Help in solving a question to trace the perimeter of a face in a polyhedron
$begingroup$
I have the following question with me:
"We assign an arrow to each edge of a convex polyhedron, so that at least one arrow starts at each vertex and at least one arrow arrives. Prove that there exist 2 faces of the polyhedron, so that you can trace their perimeters in the direction of the arrow"
I will be needing a little elaborate explanation please, if possible, as I am pretty weak with the understanding of 3-D figures.
Only thing that I deduced was that I will indeed reach the vertex I started off after a finite number of steps, but I dont know what to do with this conclusion. Does it help?
graph-theory
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
"We assign an arrow to each edge of a convex polyhedron, so that at least one arrow starts at each vertex and at least one arrow arrives. Prove that there exist 2 faces of the polyhedron, so that you can trace their perimeters in the direction of the arrow"
I will be needing a little elaborate explanation please, if possible, as I am pretty weak with the understanding of 3-D figures.
Only thing that I deduced was that I will indeed reach the vertex I started off after a finite number of steps, but I dont know what to do with this conclusion. Does it help?
graph-theory
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
$endgroup$
add a comment |
$begingroup$
I have the following question with me:
"We assign an arrow to each edge of a convex polyhedron, so that at least one arrow starts at each vertex and at least one arrow arrives. Prove that there exist 2 faces of the polyhedron, so that you can trace their perimeters in the direction of the arrow"
I will be needing a little elaborate explanation please, if possible, as I am pretty weak with the understanding of 3-D figures.
Only thing that I deduced was that I will indeed reach the vertex I started off after a finite number of steps, but I dont know what to do with this conclusion. Does it help?
graph-theory
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
$endgroup$
I have the following question with me:
"We assign an arrow to each edge of a convex polyhedron, so that at least one arrow starts at each vertex and at least one arrow arrives. Prove that there exist 2 faces of the polyhedron, so that you can trace their perimeters in the direction of the arrow"
I will be needing a little elaborate explanation please, if possible, as I am pretty weak with the understanding of 3-D figures.
Only thing that I deduced was that I will indeed reach the vertex I started off after a finite number of steps, but I dont know what to do with this conclusion. Does it help?
graph-theory
graph-theory
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
asked Dec 16 '18 at 11:01
saisanjeevsaisanjeev
987212
987212
add a comment |
add a comment |
1 Answer
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We assume that the polyhedron has finitely many vertices. A not perfectly rigorous proof is the following. Start from any vertex of the polyhedron and go along edges towards arrows. Since the polyhedron has finitely many vertices, after some step we (first time) visit a vertex which we have already visited. So we have constructed a cycle $C$ directed towards arrows. The cycle $C$ divides the surface of the polyhedron into two parts $S$ and $S’$ such that $C$ is a boundary of each of these parts. We say that a cycle $C’$ contained in one of these parts bounds a face $F$ contained in the same part, if $F$ is situated inside the blue domain at the left picture. In each of the parts $S$ and $S’$ pick a cycle directed towards arrow which bounds the smallest number of faces. We claim that each of these cycles bounds exactly one face. Indeed, assume the converse, that such a cycle $C’$ bounds more than one face. Then at some vertex $v$ of $C’$ there is an edge $e$ located inside the domain bounded by $C’$ (look at the right picture for instances of $v$ and $e$). Starting from the vertex $v$ we shall go along $e$. We shall continue our walk till we visit a vertex of the cycle $C’$ or a vertex which we have already visited. When we leave an intermediate vertex, we can choose theoutgoing edge which arrow direction corresponds the to arrow direction of the incoming edge (look at the picture). When we shall stop, we obtain a cycle which bounds the smaller number of faces than $C’$ (look at the picture), a contradiction.
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
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1 Answer
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1 Answer
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$begingroup$
We assume that the polyhedron has finitely many vertices. A not perfectly rigorous proof is the following. Start from any vertex of the polyhedron and go along edges towards arrows. Since the polyhedron has finitely many vertices, after some step we (first time) visit a vertex which we have already visited. So we have constructed a cycle $C$ directed towards arrows. The cycle $C$ divides the surface of the polyhedron into two parts $S$ and $S’$ such that $C$ is a boundary of each of these parts. We say that a cycle $C’$ contained in one of these parts bounds a face $F$ contained in the same part, if $F$ is situated inside the blue domain at the left picture. In each of the parts $S$ and $S’$ pick a cycle directed towards arrow which bounds the smallest number of faces. We claim that each of these cycles bounds exactly one face. Indeed, assume the converse, that such a cycle $C’$ bounds more than one face. Then at some vertex $v$ of $C’$ there is an edge $e$ located inside the domain bounded by $C’$ (look at the right picture for instances of $v$ and $e$). Starting from the vertex $v$ we shall go along $e$. We shall continue our walk till we visit a vertex of the cycle $C’$ or a vertex which we have already visited. When we leave an intermediate vertex, we can choose theoutgoing edge which arrow direction corresponds the to arrow direction of the incoming edge (look at the picture). When we shall stop, we obtain a cycle which bounds the smaller number of faces than $C’$ (look at the picture), a contradiction.
