After many years I finally solved it.
After many years I finally solved it.
Low IQ
Pepe? Is that you?
Oh... The pic looks like a smiling pepe tbhnglno
Mathematically, the problem can be formulated in terms of graph drawings of the complete bipartite graph K3,3. The three utilities problem is the question of whether this graph is a planar graph.
It is possible to show that any bridgeless bipartite planar graph with V vertices and E edges has E ≤ 2V − 4 by combining the Euler formula V − E + F = 2 (where F is the number of faces of a planar embedding) with the observation that the number of faces is at most half the number of edges (the vertices around each face must alternate between houses and utilities, so each face has at least four edges, and each edge belongs to exactly two faces). In the utility graph, E = 9 and 2V − 4 = 8, violating this inequality, so the utility graph cannot be planar.
It is possible to show that any bridgeless bipartite planar graph with V vertices and E edges has E ≤ 2V − 4 by combining the Euler formula V − E + F = 2 (where F is the number of faces of a planar embedding) with the observation that the number of faces is at most half the number of edges (the vertices around each face must alternate between houses and utilities, so each face has at least four edges, and each edge belongs to exactly two faces). In the utility graph, E = 9 and 2V − 4 = 8, violating this inequality, so the utility graph cannot be planar.
I did it too lifefuel
Isn't this also some kind of autism? High IQ though
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