# Test case.  In this case, to simplify the ascii-graphics, I'm not
#  actually drawing the resistors; each line is a one-ohm resistor:
#
#	     a (1V)
#	    /|\
#	   / | \
#	  /  b  \
#	 /  / \  \
#	c--d   e--f
#	 \  \ /  /
#	  \  g  /
#	   \ | /
#	    \|/
#	     h (0V)
#
# By symmetry, c, d, e, and f are all at 1/2V, so the c-d and e-f
#  resistors carry no current and can be ignored.  Then the a-c-h path
#  has resistance 2 and current 1/2.  The a-f-h path likewise.  Then
#  b-d-g and b-e-g each have resistance 2, so the b-(d,e)-g diamond is
#  two two-ohm resistors in parallel and thus has resistance 1.  Then
#  a-b-(d,e)-g-h is three one-ohm resistors in series and thus has
#  resistance 3.  Then the whole network is two two-ohm and one
#  three-ohm resistors in parallel, giving it resistance
#  1/((1/2)+(1/2)+(1/3)), or 3/4.
#
# Total current is thus 4/3, with 1/2 passing through a-f-h and 1/2
#  through a-c-h and the remaining 1/3 through a-b-(d,e)-g-h: a-b and
#  g-h carry 1/3 and b-d-g and b-e-g carry half that, or 1/6.

V a 1
R a b 1
R a c 1
R a f 1
R b d 1
R b e 1
R c d 1
R e f 1
R d g 1
R e g 1
R g h 1
R c h 1
R f h 1
V h 0