Paths on a dodecahedron











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Looking at this question, I misread "dodecagon" as "dodecahedron". I think the latter is a cool problem, so I'm posing it as a question of its own :)




Starting from one vertex of a dodecahedron, an ant wants to reach the opposite vertex of the dodecahedron, moving to adjacent vertices. If $p_n$ is the number of such paths with length $n$, compute $p_1+p_2+dots+p_{12}$.











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  • $p_1=p_2=p_3=p_4=0$
    – Alexander Burstein
    Dec 23 '17 at 4:37






  • 1




    The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
    – vadim123
    Dec 23 '17 at 16:01










  • Paths or simple paths?
    – MJD
    Dec 23 '17 at 16:25










  • @vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
    – Yly
    Dec 24 '17 at 19:26










  • @MJD Paths, not simple paths.
    – Yly
    Dec 24 '17 at 19:27















up vote
7
down vote

favorite












Looking at this question, I misread "dodecagon" as "dodecahedron". I think the latter is a cool problem, so I'm posing it as a question of its own :)




Starting from one vertex of a dodecahedron, an ant wants to reach the opposite vertex of the dodecahedron, moving to adjacent vertices. If $p_n$ is the number of such paths with length $n$, compute $p_1+p_2+dots+p_{12}$.











share|cite|improve this question






















  • $p_1=p_2=p_3=p_4=0$
    – Alexander Burstein
    Dec 23 '17 at 4:37






  • 1




    The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
    – vadim123
    Dec 23 '17 at 16:01










  • Paths or simple paths?
    – MJD
    Dec 23 '17 at 16:25










  • @vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
    – Yly
    Dec 24 '17 at 19:26










  • @MJD Paths, not simple paths.
    – Yly
    Dec 24 '17 at 19:27













up vote
7
down vote

favorite









up vote
7
down vote

favorite











Looking at this question, I misread "dodecagon" as "dodecahedron". I think the latter is a cool problem, so I'm posing it as a question of its own :)




Starting from one vertex of a dodecahedron, an ant wants to reach the opposite vertex of the dodecahedron, moving to adjacent vertices. If $p_n$ is the number of such paths with length $n$, compute $p_1+p_2+dots+p_{12}$.











share|cite|improve this question













Looking at this question, I misread "dodecagon" as "dodecahedron". I think the latter is a cool problem, so I'm posing it as a question of its own :)




Starting from one vertex of a dodecahedron, an ant wants to reach the opposite vertex of the dodecahedron, moving to adjacent vertices. If $p_n$ is the number of such paths with length $n$, compute $p_1+p_2+dots+p_{12}$.








combinatorics polyhedra






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asked Dec 23 '17 at 2:49









Yly

6,52621539




6,52621539












  • $p_1=p_2=p_3=p_4=0$
    – Alexander Burstein
    Dec 23 '17 at 4:37






  • 1




    The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
    – vadim123
    Dec 23 '17 at 16:01










  • Paths or simple paths?
    – MJD
    Dec 23 '17 at 16:25










  • @vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
    – Yly
    Dec 24 '17 at 19:26










  • @MJD Paths, not simple paths.
    – Yly
    Dec 24 '17 at 19:27


















  • $p_1=p_2=p_3=p_4=0$
    – Alexander Burstein
    Dec 23 '17 at 4:37






  • 1




    The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
    – vadim123
    Dec 23 '17 at 16:01










  • Paths or simple paths?
    – MJD
    Dec 23 '17 at 16:25










  • @vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
    – Yly
    Dec 24 '17 at 19:26










  • @MJD Paths, not simple paths.
    – Yly
    Dec 24 '17 at 19:27
















$p_1=p_2=p_3=p_4=0$
– Alexander Burstein
Dec 23 '17 at 4:37




$p_1=p_2=p_3=p_4=0$
– Alexander Burstein
Dec 23 '17 at 4:37




1




1




The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
– vadim123
Dec 23 '17 at 16:01




The dodecagon, expressed as a graph, is a cycle. Hence, walks of a specific length have very clear structure, and one can exploit that structure to solve the problem. In contrast, the dodecahedron graph is a big beautiful mess. Even calculating $p_5$ is not obvious (and would make a fine question on its own). Calculating larger and larger $p_k$ can only be done by computer.
– vadim123
Dec 23 '17 at 16:01












Paths or simple paths?
– MJD
Dec 23 '17 at 16:25




Paths or simple paths?
– MJD
Dec 23 '17 at 16:25












@vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
– Yly
Dec 24 '17 at 19:26




@vadim123 $p_5=6$, $p_6=12$, and both take less than a minute to compute when you look at a picture of a dodecahedron.
– Yly
Dec 24 '17 at 19:26












@MJD Paths, not simple paths.
– Yly
Dec 24 '17 at 19:27




@MJD Paths, not simple paths.
– Yly
Dec 24 '17 at 19:27










4 Answers
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Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.



For example, there are



171619248



such paths of length 20 (not too far from MJD's guess), and



25768876036573921452762172776956776774411837488



such paths of length 100.






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    It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0leq ileq5)$ of vertices.



    As described above: there is a single central vertex C_0; arranged symmetrically around this are three vertices C_1, then six vertices C_2, six more C_3, three more C_4, and finally one C_5 which is drawn in three places, attached to the C_5, vertices, but with two in parentheses to indicate that they are not separate vertices.



    It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
    $${bf p}(n+1)=left[matrix
    {0&3&0&0&0&0cr
    1&0&2&0&0&0cr
    0&1&1&1&0&0cr
    0&0&1&1&1&0cr
    0&0&0&2&0&1cr
    0&0&0&0&3&0cr}right]>{bf p}(n) ,$$

    whereby ${bf p}(n)$ denotes the column vector of the $p_i(n)$.



    Since there are so many treatments of the problem already I stop here.






    share|cite|improve this answer



















    • 1




      I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
      – MJD
      Nov 14 at 15:45








    • 1




      @MJD: That's exactly the drawing I had. Thank you very much!!
      – Christian Blatter
      Nov 14 at 16:09










    • This is a nice observation!
      – Travis
      Nov 14 at 21:24


















    up vote
    3
    down vote













    One way to count paths on a graph $Gamma$, say, with vertices $v_i$, is to use its adjacency matrix: This is the square matrix $A_{Gamma}$ of size $|Gamma|$ for which the $(i, j)$ entry $(A_{Gamma})_{ij}$ is the number of edges with endpoints at vertices $v_i$ and $v_j$. An intuitive inductive argument then shows that the number of paths of length $n$ from $v_i$ to $v_j$ is exactly $(A_{Gamma}^n)_{ij}$.



    The SageMath code Delta = graphs.DodecahedralGraph(); Delta.plot() assigns the dodecahedral graph the name Delta and returns the following labeled plot of the graph, which we call $Delta$. NB Sage starts indexing at $0$.
    Dodecahedral graph
    A little visualization shows that, e.g., the vertex opposite $v_0$ is vertex $v_{15}$.



    Next, the Sage command A = dodecahedral.adjacency_matrix() assigns to A the adjacency matrix $A_{Delta}$ of $Delta$; it has size $|Delta| = 20$.



    $$small A_{Delta} = pmatrix{
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
    1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
    0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
    0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \
    0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
    0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
    0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \
    1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \
    0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \
    0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \
    1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
    } .$$



    Using our above combinatorial interpretation of $A_{Delta}^n$ gives that the number of paths of length $n$ from $v_0$ to the opposite vertex $v_{15}$ is $$boxed{p_n = (A_{Delta}^n)_{0, 15}}.$$ Asking Sage to compute $p_1, ldots, p_{12}$ and add them, say, using the code n = var('n'); sum([(A^n)[0, 15] for n in range(1,13)]), gives that
    $$color{#df0000}{boxed{p_1 + cdots + p_{12} = 34548}} ,$$
    which agrees with MJD's count.



