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KingGeorge
 3 years ago
[UNSOLVED] KingGeorge's Challenge of the Month/Annual 4th of July Problem!
Suppose you have an 8pointed 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.
KingGeorge
 3 years ago
[UNSOLVED] KingGeorge's Challenge of the Month/Annual 4th of July Problem! Suppose you have an 8pointed 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.

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Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0\[\binom{8}{2}\cdot \frac12\]?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.4There's a lot more than that. Remember that there are 16 edges here, not 8.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0Ohh edges. Can I substitute 16 in place of 8?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.4Nope. 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.

nbouscal
 3 years ago
Best ResponseYou've already chosen the best response.0Hrm... 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
 3 years ago
Best ResponseYou've already chosen the best response.1another 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.4It 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
 3 years ago
Best ResponseYou've already chosen the best response.0I used Burnside's Lemma from Group Theory using the group

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I used Burnside's Lemma from Group Theory using the group D8. D8 is also called D16 sometimes.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0What's Burnside's Lemma?

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0I also dont know about that : )

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.0Purple 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
 3 years ago
Best ResponseYou've already chosen the best response.4Unless you're familiar with group theory, I would encourage ignoring Burnside's Lemma, and trying to find another way.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0How would you arrange 2 things in a circle at 16 places, repetition allowed

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.4What do you mean with repetition allowed? Can you place two things at the same place? That would give you \(16^2\).

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0I meant repetition of the things, you can use them more than once. I am sorry :/

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0Ignore me. I will try to come up with concrete solution and only then reply

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0But I would like to know about the symmetries nbouscal mentioned. @nbouscal

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.4That 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
 3 years ago
Best ResponseYou've already chosen the best response.4Don'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.

Ishaan94
 3 years ago
Best ResponseYou've already chosen the best response.0@mukushla you might like to do this :)
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