Re: [math-fun] checkerboard version of an e-related dating problem--a graph theoretic approach
Concerning Allan Wechsler's approach: Note that the number of valid (jammed) configurations for n vertices on a line is the number of maximal matchings of the path graph with n vertices. And the number of singletons for a given configurations is the number of vertices not in the matching. [This same characterization holds for a grid graph or indeed any simple graph.] I don't know if there is a good algorithm in general for enumerating the set of all maximal matchings for a graph, but I wrote a crude one for this case and was able to find directly Allan's sequences a[n],b[n] for n from 0 to 20: a = 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200 b = 0, 1, 0, 2, 2, 3, 6, 7, 12, 17, 24, 36, 50, 72, 102, 143, 202, 282, 394, 549, 762 I cannot find b in the OEIS. My value for b[4] is 2 which differs from Allan's b[4] = 3. I didn't work out the recurrences, but Maple's gfun:-listtorec guessed that the recurrences for a and b are given by: {a(n+3)-a(n+1)-a(n) =0, a(0) = 1, a(1) = 1, a(2) = 1} {(n+3)*b(n+4)+(-5-n)*b(n+2)+(-6-n)*b(n+1) = 0, b(0) = 0, b(1) = 1, b(2) = 0, b(3) = 2} --Edwin On Wed, Nov 28, 2012 at 3:42 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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I found the error in my approach and recalculated the sequences through a[11] and b[11] by hand. They match Edwin's numbers perfectly now. My recurrence is exactly the one Maple found for a[n], but my formula for b[n] cheats by using a. b[n+3] = a[n] + b[n] + b[n+1]. I don't know if Maple's standalone formula for b is correct; it may well be. Anyway, for the actual problem described by Strogatz, we need different sequences. The "new a" should calculate how many ways there are to sequentially pair off adjacent members of a line of N things. a[4] should therefore be 3 instead of 2, because (xx)(xx) can be produced in two different ways. There's an analogous b-sequence, too, which aggregates the number of singletons in all those ways. These seem to be considerably harder to compute than my earlier formalization; I'm not seeing an easy recurrence, although there may be one. Here are the first 11 values for this new a and b. a: 1, 1, 1, 2, 3, 6, 12, 26, 60, 144, 366 b: 0, 1, 0, 2, 2, 6, 12, 30, 72, 192, 504 Neither of these are in OEIS. Strogatz's claim is that b/na -> e^(-2). exp(-2) is near 0.135; b[10]/10a[10] is around 0.138. On Thu, Nov 29, 2012 at 1:07 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
Concerning Allan Wechsler's approach:
Note that the number of valid (jammed) configurations for n vertices on a line is the number of maximal matchings of the path graph with n vertices. And the number of singletons for a given configurations is the number of vertices not in the matching. [This same characterization holds for a grid graph or indeed any simple graph.]
I don't know if there is a good algorithm in general for enumerating the set of all maximal matchings for a graph, but I wrote a crude one for this case and was able to find directly Allan's sequences a[n],b[n] for n from 0 to 20:
a = 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200 b = 0, 1, 0, 2, 2, 3, 6, 7, 12, 17, 24, 36, 50, 72, 102, 143, 202, 282, 394, 549, 762
I cannot find b in the OEIS.
My value for b[4] is 2 which differs from Allan's b[4] = 3. I didn't work out the recurrences, but Maple's gfun:-listtorec guessed that the recurrences for a and b are given by:
{a(n+3)-a(n+1)-a(n) =0, a(0) = 1, a(1) = 1, a(2) = 1} {(n+3)*b(n+4)+(-5-n)*b(n+2)+(-6-n)*b(n+1) = 0, b(0) = 0, b(1) = 1, b(2) = 0, b(3) = 2}
--Edwin
On Wed, Nov 28, 2012 at 3:42 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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In the 1-D case for the sequential seating problem we can just calculate the expected number s[n] of singletons (out of n original points) recursively by conditioning on which of the n-1 possible locations we choose for the first pair. Since placing the first pair at position k separates the line into two unfilled lines, one of length k-1 and one of length n-k-1, we have \[s[n] = \frac{1}{n-1} \sum_{k=1}^{n-1} (s[k-1] + s[n-k-1]) = \frac{2}{n-1} \sum_{k=0}^{n-2} s[k] = \frac{n-2}{n-1} s[n-1] + \frac{2}{n-1} s[n-2]\] with s[0] = 0 and s[1] = 1. Mathematica confirms that s[n]/n -> 1/e^2. The 2-D case does not seem to admit such an obvious recurrence. --Emma On Thu, Nov 29, 2012 at 4:07 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I found the error in my approach and recalculated the sequences through a[11] and b[11] by hand. They match Edwin's numbers perfectly now.
My recurrence is exactly the one Maple found for a[n], but my formula for b[n] cheats by using a.
b[n+3] = a[n] + b[n] + b[n+1].
