Visual example of an infinite planar graph with degree sequence $(4^4,6^infty)$












3












$begingroup$


After reading some graph theory and talking with experts, I was intrigued. I would like to construct and visualise an infinite planar graph with degree sequence:
$$D=(4^4,6^infty)$$
where the superscripts denote the number of vertices with that degree.



What is the number of graphs with this degree sequence?



Thanks in advance.










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    After reading some graph theory and talking with experts, I was intrigued. I would like to construct and visualise an infinite planar graph with degree sequence:
    $$D=(4^4,6^infty)$$
    where the superscripts denote the number of vertices with that degree.



    What is the number of graphs with this degree sequence?



    Thanks in advance.










    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      After reading some graph theory and talking with experts, I was intrigued. I would like to construct and visualise an infinite planar graph with degree sequence:
      $$D=(4^4,6^infty)$$
      where the superscripts denote the number of vertices with that degree.



      What is the number of graphs with this degree sequence?



      Thanks in advance.










      share|cite|improve this question











      $endgroup$




      After reading some graph theory and talking with experts, I was intrigued. I would like to construct and visualise an infinite planar graph with degree sequence:
      $$D=(4^4,6^infty)$$
      where the superscripts denote the number of vertices with that degree.



      What is the number of graphs with this degree sequence?



      Thanks in advance.







      graph-theory visualization infinite-graphs






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 3:49









      Blue

      47.9k870153




      47.9k870153










      asked Dec 18 '18 at 2:23









      UltradarkUltradark

      1791517




      1791517






















          4 Answers
          4






          active

          oldest

          votes


















          5












          $begingroup$

          There are infinitely many graphs with this degree sequence.



          First, here is one possible graph with degree sequence $(6^infty)$: an infinite triangular lattice.



          triangular lattice



          One way to add four vertices of degree $4$ is to pick two places where we "split up" a degree-$6$ vertex into two adjacent degree-$4$ vertices:



          modified triangular lattice



          There are other ways to make a graph with degree sequence $(4^4,6^infty)$, but this operation can already be done in infinitely many non-isomorphic ways, since we can put the two pairs of degree-$4$ vertices arbitrarily far apart in the infinite lattice.



          (And there are also other graphs we can start with that have degree sequence $(6^infty)$, this one is just the nicest.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 3:28



















          5












          $begingroup$

          It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)



          For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^infty)$. The vertex set is the set of all finite sequences $langle x_1,x_2,dots,x_nrangle$ (including the null sequence) with $x_iin{0,1,2,3,4}$, and with an edge joining $langle x_1,x_2,dots,x_nrangle$ to $langle x_1,x_2,dots,x_n,x_{n+1}rangle$. The null sequence has degree $5$, the rest have degree $6$.



          By imposing the further conditions $x_1ne0$ and $x_1in{1,2,3}implies x_2in{1,2,3}$ we get a tree with degree sequence $(4^4,6^infty)$; the vertices of degree $4$ are the null sequence and the sequences $langle1rangle$, $langle2rangle$, and $langle3rangle$.



          The number of trees with degree sequence $(4^4,6^infty)$ is $aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            This is probably the best answer and I feel guilty for making things way too complicated.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 4:11










          • $begingroup$
            Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 5:02





















          4












          $begingroup$

          If $6^infty$ means that there are $aleph_0$ vertices of degree $6$, then there are $2^{aleph_0}$ isomorphism classes of such graphs.



          There can't be more than that, because each graph with $aleph_0$ vertices can be represented as a subset of $mathbb Ntimesmathbb N$ in at least one way.



          On the other hand, we can construct that many different graphs:



          Start with an infinite tiling of the plane with equilateral triangles. That gives you an infinity of degree-6 vertices. Now pick two of them and split each of those two into two degree-4 vertices:



            /               /
          --O-- --> --O--O--
          / /


          This gives at least one graph with your degree sequence. But there are uncountably many ways we can modify it to give rise to a non-isomorphic graph. Namely, somewhere in the grid we can slice it open with a cut stretching to infinity:



             /  /  /  /  /  /  /  /
          --O---O---O---O---O---O---O---O--
          / / / / / / / /
          O---O---O---O---O---O---O---O---O
          / / /
          --O---O---O---O---O---O---O---O--
          / / / / / / / /
          O---O---O---O---O---O---O---O---O
          / / / / / / / /


          and then fill the gap with another copy of the same sliced-open grid, stitched to your existing grid along the edge. (You'll need to squeeze it to make it fit, of course -- but even if your concept of "infinite planar graph" requires that the set of vertices has no limit points in the plane, you can achieve this by squeezing most of the the new copy out towards infinity such that at every finite distance from the place we're looking at here there's a certain minimal spacing between the vertices).



