom satisfaction for match-preference operators
Clearly any axiom incompatible with chain-min cannot hold for match-preference operators. However, it is possible that the match-preference scheme is restrictive enough to rule out axioms that are compatible with chain-min.
dual: Not sure. Here’s some relevant bits of information for duality in relation to match-preference operators.
For match-preference operators we essentially just choose one of the closest chain tournaments by looking at the ‘difference’ matrices: \(D_K = \{K \oplus K' \mid K' \in \minch{K}\}\). Now, it holds that \(\dual{K} \oplus \dual{K'} = (K \oplus K')^\tr\), and recall that \(\minch{\dual{K}} = \dual{\minch{K}}\).
This means that the difference matrices for \(\dual{K}\) are just the transposes of those for \(K\), and for duality we need to make sure the matrices chosen for \(K\) and \(\dual{K}\) are transpose pairs. Looking at square tournaments therefore seems like the way to go. If there exists \(K\) such that \(D_K\) is closed under transposition but contains no symmetric matrices, then we will get a counterexample to duality. A brute force search yields no such tournaments for up to 3x3, but unfortunately 4x4 seems to already be too large to search in a reasonable time.
The two match-preference operators which seem the most ‘sensible’ are where \(\trianglelefteq\) is the lexicographic order on \(\N \times \N\) (which corresponds to row-by-row vectorisation of the difference matrices) and the ‘reverse’ lexicographic order (where \((a,b) \trianglelefteq (a',b')\) iff \((b,a)\) preceeds \((b',a')\) lexicographically; this corresponds to column-by-column vectorisation).
Neither of these satisfy duality: a counterxample is \(\left[\begin{smallmatrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{smallmatrix}\right]\). This is quite interesting since it suggests that any duality-respecting match-preference operator must have some kind of weird ordering \(\trianglelefteq\).
extension: I think it holds for all match-preference operators. In the remarks after the proof of Proposition 9 it was shows that whenever \(v\) is compatible with \({\ale_K^\phi}\) for a chain-minimal operator \(\phi\), we have
so the differences matrices for \([K \mid v]\) are of the form \([K \mid v] \oplus [K' \mid v] = [K \oplus K' \mid 0]\). This column of zeros seems to not affect the ranking after vectorisation, so that choosing \(\alpha_{\trianglelefteq}([K \mid v])\) seems to come down to the same thing as choosing \(\alpha_{\trianglelefteq}(K)\), which would guarantee extension. I need to prove this properly, though.
Note that this argument relies on the fact that the ordering for \(m \times n\) and \(m \times (n + 1)\) tournaments comes from a common ordering on \(\N \times \N\). The generalisation of match-preference operators from footnote 11 would not guarantee extension without further restrictions.
A maximum likelihood interpretation¶
Rough notes on MLE and references
Bradlet-Terry model is MLE for tournaments, but in a different noise model
Lots of work on MLE voting mechanisms in social choice (e.g. section 8 of COMSOC handbook)
Belief merging via MLE
In this section we show that chain-minimal operators arise naturally as maximum likelihood estimators in a particular probabilistic model. Throughout the section we fix \(m, n \in \N\) and write \(A = [m]\), \(B = [n]\).
In the maximum likelihood approach we take an epistemic view of tournament ranking: we assume there exists a true “state of the world” which determines the tournament results along with objective rankings of \(A\) and \(B\). A given tournament \(K\) is then seen as a noisy observation derived from the true state, and a maximum likelihood estimate is a state for which the probability of observing \(K\) is maximal.
More specifically, a state of the world is represented as a vector of skill levels for the players in \(A\) and \(B\). 14
Definition 10
A state of the world is a tuple \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\), where \(\mathbf{x} \in \R^m\) and \(\mathbf{y} \in \R^n\), satisfying the following properties:
Write \(\Theta\) for the set of all states.
For \(a \in A\), \(x_a\) is the skill level of \(a\) in state \(\theta\) (and similarly for \(y_b\)). These skill levels represent the true capabilities of the players in \(A\) and \(B\) in state \(\theta\): \(a\) is capable of defeating \(b\) if and only if \(x_a \ge y_b\). Note that condition \((*)\) simply says that \(a'\) can only be strictly more skilful than \(a\) if there is a good reason for it, i.e. if there is some \(b \in B\) which \(a'\) can defeat but \(a\) cannot (condition \((**)\) is analogous for the \(B\)s). These conditions are intuitive – and necessary for technical reasons later – if we assume that skill levels are relative to the sets \(A\) and \(B\) currently under consideration (for example, they do not reflect the abilities of players in future matches against new contenders outside of \(A\) or \(B\)).
TO COMMENT ON: Skill levels need not have the same meaning for the \(A\)s and \(B\)s in bipartite tournaments. E.g. in the educational setting, the ‘skill’ of a question \(b\) might represent its difficulty, and the skill of a student \(a\) would represent the level of the most difficult question they are capable of answering.
Now, the capabilities of players in principle may not align with what is observed in reality. Noise may be introduced in the observed tournament \(K\) via false positives (where \(a \in A\) defeats a more skilled \(b \in B\) by accident) and false negatives (where \(a \in A\) is defeated by an inferior \(b \in B\) by mistake). 15 The noise model is therefore parametrised by the false positive and false negative rates \(\mathbf{\alpha} = \tuple{\alpha_+, \alpha_-} \in [0,1]^2\), which we assume are the same for all \(a \in A\).
Definition 11
Let \(\mathbf{\alpha} = \tuple{\alpha_+, \alpha_-} \in [0,1]^2\). We define a family of probability distributions \(\{P_{\mathbf{\alpha}}(\cdot \mid \theta) \mid \theta \in \Theta\}\) over \(m \times n\) tournaments by
where \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\),
and \(\gamma = \sum_{K'}{\prod_{a, b}{p_{ab, K'}}}\) is a normalisation constant.
\(P_{\mathbf{\alpha}}(K \mid \theta)\) is the probability of observing the tournament results \(K\) when the false positive and negative rates are given by \(\alpha\) and the true state of the world is \(\theta\). Note that the four cases in \(p_{ab}\) correspond to a false positive, false negative, true positive and true negative respectively.
TODO: Do I have \(\alpha_+\) and \(\alpha_-\) the wrong way around in the last two cases?! Need to fix this here and throughout.
We can now define a maximum likelihood operator.
Definition 12
Let \(\mathbf{\alpha} \in [0,1]^2\). \(\theta\) is a maximum likelihood estimate (MLE) for an \(m \times n\) tournament \(K\) if \(\theta \in \argmax_{\theta' \in \Theta}{P_{\mathbf{\alpha}}(K \mid \theta')}\).
\(\phi\) is a maximum likelihood operator for \(m \times n\) tournaments if for all \(K\) there is an MLE \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\) such that
In order to calculate the maximum likelihood estimates for a given \(K\), consider the tournament \(K_\theta\) associated with each state \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\) given by
Note that \(K_\theta\) is the unique tournament with non-zero probability in \(P_{\tuple{0,0}}(\cdot \mid \theta)\), i.e. when there are no false positive or false negatives.
Lemma 5
Let \(K\) be an \(m \times n\) tournament, \(\mathbf{\alpha} \in [0,1]^2\) and \(\theta \in \Theta\). Then
Proof.
Expanding the product in Definition 11, we have
Let \(a \in A\). Note that \(B\) can be written as the disjoint union \(B = B_1 \cup B_2 \cup B_3 \cup B_4\), where
Recall that \(b \in K_\theta(a)\) iff \(\mathbf{x}_a \ge \mathbf{y}_b\) (where \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\)). It follows that
\(b \in B_1\) iff \(K_{ab} = 1\) and \(\mathbf{x}_a < \mathbf{y}_b\)
\(b \in B_2\) iff \(K_{ab} = 0\) and \(\mathbf{x}_a \ge \mathbf{y}_b\)
\(b \in B_3\) iff \(K_{ab} = 1\) and \(\mathbf{x}_a \ge \mathbf{y}_b\)
\(b \in B_4\) iff \(K_{ab} = 0\) and \(\mathbf{x}_a < \mathbf{y}_b\)
Note that this corresponds to exactly the four cases in the definition of \(p_{ab,K}\). Consequently
Taking the product over all \(a \in A\) we reach the desired expression for \(P_{\mathbf{\alpha}}(K \mid \theta)\).
The maximum likelihood estimates take a particularly simple form if \(\alpha_+ = \alpha_-\), i.e. if false positives and false negatives occur at the same rate.
Proposition 13
Suppose \(\mathbf{\alpha} = \tuple{\beta, \beta}\) with \(\beta < \frac{1}{2}\). Then \(\theta\) is an MLE for \(K\) if and only if \(\theta \in \argmin_{\theta' \in \Theta}{d(K, K_{\theta})}\).
Proof.
Let \(\theta \in \Theta\). From Lemma 5 we get
Note that
and so
Now, \(P_{\mathbf{\alpha}}(K \mid \theta)\) is maximal when its logarithm is. We have
Note that \(c\) does not depend on \(\theta\). Noting also that \(\beta < 1/2\) implies \(\log{\left(\frac{\beta}{1 - \beta}\right)} < 0\), it follows that for any \(\theta, \theta' \in \Theta\):
which proves the result.
Proposition 13 characterises the MLE states for \(K\) as those for which \(K_\theta\) is the closest to \(K\). As it turns out, the tournaments \(K_\theta\) that arise in this way are exactly those with the chain property.
Lemma 6
Let \(\theta = \tuple{\mathbf{x}, \mathbf{y}} \in \Theta\). Then for all \(a, a' \in A\) and \(b, b' \in B\):
\(K_\theta(a) \subseteq K_\theta(a')\) iff \(\mathbf{x}_a \le \mathbf{x}_{a'}\)
\(K_\theta^{-1}(b) \supseteq K_\theta^{-1}(b')\) iff \(\mathbf{y}_b \le \mathbf{y}_{b'}\).
Proof.
We prove (1); (2) is shown similarly.
Let \(a, a' \in A\). First suppose \(\mathbf{x}_a \le \mathbf{x}_{a'}\). Let \(b \in K_\theta(a)\). Then \(\mathbf{y}_b \le \mathbf{x}_a \le \mathbf{x}_{a'}\), so \(b \in K_\theta(a')\) also. This shows \(K_\theta(a) \subseteq K_\theta(a')\).
Now suppose \(K_\theta(a) \subseteq K_\theta(a')\). For the sake of contradiction, suppose \(\mathbf{x}_a > \mathbf{x}_{a'}\). By \((*)\) in the definition of a state (Definition 10), there is \(b \in B\) such that \(\mathbf{x}_{a'} < \mathbf{y}_b \le \mathbf{x}_{a}\). But this means \(b \in K_\theta(a) \setminus K_\theta(a')\), which contradicts \(K_\theta(a) \subseteq K_\theta(a')\). Thus (1) is proved.
Proposition 14
\(K\) has the chain property if and only if \(K = K_\theta\) for some \(\theta \in \Theta\).
Proof.
The “if” direction follows from Lemma 6 part (1): if \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\) and \(a, a' \in A\) then either \(\mathbf{x}_a \le \mathbf{x}_{a'}\) – in which case \(K_\theta(a) \subseteq K_\theta(a')\) – or \(\mathbf{x}_{a'} < \mathbf{x}_a\) – in which case \(K_\theta(a') \subseteq K_\theta(a)\). Therefore \(K_\theta\) has the chain property.
For the “only if” direction, suppose \(K\) has the chain property. Define \(\theta = \tuple{\mathbf{x}, \mathbf{y}}\) by
It is easily that since the neighbourhood-subset relation \({\anle_K}\) is a total preorder, we have \(K(a) \subseteq K(a')\) if and only if \(\mathbf{x}_a \le \mathbf{x}_{a'}\).
