some time ago, i wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
it is possible to have arbitrarily many points in the plane, no three collinear, such that all distances between them are integers, and no two such distances are equal.
there hasn't been any response, except for dan's query "why is this true?". since no one has answered, here is my solution. rational points on the unit circle x^2 + y^2 = 1 are (rationally) parametrized by (x, y) = P(t) = ((1 - t^2) / (1 + t^2), 2t / (1 + t^2)) . let A = P(0) = (1, 0), B = P(oo) = (-1, 0) and O = (0, 0) ; then the parameter t is tan(<PBA) = tan(<POA / 2) . it is easy to show that the distances PA and PB are rational if and only if t^2 + 1 is the square of a rational number, which is the case if and only if s = tan(<PBA / 2) is rational. we get a new parametrization of the unit circle by this parameter s : P(s) = ((1 - 6s^2 + s^4) / (1 + s^2)^2, 4s(1 - s^2) / (1 + s^2)^2) for rational s , such points have rational distance to both A and B , and the points obtained are dense in the unit circle. moreover, by ptolemy's theorem, any two such have rational distance to each other. this gives an infinite subset of the plane such that all the pairwise distances are rational, and one can also include the center of the circle. i do not know if this is a maximal subset of the plane with this property, although i think that it is likely. it remains to get a set with all distances different. this can be done in several ways. one way is to choose an infinite sequence {P_j} of points in the above set such that each point is on the top half of the circle, and such that <P_(j+1)OA < (<P_jOA) / 2 . for an explicit example, choose s = 2^(-j) , j = 0, 1, 2, ... to get the points P_j = ((2^(4j) + 6(2^(2j)) + 1) / (2^(2j) + 1)^2 , 4(2^j)(2^(2j) - 1) / (2^(2j) + 1)^2) . another way is to choose z to be the complex number corresponding to P(s) , and then to take the complex numbers z^(2^k) , k = 1, 2, 3, ... . (this will work as long as s is not 0, 1, -1 or oo , so that z is not a root of unity.) for a *finite* set, we can get all distances to be integers simply by scaling. i do not know if it's possible to have an infinite set, not all on a line, such that all pairwise distances are integers. i also do not know about dan's follow-up question: | And what about the higher-dimensional analogues: | | Q3: What is the maximum number of points in R^n, no hyperplane | (i.e., n-1 dimensional plane) containing n+1 of them, such that all | interpoint distances are integers? but i note that even with the weaker condition that no hyperplane contains all of the points, the question is still non-trivial! mike
On 12/12/06, Michael Reid <reid@math.ucf.edu> wrote:
some time ago, i wrote:
Before people go too far looking for such charts, I have a question: is it possible to have 4 points in a plane, no 3 colinear, so that all 6 distances are integers? And with the additional constraint that no 2 distances are equal?
it is possible to have arbitrarily many points in the plane, no three collinear, such that all distances between them are integers, and no two such distances are equal.
there hasn't been any response, except for dan's query "why is this true?". since no one has answered, here is my solution.
I developed a geometric construction, which boils down to the following: given a circle, assume that we can find a triangle with rational sides inscribed within it. Permuting two sides creates a congruent triangle, also inscribed in the circle. By Ptolemy, all distances between the four points are rational. Now continue, permuting the new triangles. True, I haven't shown that this process does not terminate. But it does apparently produce results for circles of radius sqrt(1/m) for many natural m; m = 17 is the smallest so far not observed. So I enquire Does this construction find all charts with given radius, assuming we can find an initial triangle? Can Mike Reid's algebraic construction be modified to deal with radius the reciprocal square root of arbitrary natural m? Example: for radius 1/sqrt(3) we find charts [3,5,7,7,8,3]/7, [3,5,7,7,5,8]/7, [3,5,7,8,7,7]/7, [3,7,8,8,7,5]/7, [7,8,13,13,15,7]/13, [7,8,13,13,8,15]/13, [7,8,13,15,13,13]/13, [7,13,15,15,13,8]/13, ...
i also do not know about dan's follow-up question:
| And what about the higher-dimensional analogues: | | Q3: What is the maximum number of points in R^n, no hyperplane | (i.e., n-1 dimensional plane) containing n+1 of them, such that all | interpoint distances are integers?
but i note that even with the weaker condition that no hyperplane contains all of the points, the question is still non-trivial!
