Приклади вживання Number of vertices Англійська мовою та їх переклад на Українською
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N:= number of vertices in Graph.
Print the desired number of vertices.
Find the number of vertices in the connected component where the selected vertex(including the selected vertex). .
The first line contains the number of vertices in a graph n(1≤ n≤ 100).
Find the area of a green spotthat is obtained with superposition of two spots and the form(number of vertices) of a green spot.
The size and number of vertices depends on your desire;
If G is a regular graph of degree d whose edge connectivity is at least d- 1,and G has an even number of vertices, then it has a perfect matching.
First line contains number of vertices n(n≤ 100) and edges m of undirected graph.
Every Halin graph has a Hamiltonian cycle through all its vertices, as well as cycles of almost all lengths up to their number of vertices.
The first line contains the number of vertices n(1≤ n≤ 100) in the graph.
For dense graphs, it is more efficient to use this simple algorithm which relies on matrix multiplication, since it gets the time complexity down to O(n2.373),where n is the number of vertices.
The first line contains the number of vertices n(1≤ n≤ 100000) in a tree.
A strengthened version of the conjecture states that each such graph has an equitable coloring with exactly Δ colors, with one additional exception, a complete bipartite graph in which bothsides of the bipartition have the same odd number of vertices.[1].
QGIS doesn't have a built-in function to calculate number of vertices for each feature in a layer.
The condition on the number of vertices can be omitted from this result when the degree is odd, because in that case(by the handshaking lemma) the number of vertices is always even.
This can onlyoccur when the graph has an odd number of vertices, and such a matching must be maximum.
These include the bipartite graphs, the chordal graphs, the comparability graphs, the distance-hereditary graphs(in which shortest path distances in connected induced subgraphs equal those in the whole graph),and the wheel graphs that have an odd number of vertices.
The first line contains the number of vertices n(1≤ n≤ 100) and the number of edges m of the graph.
Each step increases the number of consecutive pairs in the cycle that are adjacent in the graph, by one or two pairs(depending on whether vj and vj+ 1 are already adjacent), so the outer loop can only happen at most n times before the algorithm terminates,where n is the number of vertices in the given graph.
The first line contains the number of vertices in a graph n(1≤ n≤ 100) and the number of edges m(1≤ m≤ n(n- 1)/ 2).
By the Davenport- Schinzelargument, the lower envelope in the worst case has Θ( n α( n)){\displaystyle\Theta(n\alpha(n))} number of vertices, where α( n){\displaystyle\alpha(n)} is the inverse Ackermann function.
Like all regular polygons, their vertices lie on a circle. m also corresponds to the number of vertices around the circle to get from one end of a given edge to the other, starting at 1.
By a theorem of Descartes,this is equal to 4π divided by the number of vertices(i.e. the total defect at all vertices is 4π).
These algorithms run in O(| V|+| E|){\displaystyle O(| V|+| E|)}, or linear time,where V is the number of vertices, and E is the number of edges between vertices. .
A regular star polygon is represented by its Schläfli symbol{n/m}, where n is the number of vertices, m is the step used in sequencing the edges around it, and m and n are co-prime(i.e. have no common divisor).
An adjacency matrix is preferred if the graph is dense, that is the number of edges|E| is close to the number of vertices squared,|V|2, or if one must be able to quickly look up if there is an edge connecting two vertices.[5][6].
The following table gives the time complexity cost of performing various operations on graphs, for each of these representations,with|V| the number of vertices and|E| the number of edges.[citation needed] In the matrix representations, the entries encode the cost of following an edge.
No two adjacent vertices have the same color, and The numbers of vertices in any two color classes differ by at most one.
Pál Turán's"brick factory problem"asks more generally for a formula for the minimum number of crossings in a drawing of the complete bipartite graph Ka, b in terms of the numbers of vertices a and b on the two sides of the bipartition.
The first line contains the number of graph vertices n(1≤ n≤ 100).