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The section points are houses and lines are streets, all with one unit length. What is the fewest number of units you must travel to visit every street at least once?

a 5-unit size 30-60-30-60 rhombus with two opposite ends one triangle off, divided into one-unit triangles

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  • $\begingroup$ To "visit" a street do we need to walk its entire length or is just touching an endpoint enough? $\endgroup$ – Paul Panzer Jan 3 at 20:31
  • $\begingroup$ Visiting w street actually means just walking along a one unit long edge $\endgroup$ – balazs.com Jan 3 at 21:54
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There are 52 edges, so that is a lower bound. There are six odd-degree nodes. If you choose the two middle ones to be the endpoints of the overall path, the other four can be paired up with distance 1+1. Adding these two edges yields

52+2=54 edges.

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  • $\begingroup$ Rot13(V qryrgrq zl nafjre orpnhfr V zvffrq gur snpg gung gjb cnvef bs bqq-abqrf ner pbaarpgrq, naljnl lbh jrer nobhg 1 frpbaq orsber zr.) $\endgroup$ – Weather Vane Jan 3 at 20:21
  • $\begingroup$ @RobPratt Sorry, somebody mistakenly closed this question (First it says "off-topic" then "This looks like a puzzle you found elsewhere. For content you did not create yourself, proper attribution is required. If you have permission to repost this, please edit to include (at minimum) where it came from, then vote to reopen. Posts which use someone else's content without attribution are generally deleted." While none is the real case. Anyway, the answer is correct, congrats!! $\endgroup$ – balazs.com Jan 3 at 20:37
  • $\begingroup$ @balazs is that your own website 'not quite' readable in the bottom left corner of the image? $\endgroup$ – Weather Vane Jan 3 at 20:40
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    $\begingroup$ @balazs.com If it's not the case, you can simply edit the question accordingly and vote to reopen, as the comment says. The image clearly contains a blurry address of a website. $\endgroup$ – Lukas Rotter Jan 3 at 20:49
  • $\begingroup$ @Lukas Rotter yeah I see it now, that might be a part of the original image I used when putting up this image. But I came up with the whole problem. Anyway, the solution is correct (I mean... I have the same solution) and no need to draw the path...which is most likely very difficult :)So.. I think that was the right idea..to connect the neighbour odd nodes, that needs exactly +2 edges, which is the least number of addition to a hypothetical map that really has the Eulerian path. Just as RobPratt said. Congrats again! $\endgroup$ – balazs.com Jan 3 at 20:55

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