First we show that \(K_\theta = K\) by showing that \(K_{ab} = 1\) if and only if \([K_\theta]_{ab} = 1\). Suppose \(K_{ab} = 1\). Then \(a \in K^{-1}(b)\), so \(\mathbf{y}_b = \min\{\mathbf{x}_{a'} \mid a' \in K^{-1}(b)\} \le \mathbf{x}_a\) and consequently \([K_\theta]_{ab} = 1\).
Now suppose \([K_\theta]_{ab} = 1\). Then \(\mathbf{x}_a \ge \mathbf{y}_b\). We must have \(K^{-1}(b) \ne \emptyset\); otherwise \(\mathbf{y}_b = 1 + |A| > |A| \ge \mathbf{x}_a\). We can therefore take \(\hat{a} \in \argmin_{a' \in K^{-1}(b)}{\mathbf{x}_{a'}}\). By definition of \(\mathbf{y}_b\), \(\mathbf{x}_{\hat{a}} = \mathbf{y}_b \le \mathbf{x}_a\). But \(\mathbf{x}_{\hat{a}} \le \mathbf{x}_a\) implies \(K(\hat{a}) \subseteq K(a)\); since \(\hat{a} \in K^{-1}(b)\) this gives \(b \in K(\hat{a})\) and \(b \in K(a)\), i.e. \(K_{ab} = 1\). This completes the claim that \(K = K_\theta\).
It only remains to show that \(\theta\) satisfies conditions \((*)\) and \((**)\) of Definition 10. For \((*)\), suppose \(\mathbf{x}_a < \mathbf{x}_{a'}\). Then \(K(a) \subset K(a')\), i.e there is \(b \in K(a') \setminus K(a) = K_\theta(a') \setminus K_\theta(a)\). \(b \in K_\theta(a')\) gives \(\mathbf{y}_b \le \mathbf{x}_{a'}\), and \(b \not\in K_\theta(a)\) gives \(\mathbf{x}_a < \mathbf{y}_b\); this shows that \((*)\) holds.
For \((**)\), suppose \(\mathbf{y}_b < \mathbf{y}_{b'}\). Clearly \(K^{-1}(b) \ne \emptyset\) (otherwise \(\mathbf{y}_b = 1 + |A|\) is maximal). Thus there is \(a \in K^{-1}(b)\) such that \(\mathbf{y}_b = \mathbf{x}_a\). This of course means \(\mathbf{x}_a < \mathbf{y}_{b'}\); in particular we have \(\mathbf{y}_b \le \mathbf{x}_a < \mathbf{y}_{b'}\) as required for \((**)\).
We have shown that \(K = K_\theta\) and that \(\theta \in \Theta\), and the proof is complete.
Corollary 1
If \(\mathbf{\alpha} = \tuple{\beta, \beta}\) for some \(\beta < \frac{1}{2}\), then \(\theta\) is an MLE for \(K\) if and only if \(K_\theta \in \minch{K}\).
Proof. Write \(\K_\Theta = \{K_\theta \mid \theta \in \Theta\}\). By Proposition 13, \(\theta\) is an MLE if and only if \(d(K, K_\theta) \le d(K, K_{\theta'})\) for all \(\theta' \in \Theta\), i.e. \(K_\theta \in \argmin_{K' \in \K_\Theta}{d(K, K')}\).
But by Proposition 14 \(\K_\Theta = \ch_{m,n}\), i.e. the set of all \(m \times n\) tournaments with the chain property. We see that \(\argmin_{K' \in \K_\Theta}{d(K, K')} = \argmin_{K' \in \ch_{m,n}}{d(K, K')} = \minch{K}\) by definition of \(\minch{K}\). This proves the result.
TODO:
Go back through this section and remove dependence on a specific \(m\) and \(n\): have definitions be with respect to some \(m, n\), and an operator has the MLE property iff it gives an \((m,n)\)-MLE ranking for every \(m \times n\) tournament, for any \(m, n \in \N\). Then we can say that the MLE property is equivalent to chain-min.
Give the other results about chain-addition and chain-deletion
Relaxing chain-min in the name of symmetry¶
Chain-minimal operators have intuitive support via the maximum-likelihood characterisation, but are not suitable when symmetry is an absolute requirement (e.g. when no tie-breaking mechanism is available). The goal of this section is to investigate how chain-min can be relaxed so that sym can be satisfied, while staying close to the spirit of ranking via closest chain tournaments.
Symmetry and chain rankings¶
Before actually considering relaxations of chain-min, it will be useful to collect some symmetry-related properties of chain tournaments. Indeed, while there is no chain-minimal operator satisfying symmetry, the set of chain tournaments \(\minch{K}\) (and the induced chain-definable rankings) do satisfy the expected symmetry properties.
Notation. For a relation \(R \subseteq X \times X\) and a permutation \(\tau: X \to X\), write \(\tau(R) = \{(\tau(x), \tau(y)) \mid (x, y) \in R\}\).
Lemma 7
Let \(K\) be a tournament and \(\sigma: A \to A\), \(\pi: B \to B\) permutations. Write \(K' = \pi(\sigma(K))\). Then
\({\anle_{K'}} = \sigma({\anle_K})\) and \({\bnle_{K'}} = \pi({\bnle_K})\)
\(K\) has the chain property if and only if \(K'\) does
\(\minch{K'} = \{\pi(\sigma(\tilde{K})) \mid \tilde{K} \in \minch{K}\}\)
\(\dual{K'} = \pi(\sigma(\dual{K}))\)
Proof.
Let \(a \in A\). Note that
\[\begin{aligned} b \in K(a) &\iff K_{a,b} = 1 \\ &\iff K'_{\sigma(a),\pi(b)} = 1 \\ &\iff \pi(b) \in K'(\sigma(a)) \end{aligned} \]which implies \(K(a) = \pi^{-1}(K'(\sigma(a)))\). This means
\[\begin{aligned} a \anle_{K} a' &\iff K(a) \subseteq K(a') \\ &\iff \pi^{-1}(K'(\sigma(a))) \subseteq \pi^{-1}(K'(\sigma(a'))) \\ &\iff K'(\sigma(a)) \subseteq K'(\sigma(a')) \\ &\iff \sigma(a) \anle_{K'} \sigma(a') \end{aligned}\]where the outer \(\pi^{-1}\) can be removed in the third step since \(\pi^{-1}\) is a bijection. This shows that \(a \anle_K a'\) iff \(\sigma(a) \anle_{K'} \sigma(a')\), i.e. \({\anle_{K'}} = \sigma({\anle_K})\) as required. An analogous argument shows that \({\bnle_{K'}} = \pi({\anle_K})\).
Suppose \(K\) has the chain property. Let \(a, a' \in A\). From (1) we have \(a \anle_{K'} a'\) iff \(\sigma^{-1}(a) \anle_K \sigma^{-1}(a')\). Since \(K\) has the chain property, \({\anle_K}\) is a total preorder and thus \(\sigma^{-1}(a)\) and \(\sigma^{-1}(a')\) are comparable with respect to \({\anle_K}\); this tells us that \(a\) and \(a'\) are comparable with respect to \({\anle_{K'}}\), and thus \(K'\) has the chain property.
An identical argument also proves the converse direction.
This follows from part (2) together with the fact that the Hamming distance satisfies the following symmetry property:
\[d(K, \tilde{K}) = d(\pi(\sigma(K)), \pi(\sigma(\tilde{K})))\]We have
\[\begin{aligned} [\dual{\pi(\sigma(\dual{K}))}]_{ab} &= 1 - [\pi(\sigma(\dual{K}))]_{ba} \\ &= 1 - \dual{K}_{\pi^{-1}(b), \sigma^{-1}(a)} \\ &= 1 - (1 - K_{\sigma^{-1}(a), \pi^{-1}(b)}) \\ &= K_{\sigma^{-1}(a), \pi^{-1}(b)} \\ &= K'_{ab} \end{aligned}\]so that \(K' = \dual{\pi(\sigma(\dual{K}))}\). Taking the dual of both sides yields the desired result.
Note that (3) implies symmetry within \(\minch{K}\) when \(K\) is ‘symmetric’ itself: if there is a (non-trivial) permutation which maps \(K\) onto itself, then \(\minch{K}\) is closed under this permutation. 16
This can account for the non-uniqueness of the closest chain tournament for ‘symmetric’ tournaments (such as \(K\) from the proof of Proposition 3). However, it is not the case that non-uniqueness arises only in this way. For example, consider
There is no non-trivial permutation mapping \(K\) onto itself, and yet \(|\minch{K}| = 8\). 17
Approximating chain rankings¶
The idea of this subsection is the following: instead of requiring an operator \(\phi\) to output rankings corresponding to some closest chain tournament \(K'\), let us instead only require that \(\phi\) minimises the distance from these rankings, with the minimum taken over some set of operators \(\Phi\).
If \(\Phi\) is the set of all operators then there is clearly no restriction, and we arrive back at chain-min. However, limiting \(\Phi\) to the set of symmetric operators, say, allows us to consider symmetric operators which are the ‘closest’ to satisfying chain-min.
To this end, let \(\orderings{X}\) denote the set of total preorders on a set \(X\), and let \(d_{m,n}\) be a distance metric on \(\orderings{[m]} \times \orderings{[n]}\), for each \(m,n \in \N\). We drop the subscripts when no ambiguity arises. 18 We will also assume that \(d\) satisfies the following symmetry properties:
where \(\sigma\) and \(\pi\) are permutations on \([m]\) and \([n]\) respectively.
Definition 13 (Chain error)
The chain error of an operator \(\phi\) for a tournament \(K\) is
Given a non-empty set of operators \(\Phi\), we say that \(\phi \in \Phi\) is chain-optimal in \(\Phi\) on \(K\) if \(\cherr_{\phi,K} \le \cherr_{\phi', K}\) for all \(\phi' \in \Phi\).
Note that since \(\minch{K}\) is finite and each \(\phi(K)\) takes one of finitely many values (there are a finite number of rankings of \(A\) and \(B\)), there always exists a chain-optimal operator for any non-empty set \(\Phi\).
Interestingly, we can construct a symmetric operator which is chain optimal for all \(K\), amongst those operators satisfying sym and the ‘dual’ version for the ranking on \(B\): 19
b-sym: For any permutations \(\sigma: A \to A\) and \(\pi: B \to B\), \(b \ble_K^\phi b' \iff \pi(b) \ble_{\pi(\sigma(K))}^\phi \pi(b')\)
Proposition 15
Let \(\hat{\Phi}\) denote the set of operators satisfying sym and b-sym. Then there exists an operator \(\hat{\phi}\) that is chain-optimal in \(\hat{\Phi}\) for all tournaments \(K\).
Proof.
For tournaments \(K, K'\), write \(K \sim K'\) iff \(K\) and \(K'\) are of the same size and there exist permutations \(\sigma: A \to A\), \(\pi: B \to B\) such that \(K' = \pi(\sigma(K))\). Note that \(\sim\) is an equivalence relation on the set of all tournaments \(\K\). 20 First we show that for any \(\phi \in \hat{\Phi}\) and \(K \sim K'\), \(\cherr_{\phi,K} = \cherr_{\phi,K'}\).
Indeed, suppose \(K \sim K'\), and write \(K' = \pi(\sigma(K))\). Applying Lemma 7 parts (1) and (2), we may write
Due to sym and b-sym, we have \({\ale_{K'}^\phi} = \sigma({\ale_K^\phi})\) and \({\ble_{K'}^\phi} = \pi({\ble_K^\phi})\). Combining this with the assumed symmetry properties of \(d\) from (1) and the expression for \(\cherr_{\phi,K'}\) above, we get
So \(\cherr_{\phi,{K'}} = \cherr_{\phi, K}\) as required. Since this holds for arbitrary \(\phi \in \hat{\Phi}\), it follows that \(\phi\) is chain-optimal in \(\hat{\Phi}\) on \(K\) if and only \(\phi\) is chain-optimal in \(\hat{\Phi}\) on \(K'\). That is, \(\phi\) is chain optimal in \(\hat{\Phi}\) on \(K\) if and only if it is chain-optimal on the whole equivalence class \([K]\).
