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Perform the following step as many times as possible: Among the edges (..) not yet chosen, choose the shortest edge, which does not form any loops with those edges already chosen—Kruskal, 1956
In this case, the shortest edge is the one with the lowest weight.
|AD and CE are the shortest edges, with length 5, and AD has been arbitrarily chosen, so it is highlighted.|
|CE is now the shortest edge that does not form a cycle, with length 5, so it is highlighted as the second edge.|
|The next edge, DF with length 6, is highlighted using much the same method.|
|The next-shortest edges are AB and BE, both with length 7. AB is chosen arbitrarily, and is highlighted. The edge BD has been highlighted in red, because there already exists a path (in green) between B and D, so it would form a cycle (ABD) if it were chosen.|
|The process continues to highlight the next-smallest edge, BE with length 7. Many more edges are highlighted in red at this stage: BC because it would form the loop BCE, DE because it would form the loop DEBA, and FE because it would form FEBAD.|
|Finally, the process finishes with the edge EG of length 9, and the minimum spanning tree is found.|
- Joseph. B. Kruskal (Feb 1956). "On the Shortest Spanning Subtree of a Graph and the Traveling Salesman Problem" (pdf). Proceedings of the American Mathematical Society 7 (1): 48-50. http://penguin.ewu.edu/class/cscd320/Topic/Graph/Kruskal/PAMS_1956.pdf.