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010/main.py

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+import os
+import math
+import numpy as np
+
+
+# Find the sum of all prime numbers below two million
+
+def isDivisible(n, ms):
+    res = False
+    for m in ms:
+        if n % m == 0:
+            res = True
+            break
+    return res
+
+
+def getNextPrime(ps):
+    if len(ps) == 0:
+        return [2]
+    else:
+        p = ps[-1] + 1
+        while isDivisible(p, [q for q in ps if q <= np.sqrt(p)]):
+            p = p + 1
+        ps.append(p)
+        return ps
+
+def getFirstNPrimesBad(n):
+    assert n >= 0
+
+    l = 0
+    ps = []
+
+    while l < n:
+        ps = getNextPrime(ps)
+        l = l + 1
+
+    return ps
+
+def sieve(n):
+    assert n > 1
+
+    ns = [True] * n
+
+    for i in range(2, math.ceil(np.sqrt(n))):
+        if ns[i]:
+            j = pow(i, 2)
+
+            while j < n:
+                ns[j] = False
+                j = j + i
+
+    return [i for i,val in enumerate(ns) if val][2:]
+
+
+def product(xs):
+    res = 1
+
+    for x in xs:
+        res = res * x
+
+    return res
+
+
+# Found a formula for this exact question:
+# https://oeis.org/A007504
+# a(n) = (n^2/2)*(log n + log log n - 3/2 + (log log n - 3)/log n + (2 (log log n)^2 - 14 log log n + 27)/(4 log^2 n) + O((log log n/log n)^3))
+
+# def getSumOfPrimes(n):
+#     return 0.5 * pow(n, 2) * (np.log(n) + np.log(np.log(n)) - 1.5 + (np.log(np.log(n)) - 3) / np.log(n) + 
+#         (2 * pow(np.log(np.log(n)), 2) - 14 * np.log(np.log(n)) + 27) / (4 * np.log2(n)) + some O term)
+
+# Unusable because of the o term, unfortunately. Also, I don't know exactly for which n prime numbers I want to calculate this sum.
+# We will have to do it the hard way
+
+
+def main():
+    print("Hello, this is Patrick")
+
+    ps = sieve(2000000)
+    print(sum(ps))
+
+
+if __name__ == "__main__":
+    main()