Making everything concyclic no longer helps here. The condition for n+2 points to lie on a (n-1)-sphere is that the matrix with elements (s_ij)^2 be singular, where s_ij is the distance between points i,j. For n = 1 this happens to decompose into four bilinear factors, each a version of Ptolemy's theorem with various signs; but for larger n it is irreducible. Fred Lunnon
On 12/12/06, Michael Reid <reid@math.ucf.edu> wrote:
i also do not know about dan's follow-up question:
| And what about the higher-dimensional analogues: | | Q3: What is the maximum number of points in R^n, no hyperplane | (i.e., n-1 dimensional plane) containing n+1 of them, such that all | interpoint distances are integers?
Well, I had a little look at flat rational pentatopes, or sets of 5 points in Euclidean 3-space with integer distances. They seem much easier to find than planar charts --- in particular, a dumb search program came up immediately with the following examples [AB, AC, BC, AD, BD, CD, AE, BE, CE, DE] = [3, 3, 3, 2, 2, 2, 2, 2, 2, 2] [4, 3, 2, 4, 2, 3, 2, 3, 2, 4] [4, 3, 3, 4, 2, 3, 2, 3, 3, 4] [4, 4, 4, 4, 2, 3, 2, 4, 3, 3] [4, 4, 4, 4, 4, 2, 2, 4, 3, 4] [for which independent verification would be appreciated!] All except the first are conspheric. Rejecting the 120 possible symmetries is a technical problem yet to be completely sorted; till then a longer list is not very practicable. Fred Lunnon
Isomorph rejection now accomplished for rational simplex search program, which finds 47 distinct examples of integer 3-flat 5-topes with edge at most 6; list available on request. This search cost only 1000 sec on a Mac Powerbook, and could be pushed a good way further if anybody's interested. All but 6 of these charts are conspheric. Two exceptions, having a common tetrahedron, with edges in natural order, are [6, 5, 4, 4, 5, 2, 4, 4, 3, 2] [6, 5, 4, 4, 5, 2, 6, 3, 4, 4] These looked a good bet for gluing together to form a rational 3-flat 6-tope; alas, the 15-th side turns out to be sqrt(40/7)! Fred Lunnon
On 12/16/06, Fred lunnon <fred.lunnon@gmail.com> wrote:
Two exceptions, having a common tetrahedron, with edges in natural order, are
[6, 5, 4, 4, 5, 2, 4, 4, 3, 2] [6, 5, 4, 4, 5, 2, 6, 3, 4, 4]
These looked a good bet for gluing together to form a rational 3-flat 6-tope; alas, the 15-th side turns out to be sqrt(40/7)!
But I failed to take inot account that the second simplex might alternatively be written [6, 5, 4, 4, 5, 2, 3, 6, 4, 4] Gluing this on to the first instead gives a rational 3-flat 6-tope with edges [in alphabetic order of vertices] [6, 5, 4, 4, 5, 2, 3, 6, 4, 4, 4, 4, 3, 2, 5] Fred Lunnon
More thorough investigation of the puzzling phenomenon of large numbers of conspheric rational 3-flat 5-topes reveals the all-too-predictable explanation: an unfinished test in the search program letting through large numbers of improper examples, some of whose tetrahedral cells are planar. When the dust has settled, only the following (proper) rational examples remain, up to edge 6 --- intriguingly, they are exactly those non-conspheric ones found before! [3, 3, 3, 2, 2, 2, 2, 2, 2, 2], [6, 6, 3, 5, 4, 4, 4, 5, 4, 2], [6, 5, 4, 4, 5, 2, 4, 4, 3, 2], [6, 6, 5, 4, 5, 4, 3, 6, 4, 2], [6, 6, 6, 4, 4, 4, 4, 4, 4, 4], # gcd = 2 [6, 6, 6, 5, 4, 4, 5, 4, 4, 5]. The 3-flat 6-tope discussed earlier has 4 of its 6 5-topes in this list, but the other 2 now turn out to have been improper; so it falls over as well. Notice that there are (so far) no occurrences of an edge of unit length, in either 2- or 3-space. Is there a theorem here? Fred Lunnon
On 12/17/06, Fred lunnon <fred.lunnon@gmail.com> wrote: More thorough investigation of the puzzling phenomenon of large numbers of conspheric rational 3-flat 5-topes reveals the all-too-predictable explanation: an unfinished test in the search program letting through large numbers of improper examples, some of whose tetrahedral cells are planar.
