mercredi 13 octobre 2010

The Social Network: a flawed voting scheme

As it happens, I saw The Social Network. For my defense, I should mention that this is solely the unfortunate result of my friends' perfidy: a moment of inattention and bim! you're suddenly learning about the romantic life of Mark Zuckerberg - or whatever. Nothing really made me go "wait a second", except in the very beginning of the film when our protagonist happily codes a voting scheme for ranking Harvard's girls - by "hotness" of course. As the story goes, he calls his best friend over for him to explain "the chess playing algorithm". I am clueless as to what chess has to do with this, but let's move on. (EDIT: They are probably talking about the Elo rating system which does indeed make some sense.)

Back at my university, we had the same thing on our local network periodically going on (at each beginning of the academic year - I let you guess why). The concept is fairly primitive. You are presented two girls' pictures and you click on which of the two you like best. Then two other girls are randomly paired and presented to you, and so on. In the end, after many many votes, you magically end up with some ranking of the girls. No comments on the social value of this. So, what is really going on behind the scene?

Let's get more precise. Basically, you have $n$ voters and $m$ choices. Each voter has a complete ordering over $\{1,...,m\}$, meaning that for any two choices $x,y$, one has either $x < y$ or $x > y$. Is complete ordering a realistic assumption? Well, it does not matter, as long as you are only given the $<$ or $>$ alternative in the voting scheme, so we will restict ourselves to this case. There are exactly $2^{\frac{m(m-1)}{2}$ possible distinct complete orderings, some of them are not transitive: you can have $a>b$ and $b>c$ but $a < c$. But is it possible that if I prefer $a$ to $b$ and $b$ to $c$, then I actually prefer $c$ to $a$? The answer is negative if I am being self-consistent. Thus, an opinion is a transitive complete ordering, i.e. a permutation. Therefore, voters' opinions are reasonably accounted for by permutations over the $m$ choices, and there is exactly $m!$ possible distinct opinions.

Thus, if $S_m$ denotes the group of permutations of size $m$, the voting scheme is adequately represented by a function $f$ which type is:

$$f : S_m^n \rightarrow S_m$$
Functions such as $f$ are called aggregation functions, because their aim is to aggregate voters' information into a single, hopefully fair and representative global choice. Now, unpleasant things start happening here. Even without knowing how $f$ works, we have some important limitations on what can $f$ achieve at best. Perhaps the simplest thing to think of is Condorcet's voting paradox: there exist configurations of voters' rankings such that $f$ will contradict a majority of voters' preferences.

Take 3 voters having personal preferences $a>b>c$, $b>c>a$ and $c>a>b$. We see that a strict majority of 2 voters prefer $a$ to $b$, $b$ to $c$ and $c$ to $a$. Thus, the majority ordering is not commutative, but $f$ is restricted to deliver a permutation. Thus any resulting aggregated ranking will contradict at least one of the 3 majority preferences.
Arrow's impossibility theorem extends and precises this puzzling result: basically there exists no fair (several natural requirements) aggregating functions for this problem. So, whatever the actual algorithm does, it can not claim to find the majority ranking of candidates - because no such thing exists in general. That's fairly poor for something supposed to give the general opinion of voters, right?

But in all honesty, we did not need Arrow's theorem to figure out that the idea does not work so well. Basically, the easiest way to implement this stuff is to keep track of the number of clicks on each candidate and to rank them accordingly. Thus, cheating is extremely easy by repeatedly clicking on someone's face. And the more you click, the stronger your voice is. Well, from my perspective, it seems that this is the number one reason why after a short initial enthusiastic phase, such applications invariably die out. And that's for the best :-)

2 commentaires:

  1. Choose a reasonable definition for a distance between two orderings (e.g., the smallest number of transpositions needed to go from one to another, or the number of ranks for which they disagree, or a combination of the above, or...).

    1. Is there an (efficient) way to compute the centroid of a collection of orderings if you are given all orderings?

    2. Assuming that you can probe the value of an ordering only by asking it to perform a comparison between two elements you choose, how many comparisons will be needed to get a (good) approximation of that centroid?

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  2. Indeed, a natural distance would be the number of transpositions. This is called the Kendall tau distance.

    1. It is notorious that computing the centroid in the case of Kendall's distance is NP-hard. See Bartholdi, Tovey, and Trick: Voting schemes for which it can be difficult to tell who
    won the election (1989). Love the paper name by the way...

    2. There is a bunch of approximation algorithms for the task, some randomized, others not. A general question is that of ordering under noise: every time you compare two elements, you get a wrong result with some probability e<1/2. Surprisingly, you can get a very high probability ordering in O(n log n) time: you just loose a constant factor compared with the usual sorting methods! See Feige, Raghavan, Peleg, and Upfal: Computing with noisy information (1994).

    There's also Colin Mallows model (exponential in Kendall's distance probability distributions for the votes) (still NP-Hard) studied in Meila, Phadnis, Patterson and Bilmes: Consensus ranking under the exponential model (2007).

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