Solved the problem 2

0001-3and5.py

@@ -1,4 +1,10 @@
-import math
+"""
+Problem 1
+Add all the natural numbers below one thousand that are multiples of 3 or 5.
+
+If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
+Find the sum of all the multiples of 3 or 5 below 1000.
+"""
 def bruteforce():
     sum_ = 0
     number = 1000

0002-fib-4m.py

@@ -0,0 +1,61 @@
+"""
+Problem 2
+Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
+
+1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
+
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued t
+"""
+def fibonacci():
+    """
+    Just the usual fibonacci generator which starts from 1,2,3,..
+    """
+    a = 1
+    b = 2
+
+    while True:
+        yield a
+        sum_ = a + b
+        a = b
+        b = sum_
+
+def mod_fibonacci():
+    """
+    A modified version which returns every third number from 1,1,2,3,5,8,..
+    Note: Every third number is an even number
+    """
+    a = 1
+    b = 1
+    c = a + b
+    while True:
+        yield c
+        a = b + c
+        b = c + a
+        c = a + b
+
+def traditional():
+    """
+    Function to calculate sum of the usual fibonacci
+    """
+    sum_ = 0
+    for x in fibonacci():
+        if x >= 4000000:
+            break
+        if x % 2 == 0:
+            sum_ = sum_ + x
+    return sum_
+
+def third_number():
+    """
+    Function to calculate the sum out of the modified fibonacci
+    """
+    sum_ = 0
+    for x in mod_fibonacci():
+        print x
+        if x >= 4000000:
+            break
+        sum_ = sum_ + x
+    return sum_
+
+print "Answer by traditional method:",traditional()
+print "Answer by optimized method:",third_number()