When the dust has settled, only the following (proper) rational examples remain, up to edge 6 --- intriguingly, they are exactly those non-conspheric ones found before!
This raises the question of whether there are conspheric ones with no tetrahedral cells planar or other degeneracies. Presumably there are, so what is the 'smallest' one?
[3, 3, 3, 2, 2, 2, 2, 2, 2, 2], [6, 6, 3, 5, 4, 4, 4, 5, 4, 2], [6, 5, 4, 4, 5, 2, 4, 4, 3, 2], [6, 6, 5, 4, 5, 4, 3, 6, 4, 2], [6, 6, 6, 4, 4, 4, 4, 4, 4, 4], # gcd = 2 [6, 6, 6, 5, 4, 4, 5, 4, 4, 5].
The 3-flat 6-tope discussed earlier has 4 of its 6 5-topes in this list, but the other 2 now turn out to have been improper; so it falls over as well.
Notice that there are (so far) no occurrences of an edge of unit length, in either 2- or 3-space. Is there a theorem here?
Fred Lunnon
Jim Buddenhagen
On 12/19/06, James Buddenhagen <jbuddenh@gmail.com> wrote:
This raises the question of whether there are conspheric ones with no tetrahedral cells planar or other degeneracies. Presumably there are, so what is the 'smallest' one?
In 2-space, I found 250 disitnct proper integer charts with max sum of edge lengths 111, most of which belong to the special 4-parameter (concyclic) trapezium family mentioned earlier. Excluding these there remain just 53, including 19 concyclic: so approximately one third of the non-trapezoidal total is concyclic. In 3-space, I have now searched up to edge-length 8, finding 26 proper integer charts --- list available on request. None is conspheric, and none has an edge of length unity. Draw your own conclusions! Fred Lunnon
I put in some time [too much, probably] on rational simplices over the holiday: here's a summary. Edge-lengths in order [AB,AC,BC,AD,BD,CD,AE,BE,CE,DE,...] where vertices are A,B,C,D,E,...; cases differing only by a permutation of the vertices considered equivalent. A simplex spanning n dimensions is "proper" when every subset with n vertices spans n-1 dimensions; "noncyclic" when none with n+1 lies on a (n-1)-sphere. [The latter implies the former if an extra vertex at inversive infinity be attached --- this will assumed]. 4 vertices in 2 dimensions --- For edges not exceeding 32, there are 513 distinct proper integer cases. Approx. 80 percent of these are trapezia or parallelograms, each constituting a rational 4-parameter class: quadratic for trapezia [a*b, (b*d+a*c)/2, (b*d-a*c)/2, (b*d-a*c)/2, (b*d+a*c)/2, c*d]; cubic for parallelograms [(p^2+q^2)*t+4*p*q*s, (p^2-q^2)*t+(p^2+q^2)*s, (2*p*q)*t+(p^2+q^2)*s, (2*p*q)*t+(p^2+q^2)*s, (p^2-q^2)*t+(p^2+q^2)*s, (p^2+q^2)*t+2*(p^2-q^2)*s]. The latter is a linear combination of rectangle and rhombus, both constructed from Pythagorean triples. [Care is required with both to ensure that the results are proper and integer.] 5 vertices in 2 dimensions --- All but 129 of the 513 are concyclic, and it has been earlier established that there are concyclic proper integer m-vertex configurations for any m; however, the construction does not generalise to more than 2 dimensions. This suggests exploring noncyclic cases: it transpires that there is just one such plane 5-vertex with edges up to 32: [26, 25, 9, 18, 20, 13, 13, 15, 18, 19]. 5 vertices in 3 dimensions --- for edges up to 10, there are 63 distinct proper integer cases. Just 2 of these are conspheric [with the same radius] [9, 9, 6, 8, 7, 5, 7, 8, 4, 7], [9, 8, 7, 8, 5, 4, 7, 8, 7, 9]. In contrast, there is also the single ambiguous pair [7, 6, 2, 6, 2, 3, 5, 4, 4, 4], [7, 6, 2, 6, 2, 3, 4, 5, 4, 4]. 6 vertices in 3 dimensions --- none for edges up to 10. Although there are numerous pairs which may be glued together at a common tetrahedron, either the single new edge created is irrational, or else one of the component tetrahedra turns out to be planar. Almost certainly a considerably more extensive search is required to find such an object, if it exists. No proper integer case has emerged with any edge of length unity; just one with edge 2 has been found in 2 dimensions, and only 12 out of 63 in 3 dimensions. Fred Lunnon
Fred Lunnon wrote:
I put in some time [too much, probably] on rational simplices over the holiday[...]
