Solves Problem 37 of Project Euler

Could be refactored a bit to avoid code redondance.
Uses my list of primes numbers

Performs calculations in less than one minute.

README.markdown

@@ -46,6 +46,7 @@ Should be used in order to help future reuse of code :)
 34 - Find the sum of all numbers which are equal to the sum of the factorial of their digits. - 30 sec - 16 sec  <br />
 35 - How many circular primes are there below one million? <br />
 36 - Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2. - 0.933 <br />
+37 - Find the sum of all eleven primes that are both truncatable from left to right and right to left. < 1 min <br />
 39 - If p is the perimeter of a right angle triangle, {a, b, c}, which value, for p <= 1000, has the most solutions? - 1min<br />
 42 - Using words.txt, a 16K text file containing nearly two-thousand common English words, how many are triangle words? - < 1 sec <br />
 45 - Find the next triangle number that is also pentagonal and hexagonal. - < 1 sec <br /> 
@@ -58,7 +59,6 @@ Should be used in order to help future reuse of code :)
 
 26 - Find the value of d < 1000 for which 1/d contains the longest recurring cycle. <br />
 33 - Discover all the fractions with an unorthodox cancelling method. <br />
-37 - Find the sum of all eleven primes that are both truncatable from left to right and right to left. <br />
 38 - What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ? <br />
 41 - What is the largest n-digit pandigital prime that exists? <br />
 97 - Find the last ten digits of the non-Mersenne prime: 28433 × 2^7830457 + 1.

e_37.py

@@ -11,7 +11,59 @@ Find the sum of the only eleven primes that are both truncatable from left to ri
 
 NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
 ''' 
+import pickle
+
+plist = pickle.load(open("primes_list.dup", "rb"))
+
+def is_prime(val):
+    """
+    Returns True if the number is prime
+    """
+    return (val in plist)
+
+def trunc_left(val):
+    """
+    Returns true if the prime number is truncatable from left to right
+    """
+    if not is_prime(val):
+        return False
+    if len(str(val)) == 1:
+        return True
+    else:
+        return trunc_left(int(str(val)[1:]))
+
+def trunc_right(val):
+    """
+    Returns true if the prime number is truncatable from left to right
+    """
+    if not is_prime(val):
+        return False
+    if len(str(val)) == 1:
+        return True
+    else:
+        return trunc_right(int(str(val)[:-1]))
+
+def trunc_primes():
+    """
+    Returns the complete list of primes that are both truncatable from left to right and right to left
+    """
+    # there we already have to 1 million.
+    inplist = list(plist)
+    out_list = []
+
+    for p in inplist:
+        if p > 11:
+            if (trunc_left(p) and trunc_right(p)):
+                print "Found one : %d" %(p)
+                out_list.append(p)
+                
+        if len(out_list) > 10 : #already found them all
+            return out_list
+        
+    print "End of the line. Should go higher than one million !"
+    return out_list
 
 if __name__ == '__main__':
-    print 1
-    #print "Answer : %d " % (last_ten())
+    # We know that there are only 11 of them. Lets try to find them
+    print trunc_primes()
+    print "Answer : %d " % (sum(trunc_primes()))