Sentence examples for be a graph containing from inspiring English sources

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Let G be a graph containing a pendant vertex, and let H be the induced subgraph of G obtained by deleting the pendant vertex together with the vertex adjacent to it.

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The result of HERCULE is a graph containing all the events related to a multi-step attack.

But it had been shown in [16] that if E is a graph containing at least one cycle, then L K ( E ) is not von Neumann regular, so, in particular, L K ( 1, 2 ) ≅ L K ( R 2 ) is not von Neumann regular.

Let p be an irrational infinite path in E. Then Ext L K ( E ) 1 ( V [ p ], T ) ≠ { 0 } if and only if r ( X i ( p ) ) ∩ U ( T ) ≠ ∅ for infinitely many i ≥ 0. As a consequence of Theorem 26, whenever E is a graph containing at least one cycle, then (non-projective) indecomposable L K ( E ) -modules of any desired finite length can be constructed.

First, we note that a protein's annotation (experimental or predicted) is a graph containing a subset of nodes in the ontology together with edges connecting them.

A DAG is a graph containing only directed edges and no cycles, and the skeleton of a DAG is the DAG itself where directionality has been removed.

If a network contains three edge-disjoint spanning trees, for example, information can flow in parallel along each of these trees at the same time, meaning three times more bandwidth than would be possible in a graph containing just one tree.

All Pearson correlations with r ≥0.7 were saved to a '.pearson' file and a correlation cut off of r = 0.8 was used to construct a graph containing 20,355 nodes (probesets) and 1,251,575 edges (correlations between nodes above the threshold).

Figure 11(b) shows an alternative representation of the same structure, which is restricted to use of a graph containing only atom nodes and bond edges.

That is, a 1-tree is a connected graph containing exactly one cycle.

Lemma 2.6 Let G σ be an oriented graph containing no even cycles and G 1 σ a spanning subgraph (resp. proper spanning subgraph) of G σ. Then G σ ⪰ G 1 σ (resp. G σ ≻ G 1 σ ).

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