Prove that the shortest path between two points in a Delaunay triangulation minimizes angle at each step.











up vote
1
down vote

favorite
1












Say we have a set of points $p_{k (k in K)}in P$ Poisson distributed in a real coordinate plane $X$ residing in ${rm I!R}^{2}$ and with Euclidean distance function $d$ and $K$ is a set of indices. The points form a Delaunay triangulation $D = {P, E}$ where E is the set of edges between points that obeys the empty circle property that no circumcircle of any triangle in $D$ has a point of $P$ in its interior.



Let two points $p_0$ and $p_N$ be points in $D$ that are separated by a shortest path geodesic distance of $N$ and we are interested in determining a shortest geodesic path $Lambda$ between them i.e. a set of edge tuples
begin{equation}
Lambda = {(p_1,p_0) ... (p_{N-2},p_{N-1}), (p_{N-1},p_N) } subset E.
end{equation}



At any step $i$ along the path, $vec{omega} = (p_i,p_N)$ is a vector pointing from a current vertex along the shortest path toward the destination $p_N$ and $vec{nu} = (p_i,p_{i+1})$ is a vector pointing to a next vertex along the shortest path to $p_N$, and let $vec{mu} = (p_i,p_j)$ be the vector pointing to some neighbor $p_{j (j in J)}$ geodesically adjacent to $p_i$ and $J$ is the set of indices of vertices adjacent to $p_i$.



The angle $theta_{i,j} = cos ^{-1} ( frac{ vec{ omega } cdot vec{ mu} }{|vec{ omega }| , |vec{ mu}|})$ is the angle between the vector from current point to the destination and the vector to neighbor $j$. Can it be shown that
begin{equation}
theta_{i,{i+1}} = cos ^{-1} (frac{ vec{ omega } cdot vec{ nu} }{|vec{ omega }| , |vec{ nu}|}) = min{ theta_{i,j} , forall j in J }
end{equation}



i.e. that by minimizing the angle made by vectors extending from the current vertex out to the destination and an adjacent vertex, the next point on the shortest path can be determined.






graph-theory metric-spaces euclidean-geometry planar-graph tessellations







share|cite|improve this question









New contributor




Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
























    up vote
    1
    down vote

    favorite
    1












    Say we have a set of points $p_{k (k in K)}in P$ Poisson distributed in a real coordinate plane $X$ residing in ${rm I!R}^{2}$ and with Euclidean distance function $d$ and $K$ is a set of indices. The points form a Delaunay triangulation $D = {P, E}$ where E is the set of edges between points that obeys the empty circle property that no circumcircle of any triangle in $D$ has a point of $P$ in its interior.



    Let two points $p_0$ and $p_N$ be points in $D$ that are separated by a shortest path geodesic distance of $N$ and we are interested in determining a shortest geodesic path $Lambda$ between them i.e. a set of edge tuples
    begin{equation}
    Lambda = {(p_1,p_0) ... (p_{N-2},p_{N-1}), (p_{N-1},p_N) } subset E.
    end{equation}



    At any step $i$ along the path, $vec{omega} = (p_i,p_N)$ is a vector pointing from a current vertex along the shortest path toward the destination $p_N$ and $vec{nu} = (p_i,p_{i+1})$ is a vector pointing to a next vertex along the shortest path to $p_N$, and let $vec{mu} = (p_i,p_j)$ be the vector pointing to some neighbor $p_{j (j in J)}$ geodesically adjacent to $p_i$ and $J$ is the set of indices of vertices adjacent to $p_i$.



