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The degree diameter problem seeks tight relations between the diameter, number of vertices, and degree of a graph. One way of formulating it is to ask for the largest graph with given bounds on its degree and diameter. For any fixed degree, this maximum size is exponential in the diameter, with the base of the exponent depending on the degree. [1]
The surface area, or properly the -dimensional volume, of the -sphere at the boundary of the (+) -ball of radius is related to the volume of the ball by the differential equation
In graph theory, the degree diameter problem is the problem of finding the largest possible graph for a given maximum degree and diameter.The Moore bound sets limits on this, but for many years mathematicians in the field have been interested in a more precise answer.
An example of a spherical cap in blue (and another in red) In geometry, a spherical cap or spherical dome is a portion of a sphere or of a ball cut off by a plane.It is also a spherical segment of one base, i.e., bounded by a single plane.
The size of G is bounded above by the Moore bound; for 1 < k and 2 < d, only the Petersen graph, the Hoffman-Singleton graph, and possibly graphs (not yet proven to exist) of diameter k = 2 and degree d = 57 attain the Moore bound. In general, the largest degree-diameter graphs are much smaller in size than the Moore bound.
The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m −1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
In forestry, quadratic mean diameter or QMD is a measure of central tendency which is considered more appropriate than arithmetic mean for characterizing the group of trees which have been measured. For n trees, QMD is calculated using the quadratic mean formula:
Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2) 2. Solving for r, we find the required result.