The only patterns which can't be
drawn with pencil in one continuous line so that you don't take the
pencil point off the paper are as follows: 2, 3, 4, 8, 12.
Possible solutions to the remaining patterns are shown in the
Now using Figure 4 we'd like to explain why it's impossible to draw it
with pencil in one continuous line so that you don't take the pencil
point off the paper, or why it isn't unicursal.
Figure 4 has exactly four points (nodes) where an odd number of lines
are branching out (5 at each), and one node in the center of the
pattern with an even number of branches (4).
Every time you go through a node not stopping at it you must of
necessity use a pair of its branches. Therefore at each of the four
nodes on the periphery of the pattern one branch in any case will be
alone. When we use this alone branch this means that our line either
starts from this node or just finishes at it. Thus we have FOUR points
(nodes) where the line has to start (or finish) doesn't matter how to
draw it. But a continuous line has only TWO ends, so the puzzle can't
Same proof is true and for the rest of "impossible" figures in our set
- 2, 3, 8 and 12.
ANY figure that has only TWO points (nodes) where an odd number of
lines are branching out, and ANY number of its nodes with an even
number of branches, CAN BE DRAWN in one continuous line. You just have
to start at one "odd" point and finish at the other. See Figures 6, 7,
10 and 11 (figure 11 has two "free" ends which are "odd" nodes too,
but just with one branch each).
And finally an excellent thing about all unicursal figures is
that you ALWAYS can draw in one continuous (and even closed in loop!)
line ANY pattern if ALL its nodes are "even." See Figures 1, 5 and 9.