KingGeorge
  • KingGeorge
[UNSOLVED] KingGeorge's Challenge of the Month/Annual 4th of July Problem! Suppose you have an 8-pointed star like in the picture below, and you want to color the edges of that star in either purple or green. How many ways are there to do this? Assume that two colorings are the same if the star can be rotated or reflected so that the two colorings look the same.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
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KingGeorge
  • KingGeorge
1 Attachment
anonymous
  • anonymous
\[\binom{8}{2}\cdot \frac12\]?
KingGeorge
  • KingGeorge
There's a lot more than that. Remember that there are 16 edges here, not 8.

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More answers

anonymous
  • anonymous
Ohh edges. Can I substitute 16 in place of 8?
KingGeorge
  • KingGeorge
Nope. There's still a lot more than that. I'll be gone for a while now. If it still isn't solved when I get back, I might give a hint.
anonymous
  • anonymous
Okay :|
anonymous
  • anonymous
Hrm... there are 8 rotational symmetries and 4 reflective symmetries. Ignoring symmetries, there are \(2^{16}\) patterns. Perhaps we simply divide the symmetries away, leaving us with \(2^{11}\)? Not sure, will have to think about it more.
asnaseer
  • asnaseer
another way of looking at it: Purple Edges Green Edges 16 0 15 1 14 2* 13 3* 12 4* 11 5* 10 6* 9 7* 8 8* where N* stands for how many ways we can select N unique edges - so 2* would be 8 in this case. Then multiply the whole lot by 2 as we can swap the Purple and Green edges.
KonradZuse
  • KonradZuse
256.
KonradZuse
  • KonradZuse
I win !!!
KingGeorge
  • KingGeorge
It is not 256. @nbouscal's way is closest to my original method, although I would be very interested in seeing if @asnaseer's way also pans out. I will point out that the solution is an odd number, however odd that sounds (pun intended). In asnaseer's potential solution, after you multiply by 2, you are counting 8 purple/8 green twice, so you have to subtract it (I think that should work). Hint containing my method in white \(\LaTeX\) below: \[\color{white}{\text{I used Burnside's Lemma from Group Theory using the group } D_8.}\]\[\color{white}{D_8\text{ is also called }D_{16}\text{ sometimes.}}\]
mathslover
  • mathslover
I used Burnside's Lemma from Group Theory using the group
mathslover
  • mathslover
I used Burnside's Lemma from Group Theory using the group D8. D8 is also called D16 sometimes.
anonymous
  • anonymous
What's Burnside's Lemma?
mathslover
  • mathslover
I also dont know about that : )
experimentX
  • experimentX
Purple Edges Green Edges Ways 16 0 1 15 1 1 14 2* 7 or 8 13 3* 12 4* 11 5* 10 6* 9 7* 8 8* looks like this will take a month.
KingGeorge
  • KingGeorge
Unless you're familiar with group theory, I would encourage ignoring Burnside's Lemma, and trying to find another way.
anonymous
  • anonymous
How would you arrange 2 things in a circle at 16 places, repetition allowed
KingGeorge
  • KingGeorge
What do you mean with repetition allowed? Can you place two things at the same place? That would give you \(16^2\).
anonymous
  • anonymous
I meant repetition of the things, you can use them more than once. I am sorry :/
anonymous
  • anonymous
Ignore me. I will try to come up with concrete solution and only then reply
anonymous
  • anonymous
But I would like to know about the symmetries nbouscal mentioned. @nbouscal
KingGeorge
  • KingGeorge
That is basically what Burnside's Lemma does for you. It's a little bit more complicated, but basically you look at the colorings that are fixed by certain symmetries, multiply by some things, add a bunch of these terms, and then divide by the amount of symmetries. Of which there are 16.
KingGeorge
  • KingGeorge
Don't get discouraged by Burnside's Lemma. It's a(n) (advanced) technique that trivializes the problem to one or two lines. I would love to see another way to do this problem.
anonymous
  • anonymous
@mukushla you might like to do this :-)

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