Mama snake wants to knit a blanket for little baby snake.

She is not a dissipater and wants to make the blanket of a minimal size (area). But her baby snake is quite a lively baby and it always twists and stretches and turns around while sleeping.

If the baby changes position or from, mama takes the blanket and puts it over the baby again.

What size (and form) is the blanket ?

  • The baby snake has the length of 1
  • While sleeping the baby lays on the bed (it's a 2D-puzzle, no 3D)
  • The baby snake can take any shape it wants
    • it can have bends, even sharp bends like 120°
    • it could also be shaped like a wave, a ring or a U-form
    • it could be any shape you can imagine (even a spiral or square)
  • The blanket must always cover any shape of the snake
  • Mama turns and moves the blanket if necessary

mama snake baby snake

Mama snake and her little baby snake



This is biggest solution (from Miniman):

enter image description here

It's a circle with diameter 1 (you can see the red baby snake stretched out totally - and you can also see the baby snake performing a circle).

Semicircle small

A smaller version would be a blanket of this size and shape (a semicircle).


You see how the snake performs a bending like a U (red outline) and the resulting size of the blanket (black line). Unluckily this blanket would be too small, because if the snake bends into a 90°L-form it wouldn't fit into the blanket anymore.

Semicircle big


That great idea comes from Joe. Thanks for providing that idea! a semicircle with the diameter of 1... This definitely covers the baby snake but as you can see could easily be cut of on several parts...

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    $\begingroup$ it's not a 'rubber-blanket' @JanDvorak . it can't be stretched and it can't be shrinked, not by any purpose... $\endgroup$ – Martin Frank Oct 9 '14 at 9:03
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    $\begingroup$ I think it would be better to just let answers be answers, instead of updating the post to incorporate them all. $\endgroup$ – SQB Oct 9 '14 at 11:01
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    $\begingroup$ You could edit the answers and include the pictures there. $\endgroup$ – Moyli Oct 9 '14 at 11:04
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    $\begingroup$ This now has the distinct feel that it belongs on Math.SE, and the "logic puzzle" tag is incorrect. It's just math and geometry $\endgroup$ – Joe Oct 9 '14 at 21:59
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    $\begingroup$ It's an unsolved mathematical problem. See Wikipedia and Mathoverflow. $\endgroup$ – sth Oct 10 '14 at 21:06

17 Answers 17


As remarked by sth, this is an open problem, a topic of ongoing research. It is known as Moser's worm problem. The Math Overflow thread Smallest area shape that covers all unit length curve cites some recent result.

A known result is that the smallest possible convex blanket has an area between 0.227498 to 0.231999. (In comparison, a half-circle of diameter 1 has an area of about 0.3927.)

Reference: Tirasan Khandhawit, Dimitrios Pagonakis, Sira Sriswasdi. Lower Bound for Convex Hull Area and Universal Cover Problems. Int.J.Comput.Geom.Appl. 23 (2013) 197-212. arXiv:1101.5638. DOI: 10.1142/S0218195913500076. PDF

If the blanket is not required to be convex, smaller blankets may be possible. A special case of interest is if the snake only consists of two straight segments with a bend at a position that may vary, in which case the blanket can be a Kakeya set which can have zero area. It is even true if the snake consists of finitely many straight segments, see the works of Ward and of Davies. For real-analytic curves, the area cannot be zero by a work of Marstrand, however it appears to be not known if it can be arbitrarily small.

  • $\begingroup$ Now I know why I can't remember the solution! Oh I wish I had known, and not asked that question here :-$ I really feel bad somehow now :-( $\endgroup$ – Martin Frank Oct 13 '14 at 4:35
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    $\begingroup$ I would like to add that the lower bound quoted is for the problem where the blanket is convex. I am not sure whether existence of a zero-area set has been ruled out. (Indeed, if the baby snake is straight apart from finitely many bends, then answer is zero. It is a basically the famous Kakeya problem.) $\endgroup$ – Boris Bukh Oct 26 '14 at 0:15
  • $\begingroup$ @BorisBukh How does a solution based on a Kakeya set work for, say, three segments? Note that the length of the middle segment varies continuously. By the way, this answer is community wiki, feel free to edit it directly. $\endgroup$ – Gilles 'SO- stop being evil' Oct 26 '14 at 14:21
  • $\begingroup$ @Gilles. Edit done. ("Basically" meant that with some work one can modify the known constructions of Kakeya sets.) $\endgroup$ – Boris Bukh Oct 26 '14 at 15:17
  • $\begingroup$ A zero area blanket? Your kids must love story time. =( $\endgroup$ – corsiKa Nov 3 '14 at 18:26

EDIT: After some reflection and help/counter-examples seen in some other answers, I think this is either very close to, or is the correct answer.