          You can keep making such cuts and filling them in at different spacings. The endpoints of the original cuts will be topologically distinguishable from the rest of the graph, because they will be the only places in the graph there's a cycle of degree-6 vertices of length more than 3 where the shortest path between any two vertices in the cycle is along the cycle. We can encode an arbitrary subset of $mathbb N$, for example, as the distances between those cut endpoints.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 3:28



















          0












          $begingroup$

          In this graph I started with four vertices in the shape of a square, then kept connecting vertices in the interior to match the desired degree sequence.



          enter image description here






          share|cite|improve this answer









          $endgroup$













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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            5












            $begingroup$

            There are infinitely many graphs with this degree sequence.



            First, here is one possible graph with degree sequence $(6^infty)$: an infinite triangular lattice.



            triangular lattice



            One way to add four vertices of degree $4$ is to pick two places where we "split up" a degree-$6$ vertex into two adjacent degree-$4$ vertices:



            modified triangular lattice



            There are other ways to make a graph with degree sequence $(4^4,6^infty)$, but this operation can already be done in infinitely many non-isomorphic ways, since we can put the two pairs of degree-$4$ vertices arbitrarily far apart in the infinite lattice.



            (And there are also other graphs we can start with that have degree sequence $(6^infty)$, this one is just the nicest.)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 3:28
















            5












            $begingroup$

            There are infinitely many graphs with this degree sequence.



            First, here is one possible graph with degree sequence $(6^infty)$: an infinite triangular lattice.



            triangular lattice



            One way to add four vertices of degree $4$ is to pick two places where we "split up" a degree-$6$ vertex into two adjacent degree-$4$ vertices:



            modified triangular lattice



            There are other ways to make a graph with degree sequence $(4^4,6^infty)$, but this operation can already be done in infinitely many non-isomorphic ways, since we can put the two pairs of degree-$4$ vertices arbitrarily far apart in the infinite lattice.



            (And there are also other graphs we can start with that have degree sequence $(6^infty)$, this one is just the nicest.)






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 3:28














            5












            5








            5





            $begingroup$

            There are infinitely many graphs with this degree sequence.



            First, here is one possible graph with degree sequence $(6^infty)$: an infinite triangular lattice.



            triangular lattice



            One way to add four vertices of degree $4$ is to pick two places where we "split up" a degree-$6$ vertex into two adjacent degree-$4$ vertices:



            modified triangular lattice



            There are other ways to make a graph with degree sequence $(4^4,6^infty)$, but this operation can already be done in infinitely many non-isomorphic ways, since we can put the two pairs of degree-$4$ vertices arbitrarily far apart in the infinite lattice.



            (And there are also other graphs we can start with that have degree sequence $(6^infty)$, this one is just the nicest.)






            share|cite|improve this answer











            $endgroup$



            There are infinitely many graphs with this degree sequence.



            First, here is one possible graph with degree sequence $(6^infty)$: an infinite triangular lattice.



            triangular lattice



            One way to add four vertices of degree $4$ is to pick two places where we "split up" a degree-$6$ vertex into two adjacent degree-$4$ vertices:



            modified triangular lattice



            There are other ways to make a graph with degree sequence $(4^4,6^infty)$, but this operation can already be done in infinitely many non-isomorphic ways, since we can put the two pairs of degree-$4$ vertices arbitrarily far apart in the infinite lattice.