Just the opposite here: I was hoping to rewrite my (slow Mma-based) search in gp/pari, but never found the time. (Surely there is more than one ambiguous hexad!) The real problem with not being a math professor any more: no grad students to run off and do this kind of thing for you.
5 vertices in 3 dimensions [...] there is also the single ambiguous pair
[7, 6, 2, 6, 2, 3, 5, 4, 4, 4], [7, 6, 2, 6, 2, 3, 4, 5, 4, 4].
Very nice to see! As with the only 4-points-in-2d example we have, this is a proper simplex plus one point with ambiguous placement. This seems like the odds-on bet for where to look for ambiguous cases, especially ones with more than two configurations. I wonder if there's a good way to build a search that looks for that situation specifically? In particular, given a proper simplex (with, say, integer edge lengths), is there an easy (eg finite) way to tell whether there is a set of (let's say rational) distances from its vertices resulting in ambiguous placement of the last point?
6 vertices in 3 dimensions --- none for edges up to 10. Although there are numerous pairs which may be glued together at a common tetrahedron, either the single new edge created is irrational, or else one of the component tetrahedra turns out to be planar. Almost certainly a considerably more extensive search is required to find such an object, if it exists.
My intuition is that this, like the abiguous-planar case, would be better suited to a search targeted at finding those pairs to glue, rather than looking for an example with small edge lengths. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 1/4/07, Michael Kleber <michael.kleber@gmail.com> wrote:
My intuition is that this, like the abiguous-planar case, would be better suited to a search targeted at finding those pairs to glue, rather than looking for an example with small edge lengths.
In 2 dimensions, a direct search for 6-vertex configurations became tedious at around edge-length 12; in contrast, gluing together pairs of (previously computed) 5-vertex cases was good for edge 32 --- and struck oil! It could well be that an "intelligent" construction, similar to that of the planar ambiguous case, is the way to go in 3-D. So --- any ideas, anybody? Fred Lunnon
Aaarghhh! Last message should have read as follows: On 1/4/07, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 1/4/07, Michael Kleber <michael.kleber@gmail.com> wrote:
My intuition is that this, like the abiguous-planar case, would be better suited to a search targeted at finding those pairs to glue, rather than looking for an example with small edge lengths.
In 2 dimensions, a direct search for 5-vertex configurations became tedious at around edge-length 12; in contrast, gluing together pairs of (previously computed) 4-vertex cases was good for edge 32 --- and struck oil!
It could well be that an "intelligent" construction, similar to that of the planar ambiguous case, is the way to go in 3-D. So --- any ideas, anybody?
Fred Lunnon
The other day, Fred wrote:
No proper integer case has emerged with any edge of length unity; just one with edge 2 has been found in 2 dimensions, and only 12 out of 63 in 3 dimensions.