    One can derive this in a more illuminating way. To compute $A_{Gamma}^n$ efficiently, it is useful to use the Jordan decomposition $A_{Gamma} = Q J Q^{-1}$. Then, $A_{Gamma}^n = (Q J Q^{-1})^n = Q J^n Q^{-1}$. Since an adjacency matrix (of an undirected graph) is symmetric, $A_{Gamma}$ is diagonalizable and thus $J$ is diagonal, say, $J = operatorname{diag}(lambda_i)$, and $J^n = operatorname{diag}(lambda_i^n)$.



    Now, the number of paths of length $n$ from $v_i$ to $v_j$ is $$(A_{Delta}^n)_{ij} = (Q J^n Q^{-1})_{ij} = sum_{k, ell} Q_{ik} (J^n)_{k ell} Q^{-1}_{ell j} ,$$ and since $J^n$ is diagonal with entries $lambda_a^n$, terms only contribute to the sum when $ell = k$, leaving $$(A_{Delta}^n)_{ij} = sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n .$$ The number of paths from $v_i$ to $v_j$ of length $leq m$ is then the partial sum
    $$sum_{n = 0}^m (A_{Delta}^n)_{ij}
    =sum_{n = 0}^m sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n
    =sum_k (Q_{ik} Q^{-1}_{kj}) sum_{n = 0}^m lambda_k^n
    =sum_k (Q_{ik} Q^{-1}_{kj}) frac{lambda_k^{m + 1} - 1}{lambda_k - 1},$$

    where we interpret $frac{lambda_k^{m + 1} - 1}{lambda_k - 1}$ as $m + 1$ if $lambda_k = 1$.



    For the dodecahedral graph $Delta$, the Jordan decomposition (produced, for example, using A.change_ring(QQ[sqrt(5)]).jordan_form(transformation=True)), is given by



    $$J = operatorname{diag}(3, sqrt{5}, sqrt{5}, sqrt{5}, -sqrt{5}, -sqrt{5}, -sqrt{5}, 0, 0, 0, 0, -2, -2, -2, -2, 1, 1, 1, 1, 1) ,$$
    begin{multline*}Q \ = scriptsize pmatrix{
    1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
    1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
    1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
    1 & 1 & -phi & phi & 1 & phi^{-1} & -phi & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
    1 & phi^{-1} & -2 & phi & -phi & -2 & -phi & 1 & 1 & 0 & 1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
    1 & -phi & -phi & 1 & phi & phi & 1 & -1 & 0 & -1 & -1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
    1 & -1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & 0 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
    1 & -phi & 2 & -phi & phi^{-1} & 2 & phi & 1 & 0 & 0 & 1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
    1 & -1 & sqrt{5} & -1 & -1 & -sqrt{5} & -1 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
    1 & -phi & 2 & -phi & phi & 2 & phi^{-1} & -1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
    1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & -1 & 1 & 0 & 1 & 1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
    1 & 1 & -phi & -phi & 1 & phi & phi & 0 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
    1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
    1 & -1 & phi & -phi & -1 & -phi & phi^{-1} & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
    1 & -phi & phi & -1 & phi^{-1} & -phi & -1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1 \
    1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
    1 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
    1 & 1 & -sqrt{5} & 1 & 1 & sqrt{5} & 1 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
    1 & phi & -2 & phi^{-1} & -phi & -2 & -phi & -1 & 0 & 0 & -1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
    1 & phi & -phi & 1 & -phi & phi^{-1} & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1} .end{multline*}

    Here, $phi := frac{1}{2}(1 + sqrt{5})$ is the Golden Ratio.



    Taking $i = 0, j = 15$ and substituting (or just executing (A^n)[0, 15]) gives
    $$boxed{p_n = (A_{Delta}^n)_{0, 15} = frac{1}{20} cdot 3^n - frac{3}{20} (1 + (-1)^n) (sqrt{5})^n + frac{1}{5} , left(-2right)^n + frac{1}{4}}.$$
    (This agrees with the formula given in OEIS A054883, Number of walks of length n along the edges of a dodecahedron between two opposite vertices.)
    Then, the number of such paths of length $leq m$ is the partial sum $s_m = sum_{n = 0}^m p_n$, and indeed, $s_{12} = 34548$ as claimed.



    For large $n$, the formula for $p_n$ is dominated by the first term, so $p_n sim frac{1}{20} cdot 3^n$; so, for large $m$, $s_m sim frac{1}{40} cdot 3^{m + 1}$.



    Edit A Christian Blatter observed in his answer, exploiting the symmetry of the dodecahedron translates the problem into a path-counting problem on the digraph $Delta'$ of classes $C_0, ldots, C_5$ of vertices, where $C_k$ consists of the vertices of distance $k$ from $v_0$. This yields the adjacency matrix
    $$A_{Delta'} = pmatrix{
    0&3&0&0&0&0cr
    1&0&2&0&0&0cr
    0&1&1&1&0&0cr
    0&0&1&1&1&0cr
    0&0&0&2&0&1cr
    0&0&0&0&3&0cr}.$$

    The number of paths of length $n$ from $C_0$ (which contains only a single vertex) to $C_5$ (which contains its antipodal vertex) is then $(A_{Delta'}^n)_{05}$. As before, we can compute this efficiently with the Jordan decomposition $A_{Delta'} = Q' J' (Q')^{-1}$, where
    $$J' = operatorname{diag}(3, sqrt{5}, 1, 0, -2, -sqrt{5}), qquad Q' = pmatrix{
    1 & 1 & 1 & 1 & 1 & 1 \
    1 & frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & -frac{1}{3} sqrt{5} \
    1 & frac{1}{3} & -frac{1}{3} & -frac{1}{2} & frac{1}{6} & frac{1}{3} \
    1 & -frac{1}{3} & -frac{1}{3} & frac{1}{2} & frac{1}{6} & -frac{1}{3} \
    1 & -frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & frac{1}{3} sqrt{5} \
    1 & -1 & 1 & -1 & 1 & -1 .
    } .$$






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      I'm still working out a combinatorial method of calculating the answer, and I may not be successful. In the meantime, here are the results from a computer enumeration of all paths:
      $$
      begin{array}{rrr}
      text{Length} & text{Simple paths} & text{All paths} \
      5 & 6 & 6 \
      6 & 12 & 12 \
      7 & 6 & 84 \
      8 & 12 & 192 \
      9 & 30 & 882 \
      10 & 24 & 2;220 \
      11 & 42 & 8;448 \
      12 & 84 & 22;704 \
      13 & 96 & 78;078 \
      14 & 132 & 218;988 \
      15 & 150 & 710;892 \
      16 & 72 & 2;048;256 \
      17 & 48 & 6;430;794 \
      18 & 60 & 18;837;516\
      19 & 6 & 58;008;216\
      hline
      text{Total} & 780 & 86;367;288
      end{array}
      $$



      The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed. For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes. (After computing the $1;042;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84;442;986$, which is quite close to the correct result.)






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      • Out of curiosity, did you ever find a combinatoral method?
        – tox123
        Nov 4 at 21:56






      • 1




        No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
        – MJD
        Nov 4 at 23:15






      • 1




        @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
        – MJD
        Nov 7 at 15:10












      • interestingly, the OEIS doesn't even have the cube sequence.
        – tox123
        Nov 11 at 4:39






      • 1




        @tox123 oeis.org/A054880
        – MJD
        Nov 11 at 5:52











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      up vote
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      Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.