What about half of a square with a diagonal of $1$, cut along the diagonal?

For example:

Square Diagonal

Using one of the triangles as the blanket, the long side (i.e. the diagonal of the square) is of length $1$, and the other two sides are length $\sqrt{\frac{1}{2}}$.

The area of this is $\frac{(\sqrt{\frac{1}{2}})^2}{2}=\frac{1}{4}$, and I think all shapes of snakes would fit in it.

Some common snake shapes that will fit:

  • A square of side length $\tfrac{1}{4}$
  • A circle
  • A "U" shape (easy to prove if it is symmetrical, and if it is not symmetrical, then it is covered by a smaller area)
    • e.g. where all three sides are length $\frac{1}{3}$
  • An "S" shape - head goes left 90 degrees, tail goes right 90 degrees

As pointed out by Mooing Duck, if you orient this triangle with its long edge down as follows: enter image description here

then its height is $\tfrac{1}{2}$, but you don't need all that height. The tallest shape the snake can make is when it forms 2 sides of an equilateral triangle. So, if we crop off that portion of the triangle, we end up with a truncated triangle (looks similar to the half hexagon in DenDenDo's answer!). It looks like follows: enter image description here

Its area is now $\tfrac{1}{4}$ minus the amount we just removed. The height of an equilateral triangle of side length $\tfrac{1}{2}$ is $0.433$ ($\sqrt{\tfrac{1}{2}^2-\tfrac{1}{4}^2}$), so the triangle we just removed has height $0.067$. The area of that triangle is $0.0045$.

Thus the area of the remaining blanket is ~ $0.2455127...$


After some more reflection, I wonder if this shape needs to be so symmetrical. You can always flip the blanket over. For example, the left side of the above blanket can cover a snake bent into 2 sides of an equilateral triangle, but the right side isn't used. If you make a "U" shape from the snake with the sides of length $\frac{1}{3}$, then you only need a portion of the right side.

For example, the following blanket still covers both shapes: enter image description here

But the upper right portion doesn't seem necessary. Can it be simply removed? Here is the shape again with some other snakes covered by it.
enter image description here

NOTE: Does the blue equilateral snake need to be oriented that direction? If we orient it so that the missing third line is on the right, can we recover some of the blanket?

Likewise, the brown "U" shape could have the opening facing up, so we could recover some of the interior of the "U".

Also, the green snake is $3$ segments each of length $\frac{1}{3}$. If the middle segment is longer, then the overall shape isn't as tall. If the middle segment is shorter, then it also begins to get shorter, so this one is the tallest of these types, so they will never use the space above the brown "U".

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    $\begingroup$ What does "half a square with a diagonal of 1" mean exactly? I can't think of an interpretation with an area of 1/4 that also covers a snake in a straight line. $\endgroup$ – Mooing Duck Oct 9 '14 at 23:08
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    $\begingroup$ @MooingDuck His "half of a square with diagonal 1" is a "right isosceles triangle with hypotenuse of length 1". $\endgroup$ – Alexander Oct 10 '14 at 7:43
  • $\begingroup$ I agree with the triangle idea. If the snake bends, any bend will reduce it's length by at least twice the length of the side protrusion, so you can move the baby away from the longest border and he's still covered $\endgroup$ – algiogia Oct 10 '14 at 8:11
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    $\begingroup$ you say the two sides are $\tfrac{1}{2}$ but they are actually $\sqrt{\tfrac{1}{2}}$ $\endgroup$ – Ivo Beckers Oct 10 '14 at 13:23
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    $\begingroup$ How does your triangle cover a snake like this, with three segments and angles close to 180 degrees? $\endgroup$ – Lopsy Oct 10 '14 at 18:58

What about a rectangular blanket that is 1 x circumference of baby snake?