Now, since each tournament has at least one chain-optimal \(\hat{\Phi}\)-operator, we can choose an operator \(\phi_{[K]} \in \hat{\Phi}\) for each equivalence class \([K]\) such that \(\phi_{[K]}\) is chain-optimal on all \(K' \in [K]\). Finally, define
Clearly \(\hat{\phi}\) satisfies sym and b-sym: if \(K' = \pi(\sigma(K))\) we have
where sym and b-sym of \(\phi_{[K]}\) is applied in the fourth step. This shows \(\hat{\phi} \in \hat{\Phi}\).
For chain-optimality, note that for any tournament \(K\) we have \(\cherr_{\hat{\phi}, K} = \cherr_{\phi_{[K]}, K}\); since \(\phi_{[K]}\) is chain-optimal in \(\hat{\Phi}\) on \(K\), so too is \(\hat{\phi}\). This completes the proof.
Unfortunately, Proposition 15 is another non-constructive proof which does not help us single out a unique sym-operator. The problem still remains that we need to choose which closest chain tournament \(K'\) we want to approximate with a symmetry-respecting ranking. It would be interesting to see if a chain choice function \(\alpha\) is sufficient for this purpose, i.e. whether we can construct a chain-optimal operator \(\hat{\phi} \in \hat{\Phi}\) as above such that the chain error \(\cherr_{\phi,K}\) is minimised at \(\alpha(K)\), for each \(K\).
Removing symmetry in K¶
Note that sym only places a restriction on the rankings \(\phi(K)\) when \(K\) itself has some kind of symmetry (i.e. there exists a non-trivial permutation mapping \(K\) onto itself). One possibility for ensuring sym, then, is to ‘remove’ this symmetry by merging together \(A\)s and \(B\)s that necessarily rank equally, thus reducing \(K\) to a (smaller) tournament \(\hat{K}\). One could then apply a chain-minimal operator to \(\hat{K}\) and unmerge the rows and columns of \(\hat{K}\) to get a ranking of \(K\).
A motivating example¶
For example, consider
The permutations that map \(K\) to itself are \(\sigma = (1\ 2)\) and \(\pi = (3\ 4)\). That is, swapping the first and second rows and the third and fourth columns brings us back to \(K\). For any operator \(\phi\) satisfying sym (and the dual version b-sym), it must therefore hold that \(1 \aeq_K^\phi 2\) in \(A\) and \(3 \beq_K^\phi 4\) in \(B\). Now, how can \(1, 2 \in A\) and \(3, 4 \in B\) be merged? Let us rewrite \(K\) with the relevant rows and columns grouped together.
Observe that only the top-right block poses any problem for reducing to a \(3 \times 3\) tournament; the other blocks contain only a single number. However, the symmetry in \(K\) dictates that \(K_{13} = K_{\sigma(1),\pi(3)}=K_{24}\) and \(K_{23} = K_{\sigma(2),\pi(3)}=K_{14}\): there is symmetry within the top \(2 \times 2\) block itself. It therefore seems reasonable to represent the block with the number \(1 / 2\): we obtain the non-integer tournament
as the reduction of \(K\). Note that \(\hat{K}\) does not display the symmetry that \(K\) did, and therefore any ranking is permissible from the point of view of sym. In particular we could consider the closest chain graphs to \(\hat{K}\). For example, using the L₁ distance as a natural generalisation of the Hamming distance to non-integer tournaments, we get closest chain tournaments 21
each with a distance of \(1.5\) from \(\hat{K}\). Any mechanism for choosing a closest chain tournament can be applied here. The resulting rankings can be re-interpreted as rankings for \(K\) by ‘unmerging’ \(1, 2 \in A\) and \(3, 4 \in B\), and finally we obtain an operator which satisfies sym which (indirectly) uses chain-minimisation to rank the tournament.
Generalising this approach¶
I think this approach can be generalised beyond the simple example above. Let \(K\) be an arbitrary tournament, and suppose there are permutations \(\sigma: A \to A\), \(\pi: B \to B\) such that \(\pi(\sigma(K)) = K\). Since any permutation on a finite set has a unique cycle decomposition, we may write
where the \(A_1,\ldots,A_k\) and \(B_1,\ldots,B_l\) are disjoint cycles.
Now, take any sub-cycle \(A_i = (a_1,\ldots,a_p)\). For any operator satisfying sym we must have
It can be seen by totality and transitivity of \({\ale_K^\phi}\) that the ranking must in fact be flat on the whole cycle: \(a_1 \aeq_K^\phi \cdots \aeq_K^\phi a_p\). An identical argument shows that \({\ble_K^\phi}\) must be flat on the sub-cycles \(B_j\) whenever \(\phi\) satisfies b-sym.
We therefore aim to reduce \(K\) to a \(k \times l\) matrix \(\hat{K}\), with entry \(\hat{K}_{ij}\) obtained from the sub-matrix of \(K\) induced by \(A_i\) and \(B_j\).
To this end, fix some \(i\) and \(j\), and enumerate \(A_i = (a_1,\ldots,a_p)\), \(B_j = (b_1,\ldots,b_q)\). Since \(K = \pi(\sigma(K))\), we have
so that, at least initially, the entries of the sub-matrix are equal along the leading diagonal. Things get interesting when we exhaust the shortest cycle and ‘wrap around’. To illustrate, suppose \(p = 4\) and \(q = 6\). The red entries in the following matrix shows the entries of the sub-matrix which must be equal due to chain of equalities in (2), together with those that follow after the wrap-around:
But the argument in (2) can also be applied starting with \(a_2\) and \(b_1\): we get \(K_{a_2,b_1} = K_{a_3,b_2} = K_{a_4,b_3}\) etc, and so the blue entries are also all equal:
Clearly this has now covered all the entries of the sub-matrix. Consequently the entries can be partitioned into two sets arranged in this chequerboard-like pattern, such that the corresponding values in \(K\) are constant within each partition.
In fact, this pattern seems to occur for any \(p\) and \(q\). Indeed, the matrix on the left below shows the pattern for \(p = 3\), \(q=6\), and the matrix on the right shows \(p = 3\), \(q = 7\)
I conjecture that the number of partitions (i.e. number of colours required to cover the matrix) is \(\operatorname{gcd}(p, q)\). This must surely be a consequence of some known result.
As a special case, note that if \(A_i\) is a singleton cycle \(A_i = (a)\) (i.e. \(\sigma(a) = a\)) then the sub-matrix is of size \(1 \times q\) and the number of partitions is \(\operatorname{gcd}(1,q)= 1\), i.e. all entries are equal. This matches what was seen in the motivating example in the previous subsection.
Now, if this pattern does indeed hold for all \(p\) and \(q\), each row and column has the same number of 1s, just in different positions. It seems reasonable to represent the collective performance of the \(A_i\) against the \(B_j\) by the proportion of partitions consisting of 1s. Note this is exactly how I arrived at the value of \(1/2\) for the motivating example above.
In summary, then, the procedure for reducing a symmetric tournament \(K\) to a non-symmetric tournament \(\hat{K}\) is (roughly) as follows:
Write \(K = \pi(\sigma(K))\) for permutations \(\sigma: A \to A\) and \(\pi: B \to B\).
Compute the cycle decompositions \(\sigma = A_1 \cdots A_k\), \(\pi = B_1 \cdots B_l\), so that the \(A_1,\ldots,A_k\) are disjoint and cover \(A\) (and similarly for \(B\)).
For each \((A_i, B_j)\) pair:
Partition the set of entries \([|A_i|] \times [|B_i|]\) according to the equations in (2) and the discussion that follows (i.e. compute the chequerboard pattern).
Let \(y_0, y_1\) be the number of partitions such that the corresponding entries in \(K\) are 0 and 1 respectively.
Set \(\hat{K}_{ij} = y_1 / (y_0 + y_1)\).
To use this reduction together with a non-integer tournament operator \(\hat{\phi}\) to define a new operator \(\phi\):
For each \(a, a' \in A\):
Find the unique \(i, i'\) such that \(a \in A_{i}\) and \(a' \in A_{i'}\)
Set \(a \ale_K^\phi a'\) iff \(i \ale_{\hat{K}}^{\hat{\phi}} i'\)
Construct the \(B\) ranking similarly.
Problems with this idea:
Is it even well-defined? In particular, what do we do in step (1) above if there are multiple \(\sigma\) and \(\pi\) that map \(K\) onto itself? Intuitively we want to choose the ‘maximal’ such permutations which give the largest cycles. But the details would need to be worked out.
Can we be sure that \(\hat{K}\) is non-symmetric? Might we have to repeat the procedure multiple times?
It opens up a whole new question on how to rank non-integer tournaments. That might be something for another day.
Is it feasible computationally? Finding whether non-trivial \(\sigma\) and \(\pi\) even exist for a given \(K\) is equivalent to the graph automorphism problem. The linked Wikipedia article suggests that this problem is in \(\mathsf{NP}\) but it is not known if it is in \(\mathsf{P}\), although efficient implementations exist in practise.
Some concrete operators¶
This subsection puts forth some concrete proposals for symmetric operators which are somehow inspired by chain-minimality.
Averaging chain-minimal rankings¶
Notation. For a total preorder \({\le}\) on a finite set \(X\), write \(\position{x}{\le} \in \N\) for the position of \(x \in X\), where \(\position{x}{\le} = 1\) for minimal \(x\).
Definition 14 (Average-chain operator)
The average-chain operator \(\phi_{\avgmin}\) is defined according to the rating vectors
i.e. the rating for \(a\) is simply the average position of \(a\) in the neighbourhood-subset rankings \({\anle_{K'}}\) across the closest chain tournaments \(K'\).
Note that the factor of \(\frac{1}{|\minch{K}|}\) at the front is not really necessary; it does not affect the rankings.
Example 4. Consider
\(K\) has four closest chain tournaments \(\minch{K} = \{K_1,\ldots,K_4\}\), where
Letting \(\mathbf{p}_a = (\position{a}{\anle_{K_1}},\ldots,\position{a}{\anle_{K_4}})\) be the vector of positions of \(a\) in each \(K_i\), we have
Averaging each vector, we find that the ratings \(r_{\phi_\avgmin}(K)\) are given by
and the ranking on \(A\) is therefore
Is there some link kind of between \(\phi_\avgmin\) and the count operator \(\phi_\text{count}\)? Example 4 already shows that the two operators do not always agree, but note that the count-based ranking of \(K\) in Example 4 is a refinenment of the average-chain ranking: the \(\phi_\text{count}\) ranking just removes \((3,4)\) from the order relation, so \({\ale_K^{\phi_\text{count}}} \subset {\ale_K^{\phi_\avgmin}}\). Interesting, for all the \(K\) I have searched so far such that neither \({\ale_K^{\phi_\text{count}}}\) nor \({\ale_K^{\phi_\avgmin}}\) is a subset of the other, \(K\) has a unique closest chain graph.
We move on to look at properties of \(\phi_\avgmin\).
Proposition 16
\(\phi_\avgmin\) satisifes sym and dual.
If \(a \anle_{K'} a'\) for all \(K' \in \minch{K}\), then \(a \ale_K^{\phi_\avgmin} a'\).
\(\phi_\avgmin\) does not satisfy chain-def
Proof.
For sym, let \(K\) be a tournament and \(K' = \pi(\sigma(K))\) for some permutations \(\sigma: A \to A\), \(\pi: B \to B\).