Suppose a proper set of points P has all pairwise distances integers, and P contains A,B with d(A,B)=1. Then for all C, d(C,A)=d(C,B), since by the proper triangle inequality d(C,A) - d(C,B) < 1. So the n-2 points in P other than {A,B} all live in some n-3-dimensional subspace, and the unit-length line segment AB has that plane as its perpendicular bisector. That makes it easy to see that there will be no 4-point solution. Let A=(0,1/2), B=(0,-1/2), then we would need two more points C, D on the x-axis, say at respective distances c and d from A. The x-coord of C is then +-sqrt( c^2 - 1/4 ) = +-sqrt(4c^2-1)/2, and likewise D is at +-sqrt(4d^2-1)/2. That makes it clear that the x-coords of C and D are both "irrational but sqrt(rational)", and the sum or difference of two such numbers is always irrational (just look at (sqrt(n)+m)^2 to see why). I have a niggling feeling that a decent number theorist might immediately see how to generalize that argument to higher dimensions, but perhaps I'm being (sorry) irrational. Even if that doesn't lead to a proof of impossibility, it certainly means it's not surprising that a search for general integer-separation point sets would have a hard time finding ones with a unit distance: almost none of the sets you consider will meet the distance pairing constraint. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 1/5/07, Michael Kleber <michael.kleber@gmail.com> wrote:
... Even if that doesn't lead to a proof of impossibility, it certainly means it's not surprising that a search for general integer-separation point sets would have a hard time finding ones with a unit distance: almost none of the sets you consider will meet the distance pairing constraint.
Oddly enough, for 5 vertices in 3 dimensions, the very first case [3, 3, 3, 2, 2, 2, 2, 2, 2, 2] does satisfy the above constraint! However, no other with edge up to 10 does so. WFL
On 1/5/07, Fred lunnon <fred.lunnon@gmail.com> wrote:
Oddly enough, for 5 vertices in 3 dimensions, the very first case
[3, 3, 3, 2, 2, 2, 2, 2, 2, 2]
does satisfy the above constraint! However, no other with edge up to 10 does so. WFL
Er, except for multiples of the first case, that is ... WFL
I suppose, having written
Even if that doesn't lead to a proof of impossibility, it certainly means it's not surprising that a search for general integer-separation point sets would have a hard time finding ones with a unit distance: almost none of the sets you consider will meet the distance pairing constraint.
, that I was more or less obligated to do the simple search that takes that constraint into account. With 10 mintues each of coding time (I think I got all the properness constraints in there) and run time, I find the following two examples: [1, 24, 23, 16, 24, 23, 16, 46, 32, 30] [1, 25, 25, 2, 25, 25, 2, 42, 24, 24] That second one has a second plane of mirror symmetry, apart from the necessary symmetry which just swaps the two distance-1 points. The search had no cleverness involved, beyond the planarity constraint that I mentioned this morning. Brute-force search over the 5-dimensional space where first you specify the distance from the two sep-1 points to the other three points (say x,y,z, constraining each of them to lie on one of three circles, all coplanar and concentric) and further specify d(x,y) and d(x,z). You can write down explicit coordinates for x, y WLOG, and there are only two choices for z, and the two possible distances between y and z can be written with square roots nested only two deep. Then just search in lex order until you happen across an integer for the one unspecified separation. Er, since I'm searching first based on the three circle radii, it's still possible that there is a solution with maximal edge length smaller than the 42 above. By the way, in this morning's I forgot to mention the observation that this problem (two points at unit separation) is exactly the same as looking for a single proper simplex with integer edge-lengths and an altitude equal to 1/2, which is perhaps a nicer formulation than the one about the perpendicular bisector plane. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 1/5/07, Michael Kleber <michael.kleber@gmail.com> wrote:
... I find the following two examples:
[1, 24, 23, 16, 24, 23, 16, 46, 32, 30] [1, 25, 25, 2, 25, 25, 2, 42, 24, 24]
That second one has a second plane of mirror symmetry, apart from the necessary symmetry which just swaps the two distance-1 points.
