[英]puzzle function with fold in haskell?
我很難弄清楚該功能的工作原理,我需要專家的解釋。 感謝任何幫助!
puzzle n x = scanr (\y acc -> (acc + y/acc)/2) 1 (replicate n x)
我嘗試運行這些:
--puzzle 10 2
--puzzle 10 5
--puzzle 10 36
它分別給我這個輸出:
[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]
它列出了n個x-es的列表,例如[x,x,x,x,x,..](n次)
然后它的作用是:
x1 = ( 1 + x/1)/2
x2 = (x1 + x/x1)/2
x3 = (x2 + x/x2)/2
x4 = (x3 + x/x3)/2
結果是[xn,x(n-1),...,x2,x1]
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