"That leaves the remaining six peers with congestion on almost all of the interconnect ports between us. Congestion that is permanent, has been in place for well over a year and where our peer refuses to augment capacity. They are deliberately harming the service they deliver to their paying customers. They are not allowing us to fulfil the requests their customers make for content.
Five of those congested peers are in the United States and one is in Europe. There are none in any other part of the world. All six are large Broadband consumer networks with a dominant or exclusive market share in their local market. In countries or markets where consumers have multiple Broadband choices (like the UK) there are no congested peers."
First, Sperner's Lemma . . . nrich.maths.org/1383 . . . Take a triangle ABC, labeled counterclockwise, and subdivide it into smaller triangles in whatever way you like. Then label all the new vertices as follows:
• vertices along AB may be labelled either A or B, but not C
• vertices along BC may be labelled either B or C, but not A
• vertices along CA may be labelled either C or A, but not B
• vertices inside triangle ABC may be labelled A or B or C.
Now shade in every small triangle that has three different labels. Use two different shadings to distinguish the triangles which have been labeled counterclockwise (i.e. in the same sense as triangle ABC) from the triangles which have been labelled clockwise (i.e. in the sense opposite to that of as triangle ABC).
Then Sperner's Lemma says there will be exactly one more counterclockwise triangle than clockwise triangles. In particular, the number of shaded triangles will be odd. And not to put too fine a point on it, but even more particularly, the number of shaded triangles will be nonzero!
The web page I mentioned at the top of this post, nrich.maths.org/1383, has a great proof of Sperner's Lemma, which works like this: Inside each little triangle, label each edge 0 if its endpoints are labeled with the same letter, 1 if the edge is labeled with different letters in counterclockwise order (like the big outer triangle), and −1 if the edge is labeled with different letters in clockwise order. Then in a circle inside each of the little triangles, write the sum of the 3 edge labels −− it will be +3 or −3, which you can verify yourself by considering all the possible labelings. Now, observe that among the little edges, there are "exterior edges" along lines AB, BC, and CA, and "interior edges". The interior edges have labels on both sides, which sum to zero, because either both labels are zero, or else they are 1 and −1. The exterior edges along line AB must sum to 1 because there has to be one more AB edge than BA edge, since the starting point of that line is A and the endpoint of that line is B. The same is true for lines BC and CA, so the sum of the exterior edge labels is 3, hence the sum of all the edge labels is 3. The sum of the numbers in the circles in the little triangles is the same as the sum of the edge numbers, so the sum of these circled numbers in the little triangles must also be 3, proving the lemma. If this was all too fast, read the NRICH web page, which takes it a bit slower.
Now you may be wondering what use Sperner's Lemma can be put to... Take a look at this story: nytimes.com/2014/04/29/science/to-divide-the-rent-start-with-a-triangle.html
If you read the story and you still don't get how Sperner's Lemma can help 3 roommates each pick a bedroom and still divide the rent fairly −− that's how I felt after reading the story −− then here's how Sperner's Lemma applies.
Take another look at the ABC triangle I posted here. It's divided into 36 smaller triangles with 28 different vertices. Think of the vertices of the smaller triangles as points in a triangular lattice. Each lattice point is a certain distance from A, B, and C, and these distances always add up to 12. (For more on this, google "barycentric coordinates") So if the total rent is, say, $1200, then each lattice point represents a division of the rent according to which bedroom is occupied. For example, the lattice point that's directly below "A" is 2 units from A, 5 units from B, and 5 units from C, representing a rent distribution of $200 for bedroom A, and $500 each for bedrooms B and C. If the three roommates take turns labeling each price point with the room they prefer at that price point, then Sperner's lemma guarantees that at least one small triangle will be labeled with three different labels by the three roommates. (Ensuring each of the three vertices of each triangle was labeled by a different person is an exercise left for the reader!) That means that at price points differing by less than $100, each person prefers a different bedroom. Then, by subdividing that triangle the same way, and having the three roommates take turns labeling price points with the room they prefer, again Sperner's lemma guarantees that one of these even smaller triangles will be labeled differently by each of the three roommates. By repeating this process, Sperner's Lemma guarantees that at some price point plus or minus, say, $1, each of the three roommates will pick a different bedroom.
- Children's Hospital BostonStaff Scientist, present
- Harvard Medical SchoolInstructor in Surgery, present
- Ph.D.Computer Science
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