Egyptian Fracs: Algorithm for Minimizing Terms/Denominator

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Is there an algorithm which can convert any rational number to a sum of distinct unit fractions which minimizes the number of terms or the largest denominator?
 
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My hunch would be the "greedy" algoritm. Given rational x1, let n be the smallest integer such that 1/n< x1. Now repeat the process with x2= x1- 1/n.

For example to find the unit fractions for 13/17:
It is clear that 1/2< 13/17 so our first unit fraction is 1/2. 13/17- 1/2= 26/34- 17/34= 9/34. 1/3> 9/34 but 1/4< 9/34 so our second unit fraction is 1/4. 9/34- 1/4= 18/68- 17/68= 1/68 which is itself a unit fraction.
13/17= 1/2+ 1/4+ 1/68.

Is it necessary to include "which minimizes the number of terms or the largest denominator"? Isn't decomposition into unit fractions unique?
 
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The decomposition is not at all unique. 1/2=1/3+1/6 for example. See the algorithm described in

https://www.physicsforums.com/showthread.php?t=119708&highlight=egyptian


The greedy algorithm can fail to give the representation with the least number of terms and smallest denominators. See

http://www.mcs.surrey.ac.uk/Personal/R.Knott/Fractions/egyptian.html#Fibgreedy

I'm not sure of the algorithm that finds the shortest, but they have a calculator on that page that claims to, so that page is probably a good place to start.
 
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Thanks, shmoe, I was wondering about that.

I have edited my post so it makes sense now!
 
Thanks for the help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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