Solves 4 problems.

Eclipse integration for simpler use with Windows. 

Still have to perform some good work on prime numbers! 


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

README.markdown

@@ -39,13 +39,18 @@ Should be used in order to help future reuse of code :)
 24 - What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? - too long <br />
 25 - What is the first term in the Fibonacci sequence to contain 1000 digits? - 0.741 <br />
 27 - Find a quadratic formula that produces the maximum number of primes for consecutive values of n. - too long <br />
+29 - How many distinct terms are in the sequence generated by ab for 2  a  100 and 2  b  100? - < 1 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 />
+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 />
 67 - Using an efficient algorithm find the maximal sum in the triangle? - 0.027 <br />
 
 **In progress: **
 
 26 - Find the value of d < 1000 for which 1/d contains the longest recurring cycle. <br />
+30 - Find the sum of all the numbers that can be written as the sum of fifth powers of their digits. <br />
 35 - How many circular primes are there below one million? <br />
 
 **WARNING : Spoil inside for those who want to solve problems by themselves :)**

e_29.py

@@ -0,0 +1,38 @@
+#!/usr/bin/env python 
+'''
+Created on 7 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 29 of Project Euler
+Consider all integer combinations of a^b for 2<=a<=5 and 2<=b<=5:
+
+2^2=4, 2^3=8,2^4=16, 2^5=32
+3^2=9, 3^3=27, 3^4=81,3^5=243
+4^2=16, 4^3=64, 4^4=256, 4^5=1024
+5^2=25, 5^3=125, 5^4=625, 5^5=3125
+If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
+
+4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
+
+How many distinct terms are in the sequence generated by a^b for 2<=a<=100 and 2<=b<=100?
+''' 
+def integer_combinations(a_max, b_max):
+    """
+    Returns a list of all integer combinations of a^b for 2<=a<=5 and 2<=b<=5
+    """
+    olist = []
+    for a in range(2, a_max + 1):
+        for b in range(2, b_max + 1):
+            olist.append(a**b)
+
+    return olist
+
+def sort_n_short(olist):
+    """
+    Sort elements of olist and remove all duplicates.
+    """
+    return sorted(set(olist))
+
+if __name__ == '__main__':    
+    print "Answer : %d " % (len(sort_n_short(integer_combinations(100, 100))))

e_30.py

@@ -0,0 +1,29 @@
+#!/usr/bin/env python 
+'''
+Created on 7 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 30 of Project Euler
+Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
+
+1634 = 1^4 + 6^4 + 3^4 + 4^4
+8208 = 8^4 + 24 + 0^4 + 8^4
+9474 = 9^4 + 4^4 + 7^4 + 4^4
+As 1 = 1^4 is not a sum it is not included.
+
+The sum of these numbers is 1634 + 8208 + 9474 = 19316.
+
+Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
+'''
+
+
+def largest():
+    """
+    Returns
+    """
+
+    return 1
+
+if __name__ == '__main__':
+    print "Answer : %d " % (largest())

e_35.py

@@ -11,7 +11,8 @@
  How many circular primes are there below one million?
  ##---
  """
- 
+import pickle
+
 def all_permutations(seq):
     """permutate a sequence and return a list of the permutations"""
     if not seq:
@@ -52,20 +53,23 @@ def prime_list(max_val):
         
     return plist
  
-def perm_primes(max_val):
+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(list))
+        print "%d/%d" % ( cpt, len(plist))
         print circulars, circl
         perms = all_permutations(str(prime)) # finds all permutations
         
@@ -82,8 +86,47 @@ def perm_primes(max_val):
         
     return circulars, circl
  
+def perm_primes(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"))
+    #plist = prime_list(max_val)
+    checks_list = [1] * len(plist) # all primes to be tested
+    
+    for ii in range(len(plist)):
+        if (ii % 99 == 0):
+            print "%d/%d" % ( ii + 1, len(plist))
+
+        if checks_list[ii]: # if prime still to be checked
+            prime = plist[ii]
+            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):                  
+                # updating checks_list
+                for perm in perms:
+                    if not(int(perm) in circl): # avoiding duplicates
+                        circulars += 1
+                        circl.append(int(perm))
+                    checks_list[plist.index(int(perm))] = 0
+                    
+            else:
+                checks_list[ii] = 0
+    
+    return circulars, circl
+ 
 if __name__ == '__main__' :
     answer, plist = perm_primes(1000000)
     print "Answer is : %d"  % (answer)
-    print plist
+    print plist
-    raw_input()

