A blanket for my baby snake

Question

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 and her little baby snake

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Examples

Circle

This is biggest solution (from Miniman):

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...

Answer 1

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.

Answer 2

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. :)

Answer 3

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:

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:

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:

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 ~ $0.2455127...$

EDIT:

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:

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.

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".

Answer 4

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:

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

Answer 5

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.

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.

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:

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.

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.

Answer 6

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

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

Answer 7

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.

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

Answer 8

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)

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.

Answer 9

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

Answer 10

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

Answer 11

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:

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.

Answer 12

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.

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

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

Further analysis is required for additional bends.

Answer 13

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.)

Answer 14

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.

Answer 15

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:

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:

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

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

Answer 16

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.

Answer 17

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...

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...