This rectangular blanket could be wrapped around the baby snake (think tube sock) and stretch and fold in any direction with him.

The big bonus here is that mama snake doesn't have to tirelessly keep replacing the blanket, the blanket will go wherever baby snake goes. :)

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    $\begingroup$ nice thinking, but the snake lies in it's bed and should be covered by that blanket (not wrapped) :/ but i'm sure that would be a really nice baby snake in that tube =) $\endgroup$ – Martin Frank Oct 9 '14 at 11:43
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    $\begingroup$ I don't understand why the obviously wrong answer gets this many upvotes. $\endgroup$ – Moyli Oct 10 '14 at 6:32
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    $\begingroup$ Who said is wrong? I think is the best solution. When the baby bends you just bend the blanket to the same shape... $\endgroup$ – algiogia Oct 10 '14 at 8:14
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    $\begingroup$ @algiogia It is wrong because the problem specifically has to be solved in 2D, and this answer involves a Z-axis. $\endgroup$ – Kheldar Oct 10 '14 at 12:15
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    $\begingroup$ It got so many votes becasue it is thinking outside the box. Almost literally in that it solves the 2D problem in 3D $\endgroup$ – Lyndon White Oct 10 '14 at 23:26

Examining all of the answers leads to interesting insights, but my intuition is now telling me that there isn't (an obvious) closed-form solution to this puzzle.

My best attempt at a solution is the union of all line segments of length 1/2 whose endpoints touch each of the x-axis and y-axis simultaneously, and are mirrored over the y-axis. This would look like Trenin's "square-minus-quarter-circle-mirrored" answer. I am fairly sure they are the same shape, but I lack proof.

Unfortunately this solution must be incomplete. Consider if the baby looped itself into any ellipse of unit perimeter--or worse: an ellipse with a break in it. For simplicity* consider a circle. The diameter of such a circle is 1/pi. It turns out a circle of such diameter is too large to fit into the shape described by my answer.

I've used MATLAB to whip up a figure of what I mean by the above descriptions:

enter image description here

Additionally, the solution could be truncated per Mooing Duck's comment on Trenin's answer, and possibly truncated further depending on the number of bends--triangles aren't the only possible bend configuration.

*As a side note, I'll point out that determining the ratio of the radii of ellipses of constant perimeter, given the length of one radius, is quite complicated, involving the inverse complete elliptic integral of the second kind. I found a journal article and related Fortran code for the inverse if anyone is interested (links below now, thanks for the upvotes!). To prove a solution is minimal would require checking every ellipse of unit perimeter, since the circle doesn't fit. This is why I don't think there is a closed-form solution to this problem. Unfortunately both determining a minimum blanket and proving that it is minimal are beyond the time I'm willing to devote to the problem, but may be the subject of a publishable journal article for an ambitious sort.

Edit: Link to article

Link to code used by author of article, as text file

  • $\begingroup$ Nice analysis. Up voted to get your rep up! Post away! $\endgroup$ – Trenin Oct 10 '14 at 11:57
  • $\begingroup$ Awesome! This certainly appears to be the same shape Trenin and I thought of, and that graph is solid proof that that it doesn't cover the circle. $\endgroup$ – Mooing Duck Oct 10 '14 at 16:26
  • $\begingroup$ I realize this is not complete, but if the circle union this shape works, isn't that an answer? $\endgroup$ – SrJoven Oct 16 '14 at 12:08
  • $\begingroup$ en.wikipedia.org/wiki/Kakeya_set $\endgroup$ – sova Oct 27 '14 at 7:41