            (And there are also other graphs we can start with that have degree sequence $(6^infty)$, this one is just the nicest.)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 15:31

























            answered Dec 18 '18 at 3:15









            Misha LavrovMisha Lavrov

            45.1k656107




            45.1k656107












            • $begingroup$
              The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 3:28


















            • $begingroup$
              The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 3:28
















            $begingroup$
            The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 3:28




            $begingroup$
            The "rearrange the edges of a chunk of the lattice" plan is not quite that simple to execute without getting multiple edges. There are plenty of other possibilities, though.
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 3:28











            5












            $begingroup$

            It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)



            For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^infty)$. The vertex set is the set of all finite sequences $langle x_1,x_2,dots,x_nrangle$ (including the null sequence) with $x_iin{0,1,2,3,4}$, and with an edge joining $langle x_1,x_2,dots,x_nrangle$ to $langle x_1,x_2,dots,x_n,x_{n+1}rangle$. The null sequence has degree $5$, the rest have degree $6$.



            By imposing the further conditions $x_1ne0$ and $x_1in{1,2,3}implies x_2in{1,2,3}$ we get a tree with degree sequence $(4^4,6^infty)$; the vertices of degree $4$ are the null sequence and the sequences $langle1rangle$, $langle2rangle$, and $langle3rangle$.



            The number of trees with degree sequence $(4^4,6^infty)$ is $aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              This is probably the best answer and I feel guilty for making things way too complicated.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 4:11










            • $begingroup$
              Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 5:02


















            5












            $begingroup$

            It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)



            For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^infty)$. The vertex set is the set of all finite sequences $langle x_1,x_2,dots,x_nrangle$ (including the null sequence) with $x_iin{0,1,2,3,4}$, and with an edge joining $langle x_1,x_2,dots,x_nrangle$ to $langle x_1,x_2,dots,x_n,x_{n+1}rangle$. The null sequence has degree $5$, the rest have degree $6$.



            By imposing the further conditions $x_1ne0$ and $x_1in{1,2,3}implies x_2in{1,2,3}$ we get a tree with degree sequence $(4^4,6^infty)$; the vertices of degree $4$ are the null sequence and the sequences $langle1rangle$, $langle2rangle$, and $langle3rangle$.



            The number of trees with degree sequence $(4^4,6^infty)$ is $aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              This is probably the best answer and I feel guilty for making things way too complicated.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 4:11










            • $begingroup$
              Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 5:02
















            5












            5








            5





            $begingroup$

            It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)



            For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^infty)$. The vertex set is the set of all finite sequences $langle x_1,x_2,dots,x_nrangle$ (including the null sequence) with $x_iin{0,1,2,3,4}$, and with an edge joining $langle x_1,x_2,dots,x_nrangle$ to $langle x_1,x_2,dots,x_n,x_{n+1}rangle$. The null sequence has degree $5$, the rest have degree $6$.



            By imposing the further conditions $x_1ne0$ and $x_1in{1,2,3}implies x_2in{1,2,3}$ we get a tree with degree sequence $(4^4,6^infty)$; the vertices of degree $4$ are the null sequence and the sequences $langle1rangle$, $langle2rangle$, and $langle3rangle$.



            The number of trees with degree sequence $(4^4,6^infty)$ is $aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.






            share|cite|improve this answer











            $endgroup$



            It is easy to construct an infinite tree with any reasonable sequence as its degree sequence. (All ones, for example, would not be reasonable.)



            For starters, here is a tree whose degree sequence, in your notation, is $(5^1,6^infty)$. The vertex set is the set of all finite sequences $langle x_1,x_2,dots,x_nrangle$ (including the null sequence) with $x_iin{0,1,2,3,4}$, and with an edge joining $langle x_1,x_2,dots,x_nrangle$ to $langle x_1,x_2,dots,x_n,x_{n+1}rangle$. The null sequence has degree $5$, the rest have degree $6$.



            By imposing the further conditions $x_1ne0$ and $x_1in{1,2,3}implies x_2in{1,2,3}$ we get a tree with degree sequence $(4^4,6^infty)$; the vertices of degree $4$ are the null sequence and the sequences $langle1rangle$, $langle2rangle$, and $langle3rangle$.



