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The question is inspired by this previous MO question. There it was shown that it's possible to label the edges of a cube by the numbers $\{1,2,\ldots,6,8,9, \ldots, 13\}$ in such a way that:

  1. Three edges meeting at a vertex sum to 21.
  2. Four edges constituting a face sum to 28.

The solution is unique up to isometry (numbers in red correspond to invisible edges) :

enter image description here

Similary, is it possible the label the edges of a regular icosahedron (or dodecahedron) with distinct integers in such a way that the two conditions above are satisfied (with different sums of course). If yes, is the solution unique up to isometry?

Note that for the regular tetrahedron, the answer is no: a computation shows that two opposite edges must be equal.

Moreover, what would be natural generalizations of this problem for polytopes in $n$-dimensional space?

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    $\begingroup$ Much nicer to have all sums $0$ using $-6 \cdots 6$ without $0.$ Either way, the same thing gives a companion octahedron labelling. Similarly, whatever one can or can't do for a dodecahedron, one can or can't do for an icosohedron. $\endgroup$ Nov 4, 2015 at 7:52

3 Answers 3

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Alas there's no such labeling by $\{1, 2, 3, \ldots, 15, 17, 18, 19, \ldots, 31\}$, assuming I made no computational error somewhere along the way.

The vertex and face conditions give $32$ linear equations on the $30$ edge labels, but with enough redundancy that there is a $6$-dimensional space of solutions. The obvious solution is to label all the edges $1$. A less trivial solution is obtained by fixing two opposite vertices $v,v'$ of the icosahedron, and using the edges that connect a neighbor of $v$ to a neighbor of $v'$: there are $10$ such, forming a cycle, and we can label them alternately $+1$ and $-1$, and label the remaining $20$ edges zero. This gives $6$ generators (in addition to the all-$1$ vector), satisfying one relation (they sum to zero if the signs are chosen consistently), for a total dimension of $1 + (6-1) = 6$. These turn out to generate the space of solutions even over the integers; that is, every integer solution of the linear equations is an integer linear combination of the generators.

Subtracting off $16 \cdot (1,1,1,\ldots, 1)$, we are looking for a vector in the $5$-dimensional span of the remaining generators whose coordinates are a permutation of $(\pm 1, \pm 2, \pm3, \ldots, \pm15)$. There is no obvious obstruction: no two of the $30$ coordinates are identically equal (and coordinates at opposite edges are automatically each other's negatives, which should only help). It would now be feasible to choose $5$ independent coordinates, try all $30 \cdot 29 \cdot 28 \cdot 27 \cdot 26$ choices of labeling, and check whether any of them results in a solution of the desired kind. But first I tried the shortcut of looking at all solutions modulo $2$ and $3$. The former failed: there are $15$ linear combinations with the requisite $16$ odd and $14$ even entries. But modulo $3$, (unwelcome) success: none of the $121 = (3^5-1) / 2$ pairs of nonzero linear combinations has coordinates equally split among $0,1$, and $2 \bmod 3$. (The possible counts of $0 \bmod 3$ coordinates are $6, 8, 12, 14, 20$, occurring with multiplicities $15, 60, 25, 15, 6$ respectively.) Hence there is no edge labeling satisfying all the conditions, QED.

[added later: I should have noticed that none of $6,8,12,14,20$ is $1 \bmod 3$. This can be seen directly from the matrix of inner products of our five nontrivial generators, which has rank $1 \bmod 3$. So no exhaustive search is required at all, though there's still the integer linear algebra to find our generators.]

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    $\begingroup$ Thanks a lot for your computations. It is still possible that there exists a distinct labeling which works, but according to your result it won't be as symmetric as in the cube case... $\endgroup$ Nov 4, 2015 at 4:39
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    $\begingroup$ Yes, certainly distinct labelings exist, because a "random" labeling in that 6-dimensional vector space has all labels distinct. The labels just can't be as closely packed as $\{\pm1, \pm2, \ldots, \pm15\}$. $\endgroup$ Nov 4, 2015 at 4:48
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    $\begingroup$ You're right. What is the smallest $N$ such that the labels are distinct and all $\leq N$ in absolute value? I guess there are algorithms to answer this kind of question. $\endgroup$ Nov 4, 2015 at 4:55
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    $\begingroup$ Looks like it's $17$, with several solutions. More in the next edit. $\endgroup$ Nov 4, 2015 at 15:40
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This is pretty useless, definitely not an answer, because the edge labels are not distinct. The face sum is constant at $145$, and the vertex sum constant at $87$.


          DodecaMagic
          $$\textrm{Labels}=\left( 2, 2, 4, 4, 8, 8, 10, 10, 23, 27, 27, 27, 27, 29, 29, 29, 29, 31, 31, \ 31, 31, 35, 48, 48, 50, 50, 54, 54, 56, 56 \right)$$
There are likely many similar solutions. All labels distinct might be challenging.

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    $\begingroup$ Thanks. This could actually be a variant of the original puzzle: maybe there is a nice symmetric solution allowing repetitions. $\endgroup$ Nov 4, 2015 at 6:37
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If you require that your labels are the set $\{1,2,\dotsc,31\}$ with one label $k$ removed, a back-of-the-envelope calculation reveals that $k$ must be $16$. Moreover, your vertex-sums (the sums of three edges meeting at any vertex) must be 36 48, and your face-sums must be 100 80.

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    $\begingroup$ 36 and 100 can't be right: the average label is 16, so the average vertex and face sums are 48 and 80, and if all are equal then all must equal those averages. $\endgroup$ Nov 4, 2015 at 3:03
  • $\begingroup$ You're absolutely right. My mistake. $\endgroup$ Nov 4, 2015 at 12:47

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