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Math Problem - 18

cmhardw

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Thread starter #3
haha I didn't expect this problem to be so easy. I thought of it last night, with the intention to try it after work today. I'll post my answer/how I got it after looking at it later today (assuming I can figure it out lol).

Chris
 
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MTGjumper
#5
I find it funny that the person providing the answer technically has a post count of zero =P

If anybody sees this, on a 3x3x3, how many legal positions are there where each face has at least one of each of the six colours on it?
 
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cuBerBruce
#7
Well, I haven't yet seen an explanation, or any answers other than that provided by Sir E Brum. So I think this is not so easy to work out by hand.

I made a "ballpark" estimate, and I came up with something significantly different from Sir's answer. Using a computer program to perform a brute force count with respect to all corner configurations seems to confirm my estimate is not too far off.

There are 24 corner stickers, and only 4 of the 24 match each of the centers. So the approximate probability of all 24 corner stickers not matching the center on the same face is roughly (5/6)^24. Multiplying by the number of corner configurations gives (approx.) 1109224. This will be off somewhat because of simplifying assumptions.

I used a computer program to perform a brute force count of all 88179840 corner configurations to see how many satisfied the condition that none of the corner stickers match the center sticker on the same face.

The result I came up with is:

Corner configurations with no corner stickers matching the center on the same face.

even permutations of corners: 641376
odd permutations of corners: 642656

total: 1284032

So my estimate for this was of the right order of magnitude.

Doing this for the edges might take awhile. But once the same is computed for the edges, it's a straightforward calculation to come up with the answer to cmhardw's question.
 
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#8
If anybody sees this, on a 3x3x3, how many legal positions are there where each face has at least one of each of the six colours on it?
Err... I'm guessing 43 quintillion (whatever that number is) minus the answer to Chris' question? :p

Edit: Wait, duh it's obviously wrong. I've been posting lotsa stupid responses today, brain malfunction :( (I don't know the correct answer anyway...)
 

cmhardw

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I have to be honest, I'm completely stumped as to how to figure this out. I am trying an idea with:

trying to count the number of ways with all corners permuted, then 6 permuted, then 5 permuted, etc. After this I will do the same with edges and combine all the numbers appropriately to make sure that I am ensuring to count only positions with corners and edges having the same parity.

Other than that idea, I'm stumped. I'm not crying out for a hint, in fact I would prefer not to read any hints. I just wanted to let people know that in spite of thinking about this problem I'm still stumped haha.

Chris
 
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#13
OK, I've come up with an answer. (I used Lucas's method - programming. :) )
Okay, my brute force counting (via computer program) of the edge configurations is now complete. The number of edge configurations satisfying no edge sticker matching the center sticker color on the same face is:

even permutation of edges: 8224786143
odd permutation of edges: 8224786143
total: 16449572286

My earlier result for the corners was:

even permutations of corners: 641376
odd permutations of corners: 642656

So the total number of legal cube positions with no corner or edge stickers matching the center sticker on the same face, by my calculation is:

(8224786143*641376)+(8224786143*642656) = 10,560,888,600,768,576
 
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