[math-fun] Fano Plane puzzle
Let F be the Fano Plane. Let FL be the set of 7 lines in F. Let FP be the set of 7 points in F. For L in FL, let P(L) be the set of 3 points on L. Now let SP be a set of 7 points in general position in R^3. Let m : FP ó SP be a bijection. For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)). Let SL = C(FL). Let S = (SL, SP). S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points. If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
The minimum radius has no maximum. WFL On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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I realized that after I posted it. You could choose your points in R^3 slightly perturbed from seven points equally spaced on a unit circle. I think that way the circles could be made arbitrarily close to unit circles, with the smallest radius any value < 1. I'm guessing you can't model the Fano plane with seven unit circles.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Wednesday, June 17, 2015 7:56 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position. So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve. Sincerely, Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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<< The minimum radius has no maximum. >> WFL << I'm guessing you can't model the Fano plane with seven unit circles. >> DWW << This seems decidedly non-trivial to resolve. >> APG I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide. Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide. Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote. So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii. The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] . Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here? Fred Lunnon [[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]]; [[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]]; On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to: "Is the Heawood graph a unit-distance graph?" Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv: http://arxiv.org/abs/0912.5395 So yes, it is indeed possible for all circles to have unit radius. (I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...) Sincerely, Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles. >> DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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<< Does one circle really contain 4 vertices, or does it just look that way? >> DA I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0 Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure! WFL On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles. >> DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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Yes, Adam made a beeline to a proof, including the needed lemma. That paper was numerically amazing, in that for each coordinate of the 14 vertices of the Heawood graph that they determined, it was the solution of a polynomial in one variable of degree 79 that involved all 80 possible coefficients! (Or at least the one they displayed as an example did.) In fact, that polynomial was displayed 2 coefficients {c_j} to a line, which were mostly gigantic integers, and the ragged edge on the display's right-hand side came close to following a smooth curve — at least for its 40 distinct "points", which were essentially each determined by length(line_k) = floor( log_10( c_(2k) * c_(2k+1) ) ). —Dan
On Jun 22, 2015, at 6:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0 <https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0>
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
That's pretty neat, I wouldn't have thought to look for planar solutions. My original hope was a solution with unit circles in R^3 where the circles are Fano lines and circle intersection points the Fano points. You are saying that's not possible?
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Monday, June 22, 2015 9:15 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
DWW asks for a properly 3-space Fano configuration, with circles disjoint (except at Fano points). Consider some general set of 7 planes in 3-space; define the 7 Fano points as an appropriate subset of 3-plane intersections, and the 7 Fano lines as the circumcircles in each plane. Specifying each unit radius constitutes 1 constraint; so modulo isometry, solid configuration set (complex) freedom dimension = 7.3 - 6 - 7.1 = 11 ; and its real planar (presumed) limit dimension = 4 (probably). But is a planar configuration in general the limit of properly solid ones? We have to be careful here: for instance modulo isometry, set of (7-fold coincident) unit circles with 7 concyclic points is freedom-6; whence limits of (freedom-4) planar configurations necessarily constitute a proper subset. In mechanical terms, suppose that Fano points can slide along Fano circles. If a planar configuration is picked up and shaken, will it always disorganise properly into 3-space? It's a pity my points & centres relaxation fails to converge, since it could be modified trivially to solid space (albeit sans graphics). Fred Lunnon On 6/23/15, David Wilson <davidwwilson@comcast.net> wrote:
That's pretty neat, I wouldn't have thought to look for planar solutions.
My original hope was a solution with unit circles in R^3 where the circles are Fano lines and circle intersection points the Fano points. You are saying that's not possible?
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Monday, June 22, 2015 9:15 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
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Er, on second thoughts --- 7.3 - 6 - 7.1 = 8 ?! WFL On 6/23/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
DWW asks for a properly 3-space Fano configuration, with circles disjoint (except at Fano points).
Consider some general set of 7 planes in 3-space; define the 7 Fano points as an appropriate subset of 3-plane intersections, and the 7 Fano lines as the circumcircles in each plane. Specifying each unit radius constitutes 1 constraint; so modulo isometry, solid configuration set (complex) freedom dimension = 7.3 - 6 - 7.1 = 11 ; and its real planar (presumed) limit dimension = 4 (probably).