Let \(a \in A\). From Lemma 7 part (3), we know that \(\minch{K'}\) is the result of permuting each tournament in \(\minch{K}\) by \(\pi \circ \sigma\). Additionally using part (1) together with the fact that \(\position{x}{\le} = \position{\tau(x)}{\tau({\le})}\) for any total preorder \({\le}\) and permutation \(\tau\), we get
\[\begin{aligned} [r_{\phi_\avgmin}(K')]_{\sigma(a)} &= \frac{1}{|\minch{K'}|}\sum_{\tilde{K} \in \minch{K}}{ \position{\sigma(a)}{\anle_{\pi(\sigma(\tilde{K}))}} } \\ &= \frac{1}{|\minch{K}|}\sum_{\tilde{K} \in \minch{K}}{ \position{\sigma(a)}{\sigma({\anle_{\tilde{K}}})} } \\ &= \frac{1}{|\minch{K}|}\sum_{\tilde{K} \in \minch{K}}{ \position{a}{\anle_{\tilde{K}}} } \\ &= [r_{\phi_\avgmin}(K)]_a \end{aligned}\]This means that for any \(a, a' \in A\),
\[\begin{aligned} a \ale_K^{\phi_\avgmin} a' &\iff [r_{\phi_\avgmin}(K)]_a \le [r_{\phi_\avgmin}(K)]_{a'} \\ &\iff [r_{\phi_\avgmin}(K')]_{\sigma(a)} \le [r_{\phi_\avgmin}(K')]_{\sigma(a')} \\ &\iff \sigma(a) \ale_{K'}^{\phi_\avgmin} \sigma(a') \end{aligned}\]as required for sym.
For dual, it is easily observed that Proposition 2 parts (2) and (5) imply \([s_{\phi_\avgmin}(K)]_b = [r_{\phi_\avgmin}(\dual{K})]_b\), which in turn implies dual.
We have \(\position{a}{\anle_{K'}} \le \position{a'}{\anle_{K'}}\) for each \(K' \in \minch{K}\), so
\[\begin{aligned} & [r_{\phi_\avgmin}(K)]_{a'} - [r_{\phi_\avgmin}(K)]_a \\ &= \frac{1}{|\minch{K}|}\sum_{K' \in \minch{K}}{ \left( \underbrace{ \position{a'}{\anle_{K'}} - \position{a}{\anle_{K'}} }_{ \ge 0 } \right) } \\ &\ge 0 \end{aligned}\]so \([r_{\phi_\avgmin}(K)]_{a'} \ge [r_{\phi_\avgmin}(K)]_a\) and \(a \ale_K^{\phi_\avgmin} a'\).
Consider
\[K = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\]Then
\[\minch{K} = \left\{ \begin{bmatrix} 0 & 1 & 1 \\ {\color{red}0} & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} {\color{red}1} & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \right\}\]Consequently the average \(A\) ranking is \(3 \alt 2 \alt 1\), and the \(B\) ranking is the flat one \(1 \beq 2 \beq 3\). That is, \(\ale_K^{\phi_\avgmin}\) has 3 ranks and \(\ble_K^{\phi_\avgmin}\) has 1. This contradicts chain-definability according to the equivalence between (1) and (2) in Theorem 1. Hence \(\phi_\avgmin(K)\) is not chain-definable.
Interleaving operator¶
The idea of this operator is to iteratively select the maximally-ranked \(A\)s and \(B\)s in a tournament \(K\), and remove them from the tournament at each iteration.
Definition 15 (Selection functions)
Let \(f: \K \times 2^\N \times 2^\N \to 2^\N\). We say that \(f\) is an \(A\)-selection function if for any tournament \(K\) and \(A' \subseteq A\), \(B' \subseteq B\):
\(f(K, A', B') \subseteq A'\)
If \(A' \ne \emptyset\) then \(f(K, A', B') \ne \emptyset\)
\(f(K, A', \emptyset) = A'\)
Similarly, \(g: \K \times 2^\N \times 2^\N \to 2^\N\) is a \(B\)-selection function if
\(g(K, A', B') \subseteq B'\)
If \(B' \ne \emptyset\) then \(g(K, A', B') \ne \emptyset\)
\(g(K, \emptyset, B') = B'\)
In this case we call \((f, g)\) a pair of selection functions.
Definition 16 (Interleaving operator)
Let \((f, g)\) be a pair of selection functions and \(K\) be a tournament. Write
We define rating vectors for the interleaving operator \(\phi = {\phi^{\text{int}}_{f,g}}\) corresponding to \(f\) and \(g\) by
Here \(f(K, A', B')\) intuitively represents the set of \(A\)s which should rank maximally when restricting our attention to the rows and columns of \(K\) given by \(A'\) and \(B'\) respectively.
Before giving an example, we show that this recursive procedure eventually terminates (and therefore that the \(\max\) in the definition of \(r_\phi(K)\) and \(s_\phi(K)\) is well-defined).
Proposition 17
Fix a tournament \(K\) and let \(A_i, B_i\) (\(i \ge 0\)) be as in Definition 16. Then there are \(j, j' \ge 1\) such that \(A_j = \emptyset\) and \(B_{j'} = \emptyset\).
Moreover, there is \(t \ge 1\) such that both \(A_t = B_t = \emptyset\).
Proof.
Suppose \(i \ge 0\) and \(A_i \ne \emptyset\). Then properties (1) and (2) in Definition 15 imply that \(\emptyset \subset f(K, A_i, B_i) \subseteq A_i\), and consequently \(A_{i+1} = A_i \setminus f(K, A_i, B_i) \subset A_i\).
Supposing that \(A_j \ne \emptyset\) for all \(j \ge 0\), we would have \(A_0 \supset A_1 \supset A_2 \supset \cdots\) which clearly cannot be the case since each \(A_j\) lies inside \(A\) which is a finite set. Hence there is \(j \ge 1\) such that \(A_j = \emptyset\). Moreover, since \(A_j \supseteq A_{j+1} \supseteq A_{j+2} \supseteq \cdots\), we have \(A_k = \emptyset\) for all \(k \ge j\).
An identical argument with \(g\) shows that there is \(j' \ge 1\) such that \(B_{j'} = \emptyset\) and \(B_k = \emptyset\) for all \(k \ge j'\).
Taking \(t = \max\{j, j'\}\), we have \(A_t = B_t = \emptyset\) as required.
Example 5. Define the cardinality-based interleaving operator \(\phicardint = \phi_{f,g}^{\text{int}}\) where \(f\) and \(g\) are given by
so that the ‘winners’ at each iterations are the \(A\)s with the most wins, and the \(B\)s win the least losses. We take the argmin/argmax to be the emptyset whenever \(A'\) or \(B'\) is empty.
Take for example
The following table shows the iteration of the algorithm. In each row \(i\) we show \(K\) with the rows and columns of \(A \setminus A_i\) greyed out, so as to make it more clear how the \(f\) and \(g\) values are calculated. For brevity we also write \(f\) and \(g\) in place of \(f(K, A_i, B_i)\) and \(g(K, A_i, B_i)\) respectively.
The rating vectors can be read off as
giving the ranking on \(A\) as \(4 \alt 3 \aeq 5 \alt 2 \alt 1\), and the \(B\) ranking as \(3 \beq 4 \blt 2 \blt 1 \beq 5\).
Note that each \(f(K, A_i, B_i)\) is an equivalence class of the symmetric closure of \(\ale_K^\phi\) (and similar for \(g(K, A_i, B_i)\)), so the rankings can in fact be read off by looking at the \(f\) and \(g\) columns of the table.
Any interleaving operator satisfies chain-def, and in fact and the converse is also true.
Theorem 2
Let \(\phi\) be an operator. Then \(\phi\) satisfies chain-def if and only if \(\phi = \phi^{\text{int}}_{f,g}\) for some selection functions \(f, g\).
Proof.
(⇐) First we prove the “if” direction. Let \(\phi = \phi^{\text{int}}_{f,g}\) be an interleaving operator with selection functions \((f, g)\), and fix a tournament \(K\). We will show that \(\phi(K)\) is chain-definable.
As per Proposition 17, let \(j, j' \ge 1\) be the minimal integers such that \(A_j = \emptyset\) and \(B_{j'} = \emptyset\). Then we have \(A_0 \supset \cdots \supset A_{j-1} \supset A_j = \emptyset\) and \(B_0 \supset \cdots \supset B_{j'-1} \supset B_{j'} = \emptyset\).
Recall that, for \(a \in A\), we have by definition \([r_\phi(K)]_a = -\max\{i \mid a \in A_i\}\). Also note that \(k = \max\{i \mid a \in A_i\}\) is the unique integer such that \(a \in A_k \setminus A_{k+1}\). It follows that the non-empty sets \(A_0 \setminus A_1, \ldots, A_{j-1} \setminus A_j\) form the ranks of the total preorder \({\ale_K^\phi}\) (that is, the equivalence classes of the symmetric closure \({\aeq_K^\phi}\)). Thus, \({\ale_K^\phi}\) has \(j\) ranks. An identical argument shows that \({\ble_K^\phi}\) has \(j'\) ranks.
Now, the equivalance of (1) and (2) in Theorem 1 implies that \(\phi(K)\) is chain-definable if and only if \(|j - j'| \le 1\). If \(j = j'\) this is clear. Suppose \(j < j'\). Then \(A_j = \emptyset\) and \(B_j \ne \emptyset\). By property (3) for \(g\) in Definition 15, we have \(g(K, A_j, B_j) = g(K, \emptyset, B_j) = B_j\). But this means \(B_{j+1} = B_j \setminus g(K, A_j, B_j) = B_j \setminus B_j = \emptyset\). Consequently \(j' = j+1\), and \(|j - j'| = |-1| = 1\)
If instead \(j > j'\), then a similar argument using property (3) for \(f\) in Definition 15 shows that \(j = j' + 1\), and we have \(|j - j'| = |1| = 1\).
Hence \(|j - j'| \le 1\) in all cases, and \(\phi(K)\) is chain-definable as required.
(⇒) Now for the “only if” direction. Suppose \(\phi\) satisfies chain-def. We will define \(f, g\) such that \(\phi = \phi^{\text{int}}_{f,g}\). The idea behind the construction is straightforward: since \(f\) and \(g\) pick off the next-top-ranked \(A\)s and \(B\)s at each iteration, simply define \(f(K, A_i, B_i)\) as the maximal elements of \(A_i\) with respect to our existing ordering \({\ale_K^\phi}\) (and similar for \(g\)). The interleaving algorithm will then select the ranks of \({\ale_K^\phi}\) and \({\ble_K^\phi}\) one-by-one; the fact that \(\phi(K)\) is chain-definable ensures that we select all the ranks before the iterative procedure ends. The formal details follow.
Fix a tournament \(K\). By chain-definability, \(|r({\ale_K^\phi}) - r({\ble_K^\phi})| \le 1\). Taking \(t = \max\{r({\ale_K^\phi}), r({\ble_K^\phi})\}\), we can write \(X_1, \ldots, X_t \subseteq A\) and \(Y_1, \ldots, Y_t \subseteq B\) for the ranks of \({\ale_K^\phi}\) and \({\ble_K^\phi}\) respectively, possibly with \(X_1 = \emptyset\) if \(r({\ble_K^\phi}) = 1 + r({\ale_K^\phi})\) or \(Y_1 = \emptyset\) if \(r({\ale_K^\phi}) = 1 + r({\ble_K^\phi})\). Assume these sets are ordered such that \(a \ale_K^\phi a'\) iff \(i \le j\) whenever \(a \in X_i\) and \(a' \in X_j\) (and similar for the \(Y_i\)). Also note that the \(X_i \cap X_j = \emptyset\) for \(i \ne j\) (and similar for the \(Y_i\)).
Now set 22
It is not difficult to see that \(f\) and \(g\) satisfy the conditions of Definition 15 for selection functions. We claim that for all \(0 \le i \le t\):
For \(i = 0\) this is clear: since \(X_1,\ldots,X_t\) contains all ranks of \({\ale_K^\phi}\) we have \(\bigcup_{j=1}^{t-0} = X_1 \cup \cdots \cup X_t = A = A_0\) (and similar for \(B\)).
Now suppose \((*)\) holds for some \(0 \le i < t\). We will show that \(f(K, A_i, B_i) = X_{t-i}\) by considering three possible cases, at least one of which must hold.
