Challenge: Can you draw this figure in a single line?

Originally published at: Challenge: Can you draw this figure in a single line? | Boing Boing


Can you draw this figure in a single line?

Yes. I can.

(Oh, is there a video too? Perhaps I’ll watch.)

ETA Now I’ve watched, there is more than one solution.


Thanks for this - I feel better about myself after feeling like an idiot spending 4 turns on wordle today.


Yes. It’s a little hard to explain without visual cues, but what you do is you start in the middle and go zhwip zhwip zwhip, then around the thing, around the thing, but not there, then you go zoop zoop zoop and back to the other thing you didn’t do before.


Yes, and their solution isn’t even the only one. It took me longer to get my stupid pen to start actually working than to draw it.
First, you draw the center triangle, then one side of one of the petals leading outward, followed by the outer rims looping all of the way around the outside (leading away from the petal you started), then down the other side of the first petal, then back out, in, out, and in along the other two inner petals. Not only not repeating over any lines, but also not crossing any lines outside of touching vertexes.

Yes. Can I do it well? Meh.

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A better question might be: Can you NOT draw this without lifting pen or retracing, because frankly, it’s incredibly simple


So Close Nbc 90Th Special GIF by NBC


Totally. And of course I’m biased, but I thought my initial solution was more elegant. I drew the clover shape of the exterior, then followed the third circle around to completion and kept on for the other two. Had a nice flow, and made the shape easier to get symmetrical than the way they showed, at least for me.


Why would you need to cross a line? Start at the base of the top ginkgo-shaped element and draw it in three curves, then take a short trip to the base of one of the other ginkgos, repeat twice, and you’re back at the starting point.

Cool, we’re doing “You solved it wrong!” Again.


Shropshire accent “And the answer is yes. Yes we can.”

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Yep. I’m struggling to find any path through the figure that would get you stuck and unable to complete it…

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In high school, I read about how Euler tackled the “7 bridges of Königsberg” problem and invented topology in the process. When you read that story, any such problem becomes straightforward:

Any node (crossing) where an odd number of lines meet must be the start or end. (A problem like this is only impossible if there are more than two nodes that each have an odd number of lines, as was the case in Königsberg). If there are two or zero nodes with an odd number of lines, then the problem is possible.

In this case, you have six nodes (three inner, three outer) where four arcs meet. Any of them can be the start-and-end-point (as can the middle of any arc, which can be thought of as a node where two arcs meet).

For example, start at a node, and draw one entire circle, back to where you started. Then start drawing the second circle… but when it hits the third circle, draw the entire third circle, and then finish the second circle.

If you start in the middle of an arc; same idea, except you won’t finish the last bit of your first circle until the very end. While drawing the first circle, when you hit a node, go off into the second circle, then at one of the points where it hits the third circle, draw the whole third circle at once, then finish the second, then finish the first.

Edit: And that’s not even including the needlessly complicated solutions where you draw the arcs in each circle in a less consecutive way.


Draw it three times in a row and shout “When the Levee Breaks”. You will be visited by the drunk ghost of an amazing drummer.


I’ve now solved it a couple of different ways, but none of my answers look like a f’ing owl.

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All the vertices have even degree, so yes, there is a theorem that says you can do it.

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I managed in nine seconds but a different method. I’m rubbish at this stuff usually… I’m out for now

Twelve non-crossing segments, about 20 seconds solve time.

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