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
$endgroup$
add a comment |
$begingroup$
We assume that the polyhedron has finitely many vertices. A not perfectly rigorous proof is the following. Start from any vertex of the polyhedron and go along edges towards arrows. Since the polyhedron has finitely many vertices, after some step we (first time) visit a vertex which we have already visited. So we have constructed a cycle $C$ directed towards arrows. The cycle $C$ divides the surface of the polyhedron into two parts $S$ and $S’$ such that $C$ is a boundary of each of these parts. We say that a cycle $C’$ contained in one of these parts bounds a face $F$ contained in the same part, if $F$ is situated inside the blue domain at the left picture. In each of the parts $S$ and $S’$ pick a cycle directed towards arrow which bounds the smallest number of faces. We claim that each of these cycles bounds exactly one face. Indeed, assume the converse, that such a cycle $C’$ bounds more than one face. Then at some vertex $v$ of $C’$ there is an edge $e$ located inside the domain bounded by $C’$ (look at the right picture for instances of $v$ and $e$). Starting from the vertex $v$ we shall go along $e$. We shall continue our walk till we visit a vertex of the cycle $C’$ or a vertex which we have already visited. When we leave an intermediate vertex, we can choose theoutgoing edge which arrow direction corresponds the to arrow direction of the incoming edge (look at the picture). When we shall stop, we obtain a cycle which bounds the smaller number of faces than $C’$ (look at the picture), a contradiction.
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
$endgroup$
add a comment |
$begingroup$
We assume that the polyhedron has finitely many vertices. A not perfectly rigorous proof is the following. Start from any vertex of the polyhedron and go along edges towards arrows. Since the polyhedron has finitely many vertices, after some step we (first time) visit a vertex which we have already visited. So we have constructed a cycle $C$ directed towards arrows. The cycle $C$ divides the surface of the polyhedron into two parts $S$ and $S’$ such that $C$ is a boundary of each of these parts. We say that a cycle $C’$ contained in one of these parts bounds a face $F$ contained in the same part, if $F$ is situated inside the blue domain at the left picture. In each of the parts $S$ and $S’$ pick a cycle directed towards arrow which bounds the smallest number of faces. We claim that each of these cycles bounds exactly one face. Indeed, assume the converse, that such a cycle $C’$ bounds more than one face. Then at some vertex $v$ of $C’$ there is an edge $e$ located inside the domain bounded by $C’$ (look at the right picture for instances of $v$ and $e$). Starting from the vertex $v$ we shall go along $e$. We shall continue our walk till we visit a vertex of the cycle $C’$ or a vertex which we have already visited. When we leave an intermediate vertex, we can choose theoutgoing edge which arrow direction corresponds the to arrow direction of the incoming edge (look at the picture). When we shall stop, we obtain a cycle which bounds the smaller number of faces than $C’$ (look at the picture), a contradiction.
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
$endgroup$
We assume that the polyhedron has finitely many vertices. A not perfectly rigorous proof is the following. Start from any vertex of the polyhedron and go along edges towards arrows. Since the polyhedron has finitely many vertices, after some step we (first time) visit a vertex which we have already visited. So we have constructed a cycle $C$ directed towards arrows. The cycle $C$ divides the surface of the polyhedron into two parts $S$ and $S’$ such that $C$ is a boundary of each of these parts. We say that a cycle $C’$ contained in one of these parts bounds a face $F$ contained in the same part, if $F$ is situated inside the blue domain at the left picture. In each of the parts $S$ and $S’$ pick a cycle directed towards arrow which bounds the smallest number of faces. We claim that each of these cycles bounds exactly one face. Indeed, assume the converse, that such a cycle $C’$ bounds more than one face. Then at some vertex $v$ of $C’$ there is an edge $e$ located inside the domain bounded by $C’$ (look at the right picture for instances of $v$ and $e$). Starting from the vertex $v$ we shall go along $e$. We shall continue our walk till we visit a vertex of the cycle $C’$ or a vertex which we have already visited. When we leave an intermediate vertex, we can choose theoutgoing edge which arrow direction corresponds the to arrow direction of the incoming edge (look at the picture). When we shall stop, we obtain a cycle which bounds the smaller number of faces than $C’$ (look at the picture), a contradiction.
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
edited Dec 17 '18 at 17:25
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
answered Dec 17 '18 at 17:18
Alex RavskyAlex Ravsky
41.2k32282
41.2k32282
add a comment |
add a comment |
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