      For example, there are



      171619248



      such paths of length 20 (not too far from MJD's guess), and



      25768876036573921452762172776956776774411837488



      such paths of length 100.






      share|cite|improve this answer



























        up vote
        4
        down vote













        Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.



        For example, there are



        171619248



        such paths of length 20 (not too far from MJD's guess), and



        25768876036573921452762172776956776774411837488



        such paths of length 100.






        share|cite|improve this answer

























          up vote
          4
          down vote










          up vote
          4
          down vote









          Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.



          For example, there are



          171619248



          such paths of length 20 (not too far from MJD's guess), and



          25768876036573921452762172776956776774411837488



          such paths of length 100.






          share|cite|improve this answer














          Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.



          For example, there are



          171619248



          such paths of length 20 (not too far from MJD's guess), and



          25768876036573921452762172776956776774411837488



          such paths of length 100.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 14 at 4:05

























          answered Nov 14 at 3:53









          verret

          2,8701817




          2,8701817






















              up vote
              4
              down vote













              It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0leq ileq5)$ of vertices.



              As described above: there is a single central vertex C_0; arranged symmetrically around this are three vertices C_1, then six vertices C_2, six more C_3, three more C_4, and finally one C_5 which is drawn in three places, attached to the C_5, vertices, but with two in parentheses to indicate that they are not separate vertices.



              It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
              $${bf p}(n+1)=left[matrix
              {0&3&0&0&0&0cr
              1&0&2&0&0&0cr
              0&1&1&1&0&0cr
              0&0&1&1&1&0cr
              0&0&0&2&0&1cr
              0&0&0&0&3&0cr}right]>{bf p}(n) ,$$

              whereby ${bf p}(n)$ denotes the column vector of the $p_i(n)$.



              Since there are so many treatments of the problem already I stop here.






              share|cite|improve this answer



















              • 1




                I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
                – MJD
                Nov 14 at 15:45








              • 1




                @MJD: That's exactly the drawing I had. Thank you very much!!
                – Christian Blatter
                Nov 14 at 16:09










              • This is a nice observation!
                – Travis
                Nov 14 at 21:24















              up vote
              4
              down vote













              It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0leq ileq5)$ of vertices.



              As described above: there is a single central vertex C_0; arranged symmetrically around this are three vertices C_1, then six vertices C_2, six more C_3, three more C_4, and finally one C_5 which is drawn in three places, attached to the C_5, vertices, but with two in parentheses to indicate that they are not separate vertices.



              It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
              $${bf p}(n+1)=left[matrix
              {0&3&0&0&0&0cr
              1&0&2&0&0&0cr
              0&1&1&1&0&0cr
              0&0&1&1&1&0cr
              0&0&0&2&0&1cr
              0&0&0&0&3&0cr}right]>{bf p}(n) ,$$

              whereby ${bf p}(n)$ denotes the column vector of the $p_i(n)$.



              Since there are so many treatments of the problem already I stop here.






              share|cite|improve this answer



















              • 1




                I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
                – MJD
                Nov 14 at 15:45








              • 1




                @MJD: That's exactly the drawing I had. Thank you very much!!
                – Christian Blatter
                Nov 14 at 16:09










              • This is a nice observation!
                – Travis
                Nov 14 at 21:24













              up vote
              4
              down vote










              up vote
              4
              down vote









              It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0leq ileq5)$ of vertices.



              As described above: there is a single central vertex C_0; arranged symmetrically around this are three vertices C_1, then six vertices C_2, six more C_3, three more C_4, and finally one C_5 which is drawn in three places, attached to the C_5, vertices, but with two in parentheses to indicate that they are not separate vertices.



              It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
              $${bf p}(n+1)=left[matrix
              {0&3&0&0&0&0cr
              1&0&2&0&0&0cr
              0&1&1&1&0&0cr
              0&0&1&1&1&0cr
              0&0&0&2&0&1cr
              0&0&0&0&3&0cr}right]>{bf p}(n) ,$$

              whereby ${bf p}(n)$ denotes the column vector of the $p_i(n)$.



              Since there are so many treatments of the problem already I stop here.






              share|cite|improve this answer














              It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0leq ileq5)$ of vertices.



              As described above: there is a single central vertex C_0; arranged symmetrically around this are three vertices C_1, then six vertices C_2, six more C_3, three more C_4, and finally one C_5 which is drawn in three places, attached to the C_5, vertices, but with two in parentheses to indicate that they are not separate vertices.



              It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
              $${bf p}(n+1)=left[matrix
              {0&3&0&0&0&0cr
              1&0&2&0&0&0cr
              0&1&1&1&0&0cr
              0&0&1&1&1&0cr
              0&0&0&2&0&1cr
              0&0&0&0&3&0cr}right]>{bf p}(n) ,$$

              whereby ${bf p}(n)$ denotes the column vector of the $p_i(n)$.



              Since there are so many treatments of the problem already I stop here.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 14 at 15:42









              MJD

              46.6k28205388




              46.6k28205388










              answered Nov 14 at 14:52









              Christian Blatter

              170k7111324




              170k7111324








              • 1




                I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
                – MJD
                Nov 14 at 15:45








              • 1




                @MJD: That's exactly the drawing I had. Thank you very much!!
                – Christian Blatter
                Nov 14 at 16:09










              • This is a nice observation!
                – Travis
                Nov 14 at 21:24














              • 1




                I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
                – MJD
                Nov 14 at 15:45








              • 1




                @MJD: That's exactly the drawing I had. Thank you very much!!
                – Christian Blatter
                Nov 14 at 16:09










              • This is a nice observation!
                – Travis
                Nov 14 at 21:24








              1




              1




              I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
              – MJD
              Nov 14 at 15:45






              I hope it is all right with you that I added a diagram. If you want to change the design, the original SVG image is available for editing.
              – MJD
              Nov 14 at 15:45






              1




              1




              @MJD: That's exactly the drawing I had. Thank you very much!!
              – Christian Blatter
              Nov 14 at 16:09




              @MJD: That's exactly the drawing I had. Thank you very much!!
              – Christian Blatter
              Nov 14 at 16:09












              This is a nice observation!
              – Travis
              Nov 14 at 21:24




              This is a nice observation!
              – Travis
              Nov 14 at 21:24










              up vote
              3
              down vote













              One way to count paths on a graph $Gamma$, say, with vertices $v_i$, is to use its adjacency matrix: This is the square matrix $A_{Gamma}$ of size $|Gamma|$ for which the $(i, j)$ entry $(A_{Gamma})_{ij}$ is the number of edges with endpoints at vertices $v_i$ and $v_j$. An intuitive inductive argument then shows that the number of paths of length $n$ from $v_i$ to $v_j$ is exactly $(A_{Gamma}^n)_{ij}$.



              The SageMath code Delta = graphs.DodecahedralGraph(); Delta.plot() assigns the dodecahedral graph the name Delta and returns the following labeled plot of the graph, which we call $Delta$. NB Sage starts indexing at $0$.
              Dodecahedral graph
              A little visualization shows that, e.g., the vertex opposite $v_0$ is vertex $v_{15}$.



              Next, the Sage command A = dodecahedral.adjacency_matrix() assigns to A the adjacency matrix $A_{Delta}$ of $Delta$; it has size $|Delta| = 20$.



              $$small A_{Delta} = pmatrix{
              0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
              1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
              0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
              0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
              0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \
              0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
              0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
              0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \
              1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \
              0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \
              0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \
              1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
              } .$$



              Using our above combinatorial interpretation of $A_{Delta}^n$ gives that the number of paths of length $n$ from $v_0$ to the opposite vertex $v_{15}$ is $$boxed{p_n = (A_{Delta}^n)_{0, 15}}.$$ Asking Sage to compute $p_1, ldots, p_{12}$ and add them, say, using the code n = var('n'); sum([(A^n)[0, 15] for n in range(1,13)]), gives that
              $$color{#df0000}{boxed{p_1 + cdots + p_{12} = 34548}} ,$$
              which agrees with MJD's count.



