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At first sight it seems that the cross of the two
diagonals with one additional point makes the minimal network. But, in
fact, it isn't. If the side of the square is 1 then the total length
of the cross is 2sqrt2, or about 2.828. With the same side of the
square total length of the network (with two points of intersections)
shown in the illustration is only (1 + sqrt3), or about
2.732, that makes it the minimal possible network to span the four
corners of a square. |
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