Klauz Oppenheim is putting together an animation as an art project for school. He wants to portray his message using familiar objects that his audience will be able to relate to, so he bought 16 sour skittles (individually wrapped) and placed them in a square. He will take 17 images, beginning with all of the skittles and taking away one per frame until none are left.

Klauz does not want to confuse his audience by flashing a sequence of strange ugly shapes, so he would like to maximize the beauty of his animation. For any frame we can assign a number, the amount of symmetries that the frame has (the order of its symmetry group), and so to an animation we can assign a number, the sum of the values of its frames.

Each skittle has a color and so the way they are initially laid out is important. You can tell Klauz which skittle to remove and when, but unfortunately you cannot tell him how to place them on the surface. He already went and glued them to the surface, and if he buys any more skittles he'll be over his budget. This is how he arranged them.

Colorative skittls

Summarized rules:

  • There are 17 frames. There is the one pictured above, and then one skittle disappears per frame until none are left. Your answer is the sequence of skittles that you will remove.
  • Symmetry is measured per frame and then the sum is taken. For each frame the symmetry value is the order of its symmetry group, including the identity. That is, every line of symmetry gives 1 point, and every rotation gives 1 point.
  • Your goal is to achieve the best possible symmetry score (and thus the most aesthetically pleasing animation).
  • Color matters when counting symmetries.
  • Color especially matters to us because as a reward for our help, Krauz has offered to let us eat the final skittle. Every color is a different flavor and our favorite flavor is grape (purple)! The frame with 1 skittle yields 12 points for purple, 10 points for green, 8 points for red, and 2 points for blue.
  • The frame with no skittles yields 0 points.

EDIT: GentlePurpleRain points out that 1 skittle would have infinite score. For that matter, so would 0. To deal with these cases I have added a new rule and clarified 0.

  • 2
    $\begingroup$ Where the heck did he buy individually wrapped skittles? $\endgroup$ – Joe Z. Apr 20 '15 at 20:07
  • $\begingroup$ @JoeZ, at a fundraiser for a new public water fountain to replace an old one that had warm white-tinted water. $\endgroup$ – Ben Frankel Apr 20 '15 at 20:18
  • $\begingroup$ Will you score the initial frame so that I can learn the scoring scheme? $\endgroup$ – JLee Apr 20 '15 at 20:41
  • $\begingroup$ Sure. No rotation of the initial frame results in the same image except a 360 rotation, and no reflection of the initial frame results in the same image. Thus its score is 1. As another example, the shape ( ) has two reflections (horizontal, vertical) and two rotations (180, 360) and so scores 4. $\endgroup$ – Ben Frankel Apr 20 '15 at 20:47
  • 1
    $\begingroup$ Do I understand right that a 2 by 2 square of skittles of the same color gives 8 points? $\endgroup$ – xnor Apr 20 '15 at 21:00

First try. Nothing clever here, just trying to end with the purple skittle and being greedy other than that. Eat in the numbered order. For groups of skittles with the same number, eat in any order.

enter image description here

This gets $\boxed {44}$ points, if I can count:

  • After eating group 1, you get $2$ points for diagonal symmetry.
  • Same for groups 2, 3, 4, and 5.
  • After eating group 6, you're left with a cross. The symmetry group is $D_4$ which has $8$ elements.
  • After eating M&M 7, you have reflective symmetry for $2$ points.
  • After M&M 8, you get a line of three M&Ms. This has a symmetry group of $C_2 \times C_2$, which has $4$ elements.
  • After M&M 9, you have reflective symmetry for $2$ points.
  • After M&M 10, you get $12$ points for finishing with grape.
  • There were also 6 frames where you got $1$ point for no symmetry.
  • 2
    $\begingroup$ reward for grape is OP, inb4 nerf $\endgroup$ – Lopsy Apr 20 '15 at 21:53
  • 1
    $\begingroup$ I can verify that this solution gets 44 points. It may be that the reward for finishing with grape is too high.. I didn't try the problem myself before posting it. It's too late to change the rules, but I can draw another grid and check if it's interesting first before posting it. $\endgroup$ – Ben Frankel Apr 20 '15 at 22:04
  • 1
    $\begingroup$ That would be cool. I like this puzzle idea a lot. $\endgroup$ – Lopsy Apr 20 '15 at 22:14
  • 2
    $\begingroup$ I would like to point out that Klauz Oppenheim is a magician as well as an artist. He put down Skittles and, when he ate them, they turned into M&Ms. $\endgroup$ – Engineer Toast Apr 21 '15 at 12:54
  • 1
    $\begingroup$ @EngineerToast, Aren't grape M&Ms just great? $\endgroup$ – Ben Frankel Apr 21 '15 at 22:41

