| One can re-use a point, but not an edge.
Comments from Hugo van der Sanden (hv(AT)crypt.org), Sep 27 2005:
"The n X n square has #(n) = 2n(n+1) line segments and v_o(n) = 4(n-1)
vertices of odd degree. The network is traversable only when v_o(n) <= 2.
When not, we must lose sufficient line segments to reduce v_o() to 2.
For odd n >= 3, we have an even number of odd vertices along each
edge; we can pair them up and lose the single line segment joining
each pair except for our start/end points. So in this case we have
a(n) = #(n) - (v_o(n) - 2)/2 = 2n^2 + 3.
For even n >= 2, we have an odd number of odd vertices along each
edge. The best we can do is start and end at the two vertices
adjacent to one corner; pairing up the rest allows us to lose a
single line segment for each of the remaining pairs except for
the two vertices adjacent to the diagonally opposite corner, for
which we must lose 2 line segments. So in this case we have
a(n) = #(n) - ((v_o(n) - 4)/2 + 2) = 2n^2 + 2.
For n < 2, we have no vertices of odd degree, so we cannot save a
segment by starting and ending on a pair of them. So we can specify
the exact function using a couple of characteristic functions: a(n) = 2n^2 + 1 + c(n > 1) + c(n odd)
(I'm assuming a(0) = 1 here, on the grounds that there is a single zero-length path in that case.)
Note that the theoretical maximum is always achievable, e.g. using Fleury's algorithm."
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