Draw this without lifting / going on the same line the following figure.
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$\begingroup$ @PeregrineRook no , this figure solution is possible $\endgroup$– Amruth AAug 5, 2016 at 8:50
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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 RookAug 5, 2016 at 9:11
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$\begingroup$ @PeregrineRook fyi, this figure is solvable , see the solution by meta45 below. $\endgroup$– Amruth AAug 5, 2016 at 10:47
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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, 2016 at 16:31
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2 Answers
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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)
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Below here one of the many solutions:
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