Please forgive me if I didn't feel like reading a whole paper to find out what was meant, when the problem poster could have just told us the intended problem. --Dan On 2012-11-29, at 11:27 AM, Allan Wechsler wrote:
The PDF file Gary sent some of us explains this model carefully, and indeed, it resolves all the ambiguity in the original statement. In particular, it's different (at least on its face) from any of the interpretations Andy proposed in his first response. Furthermore, it definitely differs from my formulation, in which all the legal configurations are equally likely.
This difference is already present for N=4. There are two legal configurations. If we write a 0 for a pair of unjoined hands, and 1 for a pair of held hands, these two configurations can be written 101 and 010. In my formulation these two configurations are equally likely. One of them has two singletons and one has none, for an average of one singleton per trial. But in Strogatz's formulation, the first bond to settle in creates one of the three preliminary configurations, 100, 010, 001, with equal likelihood. 010 is saturated, but 100 and 001 both must progress to 101. So 101 is twice as likely to occur as 010, for an average of 2/3 singletons per trial.
At this point I'm perfectly ready to believe the e^(-2) result, and to agree that the 2-dimensional case is much, much harder to analyze. I wouldn't expect it to have a closed-form solution, but surely it's worth doing a big Monte Carlo trial on, say, a 100 by 100 grid, and feed the resulting average density to Simon's number recognizer.
On Thu, Nov 29, 2012 at 12:47 PM, Gary Antonick <gantonick@post.harvard.edu>wrote:
Andy, Allan, Dan -
As you noticed pretty quickly, I didn't state the problem correctly. Thanks for pointing out. I'm actually interested in sequential and not simultaneous seating.
The following is Strogatz' presentation from Joy of x. I'm wondering if a 2D version of this has been solved.
Imagine there’s a very popular new movie showing at the local theater. It’s a romantic comedy, and hundreds of couples (many more than the theater can accommodate) are lined up at the box office, desperate to get in. Once a lucky couple get their tickets, they scramble inside and choose two seats right next to each other. To keep things simple, let’s suppose they choose these seats at random, wherever there’s room. In other words, they don’t care whether they sit close to the screen or far away, on the aisle or in the middle of a row. As long as they’re together, side by side, they’re happy.
Also, let’s assume no couple will ever slide over to make room for another. Once a couple sits down, that’s it. No courtesy whatsoever. Knowing this, the box office stops selling tickets as soon as there are only single seats left. Otherwise brawls would ensue.
At first, when the theater is pretty empty, there’s no problem. Every couple can find two adjacent seats. But after a while, the only seats left are singles—solitary, uninhabitable dead spaces that a couple can’t use. In real life, people often create these buffers deliberately, either for their coats or to avoid sharing an armrest with a repulsive stranger. In this model, however, these dead spaces just happen by chance.
The question is: When there’s no room left for any more couples, what fraction of the theater’s seats are unoccupied?
On Wed, Nov 28, 2012 at 12:42 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Going back to trying to make sense of the 1D problem, the number of valid configurations for N people satisfies a[0]=a[1]=a[2]=1, a[n+3]=a[n]+a[n+1]. There are about half a dozen sequences in OEIS with this "tribonacci" rule, with minor differences in offset and initial conditions.
Let b[N] be the aggregate number of singletons in all valid N-person configurations. I'm getting a recurrence b[n+3] = a[n] + b[n] + b[n+1]. Now b[0] = 0, b[1] = 1, b[2] = 0, and then the recurrence gives 2, 3, 3, 6, 7, 10, 15, 19 ... The last entry diverges from A121833, but the rule is so simple that I can't believe it's not in OEIS, so I must have miscalculated.
Anyway, the wanted expected number of singletons is lim b[n]/a[n]. Peering at the small entries I realize I must have blundered somewhere. Can somebody with more attention to give it please repeat this reasoning but do it correctly?
On Wed, Nov 28, 2012 at 3:09 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Good point, but it wasn't intentional -- I was just careless.
So, I should've said "then restrict attention to only those outcomes satisfying the conditions of the problem: That each person holds at most one hand and no two singles are adjacent."
Or, put another way: After giving each pair of adjacent hands an independent ½ probability of holding each other, restrict attention to the configurations having a maximal set of adjacent pairs of gridpoints each connected by the line segment between them, such that no line segment touches another one.
(This almost suggests a way to obtain the results for 1D.)
--Dan
On 2012-11-28, at 11:44 AM, Andy Latto wrote:
On Wed, Nov 28, 2012 at 1:49 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Whether in the 1D or 2D grid case (or for that matter in nD), one way to give this problem well-defined probabilities (that I think is in some sense the most natural way) is to first give each pair of adjacent hands an independent probability of ½ as to whether they or not they hold each other -- and then restrict attention to only those outcomes satisfying the conditions of the problem: That each person holds at most one hand.
That's a different problem, because you're ignoring the second condition of the problem; no two adjacent single people. If you restrict attention to those, you have a formulation equivalent to Allan's, I think, but different from both of mine.
Andy
I'm not sure if this agrees with the probabilities in one of Andy's two methods.
The problem becomes more symmetrical if the 1D case is a circle instead of a line (and in nD a K^n grid on an n-dimensional torus)
--
but this won't affect the asymptotic expected fraction of singles.
--Dan
On 2012-11-28, at 6:49 AM, Andy Latto wrote:
> On Wed, Nov 28, 2012 at 12:17 AM, Gary Antonick > <gantonick@post.harvard.edu> wrote: >> Cornell mathematician Steven Strogatz thinks this is an unsolved problem >> but am wondering if there's something being overlooked. >> >> Here's the basic -- and solved -- problem. There's a long line of people >> standing near each other. They're told to hold hands with someone
either
>> the person to their right or to their left, if available, but not both. >> The ratio of the resulting singles to all the people in line approaches >> 1/e^2 as the line increases.
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