              One can derive this in a more illuminating way. To compute $A_{Gamma}^n$ efficiently, it is useful to use the Jordan decomposition $A_{Gamma} = Q J Q^{-1}$. Then, $A_{Gamma}^n = (Q J Q^{-1})^n = Q J^n Q^{-1}$. Since an adjacency matrix (of an undirected graph) is symmetric, $A_{Gamma}$ is diagonalizable and thus $J$ is diagonal, say, $J = operatorname{diag}(lambda_i)$, and $J^n = operatorname{diag}(lambda_i^n)$.



              Now, the number of paths of length $n$ from $v_i$ to $v_j$ is $$(A_{Delta}^n)_{ij} = (Q J^n Q^{-1})_{ij} = sum_{k, ell} Q_{ik} (J^n)_{k ell} Q^{-1}_{ell j} ,$$ and since $J^n$ is diagonal with entries $lambda_a^n$, terms only contribute to the sum when $ell = k$, leaving $$(A_{Delta}^n)_{ij} = sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n .$$ The number of paths from $v_i$ to $v_j$ of length $leq m$ is then the partial sum
              $$sum_{n = 0}^m (A_{Delta}^n)_{ij}
              =sum_{n = 0}^m sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n
              =sum_k (Q_{ik} Q^{-1}_{kj}) sum_{n = 0}^m lambda_k^n
              =sum_k (Q_{ik} Q^{-1}_{kj}) frac{lambda_k^{m + 1} - 1}{lambda_k - 1},$$

              where we interpret $frac{lambda_k^{m + 1} - 1}{lambda_k - 1}$ as $m + 1$ if $lambda_k = 1$.



              For the dodecahedral graph $Delta$, the Jordan decomposition (produced, for example, using A.change_ring(QQ[sqrt(5)]).jordan_form(transformation=True)), is given by



              $$J = operatorname{diag}(3, sqrt{5}, sqrt{5}, sqrt{5}, -sqrt{5}, -sqrt{5}, -sqrt{5}, 0, 0, 0, 0, -2, -2, -2, -2, 1, 1, 1, 1, 1) ,$$
              begin{multline*}Q \ = scriptsize pmatrix{
              1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
              1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
              1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
              1 & 1 & -phi & phi & 1 & phi^{-1} & -phi & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
              1 & phi^{-1} & -2 & phi & -phi & -2 & -phi & 1 & 1 & 0 & 1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
              1 & -phi & -phi & 1 & phi & phi & 1 & -1 & 0 & -1 & -1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
              1 & -1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & 0 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
              1 & -phi & 2 & -phi & phi^{-1} & 2 & phi & 1 & 0 & 0 & 1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
              1 & -1 & sqrt{5} & -1 & -1 & -sqrt{5} & -1 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
              1 & -phi & 2 & -phi & phi & 2 & phi^{-1} & -1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
              1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & -1 & 1 & 0 & 1 & 1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
              1 & 1 & -phi & -phi & 1 & phi & phi & 0 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
              1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
              1 & -1 & phi & -phi & -1 & -phi & phi^{-1} & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
              1 & -phi & phi & -1 & phi^{-1} & -phi & -1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1 \
              1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
              1 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
              1 & 1 & -sqrt{5} & 1 & 1 & sqrt{5} & 1 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
              1 & phi & -2 & phi^{-1} & -phi & -2 & -phi & -1 & 0 & 0 & -1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
              1 & phi & -phi & 1 & -phi & phi^{-1} & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1} .end{multline*}

              Here, $phi := frac{1}{2}(1 + sqrt{5})$ is the Golden Ratio.



              Taking $i = 0, j = 15$ and substituting (or just executing (A^n)[0, 15]) gives
              $$boxed{p_n = (A_{Delta}^n)_{0, 15} = frac{1}{20} cdot 3^n - frac{3}{20} (1 + (-1)^n) (sqrt{5})^n + frac{1}{5} , left(-2right)^n + frac{1}{4}}.$$
              (This agrees with the formula given in OEIS A054883, Number of walks of length n along the edges of a dodecahedron between two opposite vertices.)
              Then, the number of such paths of length $leq m$ is the partial sum $s_m = sum_{n = 0}^m p_n$, and indeed, $s_{12} = 34548$ as claimed.



              For large $n$, the formula for $p_n$ is dominated by the first term, so $p_n sim frac{1}{20} cdot 3^n$; so, for large $m$, $s_m sim frac{1}{40} cdot 3^{m + 1}$.



              Edit A Christian Blatter observed in his answer, exploiting the symmetry of the dodecahedron translates the problem into a path-counting problem on the digraph $Delta'$ of classes $C_0, ldots, C_5$ of vertices, where $C_k$ consists of the vertices of distance $k$ from $v_0$. This yields the adjacency matrix
              $$A_{Delta'} = pmatrix{
              0&3&0&0&0&0cr
              1&0&2&0&0&0cr
              0&1&1&1&0&0cr
              0&0&1&1&1&0cr
              0&0&0&2&0&1cr
              0&0&0&0&3&0cr}.$$

              The number of paths of length $n$ from $C_0$ (which contains only a single vertex) to $C_5$ (which contains its antipodal vertex) is then $(A_{Delta'}^n)_{05}$. As before, we can compute this efficiently with the Jordan decomposition $A_{Delta'} = Q' J' (Q')^{-1}$, where
              $$J' = operatorname{diag}(3, sqrt{5}, 1, 0, -2, -sqrt{5}), qquad Q' = pmatrix{
              1 & 1 & 1 & 1 & 1 & 1 \
              1 & frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & -frac{1}{3} sqrt{5} \
              1 & frac{1}{3} & -frac{1}{3} & -frac{1}{2} & frac{1}{6} & frac{1}{3} \
              1 & -frac{1}{3} & -frac{1}{3} & frac{1}{2} & frac{1}{6} & -frac{1}{3} \
              1 & -frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & frac{1}{3} sqrt{5} \
              1 & -1 & 1 & -1 & 1 & -1 .
              } .$$






              share|cite|improve this answer



























                up vote
                3
                down vote













                One way to count paths on a graph $Gamma$, say, with vertices $v_i$, is to use its adjacency matrix: This is the square matrix $A_{Gamma}$ of size $|Gamma|$ for which the $(i, j)$ entry $(A_{Gamma})_{ij}$ is the number of edges with endpoints at vertices $v_i$ and $v_j$. An intuitive inductive argument then shows that the number of paths of length $n$ from $v_i$ to $v_j$ is exactly $(A_{Gamma}^n)_{ij}$.



                The SageMath code Delta = graphs.DodecahedralGraph(); Delta.plot() assigns the dodecahedral graph the name Delta and returns the following labeled plot of the graph, which we call $Delta$. NB Sage starts indexing at $0$.
                Dodecahedral graph
                A little visualization shows that, e.g., the vertex opposite $v_0$ is vertex $v_{15}$.



