# Draw without lifting / going on the same line? [duplicate]

Draw this without lifting / going on the same line the following figure. • @PeregrineRook no , this figure solution is possible Aug 5, 2016 at 8:50
• You're splitting hairs.  Equivalent questions have been asked and answered before, presenting the theory of how questions like this are answered.  Nothing remains but counting, which is a math exercise and not a puzzle. Besides, your figure is not solvable because it has four odd vertices. Aug 5, 2016 at 9:11
• @PeregrineRook fyi, this figure is solvable , see the solution by meta45 below. Aug 5, 2016 at 10:47
• @AmruthA: No, it's not, since you drew it poorly. The / diagonal needs to be farther left at the bottom for it to be solvable - otherwise, you miss the middle portion between the two intersections.
– Deusovi
Aug 5, 2016 at 16:31

This is an Eulerian Path. Euler said:

Theorem: If a network has more than two odd vertices, it does not have an Euler path.

Euler also proved this:

Theorem: If a network has two or zero odd vertices, it has at least one Euler path. In particular, if a network has exactly two odd vertices, then its Euler paths can only start on one of the odd vertices, and end on the other.

Solution:

Is solvable. As this has 2 odd vertices, you should start in one of them and end on the other. (top and bottom vertices)

Below here one of the many solutions:

• hide the image with >! Aug 3, 2016 at 10:35