I am posting a new answer, building from Trevin's trapezoid answer. I drew a diagram of the answer using Creo (which has some handy-dandy constraint mechanisms built in). Creo is a CAD tool which has a constraint-and-dimension-based sketcher, which is what I am using here. Solid yellow lines are outlines of the sketch itself, and dotted lines are construction lines. Construction lines are not part of the actual part being sketched, but allow for additional constraints to be made on the sketch. Solid blue lines are dimension leaders which tell the dimension of some feature, solid red lines are locked dimensions (so I don't mess up the sketch accidentally by clicking and dragging something). Some aren't locked because they are parameterized using mathematical relationships, as with the circle diameter, which is set equal to 1/pi. Filled yellow areas are considered solid two-dimensions sketch regions, and can be used to make three-dimensional features from. Here they are used for denoting the blanket.

original answer

If we consider a square inside the triangle of side-length 1/2, this actually takes care of the circle problem as well, and can improve the trapezoid's area.

square encompassing unit circumference circle

The square is the highest a quadrilateral of perimeter 4/3 can be (4/3 = 1 for snake + 1/3 for break). If it goes taller, turn it on its side and should still fit. If we drop the height of the square, then the two vertical sides will angle outwards. If we keep doing that, eventually it will become a straight line of length 1. If instead we increase the length of the top, then the other two sides must get shorter. Either way it will fit inside of a 45-45-90 triangle of base length 1. If we trim the sides down we get:

trimmed triangle

I am fairly sure that this area is minimal for all quadrilaterals the snake can take the shape of. As a heuristic argument, I ran MATLAB code for many symmetric quadrilaterals and they converge to a 45-45-90 triangle. It can't be seen in the figure and I am hesitant to make this answer bulkier, but each stair-step is for a different iteration of the length of the vertical legs. Then they are iterated over angle between vertical and horizontal. More iterations looks more like a 45-45-90 triangle. If we trim off the top to account for turning tall quads on their sides, then we get the solution in my third image.

matlab triangle

This solution does not consider pentagons or higher-order polygons in general. It turns out if you put a pentagon with one edge along the bottom, it sticks out into the region between the square and the equilateral triangle just a bit. Hexagons and heptagons do as well. However, none of them take up the entire space there. I outlined the regions that are likely to be improved upon in some way (outline in yellow, see first paragraph):


Height is sqrt(3)/4, base is 1, top is ( 1 - 2 * ( sqrt(3)/4 ) ). Multiply height by average of bases. The area of the 45 degree trapezoid I came up with is thus about 0.2455 square units.

  • $\begingroup$ You might mention that the "likely to be improved areas" are highlighted yellow. (As opposed to the bright blue and red lines) Otherwise this answer is incredible. $\endgroup$ – Mooing Duck Oct 10 '14 at 16:42
  • $\begingroup$ Thank you for the suggestion (and compliment!), which I've edited in. I am a math and engineering person in academia so I sometimes take details for granted. My apologies for any confusion! $\endgroup$ – wwarriner Oct 10 '14 at 17:06
  • $\begingroup$ I agree you in the height of sqrt(3)/4 and the 45° angles on the sides. You probably can cut some of the yellow area. But Where would you place the snake: straight 1/3, turn 60° left, straight 1/3, 60° right, straight 1/3? I think "long S" shapes are still a problem. $\endgroup$ – Florian F Oct 11 '14 at 16:50
  • $\begingroup$ @FlorianF: both ends of the snake on the bottom edge, centered horizontally. His "height" is only 0.259 in the middle 0.33333 of the blanket, which has a height of 0.33333, so he fits with lots of room to spare. $\endgroup$ – Mooing Duck Oct 13 '14 at 23:12
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    $\begingroup$ @MooingDuck Unfortunately reading the comment by user sth on the question indicates that the problem is currently unsolved and quite difficult. Link: Wikipedia. It was fun building on each others answers and comments though, even if it is unlikely we will come up with "the" solution. $\endgroup$ – wwarriner Oct 13 '14 at 23:48

I'm not math-y enough to even begin to go about proving this, but just based on logic I believe you can just cut the circle in half to give you a semicircle with a diameter of $1$ unit, which would have an area of $\tfrac{1}{8} \pi$ (a circle with a diameter of $1$ has an area of ${\pi \cdot\tfrac{1}{2}}^2 = \tfrac{1}{4}\pi$).

This will fit the longest possible layout along the diameter, and I don't think it should be possible for the snake to lie in such a way as to break out of the semicircle.

