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Prove that the shortest path between two points in a Delaunay triangulation minimizes angle at each step.
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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
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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
asked yesterday
Entangler
64
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Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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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
asked yesterday
Entangler
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.
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
asked yesterday
Entangler
64
New contributor
Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
Entangler
64
New contributor
Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited yesterday
asked yesterday
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asked yesterday
Entangler
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asked yesterday
Entangler
64
64
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Entangler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
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Check out our Code of Conduct.
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