## 30 August 2007

### profiles of interconnection networks

Flajolet and Sedgewick's Analytic Combinatorics (link goes to 12 MB PDF of the book, which is still in progress) is a most interesting book. I took a course based on the first three chapters of the book last fall; the purpose of the class was basically to teach us how to translate combinatorial structures into generating functions, in a fairly routine way. Unfortunately, we didn't get to the part of the text where we could then tell things about the asymptotics of those combinatorial structures, which is a tremendously useful thing to do.

So I've been working through parts of the book lately. In example V.8 Flajolet and sedgewick talk about a problem considered by Lagarias, Odlyzko and Zagier in their paper On The Capacity of Disjointly Shared Networks (although not quite in the same language). Flajolet and sedgewick ask: "There are 2n points on a line, with n point-to-point connections between pairs of points. What is the probable behavior of the width of such an interconnection network?" The width is defined in the following way: imagine the connections as circular arcs between the points on a horizontal line; let a vertical line sweep from left to right; then the width is the maximum number of arcs encountered by such a line.

For example, if we have n = 6 (and thus 12 points), we might make the connections 9-12, 5-10, 1-2, 6-8, 3-11, 4-7; as the line sweeps between 6 and 7 there are four open connections (and never are there more), so the width is 4. The "profile" of this network is the sequence given by the number of open connections just after the line sweeps past 0, past 1, past 2, and so on (call these a0, a1, a2, ...); this is the sequence

0, 1, 0, 1, 2, 3, 4, 3, 2, 3, 2, 1, 0

and in general this sequence increases to somewhere around n/2 and then decreases. Flajolet and sedgewick says that Louchard proved this, but I haven't looked at that paper. What surprised me, when I saw this problem, is that the profile actually looks something like a parabola, as you can see from the simulations at the top of p. 311 in Flajolet and sedgewick. Louchard proved that the profile "conforms asymptotically to a deterministic parabola 2nx(1-x)... to which are superimposed random fluctuations of O(√n)". So, for example, in such a network with one thousand nodes (n = 500), the profile goes as high as 250 (when x = 1/2) before going back down to zero.

Why should this be?

Imagine that you're putting together a random interconnection network. The profile sequence clearly has the property that each member ak is one more or one less than the last one. It'll be one more if k is the first member of one of the pairs, and one less if it's the second member. What's the probability that k is the first (i. e. smaller) member of its pair? If k is small, it'll be large; if k is near 2n then it'll be small. In fact, if we know k is a member of some pair, the other member of the pair is equally likely to be any integer from 1 to 2n except k; thus the probability that k is the smaller member of its pair is approximately 1 - k/2n, and the probability that it's the larger member is approximately k/2n.

Thus ak = ak-1 + 1 with probability 1 - k/2n, and ak = ak-1 - 1 the rest of the time. The expected value of ak - ak-1 is approximately (1-k/2n) - (k/2n), or 1-k/n. So, for example, one-fourth of the way through the process, when k = n/2, we expect ak to increase by about one-half each time we increase k by one.

Integrating with respect to n, we get

at ≈ ∫0t (1-k/n) dk = t - t2/2n

which is just a rescaling of Louchard's parabola. Flajolet and sedgewick's book is full of examples of things like this, where some apparently random structure turns out to have a lot of order when you "zoom out".

Anonymous said...

I believe you contradicted yourself in this paragraph (probably just a simple typo):

"What's the probability that k is the first member of its pair? If k is small, it'll be small; if k is near 2n then it'll be large. In fact, if we know k is a member of some pair, the other member of the pair is equally likely to be any integer from 1 to 2n except k; thus the probability that k is the smaller member of its pair is approximately 1 - k/2n, and the probability that it's the larger member is approximately k/2n."

Assuming that the numbers are in the ordinary order, the smaller member of each pair is also the first member of each pair, correct? In that case, the probability that k is the first member of its pair is large if k is small, and small if k is large. You wrote the opposite, but then in your detailed explanation you reversed yourself

Isabel said...

anonymous,

You're right. I'm editing the post.