On the Impossibility of Certain Ranking Functions

Transcription

1 On the Impossibility of Certain Ranking Functions Jin-Yi Cai Abstract Suppose all the individuals in a field are linearly ordered. Groups of individuals form teams. Is there a perfect ranking function of each team based on the members of the team? We prove that under a very mild and reasonable set of axioms for the ranking function, no such ranking function exists. AMS Subject Classification: 68R05, 91F10, 06A05, 06A07. Keywords: Ranking function. It is a personal observation that a non-trivial percentage of academics in U. S. universities are quite concerned with the ranking of their respective universities, colleges, departments, and maybe sub-areas within the department. Full professors are concerned with the reputation of their departments and sub-areas within the department, which by extension reflects their own reputation; associate and assistant professors are equally concerned with their respective standing within the field which affects future professional growth prospects. Also frequently applicants for positions at every level are swayed by the ranking of a university and department; even the graduate admission process is heavily influenced by such rankings. Compilations and publications of university rankings, some more meaningful, some less so, such as those by Computer Sciences Department, University of Wisconsin, Madison, WI 53706, jyc@cs.wisc.edu, Research supported in part by NSF grant CCR and CCR

2 the U.S. News and World Report, are generally taken seriously, even by the leaders of higher education such as Deans and Presidents. Generally, it is doubtful any linear order of excellence makes sense for people even in a narrow sub-area of expertise. But let us suppose that in some circumstances this is possible and is already given. For example, suppose everyone working in area X has been so linearly ordered. These people belong to various universities. Suppose further that this forms a partition so that everyone in this area belongs to a unique university. Then we ask, is there any general principle by which one can come up with a ranking of the universities in area X based on the individual ranking in area X? The famous theorem by Kenneth Arrow [1] (see also [2, 3]) states that there can not exist any social choice function that satisfies every one of a set of very mild and reasonable axioms for social choice. We prove a theorem in the spirit of Kenneth Arrow s theorem in the case of ranking functions. We prove that under a very mild and reasonable set of axioms for the ranking function, no such ranking function exists. Assume we are given an infinite totally ordered set Ω. The ordering relation is denoted by <. Each team consists of a non-empty subset of Ω. Here a team is a set of individuals put together hypothetically; in particular when we discuss two teams they need not be disjoint. Let us assume there is a ranking function r which for any finite collection of distinct teams gives an ordering of the teams, i.e., r produces a permutation of the teams. If A and B are two teams, we write A B if the ranking function r puts A ahead of B. We also write B A if A B. Also we write A B if either A B or A = B. Now we give some seemingly mild and reasonable axioms that a ranking function obviously should satisfy. Axiom 1. (Foundation) The ranking function depends only on the ordering relation < (and not on the elements per se), i.e., if f is a mapping that preserves the ordering relation on X 1... X m, then the permutation of r({x 1,..., X m }) is the same as r({f(x 1 ),..., f(x m )}). Also, if x, y Ω and x < y, then {x} {y}. This Axiom states that the ranking function is native to the underlying ordering of Ω. Axiom 2. (Identical Extension) If A B and x A B, then A {x} B {x}. 2

3 This Axiom states that the ranking for any pair is unaffected by the extension of an identical element. Axiom 3. (Extremal Extension) If x z y for all z A, then A {x} A A {y}. This Axiom states that if we extend at the extremal end, we increase the team ranking if we extend at the top, and we decrease the team ranking if we extend at the bottom. All of the above Axioms are, I believe, totally unassailable. The next Axiom is more delicate. Axiom 4. (Non-interference) In this axiom we wish to say that the relative ranking between two teams should be determined by their own respective members ranking in relation to each other. As long as their team memberships for these two teams do not change, the presence or absence of any other teams should be irrelevant to the ranking of this particular pair, in particular, the induced order on the pair is independent of all other teams. To state it formally, we say that For any {A, B, X 1,..., X m }, the order on {A, B} induced from the permutation of r({a, B, X 1,..., X m }) is the same as r({a, B}). We note that by Axiom 4, the ranking function does define a total order on the non-empty subsets of Ω. We only need to check for transitivity, if A B and B C, then A C. This is quite obvious if we consider {A, B, C}. Assuming is a total order, by transitivity Axiom 3 has the following corollary, (of which Axiom 3 is also a special case). Axiom 3. (Extremal Extension) Suppose A B. If x z and z y for all z A B, then A {x} B and A B {y}. Also, note that since ranking function r is invariant under any order isomorphism, we may rename elements as long as it respects the ordering. For example, {1, 4} {2, 3} iff {2, 5} {3, 4} iff {i 1, i 4 } {i 2, i 3 } for any i 1 < i 2 < i 3 < i 4. Now we will see that two particular ranking functions satisfy Axioms 1 to 4. These ranking functions are however not particularly interesting. Definition The Lexicographic order lex is a total order defined on all non-empty subsets of Ω as follows: Let A and B be two distinct finite subsets of Ω. Enumerate 3

