Proof of binary folding algorithm equation

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Xerxes
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Proof of binary folding algorithm equation

Post by Xerxes »

http://www.langorigami.com/science/math ... ctions.pdf

In this article, there's an equation for finding the maximum number of folds n required using binary folding algorithm to fold a fraction by a given accuracy e:

n = ceiling ((log2 1/e)-1)

Does anybody know if there is a existing proof of this published by Mr. Lang?
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origamifreak_1.6180339889
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Re: Proof of binary folding algorithm equation

Post by origamifreak_1.6180339889 »

what this binary algorithm does is it essentially provides you with the approximation of a prime division of a power of 2
for example- the first few binaries for 1/3 will provide you with these divisions
1/2
1/4
3/8
5/16
11/32
21/64
the algorithm for this was reverse engineered from this method, so just folding using the binary method is pretty much the only proof of it you will get
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origamifreak_1.6180339889
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Re: Proof of binary folding algorithm equation

Post by origamifreak_1.6180339889 »

its hard to explain in words, but once you fold it, it will make perfect sense to you
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OrigamiMagiro
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Re: Proof of binary folding algorithm equation

Post by OrigamiMagiro »

Maybe it's easier to think about it in the reverse direction, i.e. given a certain number n of parallel folds, what accuracy e can you obtain?

With n=1 folds you can obtain accuracy to within e=1/4 the side of the paper.
With n=2 folds you can obtain accuracy to within e=1/8, and so on. So, by inspection...

e = 1/2^(n+1)

Picking the largest integer n bigger than the inverse of this yields the formula given.
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