Liu Hui [刘徽] was a Chinese mathematician who lived in the Wei period. One of the things he was famous for was his method of calculating 
This approach to calculate
Liu Hui found a simple formula to find the side length of an inscribed polygon of 
Here 
We start with a hexagon. The radius of the circle is 1, the area
To calculate the next
The area of a regular polygon is 
We then have the formula for the area of the polygon: 
Let’s do a few here:
To save us some tedious work, here’s a Haskell program to do the calculations for us:
k 6 = 1 k n = sqrt $ 2 + k (n/2) s n = sqrt $ 2 - k n p n = (n/2) * s n main = mapM_ (putStrLn . (\n -> show (round n) ++ " " ++ show (p n))) $ iterate (*2) 6
Here’s part of the output:
6 3.0 12 3.1058285412302498 24 3.132628613281237 48 3.139350203046872 96 3.14103195089053 192 3.1414524722853443 384 3.141557607911622 768 3.141583892148936 1536 3.1415904632367617 3072 3.1415921060430483 6144 3.1415925165881546 12288 3.1415926186407894 24576 3.1415926453212157 49152 3.1415926453212157 98304 3.1415926453212157
This is about the maximum you could get without arbitrary precision decimals: the area of a polygon with 98304 sides. This gives us 3.14159265 which is 8 digits of
With a big-decimal library, it would be possible to calculate
Happy pi day!
Lucky's Notes 
How about the error analysis? How do we know how many digits of our result are correct, given some precision on taking the square roots?
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Hmm, I actually don’t know how you would calculate that. But I imagine it converges very slowly (compared to modern algorithms), and you need to store a lot of digits for intermediate values if you want to calculate pi digits.
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Pingback: Who Discovered Pi?
Liu Hiu lived in the common era.. Not before Christ
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it would be cool if you would write this article in a way a layman would undestand
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I have a method of calculating pi which I think is new. For N operations, it produces O(N^2) accurate digits. The methods I’ve read about in the literature seem to be:
Euler’s series, so no. of digits D is about log10(N)
power series, D is proportional to N
quadratically-convergent methods, so D is about 2^N or 3^N or (other small integer)^N
My method is in between: D is proportional to N^2.
It’s a two-stage method. It uses p steps of a first algorithm, then q steps of a second algorithm. The error is about (1/2)^(pq). The number of operations is about p+q times a constant. One step of the first algorithm uses two additions, a multiply and a square root. One step of the second algorithm uses a multiply, an add and a divide. The division is by an integer.
I have a (conceptually) related algorithm for sine and cosine. Again, it’s a two-part method with similar properties of computation time and size of error.
Anyone interested?
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