These checked out OK, once I had guessed MK is using a different ordering convention, and rewritten them in mine [1, 24, 24, 23, 23, 46, 16, 16, 32, 30] [1, 25, 25, 25, 25, 42, 2, 2, 24, 24]. I wonder if the second one in particular is a member of a parametric family, analogous to the quadratic 4-parameter trapezium family in 2 dimensions? Fred Lunnon
On 1/5/07, Fred lunnon <fred.lunnon@gmail.com> wrote:
I wonder if the second one in particular is a member of a parametric family, analogous to the quadratic 4-parameter trapezium family in 2 dimensions?
Despite appearances, there are in practice rather a lot of 5-vertex 3-dim cases which satisfy MK's "isosceles" constraint --- say AD = AE, BD = BE, CD = CE: 21 of the first 63 may be cast in this form; and 12 of those have sufficient symmetry to do so in two distinct ways. However, staring hard at these 33 cases soon dispels any hope for a simple parametric formula; and the impression is reinforced by an attempt to construct one via algebraic geometry, resulting in a quartic irrational parameterisation. Fred Lunnon PS --- after laboriously transcribing between MK ordering and WFL ordering, I belatedly realised that all that was required was to reverse the vector ...
Fred said:
Despite appearances, there are in practice rather a lot of 5-vertex 3-dim cases which satisfy MK's "isosceles" constraint --- say AD = AE, BD = BE, CD = CE: 21 of the first 63 may be cast in this form; and 12 of those have sufficient symmetry to do so in two distinct ways.
Upon further reflection, I suppose these are the simplest example of the construction "glue two tetrahedra together along a shared triangle." Given any tetrahedron with an altitude that is an integer or half-integer, you can glue it to its mirror image -- I didn't intend "upon further reflection" above to be a pun, honest! -- and get an isosceles 5-vx-3-d. Is the altitude of an integer-edged tetrahedron always degree <=2 over the rationals, as for triangles? That would make clear why these are so common.
PS --- after laboriously transcribing between MK ordering and WFL ordering, I belatedly realised that all that was required was to reverse the vector ...
Oof, sorry for the sign error. (Of course you chose the convention where adding a sixth point just involves tacking five more distances onto the end; duh.) Having run overnight, I've searched through length 64 for the longest edge touching one of the separation-1 points. Only found five so far. The list (in WFL-compatible order, I think!): [30, 32, 46, 16, 23, 24, 16, 23, 24, 1] [24, 24, 42, 2, 25, 25, 2, 25, 25, 1] [40, 42, 12, 26, 31, 40, 26, 31, 40, 1] [70, 64, 54, 31, 41, 41, 31, 41, 41, 1] [86, 54, 54, 46, 46, 49, 46, 46, 49, 1] Thus I can state for sure that the 3rd one on that list, with longest edges of length 40, has the smallest possible maximal edge length. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
On 1/6/07, Michael Kleber <michael.kleber@gmail.com> wrote:
... [30, 32, 46, 16, 23, 24, 16, 23, 24, 1] [24, 24, 42, 2, 25, 25, 2, 25, 25, 1] [40, 42, 12, 26, 31, 40, 26, 31, 40, 1] [70, 64, 54, 31, 41, 41, 31, 41, 41, 1] [86, 54, 54, 46, 46, 49, 46, 46, 49, 1]
All checked for propriety.
Thus I can state for sure that the 3rd one on that list, with longest edges of length 40, has the smallest possible maximal edge length.
Don't follow --- the second has max edge 25 ?! WFL
On 1/6/07, Michael Kleber <michael.kleber@gmail.com> wrote:
Is the altitude of an integer-edged tetrahedron always degree <=2 over the rationals, as for triangles? That would make clear why these are so common.
The squared base area and the squared volume are both polynomials in the squared edges; 3x altitude equals volume divided by base. Hence altitude equals square root of a rational. WFL
Hi,
Having run overnight, I've searched through length 64 for the longest edge touching one of the separation-1 points. Only found five so far. The list (in WFL-compatible order, I think!):
[30, 32, 46, 16, 23, 24, 16, 23, 24, 1] [24, 24, 42, 2, 25, 25, 2, 25, 25, 1] [40, 42, 12, 26, 31, 40, 26, 31, 40, 1] [70, 64, 54, 31, 41, 41, 31, 41, 41, 1] [86, 54, 54, 46, 46, 49, 46, 46, 49, 1]
Thus I can state for sure that the 3rd one on that list, with longest edges of length 40, has the smallest possible maximal edge length.