    The angle $theta_{i,j} = cos ^{-1} ( frac{ vec{ omega } cdot vec{ mu} }{|vec{ omega }| , |vec{ mu}|})$ is the angle between the vector from current point to the destination and the vector to neighbor $j$. Can it be shown that
    begin{equation}
    theta_{i,{i+1}} = cos ^{-1} (frac{ vec{ omega } cdot vec{ nu} }{|vec{ omega }| , |vec{ nu}|}) = min{ theta_{i,j} , forall j in J }
    end{equation}



    i.e. that by minimizing the angle made by vectors extending from the current vertex out to the destination and an adjacent vertex, the next point on the shortest path can be determined.






    graph-theory metric-spaces euclidean-geometry planar-graph tessellations







    share|cite|improve this question









    New contributor




    Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      Say we have a set of points $p_{k (k in K)}in P$ Poisson distributed in a real coordinate plane $X$ residing in ${rm I!R}^{2}$ and with Euclidean distance function $d$ and $K$ is a set of indices. The points form a Delaunay triangulation $D = {P, E}$ where E is the set of edges between points that obeys the empty circle property that no circumcircle of any triangle in $D$ has a point of $P$ in its interior.



      Let two points $p_0$ and $p_N$ be points in $D$ that are separated by a shortest path geodesic distance of $N$ and we are interested in determining a shortest geodesic path $Lambda$ between them i.e. a set of edge tuples
      begin{equation}
      Lambda = {(p_1,p_0) ... (p_{N-2},p_{N-1}), (p_{N-1},p_N) } subset E.
      end{equation}



      At any step $i$ along the path, $vec{omega} = (p_i,p_N)$ is a vector pointing from a current vertex along the shortest path toward the destination $p_N$ and $vec{nu} = (p_i,p_{i+1})$ is a vector pointing to a next vertex along the shortest path to $p_N$, and let $vec{mu} = (p_i,p_j)$ be the vector pointing to some neighbor $p_{j (j in J)}$ geodesically adjacent to $p_i$ and $J$ is the set of indices of vertices adjacent to $p_i$.



      The angle $theta_{i,j} = cos ^{-1} ( frac{ vec{ omega } cdot vec{ mu} }{|vec{ omega }| , |vec{ mu}|})$ is the angle between the vector from current point to the destination and the vector to neighbor $j$. Can it be shown that
      begin{equation}
      theta_{i,{i+1}} = cos ^{-1} (frac{ vec{ omega } cdot vec{ nu} }{|vec{ omega }| , |vec{ nu}|}) = min{ theta_{i,j} , forall j in J }
      end{equation}



      i.e. that by minimizing the angle made by vectors extending from the current vertex out to the destination and an adjacent vertex, the next point on the shortest path can be determined.






      graph-theory metric-spaces euclidean-geometry planar-graph tessellations







      share|cite|improve this question









      New contributor




      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Say we have a set of points $p_{k (k in K)}in P$ Poisson distributed in a real coordinate plane $X$ residing in ${rm I!R}^{2}$ and with Euclidean distance function $d$ and $K$ is a set of indices. The points form a Delaunay triangulation $D = {P, E}$ where E is the set of edges between points that obeys the empty circle property that no circumcircle of any triangle in $D$ has a point of $P$ in its interior.



      Let two points $p_0$ and $p_N$ be points in $D$ that are separated by a shortest path geodesic distance of $N$ and we are interested in determining a shortest geodesic path $Lambda$ between them i.e. a set of edge tuples
      begin{equation}
      Lambda = {(p_1,p_0) ... (p_{N-2},p_{N-1}), (p_{N-1},p_N) } subset E.
      end{equation}



      At any step $i$ along the path, $vec{omega} = (p_i,p_N)$ is a vector pointing from a current vertex along the shortest path toward the destination $p_N$ and $vec{nu} = (p_i,p_{i+1})$ is a vector pointing to a next vertex along the shortest path to $p_N$, and let $vec{mu} = (p_i,p_j)$ be the vector pointing to some neighbor $p_{j (j in J)}$ geodesically adjacent to $p_i$ and $J$ is the set of indices of vertices adjacent to $p_i$.



