Methods and complexity for computing square roots

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The discussion centers on computing the square root of an integer x with n digits, focusing on the computational cost in big O notation. Newton's method is highlighted as a commonly used algorithm for this purpose, alongside a long division-like method. The complexity of Newton's method is linked to the desired precision of the result, particularly when calculating square roots of integers. Participants suggest that the complexity may vary based on the number of digits needed in the output. The conversation concludes with a suggestion to compute the square root of 2 to n-digit precision for further clarity.
geor
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Hello everybody,

Let's say we want to compute sqrt(x), where x is an integer
of n digits. Then what is the cost of the computation, in
terms of big O notation and n?

And a second question: what is the algorithm for finding the
square root that is most commonly used in computers and
calculators (just a name or a link will do)?

Thanks a lot in advance!

Yiorgos
 
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what is the algorithm for finding the
square root that is most commonly used in computers and
calculators (just a name or a link will do)?

I don't know if it's the most commonly used, but Newton's method is used a lot.
 
There is an algorithm which resembles long division (I learned it in 4th grade) for taking square roots. It should take about the same time as long division.
 
Thanks for taking the time to reply.

Do you know how much Newton's method cost?
 
Hi,
Here's link that myght help:

http://numbers.computation.free.fr/Constants/Algorithms/inverse.html"
 
Last edited by a moderator:
BobMonahon said:
Hi,
Here's link that might help:

http://numbers.computation.free.fr/Constants/Algorithms/inverse.html"

Thanks, that is helpful, indeed..
So, to compute \sqrt{A} with Newton's method, we will use the iterations:

x_{n+1} =\frac{3}{2}x_n-\frac{1}{2}A{x_n}^3

Can you help me compute the complexity of this one?
Let's say that A has n digits and let's also say it is a square
of an integer...
 
Last edited by a moderator:
The complexity of the calculation depends on the number of digits you want in the result, I think. This might be a simpler way to get started: Calculate \sqrt{2} to n-digits precision.

See if that helps.
 

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