Mixing
Discrete Spherical Symmetry Group
$begingroup$
Take two spheres each having a certain number (say 5) of identical dots on them. What is the approach to proving/disproving that they are equivalent under the set of spherical rotations?
One could label the points: say (A,B,C,D,E),(1,2,3,4,5)
Align (A,1),(X,Y) with (X,Y) being successive attempted matches, say (B,2),(B,3)... ; then check all of the rest for matching.
But this seems rather inelegant. In general, it involves 5x4 test alignments and 3x3 tests.
One would prefer some kind of rank/determinant calculation. This seems reasonable since alignments are linear when expressed in terms of cartesian coordinates with spherical rotation mappings, having only two degrees of freedom, but the formulation eludes me.
group-theory symmetry spherical-geometry
asked Jan 22 at 20:38
rrogersrrogers
518311
$endgroup$
add a comment |
$begingroup$
Take two spheres each having a certain number (say 5) of identical dots on them. What is the approach to proving/disproving that they are equivalent under the set of spherical rotations?
One could label the points: say (A,B,C,D,E),(1,2,3,4,5)
Align (A,1),(X,Y) with (X,Y) being successive attempted matches, say (B,2),(B,3)... ; then check all of the rest for matching.
But this seems rather inelegant. In general, it involves 5x4 test alignments and 3x3 tests.
One would prefer some kind of rank/determinant calculation. This seems reasonable since alignments are linear when expressed in terms of cartesian coordinates with spherical rotation mappings, having only two degrees of freedom, but the formulation eludes me.
group-theory symmetry spherical-geometry
asked Jan 22 at 20:38
rrogersrrogers
518311
$endgroup$
2
$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11
add a comment |
$begingroup$
Take two spheres each having a certain number (say 5) of identical dots on them. What is the approach to proving/disproving that they are equivalent under the set of spherical rotations?
One could label the points: say (A,B,C,D,E),(1,2,3,4,5)
Align (A,1),(X,Y) with (X,Y) being successive attempted matches, say (B,2),(B,3)... ; then check all of the rest for matching.
But this seems rather inelegant. In general, it involves 5x4 test alignments and 3x3 tests.
One would prefer some kind of rank/determinant calculation. This seems reasonable since alignments are linear when expressed in terms of cartesian coordinates with spherical rotation mappings, having only two degrees of freedom, but the formulation eludes me.
group-theory symmetry spherical-geometry
asked Jan 22 at 20:38
rrogersrrogers
518311
$endgroup$
Take two spheres each having a certain number (say 5) of identical dots on them. What is the approach to proving/disproving that they are equivalent under the set of spherical rotations?
One could label the points: say (A,B,C,D,E),(1,2,3,4,5)
Align (A,1),(X,Y) with (X,Y) being successive attempted matches, say (B,2),(B,3)... ; then check all of the rest for matching.
But this seems rather inelegant. In general, it involves 5x4 test alignments and 3x3 tests.
One would prefer some kind of rank/determinant calculation. This seems reasonable since alignments are linear when expressed in terms of cartesian coordinates with spherical rotation mappings, having only two degrees of freedom, but the formulation eludes me.
group-theory symmetry spherical-geometry
group-theory symmetry spherical-geometry
asked Jan 22 at 20:38
rrogersrrogers
518311
asked Jan 22 at 20:38
rrogersrrogers
518311
edited Jan 24 at 15:17
rrogers
asked Jan 22 at 20:38
rrogersrrogers
518311
asked Jan 22 at 20:38
rrogersrrogers
518311
asked Jan 22 at 20:38
rrogersrrogers
518311
518311
2
$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11
add a comment |
2
$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11
2
2
$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11
add a comment |
1 Answer
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Well, you could calculate the various point-to-point distances in either setting. When the arrangements would be different then the set of distances also will come out different. You not even are forced to calculate them all: the first mismatch here is enough to deside.
--- rk
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
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add a comment |
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$begingroup$
Well, you could calculate the various point-to-point distances in either setting. When the arrangements would be different then the set of distances also will come out different. You not even are forced to calculate them all: the first mismatch here is enough to deside.
--- rk
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
$endgroup$
add a comment |
$begingroup$
Well, you could calculate the various point-to-point distances in either setting. When the arrangements would be different then the set of distances also will come out different. You not even are forced to calculate them all: the first mismatch here is enough to deside.
--- rk
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
$endgroup$
add a comment |
$begingroup$
Well, you could calculate the various point-to-point distances in either setting. When the arrangements would be different then the set of distances also will come out different. You not even are forced to calculate them all: the first mismatch here is enough to deside.
--- rk
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
$endgroup$
Well, you could calculate the various point-to-point distances in either setting. When the arrangements would be different then the set of distances also will come out different. You not even are forced to calculate them all: the first mismatch here is enough to deside.
--- rk
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
answered Feb 1 at 14:18
Dr. Richard KlitzingDr. Richard Klitzing
1,76516
1,76516
add a comment |
add a comment |
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$begingroup$
With 5 points you don't need anything elegant. With 50, I'd recommend to find the principal axes of inertia. If this fails, we'd have to think of something else.
$endgroup$
– Ivan Neretin
Jan 24 at 16:03
$begingroup$
@user1729 Presently the point sets are envisioned as the result of Groebner calculations; with some mandatory multiple equivalent answers. I envision symbolic answers to find generic rules/solutions. Like I mention below:: perhaps I should expand the question to provide context and further thoughts?
$endgroup$
– rrogers
Jan 25 at 15:11