Wikipedia:Reference desk/Archives/Mathematics/2011 November 22

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November 22

Football (soccer) league problem

A football (soccer) league comprises 10 teams. The league is conducted on a single round-robin basis, which means that each team has to play against every opponent once.

The league uses the current evaluating system (3 pts for a win, 1 pt for a draw, 0 pts for a defeat), and after the completion of all matches Team A finishes first.

If the league were to use the former evaluating system (2 pts for a win, 1 pt for a draw, 0 pts for a defeat) and all results were to be repeated, Team A would finish last.

Identify the results.

It was reported in the news that this problem was given at some kind of International Mathematical Olympiad for high school students held in Nepal a few days ago. I'm very curious to know the answer, but I'm not sure I can figure it myself, so I'm asking you for assistance. --Theurgist (talk) 02:22, 22 November 2011 (UTC)

Can we assume that there is no tie with A in either first or last place ? Otherwise we could just make all games a draw and thus A would be both first and last under both systems. StuRat (talk) 02:35, 22 November 2011 (UTC)

I suppose this can be assumed. The above description is what a contestant retells in brief in this TV reportage (in Bulgarian). I guess the exact wording of the problem might have been published somewhere by the organisers of the Olympiad, but I haven't so far been able to discover it. But I personally find it unlikely that the problem should not have ruled out scenarios like all matches ending as draws and all teams being both first and last under both systems. --Theurgist (talk) 02:53, 22 November 2011 (UTC)

Write X->Y if X defeats Y and let the outcomes be

A->B->C->A

A->D->E->A

A->F->G->A

A->H->I->A

J->A

with every other game being a draw. Then I get for 3-point wins scores of A-12, J-11, others-10 so A is first; and for 2-point wins A-8, J-10, others-9, so A is last. Not sure if it is unique in any sense so I don't think it completely solves the problem, but it does show there are outcomes that satisfy the conditions.--RDBury (talk) 06:53, 22 November 2011 (UTC)

Thanks a lot, this is indeed a solution to the problem. And maybe this is the solution to the problem, as it very nicely makes use of the only essential difference between the two systems - namely that, for two teams playing against each other, the old system distributes a total of 2 points per game in all cases, while the new system distributes 3 points if either team wins and only 2 if the game is a tie. So the final standings under each of the two systems are those:

3-point wins Pos Team W D L Pts 1 Team A 4 0 5 12 2 Team J 1 8 0 11 3-10 all others 1 7 1 10

2-point wins Pos Team W D L Pts 1 Team J 1 8 0 10 2-9 all others 1 7 1 9 10 Team A 4 0 5 8

They say in the video material that all contestants had to solve this problem within 15 minutes, and that one of the boys appearing in the material was the only one who managed to do so.

As demonstrated here, the "three points for a win" system does have its cons - a team that loses more than 50% of its games can still triumph with the title eventually, which is mathematically impossible under the old system. :)

--Theurgist (talk) 02:44, 23 November 2011 (UTC)

I think I see flaws in those chart results. If team A wins 4 games, doesn't some combo of other teams need to lose those 4 games ? And if team A loses 5 games, doesn't some other combo of 5 teams need win those 5 games ? And shouldn't the total number of draws always be even ? StuRat (talk) 04:15, 23 November 2011 (UTC)

In the tables, "all others" is eight teams, not just one. So all wins are 4 + 1 + 8 × 1 = 13 and all losses are 5 + 0 + 8 × 1 = 13. The total number of draws is 8 + 8 × 7 = 64. --Theurgist (talk) 04:24, 23 November 2011 (UTC)

OK, I see. Thanks. StuRat (talk) 04:47, 23 November 2011 (UTC)

Isometry group of the k dimensional torus

Put the cubical metric on the k-torus (i.e. the obvious cover is a k-cube, not some k-cuboid.) It seems intuitively clear that the isometry group of the k-torus should arise from translations in the k directions of the k-cube cover (inducing `rotations' about the various loops in the torus), and the symmetry group of the k-cube. (In fact, the group must be a semi-direct product of these two subgroups, surely?). The trouble is, I can't seem to find this written down anywhere. Is my intuition correct? And where can I find a reference on this? Thanks for your help, Icthyos (talk) 16:40, 22 November 2011 (UTC)

I suppose it boils down to the fact that the only isometry of the k-torus which lifts to the identity on $\mathbb {R} ^{k}$ is the identity, and that the only isometries of $\mathbb {R} ^{k}$ which induce well-defined isometries of the k-torus are the ones which preserve the lattice $\mathbb {Z} ^{k}$ -- i.e. the symmetries of the k-cube. Icthyos (talk) 11:23, 24 November 2011 (UTC)

(preserve the lattice up to translation, that is.) Icthyos (talk) 11:33, 24 November 2011 (UTC)

This phenomenon is known as "rigidity of tori". I'm not certain where you can find a good account, but Google might be a good place to start. The torus is a compact reductive group, and hence a symmetric space whose isometry group is the automorphisms of the group. The outer automorphisms stabilize a point on the torus, and this stabilizer is the Weyl group of the lattice. The inner automorphisms are just the translations, so yes it's a semidirect product. Sławomir Biały (talk) 12:09, 24 November 2011 (UTC)

Thanks, this is exactly the sort of lead I was looking for. Does the argument need to be that technical, though? I was sort of hoping my approach of lifting isometries to the universal cover (from which the metric is induced) would be sufficient. Icthyos (talk) 14:48, 24 November 2011 (UTC)

Yes, that works as well since any map of the torus will lift to the universal cover. Sławomir Biały (talk) 14:59, 24 November 2011 (UTC)