在haskell中折叠的拼图功能？

Question

I am having a hard time trying to figure out how does this function works, and I need an explanation from an expert. Appreciate any help!

puzzle n x = scanr (\y acc -> (acc + y/acc)/2) 1 (replicate n x)

I tried running these:

--puzzle 10 2

--puzzle 10 5

--puzzle 10 36

and it gives me this output, respectively:

[1.414213562373095,1.414213562373095,1.414213562373095,1.414213562373095,1.414213562373095,1.414213562373095,1.4142135623746899,1.4142156862745097,1.4166666666666665,1.5,1.0]

[2.23606797749979,2.23606797749979,2.23606797749979,2.23606797749979,2.23606797749979,2.236067977499978,2.2360688956433634,2.238095238095238,2.3333333333333335,3.0,1.0]

[6.0,6.0,6.0,6.0,6.000000005333189,6.0002529841194185,6.055351744849479,6.872226737643129,10.222972972972974,18.5,1.0]

Answer 1

This function calculates a square root using Newton`s formula and stores all iteration results in list.

Here is a Newton's method on wiki.

Storing process is based on definition of scanr function:

scanr is similar to foldr, but returns a list of successive reduced values from the right

Answer 2

It makes a list of n x-es like [x,x,x,x,x,..] (n times)

and then what it does is:

x1 = ( 1 +  x/1)/2
x2 = (x1 + x/x1)/2
x3 = (x2 + x/x2)/2
x4 = (x3 + x/x3)/2

and the result is [xn,x(n-1),...,x2,x1]