Abstract:In [Adams, 1994; The Knot Book], Colin Adams states as an open question whether removing a vertex and all edges incident to that vertex from an intrinsically knotted graph must yield an intrinsically linked graph. In this paper, we exhibit an intrinsically knotted graph for which there is a vertex that can be removed, and the resulting graph is not intrinsically linked. We further show that this graph is minor minimal with respect to being intrinsically knotted. ß
“…In this case, C 3 could be (1, 2, 3), (1, 2, 4), (2, 3, 4), or (2,3,5). If C 3 is (1, 2, 3), then C 1 is (v, 4, 6, 5), and C 3 connects to 7 via edge (1, 7), which is disjoint from C 1 .…”
Section: Important Lemmasmentioning
confidence: 95%
“…C 3 uses none of the vertices in {2, 6, 7}. Here C 3 must be (1,3,4) and C 1 must be (v, 2, 5). Here, C 3 connects to 6 via (4, 6) and to 7 via (1, 7).…”
Section: Important Lemmasmentioning
confidence: 96%
“…Furthermore, there are exactly two sets of 4 vertices in H whose removal will disconnect H. Those two sets are {v, 2, 4, 6} and {v, 7, 9, 11}. The graph H − {v, 2, 4, 6} has four components; one is K 3,3 , and the other three components are all points. The graph H − {v, 7, 9, 11} has four components that are also K 3,3 and three points.…”
Section: Theorem 33 the Graph J 14 Is Intrinsically Knottedmentioning
confidence: 98%
“…We will show our graphs are intrinsically knotted by showing that every embedding of them contains a subgraph that contracts onto an embedding of the graph D 4 with lk(C 1 , C 3 ) = 0 and lk(C 2 , C 4 ) = 0. This technique was used in [2] and [3].We will also make use of the following elementary lemma, which we state without proof. 6-cycle (a, 1, b, 2, c, 3), any two vertices in the set {a, b, c} are connected to any two vertices in the set {1, 2, 3} by disjoint edges.…”
mentioning
confidence: 99%
“…We will show our graphs are intrinsically knotted by showing that every embedding of them contains a subgraph that contracts onto an embedding of the graph D 4 with lk(C 1 , C 3 ) = 0 and lk(C 2 , C 4 ) = 0. This technique was used in [2] and [3].…”
Abstract:We demonstrate four intrinsically knotted graphs that do not contain each other, nor any previously known intrinsically knotted graph, as a minor.
“…In this case, C 3 could be (1, 2, 3), (1, 2, 4), (2, 3, 4), or (2,3,5). If C 3 is (1, 2, 3), then C 1 is (v, 4, 6, 5), and C 3 connects to 7 via edge (1, 7), which is disjoint from C 1 .…”
Section: Important Lemmasmentioning
confidence: 95%
“…C 3 uses none of the vertices in {2, 6, 7}. Here C 3 must be (1,3,4) and C 1 must be (v, 2, 5). Here, C 3 connects to 6 via (4, 6) and to 7 via (1, 7).…”
Section: Important Lemmasmentioning
confidence: 96%
“…Furthermore, there are exactly two sets of 4 vertices in H whose removal will disconnect H. Those two sets are {v, 2, 4, 6} and {v, 7, 9, 11}. The graph H − {v, 2, 4, 6} has four components; one is K 3,3 , and the other three components are all points. The graph H − {v, 7, 9, 11} has four components that are also K 3,3 and three points.…”
Section: Theorem 33 the Graph J 14 Is Intrinsically Knottedmentioning
confidence: 98%
“…We will show our graphs are intrinsically knotted by showing that every embedding of them contains a subgraph that contracts onto an embedding of the graph D 4 with lk(C 1 , C 3 ) = 0 and lk(C 2 , C 4 ) = 0. This technique was used in [2] and [3].We will also make use of the following elementary lemma, which we state without proof. 6-cycle (a, 1, b, 2, c, 3), any two vertices in the set {a, b, c} are connected to any two vertices in the set {1, 2, 3} by disjoint edges.…”
mentioning
confidence: 99%
“…We will show our graphs are intrinsically knotted by showing that every embedding of them contains a subgraph that contracts onto an embedding of the graph D 4 with lk(C 1 , C 3 ) = 0 and lk(C 2 , C 4 ) = 0. This technique was used in [2] and [3].…”
Abstract:We demonstrate four intrinsically knotted graphs that do not contain each other, nor any previously known intrinsically knotted graph, as a minor.
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