                Next, the Sage command A = dodecahedral.adjacency_matrix() assigns to A the adjacency matrix $A_{Delta}$ of $Delta$; it has size $|Delta| = 20$.



                $$small A_{Delta} = pmatrix{
                0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \
                1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \
                0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \
                0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \
                1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
                } .$$



                Using our above combinatorial interpretation of $A_{Delta}^n$ gives that the number of paths of length $n$ from $v_0$ to the opposite vertex $v_{15}$ is $$boxed{p_n = (A_{Delta}^n)_{0, 15}}.$$ Asking Sage to compute $p_1, ldots, p_{12}$ and add them, say, using the code n = var('n'); sum([(A^n)[0, 15] for n in range(1,13)]), gives that
                $$color{#df0000}{boxed{p_1 + cdots + p_{12} = 34548}} ,$$
                which agrees with MJD's count.



                One can derive this in a more illuminating way. To compute $A_{Gamma}^n$ efficiently, it is useful to use the Jordan decomposition $A_{Gamma} = Q J Q^{-1}$. Then, $A_{Gamma}^n = (Q J Q^{-1})^n = Q J^n Q^{-1}$. Since an adjacency matrix (of an undirected graph) is symmetric, $A_{Gamma}$ is diagonalizable and thus $J$ is diagonal, say, $J = operatorname{diag}(lambda_i)$, and $J^n = operatorname{diag}(lambda_i^n)$.



                Now, the number of paths of length $n$ from $v_i$ to $v_j$ is $$(A_{Delta}^n)_{ij} = (Q J^n Q^{-1})_{ij} = sum_{k, ell} Q_{ik} (J^n)_{k ell} Q^{-1}_{ell j} ,$$ and since $J^n$ is diagonal with entries $lambda_a^n$, terms only contribute to the sum when $ell = k$, leaving $$(A_{Delta}^n)_{ij} = sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n .$$ The number of paths from $v_i$ to $v_j$ of length $leq m$ is then the partial sum
                $$sum_{n = 0}^m (A_{Delta}^n)_{ij}
                =sum_{n = 0}^m sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n
                =sum_k (Q_{ik} Q^{-1}_{kj}) sum_{n = 0}^m lambda_k^n
                =sum_k (Q_{ik} Q^{-1}_{kj}) frac{lambda_k^{m + 1} - 1}{lambda_k - 1},$$

                where we interpret $frac{lambda_k^{m + 1} - 1}{lambda_k - 1}$ as $m + 1$ if $lambda_k = 1$.



                For the dodecahedral graph $Delta$, the Jordan decomposition (produced, for example, using A.change_ring(QQ[sqrt(5)]).jordan_form(transformation=True)), is given by



                $$J = operatorname{diag}(3, sqrt{5}, sqrt{5}, sqrt{5}, -sqrt{5}, -sqrt{5}, -sqrt{5}, 0, 0, 0, 0, -2, -2, -2, -2, 1, 1, 1, 1, 1) ,$$
                begin{multline*}Q \ = scriptsize pmatrix{
                1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                1 & 1 & -phi & phi & 1 & phi^{-1} & -phi & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                1 & phi^{-1} & -2 & phi & -phi & -2 & -phi & 1 & 1 & 0 & 1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                1 & -phi & -phi & 1 & phi & phi & 1 & -1 & 0 & -1 & -1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                1 & -1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & 0 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                1 & -phi & 2 & -phi & phi^{-1} & 2 & phi & 1 & 0 & 0 & 1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                1 & -1 & sqrt{5} & -1 & -1 & -sqrt{5} & -1 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                1 & -phi & 2 & -phi & phi & 2 & phi^{-1} & -1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & -1 & 1 & 0 & 1 & 1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                1 & 1 & -phi & -phi & 1 & phi & phi & 0 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                1 & -1 & phi & -phi & -1 & -phi & phi^{-1} & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                1 & -phi & phi & -1 & phi^{-1} & -phi & -1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1 \
                1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                1 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                1 & 1 & -sqrt{5} & 1 & 1 & sqrt{5} & 1 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                1 & phi & -2 & phi^{-1} & -phi & -2 & -phi & -1 & 0 & 0 & -1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                1 & phi & -phi & 1 & -phi & phi^{-1} & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1} .end{multline*}

                Here, $phi := frac{1}{2}(1 + sqrt{5})$ is the Golden Ratio.



                Taking $i = 0, j = 15$ and substituting (or just executing (A^n)[0, 15]) gives
                $$boxed{p_n = (A_{Delta}^n)_{0, 15} = frac{1}{20} cdot 3^n - frac{3}{20} (1 + (-1)^n) (sqrt{5})^n + frac{1}{5} , left(-2right)^n + frac{1}{4}}.$$
                (This agrees with the formula given in OEIS A054883, Number of walks of length n along the edges of a dodecahedron between two opposite vertices.)
                Then, the number of such paths of length $leq m$ is the partial sum $s_m = sum_{n = 0}^m p_n$, and indeed, $s_{12} = 34548$ as claimed.



                For large $n$, the formula for $p_n$ is dominated by the first term, so $p_n sim frac{1}{20} cdot 3^n$; so, for large $m$, $s_m sim frac{1}{40} cdot 3^{m + 1}$.



                Edit A Christian Blatter observed in his answer, exploiting the symmetry of the dodecahedron translates the problem into a path-counting problem on the digraph $Delta'$ of classes $C_0, ldots, C_5$ of vertices, where $C_k$ consists of the vertices of distance $k$ from $v_0$. This yields the adjacency matrix
                $$A_{Delta'} = pmatrix{
                0&3&0&0&0&0cr
                1&0&2&0&0&0cr
                0&1&1&1&0&0cr
                0&0&1&1&1&0cr
                0&0&0&2&0&1cr
                0&0&0&0&3&0cr}.$$

                The number of paths of length $n$ from $C_0$ (which contains only a single vertex) to $C_5$ (which contains its antipodal vertex) is then $(A_{Delta'}^n)_{05}$. As before, we can compute this efficiently with the Jordan decomposition $A_{Delta'} = Q' J' (Q')^{-1}$, where
                $$J' = operatorname{diag}(3, sqrt{5}, 1, 0, -2, -sqrt{5}), qquad Q' = pmatrix{
                1 & 1 & 1 & 1 & 1 & 1 \
                1 & frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & -frac{1}{3} sqrt{5} \
                1 & frac{1}{3} & -frac{1}{3} & -frac{1}{2} & frac{1}{6} & frac{1}{3} \
                1 & -frac{1}{3} & -frac{1}{3} & frac{1}{2} & frac{1}{6} & -frac{1}{3} \
                1 & -frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & frac{1}{3} sqrt{5} \
                1 & -1 & 1 & -1 & 1 & -1 .
                } .$$






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                  One way to count paths on a graph $Gamma$, say, with vertices $v_i$, is to use its adjacency matrix: This is the square matrix $A_{Gamma}$ of size $|Gamma|$ for which the $(i, j)$ entry $(A_{Gamma})_{ij}$ is the number of edges with endpoints at vertices $v_i$ and $v_j$. An intuitive inductive argument then shows that the number of paths of length $n$ from $v_i$ to $v_j$ is exactly $(A_{Gamma}^n)_{ij}$.



                  The SageMath code Delta = graphs.DodecahedralGraph(); Delta.plot() assigns the dodecahedral graph the name Delta and returns the following labeled plot of the graph, which we call $Delta$. NB Sage starts indexing at $0$.
                  Dodecahedral graph
                  A little visualization shows that, e.g., the vertex opposite $v_0$ is vertex $v_{15}$.



                  Next, the Sage command A = dodecahedral.adjacency_matrix() assigns to A the adjacency matrix $A_{Delta}$ of $Delta$; it has size $|Delta| = 20$.



                  $$small A_{Delta} = pmatrix{
                  0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                  1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                  0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \
                  1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \
                  1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
                  } .$$



                  Using our above combinatorial interpretation of $A_{Delta}^n$ gives that the number of paths of length $n$ from $v_0$ to the opposite vertex $v_{15}$ is $$boxed{p_n = (A_{Delta}^n)_{0, 15}}.$$ Asking Sage to compute $p_1, ldots, p_{12}$ and add them, say, using the code n = var('n'); sum([(A^n)[0, 15] for n in range(1,13)]), gives that
                  $$color{#df0000}{boxed{p_1 + cdots + p_{12} = 34548}} ,$$
                  which agrees with MJD's count.