I don't know if Maple's standalone formula for b is correct; it may well be.
Anyway, for the actual problem described by Strogatz, we need different sequences. The "new a" should calculate how many ways there are to sequentially pair off adjacent members of a line of N things. a[4] should therefore be 3 instead of 2, because (xx)(xx) can be produced in two different ways. There's an analogous b-sequence, too, which aggregates the number of singletons in all those ways. These seem to be considerably harder to compute than my earlier formalization; I'm not seeing an easy recurrence, although there may be one. Here are the first 11 values for this new a and b.
a: 1, 1, 1, 2, 3, 6, 12, 26, 60, 144, 366 b: 0, 1, 0, 2, 2, 6, 12, 30, 72, 192, 504
Neither of these are in OEIS. Strogatz's claim is that b/na -> e^(-2). exp(-2) is near 0.135; b[10]/10a[10] is around 0.138.
On Thu, Nov 29, 2012 at 1:07 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
Concerning Allan Wechsler's approach:
Note that the number of valid (jammed) configurations for n vertices on a line is the number of maximal matchings of the path graph with n vertices. And the number of singletons for a given configurations is the number of vertices not in the matching. [This same characterization holds for a grid graph or indeed any simple graph.]
I don't know if there is a good algorithm in general for enumerating the set of all maximal matchings for a graph, but I wrote a crude one for this case and was able to find directly Allan's sequences a[n],b[n] for n from 0 to 20:
a = 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200 b = 0, 1, 0, 2, 2, 3, 6, 7, 12, 17, 24, 36, 50, 72, 102, 143, 202, 282, 394, 549, 762
I cannot find b in the OEIS.
My value for b[4] is 2 which differs from Allan's b[4] = 3. I didn't work out the recurrences, but Maple's gfun:-listtorec guessed that the recurrences for a and b are given by:
{a(n+3)-a(n+1)-a(n) =0, a(0) = 1, a(1) = 1, a(2) = 1} {(n+3)*b(n+4)+(-5-n)*b(n+2)+(-6-n)*b(n+1) = 0, b(0) = 0, b(1) = 1, b(2) = 0, b(3) = 2}
--Edwin
On Wed, Nov 28, 2012 at 3:42 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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Very nice clean reasoning! I've been using the language of expectations, but I think it's more accurate to say we seek the limit as n -> oo of the probability that a random point is a singleton. Or better, the heck with probability altogether -- we're just aiming for the limit as n -> oo of the fraction of singletons. Dan P.S. It seems plausible that the nonlinear recurrence relation Emma derives for s[n] would imply s[n] -> 1/e^2, but it would be nice to see a proof. Probably much harder: Can it be solved in closed form for s[n] as a function of n ? On 2012-11-29, at 6:16 PM, Emma Cohen wrote:
In the 1-D case for the sequential seating problem we can just calculate the expected number s[n] of singletons (out of n original points) recursively by conditioning on which of the n-1 possible locations we choose for the first pair. Since placing the first pair at position k separates the line into two unfilled lines, one of length k-1 and one of length n-k-1, we have \[s[n] = \frac{1}{n-1} \sum_{k=1}^{n-1} (s[k-1] + s[n-k-1]) = \frac{2}{n-1} \sum_{k=0}^{n-2} s[k] = \frac{n-2}{n-1} s[n-1] + \frac{2}{n-1} s[n-2]\] with s[0] = 0 and s[1] = 1. Mathematica confirms that s[n]/n -> 1/e^2.
The 2-D case does not seem to admit such an obvious recurrence.
Dan wrote:
P.S. It seems plausible that the nonlinear recurrence relation Emma derives for s[n] would imply s[n] -> 1/e^2, but it would be nice to see a proof.
