Data commit

Task/Perfect-numbers/Python/perfect-numbers-1.py

@@ -0,0 +1,6 @@
+def perf1(n):
+    sum = 0
+    for i in range(1, n):
+        if n % i == 0:
+            sum += i
+    return sum == n

Task/Perfect-numbers/Python/perfect-numbers-2.py

@@ -0,0 +1,22 @@
+from itertools import chain, cycle, accumulate
+
+def factor2(n):
+    def prime_powers(n):
+        # c goes through 2, 3, 5, then the infinite (6n+1, 6n+5) series
+        for c in accumulate(chain([2, 1, 2], cycle([2,4]))):
+            if c*c > n: break
+            if n%c: continue
+            d,p = (), c
+            while not n%c:
+                n,p,d = n//c, p*c, d + (p,)
+            yield(d)
+        if n > 1: yield((n,))
+
+    r = [1]
+    for e in prime_powers(n):
+        r += [a*b for a in r for b in e]
+    return r
+
+def perf4(n):
+    "Using most efficient prime factoring routine from: http://rosettacode.org/wiki/Factors_of_an_integer#Python"
+    return 2 * n == sum(factor2(n))

Task/Perfect-numbers/Python/perfect-numbers-3.py

@@ -0,0 +1,6 @@
+def perf2(n):
+    return n == sum(i for i in range(1, n) if n % i == 0)
+
+print (
+    list(filter(perf2, range(1, 10001)))
+)

Task/Perfect-numbers/Python/perfect-numbers-4.py

@@ -0,0 +1,35 @@
+'''Perfect numbers'''
+
+from math import sqrt
+
+
+# perfect :: Int - > Bool
+def perfect(n):
+    '''Is n the sum of its proper divisors other than 1 ?'''
+
+    root = sqrt(n)
+    lows = [x for x in enumFromTo(2)(int(root)) if 0 == (n % x)]
+    return 1 < n and (
+        n == 1 + sum(lows + [n / x for x in lows if root != x])
+    )
+
+
+# main :: IO ()
+def main():
+    '''Test'''
+
+    print([
+        x for x in enumFromTo(1)(10000) if perfect(x)
+    ])
+
+
+# GENERIC -------------------------------------------------
+
+# enumFromTo :: (Int, Int) -> [Int]
+def enumFromTo(m):
+    '''Integer enumeration from m to n.'''
+    return lambda n: list(range(m, 1 + n))
+
+
+if __name__ == '__main__':
+    main()