If anyone can provide any kind of proof (or disproof), feel free to edit my post.

Here's an illustration of the blanket, provided by @MartinFrank. The red lines indicate possible snake positions.

enter image description here

  • $\begingroup$ i've added a picture that should display your idea... but there must be a smaller one i think... $\endgroup$ – Martin Frank Oct 9 '14 at 8:32
  • $\begingroup$ It's worth mentioning that the total area of such a blanket is ~0.39269908 $\endgroup$ – Mooing Duck Oct 9 '14 at 23:06

Half-Hexagon, area <= sqrt(3)*3/16 = 0.3927

I think the optimal shape is either a half hexagon or something slightly smaller.
Assume the head of the snake is fixed and it extends to the right along the x-axis, otherwise we just rotate it.

If the snake bends, we can assume the first bend goes up, otherwise we mirror the snake (or flip the blanket over). In worst case it bends in the middle and once it bends sharper than an equilateral triangle, we can align the long side of the triangle on the x-axis and the snake is covered again. if there is only one bend, we can get away with an even smaller droplet-shaped area which is the envelope of several triangles

If the snake bends more often it will only get shorter, which makes it even easier to cover. Again we align the longest side of the bounding polygon on the x-axis and the snake is covered. It needs more area than the shape from before, but still slightly less than a full half-hexagon.

The proof to this is pure intuition and playing around with different shapes in Inkscape

enter image description here

Edit: In the first picture the snake is simply straight.
In the second it can bend once, so it goes half the distance to the right and then in any direction. On the left are the required blankets (partially hiding each other) on the right they are rotated to fit in less then a half-circle.
Same thing with three segments of length 1/3

This is the hexagon I mean, with all helper circles and shadings removed
enter image description here

  • $\begingroup$ You haven't specified the size of the hexagon. You need the long side to be $1$ to take care of a straight worm, so let's assume that. I can still take a segment of length $1$ and make two perturbations near one end, one up ad one down, and it wont fit, so at one end the shape has to meet the line at $90$ degrees. $\endgroup$ – Ross Millikan Oct 10 '14 at 3:21
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    $\begingroup$ @Ross but by bending it you make it shorter, so you dont need to put the corner of the shape into the corner of the hexagon. By bending the endpoints of length $x$ by 90° you get a bounding parallelogram with longest side $L(x)=\sqrt{(1-2x)^2+x^2}$, angle $\phi(x) = \arctan(\frac{x}{1-2x})$ and height $h(x) = x \sin(\phi (x))$. Now we need to make sure that $L(x) < W(h(x)) = 1-\frac{h(x)}{\sqrt{3}}$ where $W(h)$ is the width of the hexagon at the distance $h$ from the baseline. The equations are ugly but in a plot it is obvious that it holds true for all x $\endgroup$ – DenDenDo Oct 10 '14 at 9:17
  • $\begingroup$ @DenDenDo I think Ross's snake will fit in the 1/2 hexagon. But you'll have trouble with the almost flat S shape: 3 equal segments with an angle 180-epsilon and 180+epsilon. The angle will shorten the snake by a length O(epsion^2) but it will stray away from the axis to a distance O(epsilon) on both sides. So for a small enough epsilon you will have to place it so close to the long edge that it overlaps. $\endgroup$ – Florian F Oct 11 '14 at 20:58

EDIT: As pointed out by Falco and Ivo Becker in the comments, there are some shapes that do not fit. I will leave this answer here, however, since there is valuable info in the comments. Also, thanks to Mooing Duck for the pictures - better than the one I put up.

I have another thought.

What if you started with a unit square blanket and took away the unit circle from the middle, and then took one quarter of that, you'd have kind of a right-angle triangle, but its long side would be caved in. Now say you mirror this triangle along one of its short edges, and it would be like a bigger right angle triangle, except its two short edges are caved in while its long edge is straight. (Blanket is the red area) enter image description here

Would all snake positions be covered by this blanket?

EDIT: As pointed out by Falco and Ivo Becker in the comments, there are some shapes that do not fit.

