Solve a new problem.

Prepares some others. 

More than 35 problem solved ! 

Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>

README.markdown

@@ -42,13 +42,14 @@ Should be used in order to help future reuse of code :)
 28 - What is the sum of both diagonals in a 1001 by 1001 spiral? - < 1 sec <br />
 29 - How many distinct terms are in the sequence generated by ab for 2  a  100 and 2  b  100? - < 1 sec <br />
 30 - Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. - < 3 sec <br />
-34 - Find the sum of all numbers which are equal to the sum of the factorial of their digits. - 30 sec  <br />
+34 - Find the sum of all numbers which are equal to the sum of the factorial of their digits. - 30 sec - 16 sec  <br />
 36 - Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2. - 0.933 <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 /> 
 48 - Find the last ten digits of 1^1 + 2^2 + ... + 1000^1000. - 0.053 <br />
 52 - Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits. - 2min <br />
+53 - How many values of C(n,r), for 1 <= n <= 100, exceed one-million? - < 1 sec <br />
 67 - Using an efficient algorithm find the maximal sum in the triangle? - 0.027 <br />
 
 **In progress: **
@@ -56,7 +57,7 @@ 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 />
 31 - Investigating combinations of English currency denominations. <br />
 35 - How many circular primes are there below one million? <br />
-53 - How many values of C(n,r), for 1 <= n <= 100, exceed one-million? <br />
+97 - Find the last ten digits of the non-Mersenne prime: 28433 × 2^7830457 + 1.
 
 **WARNING : Spoil inside for those who want to solve problems by themselves :)**
 

e_2.py

@@ -33,8 +33,6 @@ def even_sum_fib(max_value):
         
     return fib_sum
     
-    
-    
 if __name__ == '__main__':
     val = 4000000
     print "Answer : %d " % (even_sum_fib(val))                        

e_34_2.py

@@ -0,0 +1,70 @@
+#!/usr/bin/env python 
+'''
+Created on 8 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 34 of Project Euler
+145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
+
+Find the sum of all numbers which are equal to the sum of the factorial of their digits.
+
+Note: as 1! = 1 and 2! = 2 are not sums they are not included.
+''' 
+import timeit
+from utils.memoize import memoize
+
+def find_limit():
+    """
+    Returns the max number of digits we have to loop through to find the solution
+    """
+    nb_digits = 2
+    max_num = fact(9) * nb_digits 
+    
+    while len(str(max_num)) >= nb_digits:
+        nb_digits += 1
+        
+    return nb_digits - 1
+
+@memoize
+def fact(value):
+    """
+    Returns value! 
+    """
+    if value == 0:
+        return 1
+    return value * fact(value - 1)      
+
+def digsum(num):
+    """
+    Returns the sum of factorial of digits of num
+    ex : digsum(123, 4) = fact(1) + fact(2) + fact(3)
+    """
+    val = 0
+    for el in str(num):
+        val += fact(int(el))
+    
+    return val
+
+def sum_fact2():
+    """
+    Finds the sum of all the numbers that can be written as the sum of the factorial of their digits.
+    """
+    max_dig = find_limit()
+    max_val = fact(9) * max_dig
+    print "Max dig is : %d" %(max_dig)
+    sumf = 0    
+    
+    cpt = 3
+    while cpt <= max_val:    
+        if cpt ==  digsum(cpt):
+            sumf += cpt
+        
+        cpt +=1
+
+    return sumf
+
+if __name__ == '__main__':
+    # The major problem in there is to find the upper limit.
+    t2 = timeit.Timer("sum_fact2()", "from __main__ import sum_fact2")
+    print t2.timeit(1)

e_35.py

@@ -52,39 +52,7 @@ def prime_list(max_val):
         cur += 1
         
     return plist
- 
-def perm_primes_old(max_val):
-    """
-    Returns the number of circular primes below max_val
-    """
-    circulars = 0
-    circl = []
-    
-    # to avoid calculating primes
-    # plist = pickle.load(open("primes_list.dup", "rb"))
-    # print plist
-    plist = prime_list(max_val)
-    print "###"
-    # stuff here
-    cpt = 0
-    for prime in plist:
-        cpt += 1
-        print "%d/%d" % ( cpt, len(plist))
-        print circulars, circl
-        perms = all_permutations(str(prime)) # finds all permutations
-        
-        # counts how many permutations are in prime list given one prime
-        p_cpt = 0
-        for perm in perms:
-            if int(perm) in plist:
-                p_cpt += 1
 
-        # if prime is circular
-        if p_cpt == len(perms):
-            circulars += 1 
-            circl.append(prime)
-        
-    return circulars, circl
  
 def perm_primes(max_val):
     """
@@ -128,5 +96,4 @@ def perm_primes(max_val):
  
 if __name__ == '__main__' :
     answer, plist = perm_primes(1000000)
-    print "Answer is : %d"  % (answer)
-    print plist
+    print "Answer is : %d"  % (answer)

e_53.py

@@ -13,16 +13,42 @@ In combinatorics, we use the notation, 5C3 = 10.
 
 In general,
 
-nCr = n! / r!(nr)! ,where r  n, n! = n(n1)...321, and 0! = 1.
+nCr = n! / r!(n - r)! ,where r <= n, n! = n * (n - 1) * ... * 3 * 2 * 1, and 0! = 1.
 It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.
 
