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Knight visits every square once in chess (i.imgur.com)
submitted 2 日 前 by CMSTerraNova
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[–]MaturinTheTurtle 31 ポイント32 ポイント33 ポイント 2 日 前
This is called a knight's tour, and you'd think there'd be only a few of them, because of the rigid constraints place on the problem. But in fact the number of knight's tours is so vast that we don't know how many there are. We haven't discovered them all yet.
[–]RubberDong 0 ポイント1 ポイント2 ポイント 2 日 前
That is a horse though.
[–]kznlol -18 ポイント-17 ポイント-16 ポイント 2 日 前
the number of knight's tours is so vast that we don't know how many there are
Uh, no. We just don't know the exact number.
It is utterly trivial to determine a range in which it could be.
[–]MaturinTheTurtle 8 ポイント9 ポイント10 ポイント 2 日 前
No, it's not. We haven't the foggiest idea how many open tours there are.
[–]GetItRightNextTime 3 ポイント4 ポイント5 ポイント 2 日 前
Is it 6?
[–]AJs_Sandshrew 0 ポイント1 ポイント2 ポイント 2 日 前
Sorry, no. You'll probably get it right next time.
[–]kznlol -9 ポイント-8 ポイント-7 ポイント 2 日 前
Yes we do.
It is without question less than the number of unique tours, unrestricted by movement rules. That number is trivial to determine.
[–]AJs_Sandshrew 5 ポイント6 ポイント7 ポイント 2 日 前
We know the number of closed tours, which is 26,534,728,821,064. A closed tour is one where the knight can end on the same position where it started and thus repeat the tour. Like this
What we don't know is the number of open tours, meaning that the knight can't end on the position where it started and thus cannot repeat the tour.
Correct me if I'm wrong, but what I think you are arguing is that we can estimate the number of open tours (which has been done and is estimated to be somewhere around 1.22 × 1015 ) Source from arxiv
I think what MaturinTheTurtle is arguing is that we don't know the exact number of open tours.
Also, if you say that the number is trivial to determine, you should probably explain why is it trivial and how you would go about determining it rather than just stating "It's trivial."
[–]kznlol -5 ポイント-4 ポイント-3 ポイント 2 日 前*
He said, quite explicitly, that the reason we don't know the exact number is that it's "so vast".
That's not the reason. The reason is that it's computationally an extremely difficult question to answer. It's very easy to show that the exact number of open tours must be less than some x.
Because its the same as assigning a unique symbol to each square on a chessboard and calculating the unique combinations of those symbols.
I'm pretty sure that amounts to solving for the product of (64 * 63 * 62 * ...)
[edit] This is quite probably not even close to being the most accurate estimate, but I do not get paid to prove mathematical claims and it was the best I could be sure about off the top of my head. The point being, after about 40 seconds of understanding the definition of an open knight's tour I had some idea of how many open tours there are.
[–]AJs_Sandshrew 0 ポイント1 ポイント2 ポイント 1 日 前
Think about what you are calculating here. I can see your thought process: the knight must visit every square once and only once, so it should just be a simple factorial, 64! = 64 * 63 * 62 ... * 1
However, this calculation assumes that the knight can move to any square from any other square which we know is not true. The restricted movement of the knight is what makes the problem so complicated. You are right when you say that "this is quite probably not even close to being the most accurate estimate" as 64! = 1.27*1089 whereas the estimate given by Cancela and Mordecki is 1.22*1015 . This is not even remotely close to the estimate of 64! (Its off by about 1074 fold!)
I found this book on Google books that addresses the theory behind determining knights tours and other related problems if you want to read a little more on the subject.
[–]kznlol 0 ポイント1 ポイント2 ポイント 1 日 前
However, this calculation assumes that the knight can move to any square from any other square which we know is not true.
Yes, obviously. This is why I can say with confidence that the number of open tours cannot exceed the factorial.
Again - I never said the problem is easy. I said that the reason it's difficult has nothing at all to do with the size of the number.
[–]MaturinTheTurtle 1 ポイント2 ポイント3 ポイント 2 日 前
Sorry, son. You're off in the tall grass this time. Better luck next time.
no argument
Easier than expected.
What?
[–]DrGiggleFairies 2 ポイント3 ポイント4 ポイント 2 日 前
Just leave him be. We'll go talk over here.
[–]MaturinTheTurtle 0 ポイント1 ポイント2 ポイント 2 日 前
Lead the way.
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π Rendered by PID 13631 on app-29 at 2014-05-25 15:14:08.240903+00:00 running 5c9ed36.
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[–]MaturinTheTurtle 31 ポイント32 ポイント33 ポイント
[–]RubberDong 0 ポイント1 ポイント2 ポイント