If you look for the smallest largest number in your lists you have found "The Ultimate Question Of Life, the Universe and Everything" :-)) Deep thought need 7.5 million years for that, see http://en.wikipedia.org/wiki/The_Answer_to_Life,_the_Universe,_and_Everythin... I know this is more fun than math, but I wanted to share it ;-) -- Christoph
Oops, quite right, it's 42, not 40, and entries #2 and #3 both attain it. --Michael On 1/6/07, Christoph Pacher <christoph.pacher@arcs.ac.at> wrote:
Hi,
Having run overnight, I've searched through length 64 for the longest edge touching one of the separation-1 points. Only found five so far. The list (in WFL-compatible order, I think!):
[30, 32, 46, 16, 23, 24, 16, 23, 24, 1] [24, 24, 42, 2, 25, 25, 2, 25, 25, 1] [40, 42, 12, 26, 31, 40, 26, 31, 40, 1] [70, 64, 54, 31, 41, 41, 31, 41, 41, 1] [86, 54, 54, 46, 46, 49, 46, 46, 49, 1]
Thus I can state for sure that the 3rd one on that list, with longest edges of length 40, has the smallest possible maximal edge length.
If you look for the smallest largest number in your lists you have found "The Ultimate Question Of Life, the Universe and Everything" :-))
Deep thought need 7.5 million years for that, see http://en.wikipedia.org/wiki/The_Answer_to_Life,_the_Universe,_and_Everythin...
I know this is more fun than math, but I wanted to share it ;-) -- Christoph
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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Update to search results: 4 vertices in 2 dimensions --- pushed up to max edge 50, finding 1579 distinct proper integer cases. Approx 450 are non-cyclic. None is ambiguous. 5 vertices in 2 dimensions --- 29 noncyclic cases found by gluing pairs of the above. [6 of these constitute an apparently promising set where each pair of members has 6 edge-lengths in common; sadly, attempting to glue these into a 6-tope fails.] 5 vertices in 3 dimensions --- pushed up to max edge 12, finding 217 cases; 5 conspheric, 2 ambiguous pairs. 6 vertices in 3 dimensions --- I earlier opined (along with MK)
Almost certainly a considerably more extensive search is required to find such an object, if it exists then went ahead and tried anyway. Mirabile dictu, gluing pairs produces just the single example (ta-DAH!) ---
[36, 33, 30, 27, 18, 15, 21, 30, 30, 27, 21, 18, 27, 18, 13] with the customary order of edges [AB,AC,BC,AD,BD,CD,AE,BE,CE,DE,AF,BF,CF,DF,EF]. I can't resist digressing on how regularly our loosely probabilistic intuitions about these objects have been confounded :--- (a) Apparently, concyclic charts should be in a small minority, since they satisfy an extra constraint. In 3-space this is indeed the case; but in 2-space the majority turn out to be concyclic [see Michael Reid, this thread Dec 12th]. (b) In 2-space, parallelograms and trapezia [two pairs of equal edges] ought also to be uncommon; but for similar reasons, they too actually preponderate [WFL, this thread Jan 3rd]. (c) The reasonable guess that "isosceles" 5-topes in 3-space [with 3 pairs equal] turns out to be similarly way off-target --- these arise in a straightforward fashion by gluing together equal tetrahedra [MK, this thread Jan 6th]. (d) Conversely, the more general "gluing" procedure, with which I optimistically set out to construct (proper) integer charts in n-space with more than n+1 vertices, produces hundreds of improper (or irrational) examples for every proper one --- despite the probability that a random n-tope has zero volume being vanishingly small. [In 2-space, these duds arise by combining a parallelogram with a triangle, so that one side of the triangle coincides with a diagonal of the p'gram, and the other vertex of the triangle lies on the other diagonal of the p'gram. If all the existing edges are integer, then the single new edge created by gluing is the difference of two integers --- but for the same reason, the new triangle it belongs to is linear!] In all the above, the reason for the dodgy insight is that some "special" case turns out far more frequent than the general. The correct approach here seems to be (somehow!) to enumerate and consider these special cases beforehand; and only risk a probabilistic argument once they have been eliminated. Fred Lunnon