                  One can derive this in a more illuminating way. To compute $A_{Gamma}^n$ efficiently, it is useful to use the Jordan decomposition $A_{Gamma} = Q J Q^{-1}$. Then, $A_{Gamma}^n = (Q J Q^{-1})^n = Q J^n Q^{-1}$. Since an adjacency matrix (of an undirected graph) is symmetric, $A_{Gamma}$ is diagonalizable and thus $J$ is diagonal, say, $J = operatorname{diag}(lambda_i)$, and $J^n = operatorname{diag}(lambda_i^n)$.



                  Now, the number of paths of length $n$ from $v_i$ to $v_j$ is $$(A_{Delta}^n)_{ij} = (Q J^n Q^{-1})_{ij} = sum_{k, ell} Q_{ik} (J^n)_{k ell} Q^{-1}_{ell j} ,$$ and since $J^n$ is diagonal with entries $lambda_a^n$, terms only contribute to the sum when $ell = k$, leaving $$(A_{Delta}^n)_{ij} = sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n .$$ The number of paths from $v_i$ to $v_j$ of length $leq m$ is then the partial sum
                  $$sum_{n = 0}^m (A_{Delta}^n)_{ij}
                  =sum_{n = 0}^m sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n
                  =sum_k (Q_{ik} Q^{-1}_{kj}) sum_{n = 0}^m lambda_k^n
                  =sum_k (Q_{ik} Q^{-1}_{kj}) frac{lambda_k^{m + 1} - 1}{lambda_k - 1},$$

                  where we interpret $frac{lambda_k^{m + 1} - 1}{lambda_k - 1}$ as $m + 1$ if $lambda_k = 1$.



                  For the dodecahedral graph $Delta$, the Jordan decomposition (produced, for example, using A.change_ring(QQ[sqrt(5)]).jordan_form(transformation=True)), is given by



                  $$J = operatorname{diag}(3, sqrt{5}, sqrt{5}, sqrt{5}, -sqrt{5}, -sqrt{5}, -sqrt{5}, 0, 0, 0, 0, -2, -2, -2, -2, 1, 1, 1, 1, 1) ,$$
                  begin{multline*}Q \ = scriptsize pmatrix{
                  1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                  1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                  1 & 1 & -phi & phi & 1 & phi^{-1} & -phi & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                  1 & phi^{-1} & -2 & phi & -phi & -2 & -phi & 1 & 1 & 0 & 1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                  1 & -phi & -phi & 1 & phi & phi & 1 & -1 & 0 & -1 & -1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                  1 & -1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & 0 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                  1 & -phi & 2 & -phi & phi^{-1} & 2 & phi & 1 & 0 & 0 & 1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                  1 & -1 & sqrt{5} & -1 & -1 & -sqrt{5} & -1 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                  1 & -phi & 2 & -phi & phi & 2 & phi^{-1} & -1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                  1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & -1 & 1 & 0 & 1 & 1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                  1 & 1 & -phi & -phi & 1 & phi & phi & 0 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                  1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                  1 & -1 & phi & -phi & -1 & -phi & phi^{-1} & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                  1 & -phi & phi & -1 & phi^{-1} & -phi & -1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1 \
                  1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  1 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                  1 & 1 & -sqrt{5} & 1 & 1 & sqrt{5} & 1 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                  1 & phi & -2 & phi^{-1} & -phi & -2 & -phi & -1 & 0 & 0 & -1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                  1 & phi & -phi & 1 & -phi & phi^{-1} & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1} .end{multline*}

                  Here, $phi := frac{1}{2}(1 + sqrt{5})$ is the Golden Ratio.



                  Taking $i = 0, j = 15$ and substituting (or just executing (A^n)[0, 15]) gives
                  $$boxed{p_n = (A_{Delta}^n)_{0, 15} = frac{1}{20} cdot 3^n - frac{3}{20} (1 + (-1)^n) (sqrt{5})^n + frac{1}{5} , left(-2right)^n + frac{1}{4}}.$$
                  (This agrees with the formula given in OEIS A054883, Number of walks of length n along the edges of a dodecahedron between two opposite vertices.)
                  Then, the number of such paths of length $leq m$ is the partial sum $s_m = sum_{n = 0}^m p_n$, and indeed, $s_{12} = 34548$ as claimed.



                  For large $n$, the formula for $p_n$ is dominated by the first term, so $p_n sim frac{1}{20} cdot 3^n$; so, for large $m$, $s_m sim frac{1}{40} cdot 3^{m + 1}$.



                  Edit A Christian Blatter observed in his answer, exploiting the symmetry of the dodecahedron translates the problem into a path-counting problem on the digraph $Delta'$ of classes $C_0, ldots, C_5$ of vertices, where $C_k$ consists of the vertices of distance $k$ from $v_0$. This yields the adjacency matrix
                  $$A_{Delta'} = pmatrix{
                  0&3&0&0&0&0cr
                  1&0&2&0&0&0cr
                  0&1&1&1&0&0cr
                  0&0&1&1&1&0cr
                  0&0&0&2&0&1cr
                  0&0&0&0&3&0cr}.$$

                  The number of paths of length $n$ from $C_0$ (which contains only a single vertex) to $C_5$ (which contains its antipodal vertex) is then $(A_{Delta'}^n)_{05}$. As before, we can compute this efficiently with the Jordan decomposition $A_{Delta'} = Q' J' (Q')^{-1}$, where
                  $$J' = operatorname{diag}(3, sqrt{5}, 1, 0, -2, -sqrt{5}), qquad Q' = pmatrix{
                  1 & 1 & 1 & 1 & 1 & 1 \
                  1 & frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & -frac{1}{3} sqrt{5} \
                  1 & frac{1}{3} & -frac{1}{3} & -frac{1}{2} & frac{1}{6} & frac{1}{3} \
                  1 & -frac{1}{3} & -frac{1}{3} & frac{1}{2} & frac{1}{6} & -frac{1}{3} \
                  1 & -frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & frac{1}{3} sqrt{5} \
                  1 & -1 & 1 & -1 & 1 & -1 .
                  } .$$






                  share|cite|improve this answer














                  One way to count paths on a graph $Gamma$, say, with vertices $v_i$, is to use its adjacency matrix: This is the square matrix $A_{Gamma}$ of size $|Gamma|$ for which the $(i, j)$ entry $(A_{Gamma})_{ij}$ is the number of edges with endpoints at vertices $v_i$ and $v_j$. An intuitive inductive argument then shows that the number of paths of length $n$ from $v_i$ to $v_j$ is exactly $(A_{Gamma}^n)_{ij}$.



                  The SageMath code Delta = graphs.DodecahedralGraph(); Delta.plot() assigns the dodecahedral graph the name Delta and returns the following labeled plot of the graph, which we call $Delta$. NB Sage starts indexing at $0$.
                  Dodecahedral graph
                  A little visualization shows that, e.g., the vertex opposite $v_0$ is vertex $v_{15}$.