But is a planar configuration in general the limit of properly solid ones? We have to be careful here: for instance modulo isometry, set of (7-fold coincident) unit circles with 7 concyclic points is freedom-6; whence limits of (freedom-4) planar configurations necessarily constitute a proper subset.
In mechanical terms, suppose that Fano points can slide along Fano circles. If a planar configuration is picked up and shaken, will it always disorganise properly into 3-space?
It's a pity my points & centres relaxation fails to converge, since it could be modified trivially to solid space (albeit sans graphics).
Fred Lunnon
On 6/23/15, David Wilson <davidwwilson@comcast.net> wrote:
That's pretty neat, I wouldn't have thought to look for planar solutions.
My original hope was a solution with unit circles in R^3 where the circles are Fano lines and circle intersection points the Fano points. You are saying that's not possible?
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Monday, June 22, 2015 9:15 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
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Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity? On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles.
DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program. But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles. WFL On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:
Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?
On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles.
DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: > Let F be the Fano Plane. > > Let FL be the set of 7 lines in F. > > Let FP be the set of 7 points in F. > > For L in FL, let P(L) be the set of 3 points on L. > > > > Now let SP be a set of 7 points in general position in R^3. > > Let m : FP ó SP be a bijection. > > For each line of L of F, let C(L) be the circle in R^3 through the 3 > points > m(P(L)). > > Let SL = C(FL). > > Let S = (SL, SP). > > S is then a model of the Fano Plane in R^3 with circles for Fano > lines > and > points for Fano points. > > > > If we scale S so that the largest circle has radius 1, how large can > we > make > the radius of the smallest circle of S by judicious choice of SP? > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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This time there's no ambiguity; radii now within 1.5e-8 of unity --- https://www.dropbox.com/s/exsev6qcjroo2dx/fano7pt7rg_2%20copy.gif?dl=0 Coordinates posted on request. WFL On 6/23/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program.
But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles.
WFL
On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:
Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?
On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles.
DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
> Sent: Thursday, June 18, 2015 at 12:56 AM > From: "Fred Lunnon" <fred.lunnon@gmail.com> > To: math-fun <math-fun@mailman.xmission.com> > Subject: Re: [math-fun] Fano Plane puzzle > > The minimum radius has no maximum. WFL > > > > On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: > > Let F be the Fano Plane. > > > > Let FL be the set of 7 lines in F. > > > > Let FP be the set of 7 points in F. > > > > For L in FL, let P(L) be the set of 3 points on L. > > > > > > > > Now let SP be a set of 7 points in general position in R^3. > > > > Let m : FP ó SP be a bijection. > > > > For each line of L of F, let C(L) be the circle in R^3 through > > the 3 > > points > > m(P(L)). > > > > Let SL = C(FL). > > > > Let S = (SL, SP). > > > > S is then a model of the Fano Plane in R^3 with circles for Fano > > lines > > and > > points for Fano points. > > > > > > > > If we scale S so that the largest circle has radius 1, how large can > > we > > make > > the radius of the smallest circle of S by judicious choice of > > SP? > > > > _______________________________________________ > > math-fun mailing list > > math-fun@mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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Just clicked on this but got a 404 "The file you are looking for has either moved or been deleted: error." —Dan
On Jun 23, 2015, at 1:33 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
This time there's no ambiguity; radii now within 1.5e-8 of unity ---
https://www.dropbox.com/s/exsev6qcjroo2dx/fano7pt7rg_2%20copy.gif?dl=0
Coordinates posted on request.
WFL
On 6/23/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program.
But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles.
WFL
On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:
Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?
On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles.
DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote: > We can get arbitrarily close to 1 by beginning with SP being the > vertices of a regular 7-gon of unit circumradius, and perturbing > the points by epsilon so that they're in general position. > > So the question is: can we get equality to hold? That is to say, > is it possible for all circles to have unit radius? This seems > decidedly non-trivial to resolve. > > > Sincerely, > > > Adam P. Goucher > >> Sent: Thursday, June 18, 2015 at 12:56 AM >> From: "Fred Lunnon" <fred.lunnon@gmail.com> >> To: math-fun <math-fun@mailman.xmission.com> >> Subject: Re: [math-fun] Fano Plane puzzle >> >> The minimum radius has no maximum. WFL >> >> >> >> On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: >>> Let F be the Fano Plane. >>> >>> Let FL be the set of 7 lines in F. >>> >>> Let FP be the set of 7 points in F. >>> >>> For L in FL, let P(L) be the set of 3 points on L. >>> >>> >>> >>> Now let SP be a set of 7 points in general position in R^3. >>> >>> Let m : FP ó SP be a bijection. >>> >>> For each line of L of F, let C(L) be the circle in R^3 through >>> the 3 >>> points >>> m(P(L)). >>> >>> Let SL = C(FL). >>> >>> Let S = (SL, SP). >>> >>> S is then a model of the Fano Plane in R^3 with circles for Fano >>> lines >>> and >>> points for Fano points. >>> >>> >>> >>> If we scale S so that the largest circle has radius 1, how large can >>> we >>> make >>> the radius of the smallest circle of S by judicious choice of >>> SP? >>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun@mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> >> _______________________________________________ >> math-fun mailing list >> math-fun@mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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By the way, is this 4th point's closeness really a problem? And, is it really close in R^3 or only close in the displayed projection? —Dan
On Jun 23, 2015, at 12:37 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program.
But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles.
WFL
On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:
Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?
On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< Does one circle really contain 4 vertices, or does it just look that way? >> DA
I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0
Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!
WFL
On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:
"Is the Heawood graph a unit-distance graph?"
Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:
http://arxiv.org/abs/0912.5395
So yes, it is indeed possible for all circles to have unit radius.
(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)
Sincerely,
Adam P. Goucher
Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles.
DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
> Sent: Thursday, June 18, 2015 at 12:56 AM > From: "Fred Lunnon" <fred.lunnon@gmail.com> > To: math-fun <math-fun@mailman.xmission.com> > Subject: Re: [math-fun] Fano Plane puzzle > > The minimum radius has no maximum. WFL > > > > On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: >> Let F be the Fano Plane. >> >> Let FL be the set of 7 lines in F. >> >> Let FP be the set of 7 points in F. >> >> For L in FL, let P(L) be the set of 3 points on L. >> >> >> >> Now let SP be a set of 7 points in general position in R^3. >> >> Let m : FP ó SP be a bijection. >> >> For each line of L of F, let C(L) be the circle in R^3 through the 3 >> points >> m(P(L)). >> >> Let SL = C(FL). >> >> Let S = (SL, SP). >> >> S is then a model of the Fano Plane in R^3 with circles for Fano >> lines >> and >> points for Fano points. >> >> >> >> If we scale S so that the largest circle has radius 1, how large can >> we >> make >> the radius of the smallest circle of S by judicious choice of SP? >> >> _______________________________________________ >> math-fun mailing list >> math-fun@mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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On 6/23/15, Dan Asimov <asimov@msri.org> wrote:
By the way, is this 4th point's closeness really a problem?
It's largely an aesthetic matter: such collisions do make the diagram difficult to interpret.
And, is it really close in R^3 or only close in the displayed projection?
Note that all configurations so far constructed by myself are _planar_! [ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
Just clicked on this but got a 404 "The file you are looking for has either moved or been deleted: error."
Synching error apparently --- try instead https://www.dropbox.com/s/jeyqj76ng4z42s7/fano7pt7rg_2.gif?dl=0 WFL
That's pretty convincing. On Tue, Jun 23, 2015 at 6:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 6/23/15, Dan Asimov <asimov@msri.org> wrote:
By the way, is this 4th point's closeness really a problem?
It's largely an aesthetic matter: such collisions do make the diagram difficult to interpret.
And, is it really close in R^3 or only close in the displayed projection?
Note that all configurations so far constructed by myself are _planar_!
[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
Just clicked on this but got a 404 "The file you are looking for has either moved or been deleted: error."
Synching error apparently --- try instead https://www.dropbox.com/s/jeyqj76ng4z42s7/fano7pt7rg_2.gif?dl=0
WFL
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I like that one. The little sail-shaped portion of the figure on which the points lie is clearly isomorphic to the standard triangular picture of the Fano plane.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Allan Wechsler Sent: Tuesday, June 23, 2015 6:25 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
That's pretty convincing.
On Tue, Jun 23, 2015 at 6:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 6/23/15, Dan Asimov <asimov@msri.org> wrote:
By the way, is this 4th point's closeness really a problem?
It's largely an aesthetic matter: such collisions do make the diagram difficult to interpret.
And, is it really close in R^3 or only close in the displayed projection?
Note that all configurations so far constructed by myself are _planar_!
[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
Just clicked on this but got a 404 "The file you are looking for has either moved or been deleted: error."