            The number of trees with degree sequence $(4^4,6^infty)$ is $aleph_0$; each such tree is determined up to isomorphism by the smallest (finite) subtree containing all the vertices of degree $4$, those vertices being marked.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 18 '18 at 9:44

























            answered Dec 18 '18 at 4:03









            bofbof

            51.3k457120




            51.3k457120








            • 1




              $begingroup$
              This is probably the best answer and I feel guilty for making things way too complicated.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 4:11










            • $begingroup$
              Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 5:02
















            • 1




              $begingroup$
              This is probably the best answer and I feel guilty for making things way too complicated.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 4:11










            • $begingroup$
              Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
              $endgroup$
              – Henning Makholm
              Dec 18 '18 at 5:02










            1




            1




            $begingroup$
            This is probably the best answer and I feel guilty for making things way too complicated.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 4:11




            $begingroup$
            This is probably the best answer and I feel guilty for making things way too complicated.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 4:11












            $begingroup$
            Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 5:02






            $begingroup$
            Hmm, yes, and you can get all the way up to $2^{aleph_0}$, say, by allowing the "tree" to have triangles in it here and there...
            $endgroup$
            – Henning Makholm
            Dec 18 '18 at 5:02













            4












            $begingroup$

            If $6^infty$ means that there are $aleph_0$ vertices of degree $6$, then there are $2^{aleph_0}$ isomorphism classes of such graphs.



            There can't be more than that, because each graph with $aleph_0$ vertices can be represented as a subset of $mathbb Ntimesmathbb N$ in at least one way.



            On the other hand, we can construct that many different graphs:



            Start with an infinite tiling of the plane with equilateral triangles. That gives you an infinity of degree-6 vertices. Now pick two of them and split each of those two into two degree-4 vertices:



              /               /
            --O-- --> --O--O--
            / /


            This gives at least one graph with your degree sequence. But there are uncountably many ways we can modify it to give rise to a non-isomorphic graph. Namely, somewhere in the grid we can slice it open with a cut stretching to infinity:



               /  /  /  /  /  /  /  /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / / / / / / /


            and then fill the gap with another copy of the same sliced-open grid, stitched to your existing grid along the edge. (You'll need to squeeze it to make it fit, of course -- but even if your concept of "infinite planar graph" requires that the set of vertices has no limit points in the plane, you can achieve this by squeezing most of the the new copy out towards infinity such that at every finite distance from the place we're looking at here there's a certain minimal spacing between the vertices).



            You can keep making such cuts and filling them in at different spacings. The endpoints of the original cuts will be topologically distinguishable from the rest of the graph, because they will be the only places in the graph there's a cycle of degree-6 vertices of length more than 3 where the shortest path between any two vertices in the cycle is along the cycle. We can encode an arbitrary subset of $mathbb N$, for example, as the distances between those cut endpoints.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 3:28
















            4












            $begingroup$

            If $6^infty$ means that there are $aleph_0$ vertices of degree $6$, then there are $2^{aleph_0}$ isomorphism classes of such graphs.



            There can't be more than that, because each graph with $aleph_0$ vertices can be represented as a subset of $mathbb Ntimesmathbb N$ in at least one way.



            On the other hand, we can construct that many different graphs:



            Start with an infinite tiling of the plane with equilateral triangles. That gives you an infinity of degree-6 vertices. Now pick two of them and split each of those two into two degree-4 vertices:



              /               /
            --O-- --> --O--O--
            / /


            This gives at least one graph with your degree sequence. But there are uncountably many ways we can modify it to give rise to a non-isomorphic graph. Namely, somewhere in the grid we can slice it open with a cut stretching to infinity:



               /  /  /  /  /  /  /  /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / / / / / / /


            and then fill the gap with another copy of the same sliced-open grid, stitched to your existing grid along the edge. (You'll need to squeeze it to make it fit, of course -- but even if your concept of "infinite planar graph" requires that the set of vertices has no limit points in the plane, you can achieve this by squeezing most of the the new copy out towards infinity such that at every finite distance from the place we're looking at here there's a certain minimal spacing between the vertices).



            You can keep making such cuts and filling them in at different spacings. The endpoints of the original cuts will be topologically distinguishable from the rest of the graph, because they will be the only places in the graph there's a cycle of degree-6 vertices of length more than 3 where the shortest path between any two vertices in the cycle is along the cycle. We can encode an arbitrary subset of $mathbb N$, for example, as the distances between those cut endpoints.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 3:28














            4












            4








            4





            $begingroup$

            If $6^infty$ means that there are $aleph_0$ vertices of degree $6$, then there are $2^{aleph_0}$ isomorphism classes of such graphs.



            There can't be more than that, because each graph with $aleph_0$ vertices can be represented as a subset of $mathbb Ntimesmathbb N$ in at least one way.



