This question already has an answer here:

Draw this without lifting / going on the same line the following figure.enter image description here


marked as duplicate by Marius, JonMark Perry, Community Aug 5 '16 at 10:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @PeregrineRook no , this figure solution is possible $\endgroup$ – Amruth A Aug 5 '16 at 8:50
  • 1
    $\begingroup$ 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. $\endgroup$ – Peregrine Rook Aug 5 '16 at 9:11
  • $\begingroup$ @PeregrineRook fyi, this figure is solvable , see the solution by meta45 below. $\endgroup$ – Amruth A Aug 5 '16 at 10:47
  • 1
    $\begingroup$ @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. $\endgroup$ – Deusovi Aug 5 '16 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.


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:


  • 1
    $\begingroup$ hide the image with >! $\endgroup$ – lois6b Aug 3 '16 at 10:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.