Synching error apparently --- try instead https://www.dropbox.com/s/jeyqj76ng4z42s7/fano7pt7rg_2.gif?dl=0
WFL
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On 6/23/2015 6:13 PM, Fred Lunnon wrote:
[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
I worked this out yesterday, but I was hoping someone else would save me the trouble of writing it up (and perhaps find something simpler to describe). There is indeed a 3-dimensional arrangement of 7 unit circles with the correct intersections, and it can be chosen to be quite symmetrical (my example has 3-fold rotational symmetry about the z-axis and bilateral symmetry in the xz-plane, so the six-element dihedral group). Start with the unit circle in the xz-plane centered at (1, 0, 0) and add two more by rotating it 120 and 240 degrees about the z-axis (I'll call these circles 1, 2 and 3). They meet at the origin, which is Fano point A. The three lowest points on the circles are points B, C and D, and lie on unit circle 4. To complete the Fano plane, each of the points B, C and D needs to be on a new circle (5, 6 or 7) that meets whichever two of circles 1, 2 and 3 that this point isn't already on; these intersections will be Fano points E, F and G. To locate F and G, for example, start with the circle in the x = 1 plane centered at (1, 0, 0). It passes through B on circle 1, but is far from circles 2 and 3. Fix this by tilting the circle toward the z-axis, keeping the bottom point at B. When it meets circles 2 and 3, those points are F and G. Their 120-degree rotational counterpart is E on circle 1, completing the set; the tilted circle and its rotated versions are circles 5, 6 and 7. (I'm sure that wasn't the easiest thing to follow. If someone who does follow would like to make a picture, that may help.) If I've done the numerical geometry correctly, point E has coordinates (.22893146917093, 0, .63675216588960), approximately. x and z should be algebraic, and in fact they appear to be the real roots of the cubic equations 13x^3 + 3x^2 + 3x - 1 = 0 and 13z^3 + 11z^2 + 5z - 11 = 0. -- Fred W. Helenius fredh@ix.netcom.com
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it! Fred Lunnon
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions: If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then e_k * e_(k+1) = e_(k+3) (indices mod 7) (e_k)^2 = -1 e_j * e_k = -e_k * e_j These rules plus the fact that the octonions are an algebra over the reals (so additively identical to R^8) are enough to multiply any two elements —Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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How about a regular heptagon (the vertices), and use off-center ellipses for the lines? This would get the rotational symmetry. --Rich Quoting Fred Lunnon <fred.lunnon@gmail.com>:
Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL
On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
?Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles. —Dan
On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL
On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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Dear all, I've been watching this with interest. I'm hesitant to rush (well, creep) in where angels fear to tread, but I wonder if there's just a chance that not everyone is familiar with the fact that the Fano configuration (due to Kirkman & Woolhouse, by the way) has two quite different descriptions. By rotating the difference set {0,1,3} mod 7. By nim-addition of the set {1,2,4}. You don't get from one to the other by ``adding 1''. You can ``see'' that the groups SL(2,7) and GL(2,3) are isomorphic, but it's a different matter to prove this. R. On Wed, 24 Jun 2015, Dan Asimov wrote:
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
—Dan
On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL
On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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Richard's comment says that if the 7 vertices ("points") of the regular 6-simplex s_6 (in R^6 if you like) are numbered by the elements of Z/7Z, then all triples of vertices {k,k+1,k+3} will form 7 "lines" such that the incidence relations of the points and lines make a Fano plane. All 168 automorphisms of the Fano plane are realizable by rigid motions of the simplex. Are they all rotations, i.e, are the permutations all even? I think so. —Dan P.S. By substituting the centroids of the 7 "lines" for the triangles, plus possibly a little tweaking, there ought to be 14 points in R^6, representing the 7 "points" and 7 "lines" connected with 21 edges (not "lines") making this into the bipartite Heawood graph, in such a way as there are rotations of R^6 that interchanges the "points" points and the "lines" points. Let's see, how many rotations of this kind are there ...
On Jun 24, 2015, at 12:50 PM, rkg <rkg@ucalgary.ca> wrote:
Dear all, I've been watching this with interest. I'm hesitant to rush (well, creep) in where angels fear to tread, but I wonder if there's just a chance that not everyone is familiar with the fact that the Fano configuration (due to Kirkman & Woolhouse, by the way) has two quite different descriptions.
By rotating the difference set {0,1,3} mod 7.