  • $\begingroup$ fascinating shape, though I had to read this several times to get it. I think it covers all snake shapes, but I can't be certain. Area is... ~0.1073009? $\endgroup$ – Mooing Duck Oct 9 '14 at 23:13
  • $\begingroup$ The more I think on it and read the other answers, the more I think you hit the right shape. I made a diagram here, blanket is the red part: i.imgur.com/eZ5Ai3P.png The key is that the snake's head is only at the Left/Right corners if he's perfectly straight, otherwise the bend is towards the top. Or maybe it needs to be a perfect triangle like your other answer says... I can't tell. $\endgroup$ – Mooing Duck Oct 9 '14 at 23:23
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    $\begingroup$ I think this will not work, if the snake tilts its head right and tail left (so a 0.9 straight line, with 0.05 rectangular bit at each end, one to the left, one to the right) $\endgroup$ – Falco Oct 10 '14 at 8:30
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    $\begingroup$ Also, when the snake is rolled up like a square (with sides 0.25) it won't fit I think $\endgroup$ – Ivo Beckers Oct 10 '14 at 9:23
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    $\begingroup$ @MooingDuck An even bigger "square" is the "U" shape with 1/3 on each side. That one won't fit. As for the "7", It is not that much shorter than the snake when it is straight, but now it must be elevated - the length of the maximum allowed long piece shrinks more rapidly than the short side grows, so it doesn't fit either. I cut the pieces out of paper, and it is not even close to fitting. $\endgroup$ – Trenin Oct 10 '14 at 16:29

Looking at Joe's answer, half the semicircle seems to be unnecessarily big. Instead, you could use a smaller area triangle of the same height and width.

If you were to take a semicircle of diameter 1 and split it in half, replacing one side with a triangle of height 0.5 and width 0.5, I think that would be the smallest area.

Area: Half the semi-circle + triangle

Circle area: pi * r^2 = pi * (1/2)^2 = pi/4

Half the semi-circle = pi/4 * 1/2 * 1/2 = pi/16

Triangle area: length * height / 2 = 1/2 * 1/2 * 1/2 = 1/8

Total area: = pi*/16 + 1/8 ~= 0.32

  • $\begingroup$ I think it can be reduced to a right-triangle with a base of 1 and a height of 0.5. Actually, it probably doesn't even need that much height. $\endgroup$ – Mooing Duck Oct 9 '14 at 23:28
  • $\begingroup$ I think the most height required is when he forms a equilateral triangle with the long edge, so... a height of 0.433012 using an area of... ~0.21650635. The more I think on it, the more I think even this has wasted cloth, but not much. I think you're very very close. $\endgroup$ – Mooing Duck Oct 9 '14 at 23:31
  • $\begingroup$ @MooingDuck Yeah - this was my first attempt. My other answer was just a simple right angle triangle (area 0.25), and my third was the funny dented triangle (which doesn't work). I've never posted more than one answer before, but I kept getting better ideas and didn't want to delete them, since they all had valid points... $\endgroup$ – Trenin Oct 10 '14 at 12:00

A blanket the same width and length of the snake would seem to be sufficient. Blankets are soft, and bend just like snakes can bend - depending upon theur construction, they may be able to bend more. In this case, let's assume a knit blanket, loose enough that it can stretch to bend just as much as the baby stretches to bend, without having to be folded over or bunched up. This has the nice effect of keeping the problem 2 dimensional. . Mommy snake can adjust the blanket at any time so that it bends to cover baby snake

  • $\begingroup$ It's sufficient, but it can be made smaller. $\endgroup$ – Joe Z. Oct 10 '14 at 23:31
  • $\begingroup$ @JoeZ, how can you cover an object with a smaller area than the object? $\endgroup$ – atk Oct 10 '14 at 23:32
  • $\begingroup$ Well, you don't need a rectangle the length and width of the snake, as some of the other solutions have shown. $\endgroup$ – Joe Z. Oct 10 '14 at 23:34
  • $\begingroup$ All the other solutions are larger than the snake... they try to address the snake's movement without bending the blanket. I take the opposite approack. If the snake is 1x0.001, then the blanket is 1x0.001. No empty space under my blanket, but empty space under everyone else's $\endgroup$ – atk Oct 10 '14 at 23:36

Assuming the line (snake) has a negligible width, this shape (blanket) will contain every possible bent/zig-zagging of a 1-unit line, assuming the shape can be rotated in three dimensions:

enter image description here

The total area adds up to a quadrant of a 1-unit square. More specifically, ¼u² or 0.25 units squared.