      The angle $theta_{i,j} = cos ^{-1} ( frac{ vec{ omega } cdot vec{ mu} }{|vec{ omega }| , |vec{ mu}|})$ is the angle between the vector from current point to the destination and the vector to neighbor $j$. Can it be shown that
      begin{equation}
      theta_{i,{i+1}} = cos ^{-1} (frac{ vec{ omega } cdot vec{ nu} }{|vec{ omega }| , |vec{ nu}|}) = min{ theta_{i,j} , forall j in J }
      end{equation}



      i.e. that by minimizing the angle made by vectors extending from the current vertex out to the destination and an adjacent vertex, the next point on the shortest path can be determined.






      graph-theory metric-spaces euclidean-geometry planar-graph tessellations




      graph-theory metric-spaces euclidean-geometry planar-graph tessellations






      share|cite|improve this question









      New contributor




      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited yesterday





















      New contributor




      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked yesterday









      Entangler

      64




      64




      New contributor




      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.



























          active

          oldest

          votes











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });






          Entangler is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004935%2fprove-that-the-shortest-path-between-two-points-in-a-delaunay-triangulation-mini%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown






























          active

          oldest

          votes













          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          Entangler is a new contributor. Be nice, and check out our Code of Conduct.










           

          draft saved


          draft discarded


















          Entangler is a new contributor. Be nice, and check out our Code of Conduct.













          Entangler is a new contributor. Be nice, and check out our Code of Conduct.












          Entangler is a new contributor. Be nice, and check out our Code of Conduct.















           


          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004935%2fprove-that-the-shortest-path-between-two-points-in-a-delaunay-triangulation-mini%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Can a sorcerer learn a 5th-level spell early by creating spell slots using the Font of Magic feature?

          Image
          28 6 $begingroup$ Per the Font of Magic feature, sorcerer can use Flexible Casting to create 5th-level spell slots at level 7, even though they are not typically available until level 9. You can transform unexpended sorcery points into one spell slot as a bonus action on your turn. The Creating Spell Slots table shows the cost of creating a spell slot of [5th level is 7] If such a sorcerer levels up while still having this 5th-level spell slot, can they choose a 5th-level spell as the spell they gain upon levelling up? The Spellcasting feature states: Additionally, when you gain a level in this class, you can choose one of the sorcerer spells you know and replace it with another spell from the sorcerer spell list, which also must be of a level for which you have spell slots.
          Read more

          Does disintegrating a polymorphed enemy still kill it after the 2018 errata?

          Image
          up vote 13 down vote favorite 1 After the 2018 errata, the disintegrate spell description now reads: A creature targeted by this spell must make a Dexterity saving throw. On a failed save, the target takes 10d6 + 40 force damage. The target is disintegrated if this damage leaves it with 0 hit points. If you were to polymorph an enemy into a rat and then disintegrate it, would the enemy be disintegrated or would it just return to its original form? I know that for Druids, it’s not an instant kill anymore, but is this the case for polymorph as well? dnd-5e spells polymorph errata share | improve this question edited 18 hours ago
          Read more

          A Topological Invariant for $pi_3(U(n))$

          Image
          3 3 $begingroup$ Recently, I saw a construction of topological invariant for $pi_3(U(n))$ with $ngeq 2$ : $$ N=frac{1}{24pi^2}int_{S^3} d^3x epsilon^{ijk} Tr[(U^{-1}partial_{x_i}U)(U^{-1}partial_{x_j}U)(U^{-1}partial_{x_k}U)] , $$ where $Uin U(n)$ depends on $boldsymbol{x}=(x_1,x_2,x_3)in S^3$ , $epsilon^{ijk}$ is the Levi-Civita symbol, $i,j,k=1,2,3$ , and the duplicated indexes are summed over. It is claimed that $N$ is an integer, but why? Update 02/02/2019 I think I got an argument for $n=2$ . In this case, $U=e^{i varphi} q$ with $qin SU(2)$ . Due to the trace and the Levi-Civita symbol in $N$ , $varphi$ does not contribute to $N$ . As $Tr[q^{dagger}partial_i q]=0$ and $(q^{dagger}partial_i q)^{dagger}=-q^{dagger}partial_i q$ , $q^{dagger}partial_i q$ in geneeral has the form $q^{dagger}partia
          Read more