Solves problem 13

Prepares problem 14 (hummm, though one it seems)

e_13.py

@@ -11,9 +11,21 @@
 """
 def ten_dig_sum(filename):
     """
+    Returns the first 10 digits of the sum of all numbers in filename.
     """
+    return int(str(sum(load_data(filename)))[0:10])
 
-    return 1
+def load_data(filename):
+    """
+    Loads the data from a file into a table
+    """
+    file = open(filename, "r")
+    data = []
+    for line in file :
+        data.append(int(line))
+    file.close()        
+    return data
 
 if __name__ == '__main__':
-    print "Answer : %d" % (fun())
+    #data = ten_dig_sum("e_13.data")
+    print "Answer : %d" % (ten_dig_sum("e_13.data"))

e_14.py

@@ -0,0 +1,34 @@
+#!/usr/bin/env python 
+"""
+ ##---
+ # jlengrand
+ #Created on : Mon Jan 16 10:22:33 CET 2012
+ #
+ # DESCRIPTION : Solves problem 14 of Project Euler
+ The following iterative sequence is defined for the set of positive integers:
+
+ n -> n/2 (n is even)
+ n -> 3n + 1 (n is odd)
+ 
+ Using the rule above and starting with 13, we generate the following sequence:
+ 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1
+ It can be seen that this sequence (starting at 13 and finishing at 1) contains 
+ 10 terms. 
+ Although it has not been proved yet (Collatz Problem), it is thought that all 
+ starting numbers finish at 1.
+
+ Which starting number, under one million, produces the longest chain?
+
+ NOTE: Once the chain starts the terms are allowed to go above one million. 
+ ##---
+"""
+def longest_chain(max_value):
+    """
+    Returns the starting number under max_value that produces the longest 
+    chain given the sequence.
+    """
+    
+    return 1
+
+if __name__ == '__main__':
+    print "Answer : %d" % (longest_chain(1000000))