Regarding the area, it cannot get any smaller because the largest bounding box for all possible bents has that specific area - it's the 90-degree mid-way bent.

  • $\begingroup$ Can you prove that it will always cover the snake and can't get any smaller? $\endgroup$ – Ben Aaronson Oct 10 '14 at 17:49
  • $\begingroup$ I could simulate a few hundreds to tens of thousands of bends and fill in a matrix. Would that suffice? $\endgroup$ – Vercas Oct 10 '14 at 18:00
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    $\begingroup$ How would you place a snake that is straight except for 1% bent at 90° at both ends? $\endgroup$ – Florian F Oct 10 '14 at 23:01
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    $\begingroup$ @Vercas I don't think it fits. The top end is still at 0.02 units from the top. At that place the shape is only 0.0004 units wide. $\endgroup$ – Florian F Oct 11 '14 at 16:21
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    $\begingroup$ @Vercas You can calculate it. 0.5 - sqrt(0.5^2-0.02^2) = 0.00040016. I am assuming your figure has 2 perfect 1/4 circles. $\endgroup$ – Florian F Oct 11 '14 at 20:41

I'm not seeing this one here yet.

a 30-60-90 triangle with sides 1/2, sqrt(3)/2, and 1. Total area =0.2165

I believe this shape allows for all single bends of the snake. triangle layout

For bends between 0 and 60 degrees, the bend should be placed in corner A: placement for bends 0 to 60 degrees

For bends greater than 60 degrees, one end should be placed at Point A:enter image description here

Further analysis is required for additional bends.

  • $\begingroup$ It might work for 1 bend. It doesn't really tell how it works if the bend is not in the middle. But with 2 bends, some won't fit. $\endgroup$ – Florian F Oct 11 '14 at 16:23

The blanket is a circle with a diameter of 1. There is no position the baby snake can take which this blanket cannot cover. (Unless the snake's width is greater than 1; this wasn't specified, but I'm hoping it's a safe assumption.)

  • $\begingroup$ i've posted your answer into my question - i think there are lot's of areas that can be clipped. I'm looking for a smaller solution! but thank you for providing this answer! good thinking and a good start!! $\endgroup$ – Martin Frank Oct 9 '14 at 4:42
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    $\begingroup$ It's trivial to show that this is not the minimal solution. Just clip a small piece off one end, say 0.1 units. If the snake bends so that its maximum length is 0.9, it can fit anywhere under the blanket. If it's longer than 0.9, the blanket can be rotated so that it fits under the uncut part. $\endgroup$ – Moyli Oct 9 '14 at 7:13
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    $\begingroup$ @Moyli yes, you are right... don't be to harsh, i think that is still a agood start of thinking... honestly i think this is a quite very hard puzzle... $\endgroup$ – Martin Frank Oct 9 '14 at 7:43
  • $\begingroup$ but still is has funny snakes ^^ $\endgroup$ – Martin Frank Oct 9 '14 at 7:43

I think the best solution is a 1*1/2 rectangular blanket. If the baby is stretched out, you can cover him entirely. if he's in a non-stretched out pose, there cannot be more than 1 part sticking out further than 1/2 from the rest of the body.

edit: to reply to the remark: instead of a rectangle with size 1*1/2, use a rectangle with diagonal length 1, height 0.894 and width 0.447.