e_42.py

@@ -0,0 +1,100 @@
+#!/usr/bin/env python 
+'''
+Created on 7 feb. 2012
+
+@author: Julien Lengrand-Lambert
+
+DESCRIPTION: Solves problem 42 of Project Euler
+The nth term of the sequence of triangle numbers is given by, t_n = 1/2 * n * (n+1); 
+so the first ten triangle numbers are:
+1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
+
+By converting each letter in a word to a number corresponding 
+to its alphabetical position and adding these values we form a word value. 
+For example, the word value for SKY is 19 + 11 + 25 = 55 = t_10. 
+If the word value is a triangle number then we shall call the word a triangle word.
+
+Using e_42.data (right click and 'Save Link/Target As...'), 
+a 16K text file containing nearly two-thousand common English words, 
+how many are triangle words?
+''' 
+import string
+
+def load_data(filename):
+    """
+    Loads the data from a name file into a table
+    """
+    my_file = open(filename, "r")
+    line = my_file.readline()
+    data = [el for el in line.split("\"")][1::2]
+    my_file.close()        
+    return data
+
+def dict_alphabet():
+    """
+    Returns a dict of the alphabet, with value for each letter.
+    """
+    alpb= dict()
+    letters = string.ascii_uppercase
+    inc = 1
+    for j in range(len(letters)): 
+        alpb[letters[j]] = inc
+        inc += 1
+
+    return alpb
+
+def names2scores(data):
+    """
+    Returns a list of scores given a list of names.
+    Scores are calculated with word values, (A = 1, B = 2, ...)
+    """
+    scores = []
+    alpb = dict_alphabet()
+    for name in data:
+        score = 0
+        for jj in range(len(name)):
+            score += (alpb[name[jj]])
+    
+        scores.append(score)
+    
+    return scores
+
+def triangle_list(max_val):
+    """
+    Returns a list of triangle numbers up to max_val.
+    triangle_numbers is an increasing function.
+    """
+    trilist = []
+    ptr = 1
+    val = 1
+    while val < max_val:
+        val = trinum(ptr)
+        trilist.append(val)
+        ptr += 1 
+    
+    return trilist
+
+def trinum(val):
+    """
+    Returns the triangle number for a given value
+    """
+    return val * (val + 1) / 2
+
+def triangle_words(filename):
+    """
+    Returns the number of triangle words in filename
+    """
+    data = load_data(filename)
+    scores = names2scores(data)
+    trilist = triangle_list(max(scores))
+    
+    # checking if words are triangle
+    cpt = 0
+    for score in scores:
+        if score in trilist:
+            cpt += 1
+            
+    return cpt
+    
+if __name__ == '__main__':    
+    print "Answer : %d " % (triangle_words("e_42.data"))

e_45.py

@@ -0,0 +1,75 @@
+'''
+Created on 7 feb. 2012
+
+@author: Julien Lengrand-Lambert
+DESCRIPTION : Solves problem 45 of Project Euler
+Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
+
+Triangle         Tn=n(n+1)/2       1, 3, 6, 10, 15, ...
+Pentagonal       Pn=n(3(n-1))/2       1, 5, 12, 22, 35, ...
+Hexagonal        Hn=n(2(n-1))         1, 6, 15, 28, 45, ...
+It can be verified that T285 = P165 = H143 = 40755.
+
+Find the next triangle number that is also pentagonal and hexagonal.
+'''
+
+def tn(value):
+    """
+    Returns the triangle number of the given value
+    """
+    return value * (value + 1) / 2
+
+def pn(value):
+    """
+    Returns the Pentagonal number of the given value
+    """
+    return value * ((3 * value) - 1) / 2
+
+def hn(value):
+    """
+    Returns the Hexagonal number of the given value
+    """
+    return value * ((2 * value) - 1)
+
+def tripenhex(value):
+    """
+    Returns True of the triangle, pentagonal and hexagonal number of value are the same
+    """
+    return (tn(value) == pn(value) == hn(value))
+
+def big_while():
+    """
+    Returns the value for which tm = pn = ho
+    """
+    cpt_t = 2
+    cpt_p = 2
+    cpt_h = 2
+    val_t = tn(cpt_t)
+    val_p = pn(cpt_p)
+    val_h = hn(cpt_h)
+    
+    res = []
+    while len(res) < 2: # to get the two first values
+        while not (val_t == val_p == val_h):
+            
+            cpt_t += 1
+            val_t = tn(cpt_t)
+            
+            # updating val_p
+            while val_p < val_t:
+                cpt_p += 1
+                val_p = pn(cpt_p)
+                
+            # updating val_h
+            while val_h < val_t:
+                cpt_h += 1
+                val_h = hn(cpt_h)    
+        res.append(val_t)
+        cpt_t += 1
+        val_t = tn(cpt_t)
+        
+    return res[1] # outputs the second value
+
+if __name__ == '__main__':
+    
+    print "Answer : %d " % (big_while())  

e_52.py

@@ -0,0 +1,64 @@
+'''
+Created on 7 feb. 2012
+
+@author: Julien Lengrand-Lambert
+DESCRIPTION : Solves problem 52 of Project Euler
+It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.
+
+Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.
+'''
+def smallest_match(max_mul):
+    """
+    Find the smallest positive integer, x, such that ..., till max_mul*x contain the same digits.
+    """
+    cpt = 2
+    nlist = [(i + 1) * cpt for i in range(max_mul)]
+    
+    while(1):
+        if cpt % 10000 == 0:
+            print cpt
+        if same_size(nlist):
+            if same_digits(nlist):
+                print cpt, nlist
+                return cpt
+
+        cpt += 1
+        nlist = [(i + 1) * cpt for i in range(max_mul)]
+
+def all_permutations(seq):
+    """permutate a sequence and return a list of the permutations"""
+    if not seq:
+        return [seq]  # is an empty sequence
+    else:
+        temp = []
+        for k in range(len(seq)):
+            part = seq[:k] + seq[k+1:]
+            for m in all_permutations(part):
+                temp.append(seq[k:k+1] + m)
+        return temp
+
+
+def same_digits(nlist):
+    """
+    Returns True if all numbers in nlist contain the same digits. 
+    Which means they are all permutations of the first value 
+    """
+    perms = all_permutations(str(nlist[0])) # finds all permutations
+    for el in nlist:
+        if not(str(el) in perms):
+            return False      
+    
+    return True
+
+def same_size(nlist):
+    """
+    Returns True if all numbers in nlist have the same number of digits
+    """
+    nlen = len(str(nlist[0]))
+    for el in nlist :
+        if len(str(el)) != nlen:
+            return False
+    return True
+
+if __name__ == '__main__':
+    print "Answer : %d " % (smallest_match(6))