Fred lunnon wrote:
Update to search results:
Oh right! I left running my dumb search for 5 points in dim 3 including a pair with unit separation. Today it reached the nice round stopping number of 100 (maximum length of an edge indicent to the unit-separation pair). Time to kill that search, though -- it's now taking about 8 hours for each increment of that radius. That's what O(n^5) gets ya. For the record, here are the 18 instances it found: 1 [1,24,24,23,23,46,16,16,32,30] 2 [1,25,25,25,25,42,2,2,24,24] 3 [1,40,40,31,31,12,26,26,42,40] 4 [1,41,41,41,41,54,31,31,64,70] 5 [1,49,49,46,46,54,46,46,54,86] 6 [1,68,68,58,58,36,32,32,39,45] 7 [1,70,70,12,12,71,8,8,68,5] 8 [1,74,74,52,52,84,52,52,105,27] 9 [1,74,74,74,74,147,52,52,84,105] 10 [1,75,75,59,59,21,55,55,120,109] 11 [1,78,78,28,28,85,10,10,71,36] 12 [1,79,79,79,79,134,6,6,76,76] 13 [1,85,85,40,40,93,23,23,96,21] 14 [1,85,85,85,85,96,13,13,96,96] 15 [1,85,85,85,85,156,13,13,84,96] 16 [1,86,86,40,40,84,40,40,125,57] 17 [1,90,90,38,38,56,10,10,84,36] 18 [1,95,95,70,70,47,37,37,96,89] All of these are, of necessity, "isosceles" in that they have the mirror symmetry that exchanges the two points at unit distance; three of them (2, 5, 14) are doubly isosceles, having a second plane of mirror symmetry as well. Number 14 seem particularly cool: two points have separation 1, and the other three form an equilateral triangle with edge length 96. I wonder if there's an infinite family of similar examples? And Fred, congratulations on finding
[36, 33, 30, 27, 18, 15, 21, 30, 30, 27, 21, 18, 27, 18, 13] You should make a model of it!
--Michael Kleber ps: it occurred to me the other day that every Euler brick leads to a set of seven integer-separation points in three dimensions -- the origin and six more arranged as the vertices of an octahedron. But of course this isn't proper. We could drop the origin and get six points with no three collinar, but the requirement that no four be coplanar shoots this out of the water. (Surely Fred Helenius realized this as soon as the discussion started, but some of us are a little slow...) -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Crossing number 1 -- Utility graph -- 6 vertices Crossing number 2 -- Petersen graph -- 10 vertices Crossing number 3 -- Heawood graph -- 14 vertices Crossing number 8 -- Mcgee graph -- 24 vertices. What are the simplest cubic graphs with crossing number 4-7 ? As a warm-up exercise to how tricky this problem is, show that a even n-gon with opposite corners connected has crossing number 1. (No need to post a proof here.) --Ed Pegg Jr
Ed et Al, I won't post the answer, but see Guy & Harary, On the M"obius ladders [also available in Hungarian!] MR 37 #98, #2627 R. On Wed, 28 Feb 2007, Ed Pegg Jr wrote:
Crossing number 1 -- Utility graph -- 6 vertices Crossing number 2 -- Petersen graph -- 10 vertices Crossing number 3 -- Heawood graph -- 14 vertices
Crossing number 8 -- Mcgee graph -- 24 vertices.
What are the simplest cubic graphs with crossing number 4-7 ?
As a warm-up exercise to how tricky this problem is, show that a even n-gon with opposite corners connected has crossing number 1. (No need to post a proof here.)
--Ed Pegg Jr
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participants (7)
-
Christoph Pacher -
Ed Pegg Jr -
Fred lunnon -
James Buddenhagen -
Michael Kleber -
Michael Reid -
Richard Guy