By nim-addition of the set {1,2,4}.
You don't get from one to the other by ``adding 1''.
You can ``see'' that the groups SL(2,7) and GL(2,3) are isomorphic, but it's a different matter to prove this. R.
On Wed, 24 Jun 2015, Dan Asimov wrote:
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
Sadly, it would difficult to make an office toy in R^6.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
—Dan
On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL
On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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I wasn't happy with the known unit distance embeddings of the Heawood graph, so I made one of my own. --Ed Pegg Jr On Wed, Jun 24, 2015 at 5:22 PM, David Wilson <davidwwilson@comcast.net> wrote:
Sadly, it would difficult to make an office toy in R^6.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
—Dan
On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL
On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:
So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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On Jun 24, 2015, at 3:22 PM, David Wilson <davidwwilson@comcast.net> wrote:
Sadly, it would difficult to make an office toy in R^6.
Not at all. In R^6 they have even better machine shops than we do. —Dan
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com <mailto:math-fun-bounces@mailman.xmission.com>] Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
On 2015-06-24 17:07, Dan Asimov wrote:
On Jun 24, 2015, at 3:22 PM, David Wilson <davidwwilson@comcast.net> wrote:
Sadly, it would difficult to make an office toy in R^6.
Not at all. In R^6 they have even better machine shops than we do.
—Dan
And the wealth to operate them. Like I was just telling the kids, On Wed, Jun 24, 2015 at 7:19 AM, Bill Gosper <billgosper@gmail.com> wrote: Subj: Scrooge McDuck was said to have three cubic acres of money. How much is that in square gallons? In[169]:= UnitConvert[Quantity[3, "Acres"]^3, Quantity["Gallons"]^2] Out[169]= Quantity[95610810544128000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] Oops, that's 27 cubic acres! Should have been In[170]:= UnitConvert[Quantity[3, "Acres"^3], Quantity["Gallons"]^2] Out[170]= Quantity[10623423393792000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] I hate it when ducks exaggerate. --Bill
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com <mailto:math-fun-bounces@mailman.xmission.com>] Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
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The Sierra Club has you beaten. They've quoted the volume of nuclear waste in cubic liters. -- Gene From: rwg <rwg@sdf.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, June 24, 2015 6:28 PM Subject: Re: [math-fun] Fano Plane puzzle On 2015-06-24 17:07, Dan Asimov wrote:
On Jun 24, 2015, at 3:22 PM, David Wilson <davidwwilson@comcast.net> wrote:
Sadly, it would difficult to make an office toy in R^6.
Not at all. In R^6 they have even better machine shops than we do.
—Dan
And the wealth to operate them. Like I was just telling the kids, On Wed, Jun 24, 2015 at 7:19 AM, Bill Gosper <billgosper@gmail.com> wrote: Subj: Scrooge McDuck was said to have three cubic acres of money. How much is that in square gallons? In[169]:= UnitConvert[Quantity[3, "Acres"]^3, Quantity["Gallons"]^2] Out[169]= Quantity[95610810544128000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] Oops, that's 27 cubic acres! Should have been In[170]:= UnitConvert[Quantity[3, "Acres"^3], Quantity["Gallons"]^2] Out[170]= Quantity[10623423393792000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] I hate it when ducks exaggerate. --Bill
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com <mailto:math-fun-bounces@mailman.xmission.com>] Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM
But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.
I prefer shifting the indices of either the points or the lines so that {0,1,3} becomes {1,2,4}. Then we have: P_i meets L_j if and only if (i-j) is a quadratic residue mod 7; L_i meets P_j if and only if (i-j) is a quadratic nonresidue mod 7. Sincerely, Adam P. Goucher
Sent: Wednesday, June 24, 2015 at 2:53 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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Also the offsets are the powers of 2 (mod 7). I like it! WFL On 6/24/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
I prefer shifting the indices of either the points or the lines so that {0,1,3} becomes {1,2,4}. Then we have:
P_i meets L_j if and only if (i-j) is a quadratic residue mod 7; L_i meets P_j if and only if (i-j) is a quadratic nonresidue mod 7.
Sincerely,
Adam P. Goucher
Sent: Wednesday, June 24, 2015 at 2:53 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:
If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then
e_k * e_(k+1) = e_(k+3) (indices mod 7)
(e_k)^2 = -1
e_j * e_k = -e_k * e_j
These rules plus the fact that the octonions are an algebra over the reals
(so additively identical to R^8) are enough to multiply any two elements
—Dan
On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!
