Four Dragon curves generating outwards from the same vertex will traverse every edge of a grid exactly once (and as a consequence will be plane-tiling as well).

enter image description here I am captivated by this fact, and somewhat wishfully suspect that there is a simple and illuminating explanation for it. If this is not the case, however, can anybody direct me to a resource where this is discussed in detail?...I cannot find the original articles by Chandler and Donald J. Knuth. "Number representations and dragon curves", on jstor.org :(

Here are a few quick links for reference:



Historically, Dragon Curves were discovered in 1969 by two NASA engineers who were interested in the pattern produced by folding a piece of paper repeatedly and then unfurling it such that all the folds were manifested as right angles, a fundamental construction which I believe to be part of the key to intuitively understanding all these properties...

  • $\begingroup$ Very interesting! But could you include a link or a definition of "Dragon Curves" $\endgroup$
    – Mike
    Jan 6, 2019 at 20:28
  • $\begingroup$ Certainly! I'll add it right now, my preferred definition is the one from Wolfram which describes concatenation of binary strings, but there are many equivalent definitions as well @Mike $\endgroup$ Jan 6, 2019 at 20:35
  • $\begingroup$ A paper by Christoph Bandt explains how these types of constructions arise from expanding integer matrices. $\endgroup$ Jan 7, 2019 at 1:04

1 Answer 1


Here’s the substitution rule that generates the dragon curve. It replaces each arrow with two smaller rotated arrows with particular orientations.

figure 1

If we apply the same substitution rule to this infinite grid of arrows, we happen to get a smaller rotated copy of the same grid. After seeing it happen once, it’s obvious that it will continue to happen infinitely many times.

figure 2

Now we can color four arrows of the initial grid and watch them grow into four dragons:

figure 3

We’ve generated exactly the figure that you posted. But this construction makes it clear why they have to fit together this way—each arrow in the grid must be used exactly once.

  • $\begingroup$ This is exactly the concise and insightful explanation I hoped to find! Thank you :) Could elaborate just a bit more on why this proves that every arrow in the grid must be used exactly once? I am almost but not quite understanding $\endgroup$ Jan 8, 2019 at 0:32
  • 1
    $\begingroup$ You can see that each arrow is used exactly once in the black and white grid, just by applying the substitution rule and seeing what happens. All we’ve done to get from the black and white grid to the colored picture is recolor some arrows. Does this make sense? Which part are you stuck on? $\endgroup$ Jan 8, 2019 at 1:51
  • $\begingroup$ The part I was stuck on was seeing why every arrow in the grid ends up colored (why we don't ever "miss" an arrow as the curves grow). I think that I understand now that this is because all four arrows in the starting position form a complete, bijective coloring of the interior of the initial 2x2 unit square. Because of the scaling action of our iterative process, this square grows to be infinitely large while retaining its complete (interior) coloring? Thanks for bearing with me – Math Enthusiast Jan 8 at 6:30 $\endgroup$ Jan 13, 2019 at 22:14
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    $\begingroup$ @MathEnthusiast Yeah…is that a question? I drew these pictures at the same scale, with the individual arrows getting smaller rather than the initial square getting bigger, but I suppose you could think of them either way. $\endgroup$ Jan 13, 2019 at 22:42
  • $\begingroup$ Ok thanks :) I thought you didn't see it sorry $\endgroup$ Jan 15, 2019 at 1:11

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