Case 1: (\(A_i \ne \emptyset\), \(B_i \ne \emptyset\)). Here we have
since the \(X_j\) form (disjoint) ranks of \({\ale_K^\phi}\) with \(X_{j} \alt X_{k}\) for \(j < k\).
Case 2: (\(B_i = \emptyset\)). Here we have \(\bigcup_{j=1}^{t-i}{Y_j} = \emptyset\). Since \(t - i > 0\) and \(Y_j \ne \emptyset\) for \(j > 1\), it must be the case that \(B_i = Y_1 = \emptyset\). Consequently \(t - i = 1\) and \(A_i = \bigcup_{j=1}^{1}{X_j} = X_1\). Therefore
Case 3: (\(A_i = \emptyset\)). By a similar argument as in case 2, we must have \(t - i = 1\) and \(A_i = X_1 = \emptyset\). Hence
We see that, in any case, \(f(K, A_i, B_i) = X_{t-i}\). Consequently, using again the fact that the \(X_j\) are disjoint,
as required. By almost identical arguments we can show that \(g(K, A_i, B_i) = Y_{t-i}\), and thus \(B_{i+1} = \bigcup_{j=1}^{t-(i+1)}{Y_j}\) also. By induction, \((*)\) holds for all \(0 \le i \le t\).
It remains to show that \(a \ale_K^\phi a'\) iff \(a \ale_K^{\phi^{\text{int}}_{f,g}} a'\) and that \(b \ble_K^\phi b'\) iff \(b \ble_K^{\phi^{\text{int}}_{f,g}} b'\).
For \(a \in A\), let \(p(a)\) be the unique integer such that \(a \in X_{p(a)}\), i.e. \(p(a)\) is the index of the rank of \(a\) in the ordering \({\ale_K^\phi}\). Note that we have
and therefore
Using the fact that \(X_i \alt X_j\) for \(i < j\), we get
Again, a similar argument shows that \(b \ble_K^\phi b'\) iff \(b \ble_K^{\phi^{\text{int}}_{f,g}} b'\) for any \(b, b' \in B\). Since \(K\) was arbitrary, we have shown that \(\phi = \phi^{\text{int}}_{f,g}\) as required.
We now present some sufficient conditions for interleaving operators to satisfy some of the other properties under consideration in this document, before taking a look at the cardinality-based interleaving operator \(\phicardint\) specifically.
Lemma 8
Let \(\phi = \phi_{f,g}^{\text{int}}\) be an interleaving operator.
If for any tournament \(K\), \(A' \subseteq A\), \(B' \subseteq B\) and for any pair of permutations \(\sigma: A \to A\) and \(\pi: B \to B\) we have
\[\begin{aligned} f(\pi(\sigma(K)), \sigma(A'), \pi(B')) &= \sigma(f(K, A', B')) \\ g(\pi(\sigma(K)), \sigma(A'), \pi(B')) &= \pi(g(K, A', B')) \end{aligned}\]then \(\phi\) satisfies sym and b-sym.
If for any tournament \(K\) and \(A' \subseteq A\), \(B \subseteq B\) we have
\[g(K, A', B') = f(\dual{K}, B', A')\]then \(\phi\) satisfies dual.
If for any tournament \(K\), \(A' \subseteq A\), \(B' \subseteq B\) and \(a, a' \in A'\) we have
\[K(a) \subseteq K(a') \implies a \not\in f(K, A', B') \text{ or } a' \in f(K, A', B')\]then \(\phi\) satisfies mon.
Proof.
Let \(K\) be a tournament. For brevity, write \(K' = \pi(\sigma(K))\). Let us write \(A_i, B_i\) and \(A_i', B_i'\) \((i \ge 0)\) for the sets defined in Definition 16 for \(K\) and \(K'\) respectively. We claim that for all \(i \ge 0\):
\[A_i' = \sigma(A_i), \quad B_i' = \pi(B_i) \quad \quad (*)\]For \(i = 0\) this is trivial since \(A_0' = A = \sigma(A) = \sigma(A_0)\) since \(\sigma\) is a bijection. The fact that \(B_0' = \pi(B_0)\) is shown similarly.
Suppose that \((*)\) holds for some \(i \ge 0\). Then applying our assumption on \(f\):
\[\begin{aligned} A_{i+1}' &= A_i' \setminus f(K', A_i', B_i') \\ &= \sigma(A_i) \setminus f(K', \sigma(A_i), \pi(B_i)) \\ &= \sigma(A_i) \setminus \sigma(f(K, A_i, B_i)) \\ &= \sigma(A_i \setminus f(K, A_i, B_i)) \\ &= \sigma(A_{i+1}) \end{aligned}\](note that \(\sigma(X) \setminus \sigma(Y) = \sigma(X \setminus Y)\) holds for any sets \(X, Y\) due to injectivity of \(\sigma\)). Using the assumption on \(g\) we can show that \(B_{i+1}' = \pi(B_{i+1})\) in a similar manner. Therefore, by induction, \((*)\) holds for all \(i \ge 0\).
This means that for any \(a \in A\) we have
\[\sigma(a) \in A_i' \iff \sigma(a) \in \sigma(A_i) \iff a \in A_i\]and therefore
\[\begin{aligned} [r_\phi(K')]_{\sigma(a)} &= -\max\{i \mid \sigma(a) \in A_i'\} \\ &= -\max\{i \mid a \in A_i\} \\ &= [r_\phi(K)]_a \end{aligned}\]From this it easily follows that \(a \ale_K^\phi a'\) iff \(\sigma(a) \ale_{K'}^\phi \sigma(a')\), i.e. \(\phi\) satisfies sym. b-sym follows from a near identical argument using the fact that \(\pi(b) \in B_i'\) iff \(b \in B_i\).
Once again, fix a tournament \(K\) and let \(A_i, B_i\) and \(A_i', B_i'\) denote the sets from Definition 16 for \(K\) and \(\dual{K}\) respectively. It is easy to show by induction that for all \(i \ge 0\) we have \(A'_i = B_i\) and \(B_i' = A_i\). This means that for any \(b \in B_K\):
\[\begin{aligned} [s_{\phi}(K)]_b &= -\max\{i \mid b \in B_i\} \\ &= -\max\{i \mid b \in A_i'\} \\ &= [r_{\phi}(\dual{K})]_b \end{aligned}\]and hence \(b \ble_K^\phi b'\) iff \(b \ale_{\dual{K}}^\phi b'\), as required for dual.
Let \(K\) be a tournament and \(a, a' \in A\) such that \(K(a) \subseteq K(a')\). We must show that \(a \ale_K^\phi a'\).
Suppose otherwise, i.e. \(a' \alt_K^\phi a\). Write \(k = \max\{i \mid a \in A_i\}\). Then \(a \in A_k \setminus A_{k+1} = f(K, A_k, B_k)\). Since \(a' \alt_K^\phi a\), we have \(\max\{i \mid a' \in A_i\} > k\). But then since \(A_k \supseteq A_{k+1} \supseteq A_{k+2}\) and so on, this means \(a' \in A_{k+1} \subseteq A_k\). In particular, \(a' \not\in f(K, A_k, B_k)\).
Piecing this all together, we have both \(a, a' \in A_k\), \(K(a) \subseteq K(a')\), \(a \in f(K, A_k, B_k)\) and \(a' \not\in f(K, A_k, B_k)\). But this directly contradicts our assumption on \(f\), so we are done.
Remark. Lemma 8 part (2) provides an easy way to construct an interleaving operator satisfying dual: take any \(A\)-selection function \(f\) and define a \(B\)-selection function \(g\) by \(g(K, A', B') = f(\dual{K}, B', A')\).
Proposition 18
Let \(\phicardint\) be defined as in Example 5. Then \(\phicardint\) satisfies sym, dual and mon. \(\phicardint\) does not satisfy pos-resp, IIM, extension or inv-rev.
Proof.
We take each property in turn.
sym. Let \(K\) be a tournament and let \(\sigma: A \to A\) and \(\pi: B \to B\) be bijective mappings. Write \(K' = \pi(\sigma(K))\). We will show that the conditions on \(f\) and \(g\) in Lemma 8 part (1) are satisfied.
Let \(A' \subseteq A\) and \(B' \subseteq B\). We have
where we make the “substitution” \(a = \sigma^{-1}(\hat{a})\). Using the defintion of \(K' = \pi(\sigma(K))\) it is easily seen that \(K'(\sigma(a)) = \pi(K(a))\). Also, since \(\pi\) is a bijection we have \(\pi(X) \cap \pi(Y) = \pi(X \cap Y)\) for any sets \(X\) and \(Y\), and \(|\pi(X)| = |X|\). Thus
as required. The result for \(g\) follows by a near-identical argument. Thus \(\phicardint\) satisfies sym by Lemma 8 part (1).
dual. Fix a tournament \(K\) and let \(A' \subseteq A\), \(B' \subseteq B\). Note that for \(b \in B'\) we have
Consequently
and, by Lemma 8 part (2), \(\phicardint\) satisfies dual.
mon. Once again, we use Lemma 8. Let \(K\) be a tournament and \(A' \subseteq A\), \(B' \subseteq B\). Suppose \(a, a' \in A'\) with \(K(a) \subseteq K(a')\). We need to show that either \(a \not\in f(K, A', B')\) or \(a' \in f(K, A', B')\)
Suppose \(a \in f(K, A', B')\). Then \(a \in \argmax_{\hat{a} \in A'}{|K(\hat{a}) \cap B'|}\), so \(|K(a) \cap B'| \ge |K(a') \cap B'|\). On the other hand \(K(a) \cap B' \subseteq K(a') \cap B'\), so \(|K(a) \cap B'| \le |K(a') \cap B'|\). Consequently \(|K(a) \cap B'| = |K(a') \cap B'|\), and so \(a' \in f(K, A', B')\). This shows the desired property, which means \(\phicardint\) satisfies mon by Lemma 8 part (3).
pos-resp. Consider
Then \(3 \aeq_K^\phicardint 4 \alt_K^\phicardint 2 \aeq_K^\phicardint 1\). If \(\phicardint\) satisfied pos-resp then we would have \(3 \alt_{K + \mathbf{1}_{4,3}}^\phicardint 4\). However, it is easily verified that \(\phicardint\) gives the same rankings in \(K + \mathbf{1}_{4,3}\) as in \(K\). Therefore pos-resp cannot hold.
IIM. Take \(K_1\) and \(K_2\) from the proof of Proposition 8. Note that the first and second rows of each tournament are identical, so IIM would imply \(1 \ale_{K_1}^\phicardint 2\) iff \(1 \ale_{K_2}^\phicardint 2\). However, it is easily verified that \(1 \alt_{K_1}^\phicardint 2\) whereas \(2 \alt_{K_2}^\phicardint 1\). Therefore \(\phicardint\) cannot satisfy IIM.
extension. Write
We have \(1 \aeq_K^\phicardint 2 \alt_K^\phicardint 3\). Write \(v = \tuple{0, 1, 1}\). Note that \(v\) is compatible with \({\ale_K^\phicardint}\). The extended tournament \([K \mid v]\) is
Now, if extension held then we should have \(1 \alt_{[K \mid v]}^\phicardint 2\) (since \(v\) “splits” the bottom rank of \({\ale_K^\phicardint}\). However, it is easily seen that we still have \(1 \aeq_{[K \mid v]}^\phicardint 2\), contradicting extension.
inv-rev. Take
The \(A\) ranking is \(1 \aeq_K^\phicardint 2 \alt_K^\phicardint 3\). But for \(\mathbf{1} - K = \left[\begin{smallmatrix}0&1&0\\1&0&1\\0&0&0\end{smallmatrix}\right]\), the \(A\) ranking is \(1 \aeq_{\mathbf{1} - K}^\phicardint 3 \alt_{\mathbf{1} - K}^\phicardint 2\). In particular, \(1 \ale_K^\phicardint 2\) but \(2 \not\ale_{\mathbf{1} - K}^\phicardint 1\), which contradicts inv-rev.
