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In graph theory, the Graham–Pollak theorem states that the edges of an -vertex complete graph cannot be partitioned into fewer than complete bipartite graphs. [1] It was first published by Ronald Graham and Henry O. Pollak in two papers in 1971 and 1972 (crediting Hans Witsenhausen for a key lemma), in connection with an application to ...
The star graphs K 1,3, K 1,4, K 1,5, and K 1,6. A complete bipartite graph of K 4,7 showing that Turán's brick factory problem with 4 storage sites (yellow spots) and 7 kilns (blue spots) requires 18 crossings (red dots) For any k, K 1,k is called a star. [2] All complete bipartite graphs which are trees are stars.
The coefficients found by Fehlberg for Formula 1 (derivation with his parameter α 2 =1/3) are given in the table below, using array indexing of base 1 instead of base 0 to be compatible with most computer languages:
In numerical analysis, the Runge–Kutta methods (English: / ˈ r ʊ ŋ ə ˈ k ʊ t ɑː / ⓘ RUUNG-ə-KUUT-tah [1]) are a family of implicit and explicit iterative methods, which include the Euler method, used in temporal discretization for the approximate solutions of simultaneous nonlinear equations. [2]
An interpretation of Plato's number is a solution for k = 3. For this special case of m = 1, some of the known solutions satisfying the proposed constraint with n ≤ k, where terms are positive integers, hence giving a partition of a power into like powers, are: [3] k = 3 3 3 + 4 3 + 5 3 = 6 3 k = 4 95800 4 + 217519 4 + 414560 4 = 422481 4 ...
Kotzig's conjecture was first listed as an open problem by Bondy & Murty in 1976, [3] attributed to Kotzig and dated to 1974. [1] Kotzig's first own writing on the conjecture appeared in 1979. [ 4 ] He later verified the conjecture for k ≤ 8 {\displaystyle k\leq 8} and claimed solution, though unpublished, for k ≤ 9 {\displaystyle k\leq 9 ...
where a i ≠ b j are positive integers for all 1 ≤ i ≤ n and 1 ≤ j ≤ m, then m + n ≥ k. In the special case m = 1, the conjecture states that if = = (under the conditions given above) then n ≥ k − 1. The special case may be described as the problem of giving a partition of a perfect power into few like powers. For k = 4, 5, 7, 8 ...
G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [17] the largest number now known not to be a sum of ...