No proof here, but it seems intuitively to me that this is the best answer. If someone can provide a proof to the contrary (or to back me up), please do so.

I will refer to the skittles by letter as follows:


I will score a stage immediately before I remove a skittle. So when I display the state of the "board", that state has not yet been scored.

It seems to me that there is no way to achieve any symmetry until you have removed at least 4 skittles.

If you remove B,E,L,O, you will have reflective symmetry along the diagonal from bottom-left to top-right.

Each of those first four steps have only 360° rotational symmetry, so they each score a single point.

Points: 4

R   B P
  B P B
B B   R

After this, you can remove D,G,J,M, one-by-one, to maintain the reflective symmetry. Each of these four steps provide 360° rotational symmetry and reflective symmetry along the bottom-left-to-top-right diagonal, resulting in 2 points for each step.

Points: 4 + 2 + 2 + 2 + 2 = 12 total

R   B
  B   B
B   B
  B   R

Now you have 180° rotational symmetry as well, and reflection along the top-left-to-bottom-right diagonal, so this board scores 4 points.

After this, we can remove skittles in pairs.
We start with the red ones. Removing one of them leaves only 1 reflection, and the 360° rotation, so the next board scores 2 points.

Points: 12 + 4 + 2 = 18 total

  B   B
B   B

Now we want to try to get a diamond of four, because that will give us maximum symmetry. That means we need to remove one of the top or bottom skittles. That will destroy all symmetry, but removing its reflection will restore 4-way symmetry. Thus these 2 steps yield 5 points (4 for the current state, 1 for the next).

Points: 18 + 4 + 1 = 23 total

B   B

Now we have 8-way symmetry (four rotations, four reflections). As above, removing a single skittle will destroy most of the symmetry, but the triangle shape that is left will still have a single line of reflectional symmetry.

We remove 1 skittle (it doesn't matter which one at this point):

Points: 23 + 8 = 31 total

B   B

We remove 1 more skittle:

Points: 31 + 2 = 33 total


Now we have 2 rotations and 2 reflections, for a total of 4 more points. We remove 1 more skittle:

Points: 33 + 4 = 37 total

We are now down to a single blue skittle, which is worth 2 points, making our final total

39 points.

*Note that if we remove D,G,J,M, and then B,E,L,O, we get the same result. It just means we achieve rotational symmetry before reflective symmetry instead of vice versa.

  • $\begingroup$ Your counting is a bit off between 12 and 20 points. $\endgroup$ – Ben Frankel Apr 20 '15 at 21:55
  • $\begingroup$ No, I think it's correct. What is wrong with it, specifically? $\endgroup$ – GentlePurpleRain Apr 20 '15 at 21:57
  • $\begingroup$ No frame has 0 points, in particular. $\endgroup$ – Ben Frankel Apr 20 '15 at 22:00
  • $\begingroup$ Gotcha. Updated the answer. Total is now 37 points, which still doesn't beat @Lopsy. :( $\endgroup$ – GentlePurpleRain Apr 20 '15 at 22:09
  • $\begingroup$ You still have one counting error. When you have 3 skittles left there is a reflectional symmetry so it scores 2. Also, you can slightly modify one of the steps to achieve another extra 1 point. So in total your score should be 39. $\endgroup$ – Ben Frankel Apr 21 '15 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.