                  Next, the Sage command A = dodecahedral.adjacency_matrix() assigns to A the adjacency matrix $A_{Delta}$ of $Delta$; it has size $|Delta| = 20$.



                  $$small A_{Delta} = pmatrix{
                  0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                  1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \
                  0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                  0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \
                  1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \
                  0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \
                  0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \
                  1 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
                  } .$$



                  Using our above combinatorial interpretation of $A_{Delta}^n$ gives that the number of paths of length $n$ from $v_0$ to the opposite vertex $v_{15}$ is $$boxed{p_n = (A_{Delta}^n)_{0, 15}}.$$ Asking Sage to compute $p_1, ldots, p_{12}$ and add them, say, using the code n = var('n'); sum([(A^n)[0, 15] for n in range(1,13)]), gives that
                  $$color{#df0000}{boxed{p_1 + cdots + p_{12} = 34548}} ,$$
                  which agrees with MJD's count.



                  One can derive this in a more illuminating way. To compute $A_{Gamma}^n$ efficiently, it is useful to use the Jordan decomposition $A_{Gamma} = Q J Q^{-1}$. Then, $A_{Gamma}^n = (Q J Q^{-1})^n = Q J^n Q^{-1}$. Since an adjacency matrix (of an undirected graph) is symmetric, $A_{Gamma}$ is diagonalizable and thus $J$ is diagonal, say, $J = operatorname{diag}(lambda_i)$, and $J^n = operatorname{diag}(lambda_i^n)$.



                  Now, the number of paths of length $n$ from $v_i$ to $v_j$ is $$(A_{Delta}^n)_{ij} = (Q J^n Q^{-1})_{ij} = sum_{k, ell} Q_{ik} (J^n)_{k ell} Q^{-1}_{ell j} ,$$ and since $J^n$ is diagonal with entries $lambda_a^n$, terms only contribute to the sum when $ell = k$, leaving $$(A_{Delta}^n)_{ij} = sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n .$$ The number of paths from $v_i$ to $v_j$ of length $leq m$ is then the partial sum
                  $$sum_{n = 0}^m (A_{Delta}^n)_{ij}
                  =sum_{n = 0}^m sum_k (Q_{ik} Q^{-1}_{kj}) lambda_k^n
                  =sum_k (Q_{ik} Q^{-1}_{kj}) sum_{n = 0}^m lambda_k^n
                  =sum_k (Q_{ik} Q^{-1}_{kj}) frac{lambda_k^{m + 1} - 1}{lambda_k - 1},$$

                  where we interpret $frac{lambda_k^{m + 1} - 1}{lambda_k - 1}$ as $m + 1$ if $lambda_k = 1$.



                  For the dodecahedral graph $Delta$, the Jordan decomposition (produced, for example, using A.change_ring(QQ[sqrt(5)]).jordan_form(transformation=True)), is given by



                  $$J = operatorname{diag}(3, sqrt{5}, sqrt{5}, sqrt{5}, -sqrt{5}, -sqrt{5}, -sqrt{5}, 0, 0, 0, 0, -2, -2, -2, -2, 1, 1, 1, 1, 1) ,$$
                  begin{multline*}Q \ = scriptsize pmatrix{
                  1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                  1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                  1 & 1 & -phi & phi & 1 & phi^{-1} & -phi & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                  1 & phi^{-1} & -2 & phi & -phi & -2 & -phi & 1 & 1 & 0 & 1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                  1 & -phi & -phi & 1 & phi & phi & 1 & -1 & 0 & -1 & -1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                  1 & -1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & 0 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                  1 & -phi & 2 & -phi & phi^{-1} & 2 & phi & 1 & 0 & 0 & 1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                  1 & -1 & sqrt{5} & -1 & -1 & -sqrt{5} & -1 & -1 & 0 & -1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                  1 & -phi & 2 & -phi & phi & 2 & phi^{-1} & -1 & -1 & 0 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 & 0 & 1 \
                  1 & phi^{-1} & phi^{-1} & -1 & -phi & -phi & -1 & 1 & 0 & 1 & 1 & -1 & 0 & 1 & 1 & 1 & -1 & 1 & -1 & 1 \
                  1 & 1 & -phi & -phi & 1 & phi & phi & 0 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 \
                  1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \
                  1 & -1 & phi & -phi & -1 & -phi & phi^{-1} & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 \
                  1 & -phi & phi & -1 & phi^{-1} & -phi & -1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1 \
                  1 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \
                  1 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & -1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \
                  1 & 1 & -sqrt{5} & 1 & 1 & sqrt{5} & 1 & 1 & 0 & 1 & 0 & -1 & -2 & -1 & 0 & -1 & 1 & -1 & 0 & 0 \
                  1 & phi & -2 & phi^{-1} & -phi & -2 & -phi & -1 & 0 & 0 & -1 & 1 & 2 & 2 & 1 & -1 & 0 & -1 & 0 & -1 \
                  1 & phi & -phi & 1 & -phi & phi^{-1} & 1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & -1 & 0 & 0 & -1 & 1 & -1} .end{multline*}

                  Here, $phi := frac{1}{2}(1 + sqrt{5})$ is the Golden Ratio.



                  Taking $i = 0, j = 15$ and substituting (or just executing (A^n)[0, 15]) gives
                  $$boxed{p_n = (A_{Delta}^n)_{0, 15} = frac{1}{20} cdot 3^n - frac{3}{20} (1 + (-1)^n) (sqrt{5})^n + frac{1}{5} , left(-2right)^n + frac{1}{4}}.$$
                  (This agrees with the formula given in OEIS A054883, Number of walks of length n along the edges of a dodecahedron between two opposite vertices.)
                  Then, the number of such paths of length $leq m$ is the partial sum $s_m = sum_{n = 0}^m p_n$, and indeed, $s_{12} = 34548$ as claimed.



                  For large $n$, the formula for $p_n$ is dominated by the first term, so $p_n sim frac{1}{20} cdot 3^n$; so, for large $m$, $s_m sim frac{1}{40} cdot 3^{m + 1}$.



                  Edit A Christian Blatter observed in his answer, exploiting the symmetry of the dodecahedron translates the problem into a path-counting problem on the digraph $Delta'$ of classes $C_0, ldots, C_5$ of vertices, where $C_k$ consists of the vertices of distance $k$ from $v_0$. This yields the adjacency matrix
                  $$A_{Delta'} = pmatrix{
                  0&3&0&0&0&0cr
                  1&0&2&0&0&0cr
                  0&1&1&1&0&0cr
                  0&0&1&1&1&0cr
                  0&0&0&2&0&1cr
                  0&0&0&0&3&0cr}.$$

                  The number of paths of length $n$ from $C_0$ (which contains only a single vertex) to $C_5$ (which contains its antipodal vertex) is then $(A_{Delta'}^n)_{05}$. As before, we can compute this efficiently with the Jordan decomposition $A_{Delta'} = Q' J' (Q')^{-1}$, where
                  $$J' = operatorname{diag}(3, sqrt{5}, 1, 0, -2, -sqrt{5}), qquad Q' = pmatrix{
                  1 & 1 & 1 & 1 & 1 & 1 \
                  1 & frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & -frac{1}{3} sqrt{5} \
                  1 & frac{1}{3} & -frac{1}{3} & -frac{1}{2} & frac{1}{6} & frac{1}{3} \
                  1 & -frac{1}{3} & -frac{1}{3} & frac{1}{2} & frac{1}{6} & -frac{1}{3} \
                  1 & -frac{1}{3} sqrt{5} & frac{1}{3} & 0 & -frac{2}{3} & frac{1}{3} sqrt{5} \
                  1 & -1 & 1 & -1 & 1 & -1 .
                  } .$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 14 at 21:51

























                  answered Nov 14 at 5:34









                  Travis

                  58.7k765142




                  58.7k765142






















                      up vote
                      2
                      down vote













                      I'm still working out a combinatorial method of calculating the answer, and I may not be successful. In the meantime, here are the results from a computer enumeration of all paths:
                      $$
                      begin{array}{rrr}
                      text{Length} & text{Simple paths} & text{All paths} \
                      5 & 6 & 6 \
                      6 & 12 & 12 \
                      7 & 6 & 84 \
                      8 & 12 & 192 \
                      9 & 30 & 882 \
                      10 & 24 & 2;220 \
                      11 & 42 & 8;448 \
                      12 & 84 & 22;704 \
                      13 & 96 & 78;078 \
                      14 & 132 & 218;988 \
                      15 & 150 & 710;892 \
                      16 & 72 & 2;048;256 \
                      17 & 48 & 6;430;794 \
                      18 & 60 & 18;837;516\
                      19 & 6 & 58;008;216\
                      hline
                      text{Total} & 780 & 86;367;288
                      end{array}
                      $$