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-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred W. Helenius Sent: Tuesday, June 23, 2015 7:14 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle
On 6/23/2015 6:13 PM, Fred Lunnon wrote:
[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]
I worked this out yesterday, but I was hoping someone else would save me the trouble of writing it up (and perhaps find something simpler to describe). There is indeed a 3-dimensional arrangement of 7 unit circles with the correct intersections, and it can be chosen to be quite symmetrical (my example has 3-fold rotational symmetry about the z-axis and bilateral symmetry in the xz- plane, so the six-element dihedral group).
Start with the unit circle in the xz-plane centered at (1, 0, 0) and add two more by rotating it 120 and 240 degrees about the z-axis (I'll call these circles 1, 2 and 3). They meet at the origin, which is Fano point A. The three lowest points on the circles are points B, C and D, and lie on unit circle 4.
To complete the Fano plane, each of the points B, C and D needs to be on a new circle (5, 6 or 7) that meets whichever two of circles 1, 2 and 3 that this point isn't already on; these intersections will be Fano points E, F and G. To locate F and G, for example, start with the circle in the x = 1 plane centered at (1, 0, 0). It passes through B on circle 1, but is far from circles 2 and
Thanks, Dr. Helenius. I was always dissatisfied with the standard 2d picture because the lines have spurious R^2 intersection points that aren't Fano points. Your construction provides a neat model of with congruent lines and Fano topology. 3. Fix this
by tilting the circle toward the z-axis, keeping the bottom point at B. When it meets circles 2 and 3, those points are F and G. Their 120-degree rotational counterpart is E on circle 1, completing the set; the tilted circle and its rotated versions are circles 5, 6 and 7.
(I'm sure that wasn't the easiest thing to follow. If someone who does follow would like to make a picture, that may help.)
If I've done the numerical geometry correctly, point E has coordinates (.22893146917093, 0, .63675216588960), approximately. x and z should be algebraic, and in fact they appear to be the real roots of the cubic equations 13x^3 + 3x^2 + 3x - 1 = 0 and 13z^3 + 11z^2 + 5z - 11 = 0.
-- Fred W. Helenius fredh@ix.netcom.com
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On 6/23/2015 10:36 PM, David Wilson wrote:
Thanks, Dr. Helenius.
I was always dissatisfied with the standard 2d picture because the lines have spurious R^2 intersection points that aren't Fano points. Your construction provides a neat model of with congruent lines and Fano topology.
You're welcome, but I'm afraid my construction is defective: it's very symmetrical, and the symmetry introduces three spurious intersections. As circles 5 and 6, for example, (the one with center on the x-axis and the one at theta = 120 degrees) tilt inward, they eventually cross the theta = 60 (or 240) degree plane where they will meet on circle 3. But in their final position they cross that plane twice, and they meet at both points. Perhaps this can be fixed by altering the construction slightly; say by giving the initial three circles a common twist (so the propeller shape they make actually works as a propeller), destroying the bilateral symmetry. But that's a lot more complicated to work out and I literally do have a plane to catch. -- Fred W. Helenius fredh@ix.netcom.com
Nice!!! Does one circle really contain 4 vertices, or does it just look that way? —Dan
On Jun 22, 2015, at 4:10 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< The minimum radius has no maximum. >> WFL
<< I'm guessing you can't model the Fano plane with seven unit circles. >> DWW
<< This seems decidedly non-trivial to resolve. >> APG
I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.
Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.
Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.
So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.
The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .
Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?
Fred Lunnon
[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];
[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];
On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.
So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.
Sincerely,
Adam P. Goucher
Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle
The minimum radius has no maximum. WFL
On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:
Let F be the Fano Plane.
Let FL be the set of 7 lines in F.
Let FP be the set of 7 points in F.
For L in FL, let P(L) be the set of 3 points on L.
Now let SP be a set of 7 points in general position in R^3.
Let m : FP ó SP be a bijection.
For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).
Let SL = C(FL).
Let S = (SL, SP).
S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.
If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?
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participants (12)
-
Adam P. Goucher -
Allan Wechsler -
Dan Asimov -
Dan Asimov -
David Wilson -
Ed Pegg Jr -
Eugene Salamin -
Fred Lunnon -
Fred W. Helenius -
rcs@xmission.com -
rkg -
rwg