-How many, not necessarily distinct, values of  nCr, for 1  n  100, are greater than one-million?
+How many, not necessarily distinct, values of  nCr, for 1 <=  n <= 100, are greater than one-million?
 ''' 
-def aaa():
+from utils.memoize import memoize
+
+@memoize
+def fact(value):
     """
-    Returns 
+    Returns value! 
     """
-    return 1
+    if value == 0:
+        return 1
+    return value * fact(value - 1)      
+
+def comb(n, r):
+    """
+    Returns nCr = n! / r!(n - r)!,where r <= n
+    """
+    if r > n :
+        raise ValueError("n should be >= r !")
+    else:
+        return (fact(n) / (fact(r) * fact(n - r)))
+
+def upper_comb(max_n, val):
+    """
+    Returns the number of values of nCr, for 1 <=  n <= max_n that are greater than value?
+    """
+    nb = 0
+    for ns in range(1, max_n + 1):
+        for nr in range(1, ns + 1):
+            if comb(ns, nr) > val:
+                nb += 1
+    
+    return nb
 
 if __name__ == '__main__':
-    print "Answer : %d " % (1)
+    print "Answer : %d " % (upper_comb(100, 1000000))

e_97.py

@@ -0,0 +1,24 @@
+#!/usr/bin/env python 
+'''
+Created on 10 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 31 of Project Euler
+The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 2^6972593 - 1.
+it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2^p - 1, have been found which contain more digits.
+
+However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433 * 2^7830457 + 1.
+
+Find the last ten digits of this prime number.
+''' 
+def last_ten():
+    """
+    Returns the last ten digits of  28433 * 2^7830457 + 1
+    """
+    val = 28433 * pow(2, 7830457) + 1
+    return str(val)[-10:]
+
+if __name__ == '__main__':
+    #print "Answer : %d " % (last_ten())
+    print  pow(2, 7830457)

utils/Memoized.py

@@ -0,0 +1,34 @@
+'''
+Created on 
+
+@author: airballman
+@contact: julien@lengrand.fr
+'''
+import functools
+
+class Memoized(object):
+    """
+    Decorator that caches a function's return value each time it is called.
+    If called later with the same arguments, the cached value is returned, and
+    not re-evaluated.
+    """
+    def __init__(self, func):
+        self.func = func
+        self.cache = {}
+    def __call__(self, *args):
+        try:
+            return self.cache[args]
+        except KeyError:
+            value = self.func(*args)
+            self.cache[args] = value
+            return value
+        except TypeError:
+            # uncachable -- for instance, passing a list as an argument.
+            # Better to not cache than to blow up entirely.
+            return self.func(*args)
+    def __repr__(self):
+        """Return the function's docstring."""
+        return self.func.__doc__
+    def __get__(self, obj, objtype):
+        """Support instance methods."""
+        return functools.partial(self.__call__, obj)

utils/__init__.py

@@ -0,0 +1,6 @@
+'''
+Created on 
+
+@author: airballman
+@contact: julien@lengrand.fr
+'''

utils/fact_memoization.py

@@ -0,0 +1,43 @@
+#!/usr/bin/env python 
+'''
+Created on 17 feb. 2012
+
+@author: Julien Lengrand-Lambert
+@contact: julien@lengrand.fr
+''' 
+import timeit
+from memoize import memoize
+from Memoized import Memoized
+
+def fact1(value):
+    """
+    Returns value! 
+    """
+    if value == 0:
+        return 1
+    return value * fact1(value - 1)      
+    
+@memoize
+def fact2(value):
+    """
+    Returns value!, using memoization 
+    """
+    if value == 0:
+        return 1
+    return value * fact2(value - 1)     
+
+@Memoized
+def fact3(value):
+    """
+    Returns value!, using memoization 
+    """
+    if value == 0:
+        return 1
+    return value * fact3(value - 1)   
+    
+if __name__ == '__main__':
+    for i in range(3):
+        t1 = timeit.Timer("fact1(150)", "from __main__ import fact1")
+        t2 = timeit.Timer("fact2(150)", "from __main__ import fact2")
+        t3 = timeit.Timer("fact3(150)", "from __main__ import fact3")
+        print t1.timeit(1), t2.timeit(1), t3.timeit(1)

utils/fib_memoization.py

@@ -0,0 +1,40 @@
+#!/usr/bin/env python 
+'''
+Created on 17 feb. 2012
+
+@author: Julien Lengrand-Lambert
+@contact: julien@lengrand.fr
+''' 
+import timeit
+from memoize import memoize
+from Memoized import Memoized
+
+def fib1(n):
+    """Returns the n st value of Fibonnacci list"""
+    if n < 2:
+        return n
+    else:
+        return fib1(n-1) + fib1(n - 2)
+    
+@memoize
+def fib2(n):
+    """Returns the n st value of Fibonnacci list, using memoization"""
+    if n < 2:
+        return n
+    else:
+        return fib2(n-1) + fib2(n - 2)
+
+@Memoized
+def fib3(n):
+    """Returns the n st value of Fibonnacci list, using memoization"""
+    if n < 2:
+        return n
+    else:
+        return fib3(n-1) + fib3(n - 2)
+    
+if __name__ == '__main__': 
+    for i in range(3):
+        t1 = timeit.Timer("fib1(20)", "from __main__ import fib1")
+        t2 = timeit.Timer("fib2(20)", "from __main__ import fib2")
+        t3 = timeit.Timer("fib3(20)", "from __main__ import fib3")
+        print t1.timeit(1), t2.timeit(1), t3.timeit(1)    

utils/memoize.py

@@ -0,0 +1,19 @@
+#!/usr/bin/env python 
+'''
+Created on 17 feb. 2012
+
+@author: Julien Lengrand-Lambert
+@contact: julien@lengrand.fr
+''' 
+
+def memoize(function):
+    """Defines a memoization scheme"""
+    cache = {}
+    def decorated_function(*args):
+        if args in cache:
+            return cache[args]
+        else:
+            val = function(*args)
+            cache[args] = val
+            return val
+    return decorated_function