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Let us consider the vertex star.
Algorithm 2 (Fig. 3 ▶): Computation of the vertex star of a given vertex i in the -graphs.
Let and let us consider its vertex star in the corresponding -graph, i.e. A comparison of Tables 5 ▶ and 6 ▶ shows that [ i.e. the number of subgroups isomorphic to in, cf. equation (14)] and therefore, since the graph is regular,.
However, since has exactly different subgroups isomorphic to, then at least two vertices in the vertex star, say and, are connected by the same subgroup isomorphic to, which we denote by Q. Therefore we have This implies that, and form a triangular circuit in the graph, which is a contradiction due to Lemma 5.3, hence the result is proved.
This suggests that there is a one-to-one correspondence between elements of the vertex star of and subgroups of isomorphic to ; in other words, if we fix any subgroup P of isomorphic to, then P 'connects' with exactly another representation.
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Each of them has two single 1-simplices {12} and its copy) with common vertex: a star.
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There are only two Penrose tilings (of each type) with global pentagonal symmetry: for the P2 tiling by kites and darts, the center point is either a "sun" or "star" vertex.
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A star at a vertex p (in an ≤n-complex K ) is the ≤n-complex s t K ( p ) = { S ̄ : p ∈ S ∈ K } ; the vertex p is also called a center of a star.
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