                      The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed. For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes. (After computing the $1;042;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84;442;986$, which is quite close to the correct result.)






                      share|cite|improve this answer























                      • Out of curiosity, did you ever find a combinatoral method?
                        – tox123
                        Nov 4 at 21:56






                      • 1




                        No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                        – MJD
                        Nov 4 at 23:15






                      • 1




                        @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                        – MJD
                        Nov 7 at 15:10












                      • interestingly, the OEIS doesn't even have the cube sequence.
                        – tox123
                        Nov 11 at 4:39






                      • 1




                        @tox123 oeis.org/A054880
                        – MJD
                        Nov 11 at 5:52















                      up vote
                      2
                      down vote













                      I'm still working out a combinatorial method of calculating the answer, and I may not be successful. In the meantime, here are the results from a computer enumeration of all paths:
                      $$
                      begin{array}{rrr}
                      text{Length} & text{Simple paths} & text{All paths} \
                      5 & 6 & 6 \
                      6 & 12 & 12 \
                      7 & 6 & 84 \
                      8 & 12 & 192 \
                      9 & 30 & 882 \
                      10 & 24 & 2;220 \
                      11 & 42 & 8;448 \
                      12 & 84 & 22;704 \
                      13 & 96 & 78;078 \
                      14 & 132 & 218;988 \
                      15 & 150 & 710;892 \
                      16 & 72 & 2;048;256 \
                      17 & 48 & 6;430;794 \
                      18 & 60 & 18;837;516\
                      19 & 6 & 58;008;216\
                      hline
                      text{Total} & 780 & 86;367;288
                      end{array}
                      $$



                      The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed. For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes. (After computing the $1;042;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84;442;986$, which is quite close to the correct result.)






                      share|cite|improve this answer























                      • Out of curiosity, did you ever find a combinatoral method?
                        – tox123
                        Nov 4 at 21:56






                      • 1




                        No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                        – MJD
                        Nov 4 at 23:15






                      • 1




                        @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                        – MJD
                        Nov 7 at 15:10












                      • interestingly, the OEIS doesn't even have the cube sequence.
                        – tox123
                        Nov 11 at 4:39






                      • 1




                        @tox123 oeis.org/A054880
                        – MJD
                        Nov 11 at 5:52













                      up vote
                      2
                      down vote










                      up vote
                      2
                      down vote









                      I'm still working out a combinatorial method of calculating the answer, and I may not be successful. In the meantime, here are the results from a computer enumeration of all paths:
                      $$
                      begin{array}{rrr}
                      text{Length} & text{Simple paths} & text{All paths} \
                      5 & 6 & 6 \
                      6 & 12 & 12 \
                      7 & 6 & 84 \
                      8 & 12 & 192 \
                      9 & 30 & 882 \
                      10 & 24 & 2;220 \
                      11 & 42 & 8;448 \
                      12 & 84 & 22;704 \
                      13 & 96 & 78;078 \
                      14 & 132 & 218;988 \
                      15 & 150 & 710;892 \
                      16 & 72 & 2;048;256 \
                      17 & 48 & 6;430;794 \
                      18 & 60 & 18;837;516\
                      19 & 6 & 58;008;216\
                      hline
                      text{Total} & 780 & 86;367;288
                      end{array}
                      $$



                      The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed. For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes. (After computing the $1;042;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84;442;986$, which is quite close to the correct result.)






                      share|cite|improve this answer














                      I'm still working out a combinatorial method of calculating the answer, and I may not be successful. In the meantime, here are the results from a computer enumeration of all paths:
                      $$
                      begin{array}{rrr}
                      text{Length} & text{Simple paths} & text{All paths} \
                      5 & 6 & 6 \
                      6 & 12 & 12 \
                      7 & 6 & 84 \
                      8 & 12 & 192 \
                      9 & 30 & 882 \
                      10 & 24 & 2;220 \
                      11 & 42 & 8;448 \
                      12 & 84 & 22;704 \
                      13 & 96 & 78;078 \
                      14 & 132 & 218;988 \
                      15 & 150 & 710;892 \
                      16 & 72 & 2;048;256 \
                      17 & 48 & 6;430;794 \
                      18 & 60 & 18;837;516\
                      19 & 6 & 58;008;216\
                      hline
                      text{Total} & 780 & 86;367;288
                      end{array}
                      $$



                      The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed. For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes. (After computing the $1;042;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84;442;986$, which is quite close to the correct result.)







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Feb 3 at 18:26


























                      community wiki





                      2 revs
                      MJD













                      • Out of curiosity, did you ever find a combinatoral method?
                        – tox123
                        Nov 4 at 21:56






                      • 1




                        No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                        – MJD
                        Nov 4 at 23:15






                      • 1




                        @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                        – MJD
                        Nov 7 at 15:10












                      • interestingly, the OEIS doesn't even have the cube sequence.
                        – tox123
                        Nov 11 at 4:39






                      • 1




                        @tox123 oeis.org/A054880
                        – MJD
                        Nov 11 at 5:52


















                      • Out of curiosity, did you ever find a combinatoral method?
                        – tox123
                        Nov 4 at 21:56






                      • 1




                        No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                        – MJD
                        Nov 4 at 23:15






                      • 1




                        @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                        – MJD
                        Nov 7 at 15:10












                      • interestingly, the OEIS doesn't even have the cube sequence.
                        – tox123
                        Nov 11 at 4:39






                      • 1




                        @tox123 oeis.org/A054880
                        – MJD
                        Nov 11 at 5:52
















                      Out of curiosity, did you ever find a combinatoral method?
                      – tox123
                      Nov 4 at 21:56




                      Out of curiosity, did you ever find a combinatoral method?
                      – tox123
                      Nov 4 at 21:56




                      1




                      1




                      No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                      – MJD
                      Nov 4 at 23:15




                      No, I moved on to other things. Thanks for the reminder; I might take it up again. I love thinking about dodecahedra.
                      – MJD
                      Nov 4 at 23:15




                      1




                      1




                      @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                      – MJD
                      Nov 7 at 15:10






                      @tox123I took another shot at it this morning. I did find a method that works well for the tetrahedron and the cube. (For a cube, I think there are 6, 0, 60, 0, 546 paths of length 3, 4, 5, 6, 7 between a vertex and its antipode.) But the method I was using turns out to be more difficult for the dodecahedron. I will consider further.
                      – MJD
                      Nov 7 at 15:10














                      interestingly, the OEIS doesn't even have the cube sequence.
                      – tox123
                      Nov 11 at 4:39




                      interestingly, the OEIS doesn't even have the cube sequence.
                      – tox123
                      Nov 11 at 4:39




                      1




                      1




                      @tox123 oeis.org/A054880
                      – MJD
                      Nov 11 at 5:52




                      @tox123 oeis.org/A054880
                      – MJD
                      Nov 11 at 5:52


















                       

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