This isn't a proof but suggests how one might get one: if I put the first few values of (n-1)!s[n] into OEIS, it tells me they're the coefficients of the e.g.f. of exp(-2x)/(1-x)^2. Well, OK, so let's just do it. (From first principles; an actual expert on this stuff would take much bigger steps.) Step 1: Exploit the correspondence between recurrence relations and differential equations. First of all, write t[n] = (n-1)!s[n]. Then the recurrence becomes t[n+2] = n t[n+1] + 2n t[n]. Now write y = sum t[n] x^n/n!. Then we have y' = sum t[n+1] x^n/n! y'' = sum t[n+2] x^n/n! xy' = sum nt[n] x^n/n! xy'' = sum nt[n+1] x^n/n! and therefore y'' = xy'' + 2xy'. Step 2: Solve the easy differential equation. Write z = y'; then z' = xz' + 2xz, or equivalently (1-x)z' = 2xz, or equivalently z'/z = 2x/(1-x), or equivalently (log z)' = 2x/(1-x) = 2/(1-x) - 2 so, integrating, log z = -2 log (1-x) - 2 x + const, or equivalently z = A exp(-2x)/(1-x)^2. (No need to get y explicitly at this point, because knowing z is just as good.) Step 3: extract the power-series coefficients and get decent estimates. We've shown that sum t[n+1] x^n/n! = A exp(-2x)/(1-x)^2, or equivalently sum s[n+1] x^n = A exp(-2x)/(1-x)^2. The constant coefficients are s[n+1] on the left and A on the right, so A=1: sum s[n+1] x^n = A exp(-2x)/(1-x)^2. OK. The (1-x)^2 is a nuisance so let's get rid of it: sum (s[n+1]-s[n]) x^n = exp(-2x)/(1-x) sum (s[n+1]-2s[n]+s[n-1]) x^n = exp(-2x) where we define s[-1] = 0. So s[n+1]-2s[n]+s[n-1] = (-2)^n/n!, so s[n+1]-s[n] = sum{0..n} (-2)^k/k! = exp(-2) + tail[k], say, so s[n] = n exp(-2) + (tail[1]+...+tail[k]) and therefore s[n]/n = exp(-2) + o(1) because the tails tend to 0. That was rather laborious, but I'm about 99% sure that to those truly skilled in the art it would take maybe three or four lines because every bit of it is standard. -- g
1) For small L > 0, Let I(L) be a random maximal collection of disjoint closed intervals each of length L and lying in [0,1] in R. Let |I(L)| denote the total length of all the intervals of I(L). It seems clear that there exists some constant C(1) such that, with probability 1, the limit of |I(L)| as L -> 0 = C. Question: Find C. ----------------- One 2D version of this is: 2) For small L > 0 let D(L) denote a random maximal collection of disjoint closed geometric disks each of diameter L and lying in [0,1]^2 in R^2. Likewise, as L -> 0 what is the limit C(2) of the total area of the disks of D(L) ? ----------------- n) The nD version: In [0,1]^n, what is C(n) for any n ? ----------------- (If you don't like edge effects, these questions can be asked for the cubical n-torus R^n / Z^n instead of [0,1]^n, with the same answer C(n).) --Dan
Renyi's parking problem: http://mathworld.wolfram.com/RenyisParkingConstants.html On Dec 2, 2012, at 8:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
1) For small L > 0, Let I(L) be a random maximal collection of disjoint closed intervals each of length L and lying in [0,1] in R.
Let |I(L)| denote the total length of all the intervals of I(L).
It seems clear that there exists some constant C(1) such that, with probability 1, the limit of |I(L)| as L -> 0 = C.
Question: Find C. -----------------
One 2D version of this is:
2) For small L > 0 let D(L) denote a random maximal collection of disjoint closed geometric disks each of diameter L and lying in [0,1]^2 in R^2.
Likewise, as L -> 0 what is the limit C(2) of the total area of the disks of D(L) ? -----------------
n) The nD version: In [0,1]^n, what is C(n) for any n ? -----------------
(If you don't like edge effects, these questions can be asked for the cubical n-torus R^n / Z^n instead of [0,1]^n, with the same answer C(n).)
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
While not wishing to be tiresome, I nevertheless feel constrained to point out that Dan's original statement again begs the question of what exactly is being "randomised". While the parking problem is surely the most natural interpretation, one might instead for instance consider a (scaled-up) process placing unit-length intervals sequentially, separated by gaps chosen uniformly in the unit interval. In this case the mean separation is obviously 1/2, and the limiting density 2/3. Any problem which imprecisely combines geometry and probability is liable to give rise to such Bertrand-style paradoxes (more properly, inconsistencies). WFL On 12/3/12, Veit Elser <ve10@cornell.edu> wrote:
Renyi's parking problem: http://mathworld.wolfram.com/RenyisParkingConstants.html
On Dec 2, 2012, at 8:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
1) For small L > 0, Let I(L) be a random maximal collection of disjoint closed intervals each of length L and lying in [0,1] in R.
Let |I(L)| denote the total length of all the intervals of I(L).
It seems clear that there exists some constant C(1) such that, with probability 1, the limit of |I(L)| as L -> 0 = C.
Question: Find C. -----------------
One 2D version of this is:
2) For small L > 0 let D(L) denote a random maximal collection of disjoint
closed geometric disks each of diameter L and lying in [0,1]^2 in R^2.
Likewise, as L -> 0 what is the limit C(2) of the total area of the disks of D(L) ? -----------------
n) The nD version: In [0,1]^n, what is C(n) for any n ? -----------------
(If you don't like edge effects, these questions can be asked for the cubical n-torus R^n / Z^n instead of [0,1]^n, with the same answer C(n).)
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (7)
-
Allan Wechsler -
Dan Asimov -
Emma Cohen -
Fred lunnon -
Gareth McCaughan -
Veit Elser -
W. Edwin Clark