In addition to the properties established in Proposition 18, \(\phicardint\) satisfies an important independence property for interleaving operators: \(f(K, A', B')\) only depends on the sub-matrix of \(K\) defined by rows \(A'\) and columns \(B'\) (and similar for \(g\)). Put another way, the greyed out rows and columns in the table from Example 5 do not affect the output of \(f\) and \(g\) (indeed, this is why they are greyed out). We state this property formally:
sub-matrix-independence: Let \(K\) be a tournament and \(A_i, B_i\) a pair of non-empty sets arising in the interleaving algorithm for \(\phi\) and \(K\) (see Definition 16). Write \(A_i = \{a_1,\ldots,a_{m'}\} \subseteq A\) and \(B_i = \{b_1,\ldots,b_{n'}\} \subseteq B\), ordered such that \(a_p < a_{p+1}\) and \(b_q < b_{q+1}\). Let \(K^-\) be the corresponding \(m' \times n'\) sub-matrix of \(K\), where \(K^-_{pq} = K_{a_p, b_q}\). Then for all \(a_p \in A_i\) and \(b_q \in B_i\):
\[\begin{aligned} a_p \in f(K, A_i, B_i) &\iff p \in f(K^-, [m'], [n']) \\ b_q \in g(K, A_i, B_i) &\iff q \in g(K^-, [m'], [n']) \end{aligned} \]
Remark. sub-matrix-independence only applies to sub-matrices \(K^-\) of \(K\) that, in a sense, arise in the interleaving procedure of Definition 16. Note that \(\phicardint\) in fact satisfies the stronger property which requires the equivalences above for any sub-matrix \(K^-\).
Proposition 19
\(\phicardint\) satisfies sub-matrix-independence.
Proof.
Let \((f, g)\) be the selection functions for \(\phicardint\), and let \(K, A_i = \{a_1,\ldots,a_{m'}\}, B_i = \{b_1,\ldots,b_{n'}\}\) and \(K^-\) be as in the statement of sub-matrix-independence. Recall that \(K^-_{pq} = K_{a_p, b_q}\), so
It follows that the mapping \(q \mapsto b_q\) is a bijection from \(K^-(p)\) to \(K(a_p) \cap B_i\), i.e. \(|K^{-1}(p)| = |K(a_p) \cap B_i|\). Consequently
as required. The required property for \(g\) is shown similarly. Hence \(\phicardint\) satisfies sub-matrix-independence.
Note that in the statement of sub-matrix-independence, \([m'] = A_{K^-}\) and \([n'] = B_{K^-}\). Consequently, sub-matrix-independence implies that the rankings of \(A\) and \(B\) are fully determined by the maximal ranks of successively smaller tournaments. In other words, if two sub-matrix-independence operators agree on the maximal \(A\) and \(B\) ranks for all tournaments \(K\), then those operators must in fact be equal.
Proposition 20
Let \(\phi, \phi'\) be two interleaving operators with corresponding selection functions \((f, g)\) and \((f', g')\) satisfying sub-matrix-independence. Additionally, suppose that for all tournaments \(K\)
Then \(\phi = \phi'\).
Proof.
Let \(K\) be a tournament. Write \(A_i, B_i\) and \(A'_i, B'_i\) (\(i \ge 0)\) for the sets defined in Definition 16 for \(\phi\) and \(\phi'\) applied to \(K\), respectively. We will show by induction that, for all \(i \ge 0\), \(A_i = A_i'\) and \(B_i = B'_i\). This will clearly imply \({\ale_K^\phi} = {\ale_K^{\phi'}}\) and \({\ble_K^\phi} = {\ble_K^{\phi'}}\); since \(K\) is arbitrary we can conclude that \(\phi = \phi'\).
For \(i = 0\) the claim is trivially true since \(A_0 = A = A'_0\) (and similar for \(B\)). Suppose that \(A_i = A_i'\) and \(B_i = B'_i\) for some \(i \ge 0\). To show \(A_{i+1} = A'_{i+1}\) it is enough to show that \(f(K, A_i, B_i) = f'(K, A'_i, B'_i)\).
If \(A_i = A'_i = \emptyset\) then \(f(K, A_i, B_i) = \emptyset = f'(K, A'_i, B'_i)\) by property (1) in the definition of an \(A\)-selection function (Definition 15). Similarly, if \(B_i = B'_i = \emptyset\) then \(f(K, A_i, B_i) = A_i = A'_i = f'(K, A'_i, B'_i)\) by property (3). Thus we may assume \(A_i \ne \emptyset\) and \(B_i \ne \emptyset\).
Write \(A_i = A'_i = \{a_1, \ldots, a_{m'}\}\) such that \(a_p < a_{p+1}\), and enumerate \(B_i = B'_i = \{b_1,\ldots,b_{n'}\}\) similarly. Let \(K^-\) be the sub-matrix of \(K\) corresponding to \(A_i\) and \(B_i\). Note that \(f(K, A_i, B_i), f'(K, A'_i, B'_i) \subseteq \{a_1, \ldots, a_{m'}\}\), and for \(1 \le p \le m'\) sub-matrix-independence for \(\phi\) gives
where we use the fact that \(f(K^-, A_{K^-}, B_{K^-})\) is the maximal rank of \({\ale_{K^-}^\phi}\). But by hypothesis, \(\max(A_{K^-}, {\ale_{K^-}^{\phi}}) = \max(A_{K^-}, {\ale_{K^-}^{\phi'}})\). Applying sub-matrix-independence for \(\phi'\) we get
i.e. \(f(K, A_i, B_i) = f'(K, A'_i, B'_i)\). Consequenly \(A_{i+1} = A'_{i+1}\). Using an almost identical argument it can be shown that \(B_{i+1} = B'_{i+1}\). By induction, the proof is complete.
From Proposition 20 it follows that \(\phicardint\) is the unique interleaving operator satisfying sub-matrix-independence and dual such that \(\max(A, {\ale_K^\phi}) = \argmax_{a \in A}{|K(a)|}\) for each \(K\). Before proving this, we characterise sub-matrix-independece interleaving operators by way of a purely ordinal property which does not refer directly to the selection functions \((f, g)\).
rank-removal: Suppose \(\max(A, {\ale_K^\phi}) \ne A\) and \(\max(B, {\ble_K^\phi}) \ne B\). Write \(A \setminus \max(A, {\ale_K^\phi}) = \{a_1,\ldots,a_{m'}\}\) and \(B \setminus \max(B, {\ble_K^\phi}) = \{b_1,\ldots,b_{n'}\}\) such that \(a_p < a_{p+1}\) and \(b_q < b_{q+1}\). Let \(K^-\) be the corresponding \(m' \times n'\) sub-matrix of \(K\). Then, for all \(a_p, a_{p'}\) and \(b_q, b_{q'}\):
\[\begin{aligned} a_p \ale_K^\phi a_{p'} &\iff p \ale_{K^-}^\phi p' \\ b_q \ble_K^\phi b_{q'} &\iff q \ble_{K^-}^\phi q' \end{aligned} \]
rank-removal says that removing the maximally ranked \(A\)s and \(B\)s from a tournament \(K\) preserves the ordering amongst the rest of \(A\) and \(B\).
Proposition 21
\(\phi\) satisfies chain-def and rank-removal if and only if \(\phi\) is an interleaving operator satisfying sub-matrix-independence.
Proof.
(⇐) First we show the “if” direction. Let \((f, g)\) be selection functions such that \(\phi = \phi_{f,g}^\text{int}\) satisfies sub-matrix-independence. By Theorem 2, \(\phi\) satisfies chain-def. To show rank-removal, let \(K\) be a tournament such that \(\max(A, {\ale_K^\phi}) \ne A\) and \(\max(B, {\ble_K^\phi}) \ne B\). Let \(m', n'\), the \(a_p, b_q\) and \(K^-\) be as in the statement of rank-removal.
Write \(A_i, B_i\) and \(A^-_i, B^-_i\) (\(i \ge 0)\) for the sets defined in the interleaving procedure for \(\phi\) applied to \(K\) and \(K^-\) respectively. We claim that for all \(i \ge 0\), \(p \in [m']\) and \(q \in [n']\) it holds that \(a_p \in A_{i+1}\) iff \(p \in A^-_i\) and \(b_q \in B_{i+1}\) iff \(q \in B^-_i\).
First take \(i = 0\). We have, by definition, \(A^-_0 = A_{K^-} = [m']\). On the other hand, \(A_1 = A \setminus f(K, A, B) = A \setminus \max(A, {\ale_K^\phi}) = \{a_1,\ldots,a_{m'}\}\). It is now clear that \(a_p \in A_1\) if and only if \(p \in [m'] = A^-_0\). Similarly, one can easily show that \(b_q \in B_1\) if and only if \(q \in B^-_0\).
Now suppose that the statement holds for some \(i \ge 0\). We consider cases.
Case 1: \(A^-_i \ne \emptyset\) and \(B^-_i \ne \emptyset\). In this case we may write \(A^-_i = \{p_1, \ldots, p_{m''}\} \subseteq A^-_0 = [m']\) and \(B^-_i = \{q_1, \ldots, q_{n''}\} \subseteq B^-_0 = [n']\), where \(m'' \le m'\) and \(n'' \le n'\). As usual, we assume an ordering such that the \(p_r\) and \(q_s\) are increasing.
Let \(K^{--}\) be the \(m'' \times n''\) sub-matrix of \(K^-\) corresponding to \(A^-_i\) and \(B^-_i\), defined by \([K^{--}]_{rs} = [K^{-}]_{p_r, q_s}\). By sub-matrix-independence, for all \(r \in [m'']\) we have
On the other hand, \(K^{--}\) is also a sub-matrix of the original tournament \(K\). Indeed, we have \([K^{--}]_{rs} = [K^{-}]_{p_r, q_s} = K_{a_{p_r}, b_{q_s}}\), so \(K^{--}\) is the sub-matrix corresponding to rows \(\{a_{p_1},\ldots,a_{p_{m''}}\}\) and columns \(\{b_{q_1},\ldots,b_{q_{n''}}\}\). But by the inductive hypothesis we have \(p \in A^-_i\) iff \(a_p \in A_{i+1}\) (\(p \in [m']\)); hence \(\{a_{p_1},\ldots,a_{p_{m''}}\} = A_{i+1}\). By a similar argument, \(\{b_{q_1},\ldots,b_{q_{n''}}\} = B_{i+1}\).
Combining this fact with another application of sub-matrix-independence we get that, for \(r \in [m'']\),
\((1)\) and \((2)\) together yield
Now, let \(p \in [m']\). If \(p \notin A^-_i\) then, by the inductive hypothesis, \(a_p \notin A_{i+1}\). Since \(A^-_{i+1} \subseteq A^-_i\) and \(A_{i+2} \subseteq A_{i+1}\), it holds that \(p \notin A^-_{i+1}\) and \(a_p \notin A_{i+2}\). In particular, it is true that \(a_p \in A_{i+2}\) if and only if \(p \in A^-_{i+1}\), as required.
If \(p \in A^-_i\), then \(p = p_r\) for some \(r \in [m'']\) and \(a_p = a_{p_r} \in A_{i+1}\). By \((3)\) we have
We have shown that \(a_p \in A_{i+2}\) if and only if \(p \in A^-_{i+1}\) in all cases, as required.
By an almost identical argument using the property of \(g\) in the definition of sub-matrix-independence, we can show that \(b_q \in B_{i+2}\) if and only if \(q \in B^-_{i+1}\).
Case 2: \(A^-_i = \emptyset\) and \(B^-_i = \emptyset\). Since \(A^-_{i+1} \subseteq A^-_i\) and \(A_{i+2} \subseteq A_{i+1}\), we have \(A^-_{i+1} = \emptyset = A_{i+2}\). It is clear that \(a_p \in A_{i+2}\) iff \(p \in A^-_{i+1}\) is vacuously true. By the same argument, \(b_q \in B_{i+2} \iff q \in B^-_{i+1}\).