4 the elements of A and B in the order of Ω, A = {a 1 < a 2 < < a n } and B = {b 1 < b 2 < < b m }, where n, m 1. Let i be the maximal index, 0 i min{n, m}, such that for all 1 j i, a j = b j. Then, if both n > i and m > i, we define A lex B iff a i+1 < b i+1 ; otherwise, we define A lex B iff n = i. In other words, we line up A and B from the leading element, and cancel the longest common prefix, if any. If both have elements left, then the next leading element determines the order of A and B. If one set is a prefix of the other, say A is a (proper) prefix of B, then A precedes B. Similarly, the Reverse-lexicographic order looks at the ordered sequence of elements of a set from the other end. Definition The Reverse-lexicographic order rlex is a total order defined on all nonempty subsets of Ω as follows: Let A and B be two distinct finite subsets of Ω. Enumerate the elements of A and B in the order of Ω from the last, A = {a n < < a 2 < a 1 } and B = {b m < < b 2 < b 1 }, where n, m 1. Let i be the maximal index, 0 i min{n, m}, such that for all 1 j i, a j = b j. Then, if both n > i and m > i, we define A rlex B iff a i+1 < b i+1 ; otherwise, we define A rlex B iff m = i. Note that in Reverse-lexicographic order if A is a (proper) postfix of B, then B precedes A. It is clear that both the Lexicographic order and the Reverse-lexicographic order satisfy Axioms 1 to 4. However, we feel that the Lexicographic order is overly elitist and thus neither realistic nor suitable for a good ranking function. Similarly the Reverselexicographic order is too conservative placing too much weight on the bottom end of the scale, and thus is similarly not suitable as a good ranking function. Thus we postulate the following Axiom 5. (Non-Extremism) The ranking function is neither the Lexicographic order nor the Reverse-lexicographic order. We will now state and prove the impossibility theorem Theorem 1 There is no ranking function that satisfies Axiom 1 to 5. We will first establish several consequences of the Axioms in 1. to 4., then prove Theorem 1 in Axiom 2 implies its converse. (Law of Cancellation): x[x A B, and A {x} B {x} = A B]. 4

5 Proof: If A B, then A {x} B {x}. 2. (Monotonicity) If A B and x < y, and x A and y B, then A {x} B {y}. More generally, 3. Axiom 2 implies a law of subadditivity. (Law of Subadditivity): For A C = B D =, A B & C D = A C B D. Moreover, if at least one of A B and C D is strict, then we also have strict A C B D. Proof: If A = B and C = D the conclusion is trivial. If A = B and C D, then we just apply Axiom of Identical Extension (Axiom 2.) A = B times to C D. Since the additional elements are all disjoint from both C and D, the application of Axiom 2 is legitimate. The case A B and C = D is symmetric. So we now assume A B and C D, both strictly so. Let = B C, and B = B and C = C. Since A C =, certainly A C =. Also by definition B C =. Thus it is legitimate to form both A C and B C by adding identical elements precisely C times. By Axiom of Identical Extension (Axiom 2), A C B C. Then from C D, since C B = D B =, we may apply Axiom 2 precisely B times, and obtain C B D B. Note that B C = C B = B C, it follows that A C B D, by transitivity of. Finally we add the elements of. We note that both (A C ) = and (B D) =, thus we may apply Axiom 2 precisely times, to obtain A C B D. 4. From Axiom of Non-interference (Axiom 4) the ranking function r induces a total order (which we denoted ), and this order uniquely determines the ranking function r. That is, satisfies the following: for any teams A and B, exactly one of the following three alternatives hold, either A B, or A = B, or B A. Moreover the binary relation is transitive and uniquely induces the ranking function r. 5. Now we are ready to prove Theorem 1. Suppose satisfies Axioms 1 to 5. Let (A, B) be a pair of distinct teams, and we suppose A B yet B lex A. By the Law of Cancellation, we may cancel the longest 5