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    $\begingroup$ It would be bigger than Joe's semicircle blanket and it would have way too much extra space. For example the snake could never touch the opposite corners of the rectangle which shows the blanket is too big to be an optimal solution. $\endgroup$ – Moyli Oct 9 '14 at 11:10
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    $\begingroup$ A rectangle with diagonal length of 1 and what would the other dimensions be? If the height would still be 0.5, it would still be bigger than a semicircle (0.43 sq vs 0.39 sq). $\endgroup$ – Moyli Oct 9 '14 at 11:34
  • $\begingroup$ @MartinFrank yes it is, just put it at an angle so that the corner is aiming towards the long side. $\endgroup$ – Nzall Oct 9 '14 at 11:47
  • $\begingroup$ ...with those measurements it's still larger than a semicircle (0.40 vs 0.39). $\endgroup$ – Moyli Oct 9 '14 at 11:55
  • $\begingroup$ if the snake would make a U (semicircle, circumfence=2, radius = 1/pi = 0.32) it would not fit under the blanket... take your diagnal line, go 0.3 into center, then make a turn of 90° and go again 0.3 units - you are outside then (freaky, i must now use my circle/compass) $\endgroup$ – Martin Frank Oct 9 '14 at 11:58

The correct answer has already been given

The SemiCircle covers everything:

First, we need to cover the full length of the snake - a line with length 1:

enter image description here

Then we need to ensure to cover - as outlined above - any angle between 0 and 180°. Worst Case: it's exactly a 50:50 split -> Thales circle:

enter image description here

This case covers everything, including the snake forming a circle itself:

enter image description here

Whatever "shapes" the snake will perform on any of the both tails (if splitting at 50%) it will NEVER have a absolute length of 0.5 therefore remaining IN the semicircle all the time.

Any solution having a streight line, but lesser area will not work, because the snake could be like v-------------------^, messing up any attempt to have a line involved just to cover the maximum length

  • $\begingroup$ I think this is the "semicircle big" that is mentioned below as being a valid- but probably not minimal- solution. I don't really understand the small semicircle myself... it's pretty trivial to see that any semicircle with diameter less than one couldn't fit a snake that was stretched completely straight, so I don't know why that much more complex diagram was included $\endgroup$ – Ben Aaronson Oct 11 '14 at 15:31
  • $\begingroup$ Well, if this is referred as semicircle big then it's the minimalisitc solution, because it needs to cover all the angles between 0 and 180°. NOTHING in the shape but the baseline could cover the snake at a streight line and every point on the circle is required per degree the snake could form. So that's minimalistic. $\endgroup$ – dognose Oct 11 '14 at 15:37
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    $\begingroup$ Not sure I buy that. For example, take any of the angles formed by the baby in your second picture. The mama snake could move the baby to the right a bit, and now he wouldn't be touching the curved edge. And with that in mind, if you point to a given part of the curved edge, I don't think it's at all clear that the mama snake couldn't get by if a little chunk was cut out of that part of the edge (with a few exceptions at certain spots) $\endgroup$ – Ben Aaronson Oct 11 '14 at 15:53

Stretch your 2D baby snake out so that it forms a straight line from nose to tale. Find the 2D silhouette that the baby now forms. Create a 2D blanket having the same shape as the silhouette. Spread glue on the blanket and attach it to the baby snake. Mama's 2D baby snake will always be covered in bed once the glue dries.

  • The blanket will have minimal size.
  • The 2D blanket will twist and stretch and turn with the 2D snake it is attached to.
  • The size of the blanket will be the area of the baby's silhouette.
  • The form or shape of the blanket will always conform to the baby's.
  • The length of the blanket will obviously be 1.
  • The 2D blanket will always take the same shape as the baby.
  • The blanket will always cover the snake since they are glued together.
  • Mama will not have to worry about the baby being covered since both are deformable.
  • $\begingroup$ that is an really good approach, i never though of this! solving the problem in 3D might really lead to an proper solution ^^ (although the snake-puzzle is supposed to be an 2D-Problem) $\endgroup$ – Martin Frank Jul 2 '15 at 5:02

well i'm not sure if i really have a valid answer, but so least blanket i could think of is shape a little weird...

enter image description here

you can put into that blanket

  • the streched snake,
  • the 90°L-baby snake,
  • the U-Shaped baby snake,
  • a snake shaped like a [___] (dunno how to call that, is bend on tail and head)...
  • a S-shaped snake would also fit in...

im not sure if I could bring in a Z-shaped baby snake if it had 2x 60° bends...

but what's worse: What is the size of this shape? I can't tell and therefor can't compare with others :-/ ... frustation ... if anyone knows sie size fo such an area let me know...


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