Case 3: \(A^-_i \ne \emptyset\) and \(B^-_i = \emptyset\). As above, \(b_q \in B_{i+2}\) iff \(a \in B^-_{i+1}\) is trivial. Note that due to property (3) in Definition 15,
Moreover, \(B^-_i = \emptyset\) implies \(B_{i+1} = \emptyset\), so
and therefore \(a_p \in A_{i+2}\) iff \(p \in A^-{i+1}\) is, again, vacuously true.
Case 4: \(A^-_i = \emptyset\) and \(B^-_i \ne \emptyset\). A near-identical argument to the one from case 3 can be applied here, using property (3) for \(g\) in Definition 15 instead.
We have now exhausted all cases. By induction, we have shown that for all \(i \ge 0\), \(p \in [m']\) and \(q \in [n']\):
This means that, for any \(a_p\):
That is, the scores of the \(a_p\) in \(K\) differ from the scores of the \(p\) in \(K^-\) only by an additive constant. Similarly, for any \(b_q\):
It is now easy to see that \(a_p \ale_K^\phi a_{p'}\) iff \(p \ale_{K^-}^\phi p'\), and \(b_q \ble_K^\phi b_{q'}\) iff \(q \ble_{K^-}^\phi q'\), and \(\phi\) therefore satisfies rank-removal.
(⇒) Next we show the “only if” statement. Suppose \(\phi\) satisfies chain-def and rank-removal. By Theorem 2, there are selection functions \((f, g)\) such that \(\phi = \phi_{f,g}^\text{int}\). We need to show that \(\phi\) satisfies sub-matrix-independence.
Let \(K, A_i, B_i\) and \(K^-\) be as in the statement of sub-matrix-independence. We also refer to \(A_j, B_j\) (\(j \ne i\)) for the other sets arising in the interleaving algorithm for \(K\) and \(\phi\).
First we claim that \(A_i = A \setminus \bigcup_{j=1}^{i}{(A_{j-1} \setminus A_j)}\). For the left-to-right inclusion, let \(a \in A_i\). Note that \(A_j \supseteq A_i\) for \(j \le i\), so \(a \in A_j\) and \(a \notin A_{j-1} \setminus A_j\) for all \(j\). Hence \(a \not\in \bigcup_{j=1}^{i}{(A_{j-1} \setminus A_j)}\) as required. We show the right-to-left inclusion by contraposition. Suppose \(a \notin A_i\). Write \(k = \min\{j \mid a \notin A_j\}\). Then \(a \in A_{k-1} \setminus A_k\). Moreover \(0 < k \le i\), so \(a \in \bigcup_{j=1}^{i}{(A_{j-1} \setminus A_j)}\) as required.
It can similarly be shown that \(B_i = B \setminus \bigcup_{j=1}^{i}{(B_{j-1} \setminus B_j)}\).
Now, recall from the proof of Theorem 2 that \(A_{j-1} \setminus A_j\) is the \(j\)-th rank of \({\ale_K^\phi}\), and \(B_{j-1} \setminus B_j\) is the \(j\)-th rank of \({\ble_K^\phi}\). Also recall that \(K^-\) is the sub-matrix of \(K\) obtained by restricting to rows \(A_i\) and columns \(B_i\). Given this and the expressions for \(A_i\) and \(B_i\) above, it follows that \(K^-\) is the sub-matrix of \(K\) resulting from the removal of the first \(i\) ranks of \({\ale_K^\phi}\) and \({\ble_K^\phi}\).
By repeatedly applying rank-removal, it can be seen that the \((i+1)\)-th rank of \({\ale_K^\phi}\) is the same as the first (maximal) rank of \({\ale_{K^-}^\phi}\), up to the appropriate reshuffling of indicies. 23
On the other hand, the \((i+1)\)-th rank of \({\ale_K^\phi}\) is \(A_i \setminus A_{i+1} = f(K, A_i, B_i)\), and the first rank of \({\ale_{K^-}^\phi}\) is \(f(K^-, A_{K^-}, B_{K^-}) = f(K^-, [m'], [n'])\). It follows that \(f(K, A_i, B_i)\) equals \(f(K^-, [m'], [n'])\) up to remapping of indicies, i.e. \(a_p \in f(K, A_i, B_i)\) iff \(p \in f(K^-, [m'], [n'])\) as required.
A similar argument shows the analogous property for \(g\). Hence \(\phi\) satisfies sub-matrix-independence.
A characterisation of \(\phicardint\) is now available.
Theorem 3
Consider the following property.
argmax: \(\max(A, {\ale_K^\phi}) = \argmax_{a \in A}{|K(a)|}\)
\(\phicardint\) is the unique operator satisying argmax, chain-def, rank-removal and dual.
Proof. First we show that \(\phicardint\) satisfies the stated properties. chain-def follows from Theorem 2 since \(\phicardint\) is an interleaving operator by definition, dual follows from Proposition 18, and argmax is evident from the definition. Finally, rank-removal follows from Proposition 19 and Proposition 21.
Now suppose some operator \(\phi\) sastisfies argmax, chain-def, rank-removal and dual. We need to show \(\phi = \phicardint\). By Proposition 21, \(\phi\) is an interleaving operator with sub-matrix-independence. Due to argmax, we also have \(\max(A, {\ale_K^\phi}) = \max(A, {\ale_K^{\phicardint}})\) for any tournament \(K\). It is easily seen that dual implies \(\max(B, {\ble_K^\phi}) = \max(B, {\ble_K^{\phicardint}})\) too. We may now conclude \(\phi = \phicardint\) by Proposition 20.
TO WRITE ABOUT:
Interleaving is a greedy algorithm for constructing a chain graph
Selection functions act as a heuristic to minimise the distance from \(K\)
We can take any social choice tournament solution (which selects a set of winners) to define \(f\) (and, by duality, \(g\))
Alternatively, could take any operator \(\phi\) and define \(f, g\) by the maximal ranks (c.f. the construction in the proof of Theorem 2. This will produce a modified operator \(\hat{\phi}\) which satisfies chain-def. Questions are: i) what properties of \(\phi\) carry over to \(\hat{\phi}\)? ii) Is it true that \(\phi = \hat{\phi}\) iff \(\phi\) already has chain-def?
Computational complexity: if \(f, g\) run in \(O(n^k)\) time then \(\phi_{f,g}^\text{int}\) can be computed in \(O(n^{k+1})\) time (roughly speaking…)
The argmax property seems to be saying that the top rank of \(\ale_K^\phi\) is chosen by the Copeland rule for tournaments [BBH16]. A characterisation is apparently given in [Hen85]. Would be good to see if the properties given in that reference can be used to axiomatically characterise argmax. Alternatively the IIM, sym and pos-resp characterisation of \(\phi_{\text{count}}\) could be adapated to act on the top rank only.
Some impossibility results¶
Whilst most of the properties listed earlier have intuitive appeal, there are combinations which are impossible to satisfy simultaneously, as the following result shows.
Theorem 4
Each set of axioms below are impossible to satisfy simultaneously:
sym, b-sym, strict-mon, chain-def
win-coherence, loss-coherence, chain-min
win-coherence, loss-coherence, dual, pos-resp
win-coherence, inv-rev, dual, pos-resp
loss-coherence, inv-rev, dual, pos-resp
TODO: (2) has now been subsumed by Proposition 10, but I refer to the proof of (2) in (3). Need to fix this…
Proof.
In each case we assume the existence of an operator \(\phi\) satisfying the stated properties, and derive a contradiction.
Write
\[K = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \\ 0 & 0 \end{bmatrix}\]It is easily seen that by taking permutations \(\sigma = \pi = (1\ 2)\) we have \(\pi(\sigma(K)) = K\). Consequently sym implies \(1 \aeq_K^\phi 2\) in \(A\), and b-sym implies \(1 \beq_K^\phi 2\) in \(B\).
Also note that \(K(4) \subset K(1), K(2) \subset K(3)\), so strict-mon implies \(4 \alt_K^\phi 1\), \(4 \alt_K^\phi 2\), \(1 \alt_K^\phi 3\) and \(2 \alt_K^\phi 3\). Both the \(A\) and \(B\) rankings are therefore fully determined as
\[\begin{array}{c|c} A & B \\ \hline 4 \alt 1 \aeq 2 \alt 3 & 1 \beq 2 \end{array}\]Note that \({\ale_K^\phi}\) has three ranks (\(\{4\}, \{1,2\}, \{3\}\)) and \({\ble_K^\phi}\) has one (\(\{1,2\}\)). But by Theorem 1, \(\phi(K) = ({\ale_K^\phi}, {\ble_K^\phi})\) is chain-definable if and only if the number of ranks differs by at most one. Therefore \(\phi(K)\) is not chain-definable, contradicting chain-def.
Consider
\[K = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]from the proof of Proposition 3. \(\minch{K}\) consists of four chain tournaments, and the corresponding rankings are
\[\begin{array}{c|c} A & B \\ \hline 1 \alt 2 & 1 \blt 2 \\ 1 \alt 2 & 2 \blt 1 \\ 2 \alt 1 & 1 \blt 2 \\ 2 \alt 1 & 2 \blt 1 \end{array}\]Note in particular that since \(\phi\) satisfies chain-min, there is necessarily a strict comparison between \(1, 2 \in B\): either \(1 \blt_K^\phi 2\) or \(2 \blt_K^\phi 1\) (and similar for the \(A\)s).
Suppose \(1 \blt_K^\phi 2\). Since \(K(1) = \{1\}\) and \(K(2) = \{2\}\), win-coherence implies \(1 \alt_K^\phi 2\). On the other hand, \(B \setminus K(1) = \{2\}\) and \(B \setminus K(2) = \{1\}\), and loss-coherence implies \(2 \alt_K^\phi 1\). Clearly this is a contradiction.
The case \(2 \blt_K^\phi 1\) is similar; either way we get the desired contradiction.
Take
\[K = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}\]Using the same argument as in part (2), the presence of win-coherence and loss-coherence together forces \(1 \beq_K^\phi 2\) in \(B\) (otherwise we would need to have both \(1 \alt_K^\phi 2\) and \(2 \alt_K^\phi 1\)). Consequently, dual implies \(1 \aeq_{\dual{K}}^\phi 2\). Note that
\[\dual{K} - \mathbf{1}_{23} = \begin{bmatrix}0&1&1\\1&0&1\end{bmatrix} - \begin{bmatrix}0&0&0\\0&0&1\end{bmatrix} = \begin{bmatrix}0&1&1\\1&0&0\end{bmatrix}\]pos-resp dictates that \(2 \in A_{\dual{K}}\) goes strictly down in ranking compared to in \(\dual{K}\), i.e. \(2 \alt_{\dual{K} - \mathbf{1}_{23}}^\phi 1\). Writing
\[K' = \dual{\dual{K} - \mathbf{1}_{23}} = \begin{bmatrix}1&0\\0&1\\0&1\end{bmatrix}\]and applying dual again, we therefore get \(1 \blt_{K'}^\phi 2\). As in part (2), win-coherence implies that \(1 \alt_{K'}^\phi 2\), whereas loss-coherence implies \(2 \alt_{K'}^\phi 1\); a contradiction.
Clearly inv-rev and dual imply b-inv-rev, which together with win-coherence implies loss-coherence. Now the impossibility follows from (3).
Identical to (4).
Note that the list above does not include 2-axiom “impossibility results” already presented in the section on chain-minimal operators.
Experimental analysis¶
Rough notes
Ideas:
How well can we efficiently approximate some aspects of chain-minimality?
Find some close chain graph (simulated annealing)
Be close to the average (will cardinality-based interleaving be good for this?)