6 common prefix of A and B. If A B and B lex A, then neither can it be the case that B is a prefix of A, nor can it be the case that A is a prefix of B. For otherwise, if B is a prefix of A, then by Axiom of Extremal Extension (Axiom 3.) we can extend the tail end of B to get A, and we would have B A. On the other hand, by the definition of the Lexicographic order, if A is a prefix of B (so a proper prefix since A B), then A lex B. Hence, after the cancellation of the longest common prefix of A and B neither is empty. We will rename A and B for what remains after this cancellation. Since B lex A, the next leading element in A B belongs to B (but not A). Call it b. Since A B, it can not be the case that there are no other elements of B besides b, for otherwise all elements of A are preceded by b, the unique element of B, but then it follows that B A by the Axiom of Extremal Extension. Thus, there are some elements both of A and of B other than b. Now move all b B other than b to somewhere after all of A, we should still have A B. More precisely, By Monotonicity, if B is obtained from B by exchanging every element of B {b} with an element of Ω which is preceded by every element of A and B, leaving b unchanged, we have B B. Since Ω is infinite, and the ranking order on teams A and B only depends on the underlying order relation and not the exact elements of A B per se (Axiom 1), we may shift the elements of A and B if necessary, and assume there are sufficiently many distinct elements in Ω after all A and B, to which place we can move B {b}. Then after a shift again (by order invariance), we may finally arrive at a pair A 1 and B 1, where we may name A 1 = {1,..., k}, and B 1 = {0, k + 1,..., k + l}, (1) where k and l are some integers 1. A 1 B 1 yet B 1 lex A 1. Next, since the order is not equal to Reverse-lexicographic order, there are pair of distinct teams C and D, where C D yet D rlex C. By a similar argument as given above for the Lexicographic order, we can produce a pair of distinct teams C 1 and D 1, with the property C 1 D 1 yet D 1 rlex C 1, and furthermore we may assume they take the following normal form D 1 = {M s,..., M 1}, and C 1 = {M (s + t),..., M (s + 1), M}, (2) where s and t are some integers 1, and M is some large integer such that M (s + t) k + l. Clearly for the pair A 1 B 1 we may extend B 1 at the tail extreme by any number of elements from Ω, and still maintain A 1 B 1 yet B 1 lex A 1. Thus, for the values 6

7 k and l in Eqn (1) we may replace an arbitrarily large l l for l for the same k. Similarly for the pair C 1 and D 1, we may extend C 1 at the leading extreme by any number of elements from Ω, and still maintain C 1 D 1 yet D 1 rlex C 1. More precisely, for the values s and t in Eqn (2) we may replace an arbitrarily large t t for t for the same s. After any chosen replacement of l by l, and t by t, we may further replace M to a larger M by a shift if necessary, so that still we may assume M (s + t ) > k + l. Now we choose l = k 2 + ts + a, and t = s 2 + kl + a, where a is an arbitrary positive integer so that we may assume l l and t t. After that we assume a suitable M is chosen so that M (s + t ) > k + l. Finally consider the following two teams X and Y. and X = {1,..., k, M (s + t ),..., M (s + 1), M}, Y = {0, k + 1,..., k + l, M s,..., M 1}. Thus, X is the (disjoint) union of A 1 and C 1, with appropriate replacement of t by t in C 1, and similarly Y is the (disjoint) union of B 1 and D 1, with appropriate replacement of l by l in B 1. Since A 1 B 1 and C 1 D 1, we have X Y, by Law of Subadditivity. However, we will decompose X and Y in a different manner to show that Y X also follows from the Axioms. Consider the following k pairs of teams and E i = {i, M (s + t ) + (i 1)l,..., M (s + t ) + (i 1)l + l 1} = {i, M (s + t ) + (i 1)l,..., M (s + t ) + il 1}, F i = {k + (i 1)k + 1,..., k + (i 1)k + k} = {ik + 1,..., (i + 1)k}, for 1 i k. Note that all E i, F i are pair-wise disjoint, and collectively k i=1 F i = {k+1,..., k+ k 2 } is part of Y, and k i=1 E i = {1,..., k, M (s + t ),..., M (s + t ) + kl 1} is part of X. The last element in k i=1 E i is (M s) s 2 a 1 by the definition of t. Each pair (F i, E i ) is order isomorphic to the pair (A 1, B 1 ), and it follows that F i E i for all 1 i k. 7

