[math-fun] The people you know...
I'm thinking of submitting the following bit of silliness to Mathematics Magazine: http://jamespropp.org/galois.pdf Comments are welcome! In particular, I'm pretty sure that statements fairly close to "The people you know are the people who know all the people who know all the people you know" already occur in print, and I'd like to give credit where credit is due (even if that prevents my article from fitting onto one page). Jim Propp
Unless "know" means something different from my intuitive notion, it is not at all clear (and in fact false) that S is a subset of K(K(S)). Let S be the singleton containing President Obama. Then K(S) is a very large subset, and I suspect K(K(S)) is empty. Are you somehow assuming that if A knows B then B knows A? --ms On 1/27/2011 1:48 PM, James Propp wrote:
I'm thinking of submitting the following bit of silliness to Mathematics Magazine: http://jamespropp.org/galois.pdf Comments are welcome!
In particular, I'm pretty sure that statements fairly close to "The people you know are the people who know all the people who know all the people you know" already occur in print, and I'd like to give credit where credit is due (even if that prevents my article from fitting onto one page).
Jim Propp
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I can't resist posing the following question: "Connect two people with a one-ohm resistor if they mutually know each other. What is the mean resistance between each pair of people?" You could, as Mike Stay suggested, replace this problem with an analogous one about a given social network. Sincerely, Adam P. Goucher
There's a nice physics problem about the equivalent resistance between two nearest points, if one is considering an infinite square lattice with 1 ohm resisters between nearest points. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Adam P. Goucher Sent: Friday, January 28, 2011 9:49 AM To: math-fun Subject: Re: [math-fun] The people you know... I can't resist posing the following question: "Connect two people with a one-ohm resistor if they mutually know each other. What is the mean resistance between each pair of people?" You could, as Mike Stay suggested, replace this problem with an analogous one about a given social network. Sincerely, Adam P. Goucher _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For the complete graph with n vertices, with 1 ohm edges, place 1 volt across two vertices. The situation is symmetrical under permutation of the remaining n-2 vertices, and under exchange of the 2 vertices with reversing the sign of the currents. So the n-2 vertices are at 1/2 volt. Then the current into the 1 volt vertex is (1 + (n-2)/2) = n/2 amps, and thus the resistance is 2/n ohms. The square lattice problem was discussed on math-fun some time ago, and a solution was presented by means of Fourier series. -- Gene ________________________________ From: "Cordwell, William R" <wrcordw@sandia.gov> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 8:51:57 AM Subject: Re: [math-fun] The people you know... There's a nice physics problem about the equivalent resistance between two nearest points, if one is considering an infinite square lattice with 1 ohm resisters between nearest points. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Adam P. Goucher Sent: Friday, January 28, 2011 9:49 AM To: math-fun Subject: Re: [math-fun] The people you know... I can't resist posing the following question: "Connect two people with a one-ohm resistor if they mutually know each other. What is the mean resistance between each pair of people?" You could, as Mike Stay suggested, replace this problem with an analogous one about a given social network. Sincerely, Adam P. Goucher _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Eugene Salamin Sent: Friday, January 28, 2011 10:06 AM To: math-fun Subject: Re: [math-fun] The people you know... For the complete graph with n vertices, with 1 ohm edges, place 1 volt across two vertices. The situation is symmetrical under permutation of the remaining n-2 vertices, and under exchange of the 2 vertices with reversing the sign of the currents. So the n-2 vertices are at 1/2 volt. Then the current into the 1 volt vertex is (1 + (n-2)/2) = n/2 amps, and thus the resistance is 2/n ohms. The square lattice problem was discussed on math-fun some time ago, and a solution was presented by means of Fourier series. -- Gene ________________________________ From: "Cordwell, William R" <wrcordw@sandia.gov> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 8:51:57 AM Subject: Re: [math-fun] The people you know... There's a nice physics problem about the equivalent resistance between two nearest points, if one is considering an infinite square lattice with 1 ohm resisters between nearest points. -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Adam P. Goucher Sent: Friday, January 28, 2011 9:49 AM To: math-fun Subject: Re: [math-fun] The people you know... I can't resist posing the following question: "Connect two people with a one-ohm resistor if they mutually know each other. What is the mean resistance between each pair of people?" You could, as Mike Stay suggested, replace this problem with an analogous one about a given social network. Sincerely, Adam P. Goucher _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2 R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580 R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789 You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2. Veit On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