            On the other hand, we can construct that many different graphs:



            Start with an infinite tiling of the plane with equilateral triangles. That gives you an infinity of degree-6 vertices. Now pick two of them and split each of those two into two degree-4 vertices:



              /               /
            --O-- --> --O--O--
            / /


            This gives at least one graph with your degree sequence. But there are uncountably many ways we can modify it to give rise to a non-isomorphic graph. Namely, somewhere in the grid we can slice it open with a cut stretching to infinity:



               /  /  /  /  /  /  /  /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / / / / / / /


            and then fill the gap with another copy of the same sliced-open grid, stitched to your existing grid along the edge. (You'll need to squeeze it to make it fit, of course -- but even if your concept of "infinite planar graph" requires that the set of vertices has no limit points in the plane, you can achieve this by squeezing most of the the new copy out towards infinity such that at every finite distance from the place we're looking at here there's a certain minimal spacing between the vertices).



            You can keep making such cuts and filling them in at different spacings. The endpoints of the original cuts will be topologically distinguishable from the rest of the graph, because they will be the only places in the graph there's a cycle of degree-6 vertices of length more than 3 where the shortest path between any two vertices in the cycle is along the cycle. We can encode an arbitrary subset of $mathbb N$, for example, as the distances between those cut endpoints.






            share|cite|improve this answer











            $endgroup$



            If $6^infty$ means that there are $aleph_0$ vertices of degree $6$, then there are $2^{aleph_0}$ isomorphism classes of such graphs.



            There can't be more than that, because each graph with $aleph_0$ vertices can be represented as a subset of $mathbb Ntimesmathbb N$ in at least one way.



            On the other hand, we can construct that many different graphs:



            Start with an infinite tiling of the plane with equilateral triangles. That gives you an infinity of degree-6 vertices. Now pick two of them and split each of those two into two degree-4 vertices:



              /               /
            --O-- --> --O--O--
            / /


            This gives at least one graph with your degree sequence. But there are uncountably many ways we can modify it to give rise to a non-isomorphic graph. Namely, somewhere in the grid we can slice it open with a cut stretching to infinity:



               /  /  /  /  /  /  /  /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / /
            --O---O---O---O---O---O---O---O--
            / / / / / / / /
            O---O---O---O---O---O---O---O---O
            / / / / / / / /


            and then fill the gap with another copy of the same sliced-open grid, stitched to your existing grid along the edge. (You'll need to squeeze it to make it fit, of course -- but even if your concept of "infinite planar graph" requires that the set of vertices has no limit points in the plane, you can achieve this by squeezing most of the the new copy out towards infinity such that at every finite distance from the place we're looking at here there's a certain minimal spacing between the vertices).



            You can keep making such cuts and filling them in at different spacings. The endpoints of the original cuts will be topologically distinguishable from the rest of the graph, because they will be the only places in the graph there's a cycle of degree-6 vertices of length more than 3 where the shortest path between any two vertices in the cycle is along the cycle. We can encode an arbitrary subset of $mathbb N$, for example, as the distances between those cut endpoints.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 29 '18 at 21:48

























            answered Dec 18 '18 at 3:25









            Henning MakholmHenning Makholm

            239k17304541




            239k17304541












            • $begingroup$
              I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 3:28


















            • $begingroup$
              I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
              $endgroup$
              – Misha Lavrov
              Dec 18 '18 at 3:28
















            $begingroup$
            I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 3:28




            $begingroup$
            I wasn't sure whether to include a discussion of cardinality or not. Thankfully, now we have one answer of each kind.
            $endgroup$
            – Misha Lavrov
            Dec 18 '18 at 3:28











            0












            $begingroup$

            In this graph I started with four vertices in the shape of a square, then kept connecting vertices in the interior to match the desired degree sequence.



            enter image description here






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              In this graph I started with four vertices in the shape of a square, then kept connecting vertices in the interior to match the desired degree sequence.



              enter image description here






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                In this graph I started with four vertices in the shape of a square, then kept connecting vertices in the interior to match the desired degree sequence.



                enter image description here






                share|cite|improve this answer









                $endgroup$



                In this graph I started with four vertices in the shape of a square, then kept connecting vertices in the interior to match the desired degree sequence.



                enter image description here







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 30 '18 at 18:24









                UltradarkUltradark

                1791517




                1791517






























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