8 and Similarly we define the following s pairs of teams G i = {(M s) s 2 a + (i 1)s,..., (M s) s 2 a + (i 1)s + (s 1)}, H i = {k + k 2 + (i 1)t + 1,..., k + k 2 + it, M s + (i 1)}, for 1 i s. Note that all G i, H i are also pair-wise disjoint (and pair-wise disjoint from all E i, F i ), and collectively s i=1 H i = {k + k 2 + 1,..., k + k 2 + st, M s,..., M 1} is part of Y, and s i=1 G i = {M s s 2 a,..., M s a 1} is part of X. Together and k s F i H i = {k + 1,..., k + k 2 + st, M s,..., M 1}, i=1 i=1 k s E i G i = {1,..., k, M (s + t ), M s a 1}. i=1 i=1 As before, each pair (H i, G i ) is order isomorphic to the pair (C 1, D 1 ), and it follows that H i G i for all 1 i s. Observe that what is left in X after taking out k i=1 E i s i=1 G i is [ k ] X s = X E i G i = {M s a,..., M s 1, M} i=1 i=1 and correspondingly what is left of Y after taking out k i=1 F i s i=1 H i is [ k ] Y s = Y F i H i = {0, k + l a + 1,..., k + l }. i=1 i=1 We can easily pair up {0} with {M}, and pair up the respective stretches of a consecutive individuals, and conclude that Y X, by Axiom 1 and Law of Subadditivity. It follows, by the decomposition of X = X k i=1 E i s i=1 G i and Y = Y ki=1 F i s i=1 H i that Y X, 8

9 by Law of Subadditivity. This contradiction shows that there can not be any ranking function which satisfies Axioms 1 to 5. The theorem is proved. 6. This theorem remains valid even if we allow the ranking function r to assign equal ranking for distinct subsets. In fact, to differ from the Lexicographic order which is a strict total order, it must differ on a pair of distinct subsets (A, B). Then without loss of generality we may assume B lex A, and yet A B. Then we can construct (A 1, B 1 ) just as before, with A 1 B 1 yet B 1 lex A 1. Similarly we can construct distinct teams C 1 and D 1, with the property C 1 D 1 yet D 1 rlex C 1. Finally X and Y are constructed in the same way and X Y. But in the decomposition of X = X k i=1 E i s i=1 G i and Y = Y k i=1 F i si=1 H i, Y X is strict, and the others pairs are at least F i E i and H i G i, and so, strictly Y X by Law of Subadditivity. Theorem 2 There is no ranking function that satisfies Axiom 1 to 5, even if it is allowed to assign equal rank on distinct subsets. 7. If the underlying order < on Ω is allowed to assign equal rank on distinct elements, and, assume such an equal rank is indeed assigned on a pair of distinct elements, then it is even easier to show that there can be no consistent ranking function for all subsets that satisfies Axiom 1 to 4. This is so even for any appropriately defined Lexicographic order or Reverse-lexicographic order. To see that, we consider any ranking function r for all multisets. We represent two individuals with equal rank as a multiple element (with multiplicity two) in the subset. Then consider how it is possible to assign the ranking between {x} and the multiset having x with multiplicity two, which we denote by {x, x}. Here we even allow r to assign equal rank on distinct multisets. Suppose {x} {x, x}. Take any y such that x y, and consider {x, y} and {x, x, y}. By Axiom of Identical Extension, But by Law of Cancellation, we get {x, y} {x, x, y}. {y} {x, y}. 9

10 However Axiom of Extremal Extension implies that {x, y} {y}. A similar contradiction ensues if we assume {x, x} {x}. This time we take a z x and consider {z, x} and {z, x, x}. Theorem 3 There is no ranking function that satisfies Axiom 1 to 4, even if it is allowed to assign equal rank on distinct sub(multi)sets. Remark: Of course academic Deans and Presidents will continue to pay close attention to their school rankings, no matter what we prove, and no matter how imperfect such rankings are. Much is the same as voting schemes, Arrow s Theorem not withstanding, imperfect schemes will continue to be used, not to mention Florida. However, one can still discuss one scheme being better than another. It would be interesting to investigate some extended notions of ranking functions, perhaps some probabilistic ones. Regarding the technical reason for the impossibility of Theorem 1, I think the real spoiler is the Axiom of Non-interference, together with the requirement that the order on subsets is invariant under order isomorphism for the underlying order of Ω. In Arrow s Theorem, a similar Axiom, called Independence of Irrelevant Alternatives is considered mostly responsible for that. References [1] K. Arrow. Social Choice and Individual Values. John Wiley and Sons (revised edition, 1963). [2] M. Balinski and H. Young. Fair Representation: Meeting the Ideal of One Man, One Vote Yale University Press, [3] W. Riker, and P. Ordeshook. An Introduction to Positive Political Theory. Prentice-Hall,