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Topic: the cube root (Read 6677 times) |
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kirru
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the cube root
« on: Aug 30th, 2013, 10:32am » |
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The cube root of a natural number n is defined as the largest natural number m such that m power 3 <=n.(n is represented in binary notation).the complexity of finding cube root of n is?
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towr
wu::riddles Moderator
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Re: the cube root
« Reply #1 on: Aug 30th, 2013, 12:12pm » |
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I think the complexity is about the same as multiplication, like it is for square root (using Newton's method).
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JohanC
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Re: the cube root
« Reply #2 on: Aug 31st, 2013, 12:56pm » |
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Wouldn't cube root (and square root) be quicker than multiplication?
For cube root you can divide your (binary or decimal) digits into groups of 3 and calculate every digit seperately. An example doing this by hand with decimal notation can be found at this url.
An interesting overview of approaches for multiplication can be found at Wikipedia.
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towr
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Re: the cube root
« Reply #3 on: Sep 1st, 2013, 7:56am » |
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on Aug 31st, 2013, 12:56pm, JohanC wrote:| For cube root you can divide your (binary or decimal) digits into groups of 3 and calculate every digit seperately. An example doing this by hand with decimal notation can be found at this url. |
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It doesn't really look faster than multiplication to me; in fact it uses quite a few, and pretty large, multiplications (especially for the last digit, since it involves the result from all previous calculations.) Just look at the last step of finding the cuberoot of 12812904: it asks us to find the largest digit an-2 such that (30*23*(230+an-2) + an-22)an-2 <= 645904
However, I did miss a part in the wikipedia article on Newton's method which discussed how it didn't work for the cube root..  So that avenue certainly doesn't work.
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| « Last Edit: Sep 1st, 2013, 7:56am by towr » |
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