How well can different operators match the ‘true’ ranking in the MLE model? In theory chain-minimal operators are the best: does this hold up when looking at the rankings out? Which tractable operators give the best performance? If approximation error compared to chain-min (see above) correlated with performance?
To do this, have a way of generating states \(\theta\); equivalently, a way of generating chain tournaments. For a bunch of sizes \(m \times n\) and \(\alpha\)s, generate a load of states \(\theta\) and sample a \(K\). Run each operator and measure distance (e.g. Kendall-tau) to the ranking from \(K_\theta\).
Existing tournament methods¶
A bipartite tournament is a special case of a partial tournament, in which players are drawn from a single set \(X\), but each \(x \in X\) does not necessarily play against all other \(x' \in X\). There already exist several ranking methods and axioms for (subclasses of) partial tournaments: our question is whether they are suitable for the bipartite case.
Definition 17 (Partial tournament [GonzalezDiazHL14])
A non-bipartite tournament is a pair \((X, T)\), where \(X = [t]\) for some \(t \in \N\) and \(T \in \R_{\ge 0}^{t \times t}\) is a non-negative \(t \times t\) matrix with \(T_{ii} = 0\) for all \(i \in [t]\).
This generalises our bipartite tournaments in two ways: i) matches need not take place in all cases, and ii) matrix entries are not restricted to lie in \(\{0,1\}\). The interpretation of the tournament matrix \(T\) is that a positive entry \(T_{ij}\) represents the score of player \(i\) against player \(j\), possibly aggregated over several matches. Note that we can have both \(T_{ij}, T_{ji} > 0\).
Given an \(m \times n\) bipartite tournament \(K\), we can form a \((m + n) \times (m + n)\) non-bipartite partial tournament \(T\) as the (anti-diagonal) block matrix
Although our setting is in principle more restricted than partial tournaments in full generality, such anti-diagonal block matrices may be reducible 24 and thus be excluded from consideration in some papers, e.g. [GonzalezDiazHL14] and [BVazquezCCDiazRGonzalezDiaz08].
The reducibility issue is side-stepped in [SV05] by decomposing \(T\) into maximally irreducible components, ranking each one separately, and combining the rankings across all components. However, their framework allows ranking methods to give partial orders on the set of players. It would be interesting to see if the particular form of \(T\) above guarantees that we can at least compare all the \(A\)s and \(B\)s (that is, the sets \(\{1,\ldots,m\}\) and \(\{m+1,\ldots,m+n\}\)). Since we always have \(K_{ab} = 1\) or \(\dual{K}_{ba} = 1\), it seems plausible that this is the case.
Existing operators¶
Write \(\mathcal{B}\) for the set of all anti-diagonal block matrices \(T\) formed from some bipartite tournament \(K\). We consider which ranking methods from [GonzalezDiazHL14] are well-defined on \(\mathcal{B}\), and what the corresponding operators look like in our restricted binary bipartite context.
Scores. This method reduces to the ‘counting’ operator \(\phi_{\text{count}}\) in our setting.
Maximum likelihood (also known as the Bradley-Terry model). It appears that the irreducibility assumption is key in the proof of existence and uniqueness of the maximum likelihood estimate [For57], so we cannot guarantee that this ranking method is well-defined on \(\mathcal{B}\).
There do exist modifications to the Bradley-Terry model which remove the irreducibility requirement, however (e.g. [Yan16]), so following these methods might be an option.
Neustadtl. Well-defined on \(\mathcal{B}\). In general this method takes into account the number of matches played by each player; since in our case this is more or less constant (\(|B|\) for the elements of \(A\) and \(|A|\) for the elements of \(B\)), the corresponding operator \(\phi_{\text{neu}}\) takes a simplified form, given by the rating vectors \(r_{\phi_{\text{neu}}}(K) = K\dual{K}\mathbf{e}\) and \(s_{\phi_{\text{neu}}}(K) = \dual{K}Ke\). That is,
\[\begin{aligned} [r_{\phi_{\text{neu}}}(K)]_{a} &= \sum_{b \in K(a)}{ |\dual{K}(b)| } \\ [s_{\phi_{\text{neu}}}(K)]_{b} &= \sum_{a \in \dual{K}(b)}{ |K(a)| } \end{aligned}\]The idea here is that when ranking the \(A\)s, for instance, it is better to have defeated those \(b \in B\) who have themselves defeated many players in \(A\).
It is clear that \(\phi_{\text{neu}}\) satisfies dual.
Fair bets. This method is not well-defined on all of \(\mathcal{B}\). Most obviously, the Fair bets ranking does not exist when there exists some \(a \in A\) with \(K(a) = B\), or some \(b \in B\) with \(K^{-1}(b) = \emptyset\). These cases are easily resolved by just rankings the relevant \(A\)s and \(B\)s maximally, however, so there is still hope for a version of this method in the bipartite context.
In [SV05] the authors generalise Fair bets to reducible tournaments using the aforementioned decomposition into maximally irreducible components, but it is possible for the resulting operator to only give partial orders.
Buchholz. Well-defined on \(\mathcal{B}\). In the full partial tournament setting this method takes into account both the number of victories and the average strength of one’s opponents. However in our case each \(a, a' \in A\) faces exactly the same opponents. This means the ranking reduces to simply counting victories, i.e. \(\phi_{\text{count}}\).
Recursive Buchholz (also known as Least squares). Again, this method coincides with \(\phi_{\text{count}}\).
Recursive performance. This method comes from [BVazquezCCDiazRGonzalezDiaz08], wherein the authors remark that it is not well-defined for anti-diagonal block tournaments (see assumption A2 in that paper).
Generalised row sum. Quite a complicated ranking method, but I think it also collapses to \(\phi_{\text{count}}\) in our setting.
Whilst it is interesting that Buchholz, Recursive Buchholz and Generalised row sum each reduce to \(\phi_{\text{count}}\) in the case of anti-diagonal block matrices, it may be unsurprising given that each method satisfies the following property:
score-consistency: ([GonzalezDiazHL14]) A ranking method should coincide with the Scores ranking for round-robin tournaments
Here the Scores ranking assigns each player \(i\) their average score \(s_i = \sum_{j}{\frac{T_{ij}}{m_i}}\) where \(m_i = \sum_{j}{(T_{ij} + T_{ji})}\) represents the matches played by player \(i\), and \(T\) is round-robin if \(T_{ij} + T_{ji} = 1\) for all \(i \ne j\).
Now our anti-diagonal block tournaments are not round-robin, but it does hold that \(T_{ij} + T_{ji} = 1\) whenever \(i \le m\) and \(j > m\). Thus there is a sort of ‘round-robinnes’ when considering matches between the \(A\)s and \(B\)s.
- 1
Indeed, if \(K = \dual{K}\) then \(K_{11} = 1 - K^\tr_{11}\) and \(K_{11} = 1/2\), which is not possible.
- 2
To borrow the terminology of González-Díaz et. al…
- 3
For an operator that satisfies dual but not inv-rev, consider \(\phi\) which always outputs the fixed rankings \(1 \alt \cdots \alt m\) and \(1 \blt \cdots \blt n\). For an operator satisfiying inv-rev but not dual, consider \(\phi\) whose output rankings are the same as \(\phi_\text{count}\) but with the \(B\) ranking reversed.
- 4
Despite the asymmetry in the definition, \({\le_1}\) is compatible with \({\le_2}\) iff \({\le_2}\) is compatible with \({\le_1}\).
- 5
Note that a full total order on \(\K\) is not necessary here; it would suffice to have a total order \(\trianglelefteq_K\) on \(\minch{K}\) for each \(K\). However, having a concrete example of a total order on \(\K\) will be useful in what follows.
- 6
Defining \(\v(K)\) as the vectorisation alone would not be injective: consider \(K_1 = \left[\begin{smallmatrix}1 & 1\end{smallmatrix}\right]\) and \(K_2 = \left[\begin{smallmatrix}1 \\ 1\end{smallmatrix}\right]\). This is why we include the size of \(K\) in \(\v(K)\).
- 7
Note that there are several choices for how vectors of different sizes should be ordered. In this application, however, \(\trianglelefteq\) is only actually used to compare between elements of \(\minch{K}\) which do have the same size, so the choice is irrelevant.
- 8
I wrote a small test in the Haskell code to try to find counterexamples to the conjecture. It seems there are none for \(|A| \le 4\) and \(|B| \le 3\).
- 9
A proof sketch: first show that \({\ale_K^\phi} = {\anle_K}\) for all \(K\) with the chain property by induction on \(|B|\). single-b covers the base case \(|B| = 1\), and Lemma 3 parts (1) and (2) together with extension provide the inductive step. Now an appeal to dual and Proposition 2 part (2) shows that \({\ble_K^\phi} = {\bnle_K}\) for all \(K\) with the chain property, so we have shown that \(\phi\) satisfies k-chain.
- 10
Although note that any tournament with \(|A| = 1\) or \(|B| = 1\) automatically has the chain property, so we must have \(\alpha_\phi(K) = K\) anyway.
- 11(1,2)
Another possible generalisation is to let \(\trianglelefteq\) depend on the size of \(K\): have a total order \(\trianglelefteq_{m,n}\) for each \(m, n \in \N\). I think this does place a restriction over chain-minimal operators.
- 12
Here \(\alpha_{\trianglelefteq_K}\) is as defined in Definition 8.
- 13
Note that we abuse notation by using the symbol \(d_W\) for all the distance functions defined for tournaments of different sizes.
- 14
For simplicity we use numerical skill levels here, although it would suffice to have a partial preorder on \(A \cup B\) such that each \(a \in A\) is comparable with every \(b \in B\).
- 15
Note that a false positive for \(a\) is a false negative for \(b\) and vice versa.
- 16
Specifically, if there are \(\sigma\) and \(\pi\) such that \(\pi(\sigma(K)) = K\) (and at least one permutation is not the identity mapping), then \(\tilde{K} \in \minch{K}\) implies \(\pi(\sigma(\tilde{K})) \in \minch{K}\) also.
- 17
It may be of minor interest to note that this \(K\) is one of the ‘smallest’ examples of ‘asymmetric’ tournaments with multiple closest chain tournaments. That is, no size smaller than \(3 \times 4\) (or \(4 \times 3\)…) exhibits such examples, and moreover there are no \(3 \times 4\) examples with fewer than 6 ones.
- 18
It should be clear from context whether \(d\) refers to the distance between pairs of rankings or the Hamming distance between tournaments.
- 19
Note that b-sym is strictly weaker than the conjunction of sym and dual.
- 20
I think we also have that \(K \sim K'\) if the bipartite graphs corresponding to \(K\) and \(K'\) are isomorphic.
- 21
In fact we could instead find the closest non-integer chain tournaments, where \(K\) has the chain property if \({\anle_K}\) is total, with \(a \anle_K a'\) iff \(K_{ab} \le K_{a',b}\) for all \(b \in B\). I think in this case this actually means \(\hat{K}\) has a unique closest chain-tournament \(\left[\begin{smallmatrix}0&1&\frac{1}{2}\\1&{\color{red}1}&1\\0&0&0 \end{smallmatrix}\right]\).
- 22
Here \(\max(Z, {\le}) = \{z \in Z \mid \not\exists z' \in Z : z < z'\}\), for any set \(Z\) and a total preorder \({\le}\) on \(Z\) (with strict part \({<}\)).
- 23
We omit a detailed proof of this, which would be similar to the “if” direction of the present proof, where sub-matrix-independence is applied to a sub-matrix \(K^{--}\) of \(K^-\) and so on.
- 24
An easy example is a tournament in which some \(a \in A\) has \(K(a) = B\); in this case \(a\) has no incoming edge in the graph corresponding to \(T\), so this graph is not strongly connected and thus \(T\) is reducible.
For a non-trivial example which does not have an immediate ‘obvious’ ranking, consider \(\left[\begin{smallmatrix}1&1&0&1\\1&0&1&1\\0&0&0&1 \\1&0&0&0\end{smallmatrix}\right]\).