I don't see the symmetry argument. Consider the 1-dimensional case; then R[1,0] = 1, not 1/2. Additionally, I would be wary of injecting current into one node without also extracting an equal current. Current must flow out to infinity. In a 2 or more dimensional lattice, how can you be sure you have a unique solution without imposing boundary conditions at infinity? -- Gene ________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 10:29:04 AM Subject: Re: [math-fun] The people you know... R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2 R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580 R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789 You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2. Veit On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
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Connect the boundary of a large hypercubic resistor grid to "ground". Ground acts as a current source at zero potential. Now apply a voltage V to the node at the origin such that 1 amp is extracted there. By symmetry, the current flowing in each of the resistors connected to the origin node is 1/(2D) amp, where D is the dimension. The potential across these resistors is 1/(2D) volt since their resistance is 1 ohm. Now consider what happens when instead, voltage -V is applied to a neighboring node, say [1,0]. We have the same symmetrical distribution of currents, only the center is shifted and the current directions are reversed. On the other hand, the voltage across the resistor R connecting [0,0] to [1,0] is the same (in magnitude and sign) as it was before. Now superimpose these two current/voltage distributions. The voltage across R will be 2/(2D) and exactly one amp will flow out of [0,0] and into [1,0]. The equivalent resistance is therefore 1/D. Veit On Jan 28, 2011, at 1:51 PM, Eugene Salamin wrote:
I don't see the symmetry argument. Consider the 1-dimensional case; then R[1,0] = 1, not 1/2. Additionally, I would be wary of injecting current into one node without also extracting an equal current. Current must flow out to infinity. In a 2 or more dimensional lattice, how can you be sure you have a unique solution without imposing boundary conditions at infinity?
-- Gene
________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 10:29:04 AM Subject: Re: [math-fun] The people you know...
R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2
R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580
R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789
You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2.
Veit
On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
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Yes, that's a good symmetry argument. I would not connect infinity to ground because in 1 or 2 dimensions, the potential drop from the injection node to infinity is infinite. Instead ground the injection node, and all is well. -- Gene ________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 11:30:05 AM Subject: Re: [math-fun] The people you know... Connect the boundary of a large hypercubic resistor grid to "ground". Ground acts as a current source at zero potential. Now apply a voltage V to the node at the origin such that 1 amp is extracted there. By symmetry, the current flowing in each of the resistors connected to the origin node is 1/(2D) amp, where D is the dimension. The potential across these resistors is 1/(2D) volt since their resistance is 1 ohm. Now consider what happens when instead, voltage -V is applied to a neighboring node, say [1,0]. We have the same symmetrical distribution of currents, only the center is shifted and the current directions are reversed. On the other hand, the voltage across the resistor R connecting [0,0] to [1,0] is the same (in magnitude and sign) as it was before. Now superimpose these two current/voltage distributions. The voltage across R will be 2/(2D) and exactly one amp will flow out of [0,0] and into [1,0]. The equivalent resistance is therefore 1/D. Veit On Jan 28, 2011, at 1:51 PM, Eugene Salamin wrote:
I don't see the symmetry argument. Consider the 1-dimensional case; then R[1,0]
= 1, not 1/2. Additionally, I would be wary of injecting current into one node
without also extracting an equal current. Current must flow out to infinity.
In a 2 or more dimensional lattice, how can you be sure you have a unique solution without imposing boundary conditions at infinity?
-- Gene
________________________________ From: Veit Elser <ve10@cornell.edu> To: math-fun <math-fun@mailman.xmission.com> Sent: Fri, January 28, 2011 10:29:04 AM Subject: Re: [math-fun] The people you know...
R[m_, n_] := Integrate[(1 - Cos[m p + n q])/(2 - Cos[p] - Cos[q]), {p, 0, 2 Pi}, {q, 0, 2 Pi}]/(2 Pi)^2
R[5,0] = (-3760 + 1203 pi)/(6 pi]) = 1.02580
R[4,3] = (48 - 5 pi)/(10 pi) = 1.02789
You can superimpose two symmetric circuit solutions, one where unit current is injected at [0,0] and the other where unit current is extracted at [1,0], to argue R[1,0] = 1/2.
Veit
On Jan 28, 2011, at 12:53 PM, Cordwell, William R wrote:
If I recall correctly, for the resistance between diagonal points, one might need Fourier series, but there is a simple method for nearest lattice points.
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participants (6)
-
Adam P. Goucher -
Cordwell, William R -
Eugene Salamin -
James Propp -
Mike Speciner -
Veit Elser