I'm publishing it now because, even though it isn't done, it will never get any more done. So that makes it done.
Problem 12: The sequence of triangle numbers is generated by adding the natural numbers. So the 7^(th) triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. What is the value of the first triangle number to have over five hundred divisors?
My first attempt was to iterate through the triangular numbers, checking their number of divisors:
Friday, December 19, 2008
Project Euler
**This was a draft written in December 2008**
I'm publishing it now because, even though it isn't done, it will never get any more done. So that makes it done.
[...] Boring as home is downloading TF2 and losing my internet to the ensuing packet deluge has made it worse. As such, I've decided to take advantage of the down time and work on some solutions to project euler problems. This post will list the problems and how I solved them.
In solving each problem, I start with whatever pops to mind first. I strive for readability and conciseness first. Then, if this doesn't work, I will implement a more complex algorithm to solve the problem. I stick to algorithms I already know (though I obviously had to look some of them up for implementation).
It appears the biggest problem here is going to be windows. GVim isn't too amused by my attempts to make it behave like normal. Apparently, they assume windows users would Also, I've decided that this will be more fun in python3k, so I need to set that up also. Windows 2003 isn't amused by this idea... Ahh, genius! obviously, I shouldn't be allowed to run a file I've pushed over a WRITE mounted samba share! In the same user's session who is sharing the directory!
Problem 1: Find the sum of all the multiples of 3 or 5 below 1000.
Kind of can't believe this didn't work on the first try (wrote and instead of or; I need to read problems more thoroughly...)
def problem1():
def problem1_generator():
for n in range(0,1000):
if n%3 == 0 or n%5 == 0:
yield n
result = sum(problem1_generator())
return result
This isn't really as pythonic as it could be though:
def problem1b():
numbers = (n for n in range(0,1000) if n%3 == 0 or n%5 == 0)
return sum(numbers)
Problem 2: sum of all even fibonacci numbers less than 4 million.
Also got this one wrong on the first try, forgot the even part...
def problem2():
def fib_generator(end=float("Infinity")):
""" return fibonacci numbers < b =" 1,2" b =" b," 2 ="="0)">
Problem 3: What is the largest prime factor of the number 600851475143 ?
A simple brute force solution.
def problem3():
from math import sqrt
def factors_generator(n, primes=[2]):
""" Due to the way python binds default method args,
primes will persist across method calls...
"""
for i in range(2, int(sqrt(n))):
if not any(p for p in primes if i%p == 0):
primes.append(i)
if n%i == 0:
n = n//i
yield i
if n == 1:
break
return sorted(list(factors_generator(600851475143)))[-1]
Problem 4: Find the largest palindrome made from the product of two 3-digit numbers.
def problem4():
numbers = list(range(100,1000))
to_test = sorted([x*y for x in numbers for y in numbers])
for n in reversed(to_test):
s = str(n)
if s == s[::-1]:
return n
Here we can see the excellent reversed built-in, added in Py3k. It's kind of unfortunate that str(reversed(s)) converts the generator itself to a string, rather than the items in it.
Problem 5: What is the smallest number that is evenly divisible by all of the numbers from 1 to 20?
I was originally doing this the hard way, writing my own custom lcm routine. Then I remembered that the lcm can be computed in terms of the gcd, and things got much easier.
def problem5():
from functools import reduce
def gcd(a,b):
while b:
a,b = b, a%b
return a
def lcm(a,b):
return a // gcd(a,b) * b
return reduce(lcm, range(1,21))
Problem 6: Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
This one is cake.
def problem6():
numbers = range(1,101)
squared = (x**2 for x in numbers)
return sum(numbers)**2 - sum(squared)
Problem 7: Find the 10001st prime.
I reuse the factors_generator method created for problem 3. This takes about 30 seconds on my machine, so it isn't winning any speed tests.
def problem7():
primes = []
i = 2
while len(primes) < 10001: for factor in factors_generator(i, primes): pass i *= 2 print(len(primes)) return primes[10000]
This is a bit entangled, so here's a standalone solution. I think it's even less code...
def problem7b():
primes = []
i = 2
while len(primes) < 10001: if not any(p for p in primes if i%p == 0): primes.append(i) i += 1 return primes[10000]
Problem 8: Find the greatest product of five consecutive digits a (given) 1000-digit number.
For the number, check the code. First problem: how the hell do you represent a 1000 digit number cleanly in code? I was forced to use a multiline string and then strip whitespace... Other than that, not much to see here.
def problem8():
from functools import reduce
from operator import mul
numberS = """
73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450
"""
number = "".join(numberS.split()) #remove whitespace
product = 0
for i in range(0, len(number)-5):
digits = (int(a) for a in number[i:i+5])
t = reduce(mul, digits)
product = max(product, t)
return product
Problem 9: There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
This is pretty simple, if you know about the modified version of Euclid's formula. This version is guaranteed to generate all Pythagorean triples (eventually).
def problem9():
from math import sqrt
def pythagorean_generator():
""" return all pairs (a,b) s.t. a**2 + b**2 = c**2
and a,b,c integers
"""
k = 1
mstart = 2
batch_size = 100
while True:
for m in range(mstart, mstart+batch_size):
for n in range(1,m):
yield ( k*2*m*n, k*(m**2-n**2) )
mstart += batch_size
k += 1
for a,b in pythagorean_generator():
c = int(sqrt(a**2 + b**2))
if a + b + c == 1000:
return a*b*c
Problem 10: Find the sum of all the primes below two million.
The naive algorithm just isn't going to be fast enough in python, clocking in around 10 minutes. Thus, it is interesting to note that in a lower level language, the naive algorithm probably would clock in at under a minute.
def problem10():
primes = [2]
i = 3
while i < 2*10**6: if not any(p for p in primes if i%p == 0): primes.append(i) i += 2 return sum(primes)
Of course, we can always try a naive implementation of the Sieve of Eratosthenes.
def problem10b():
from math import sqrt
primes = []
numbers = list(range(0, 2*10**6))
crossed = [False]*len(numbers)
for i in range(2, len(numbers)):
if not crossed[i]:
p = numbers[i]
primes.append(p)
for j in range(p, len(numbers), p):
crossed[j] = True
return sum(primes)
This gives us the answer in a couple seconds. Of course, I also managed to introduce an off-by-two error due to my indexing of the arrays (which is why numbers now includes 0 and 1, so that numbers[i] = i). I always hate these situations where you have to iterate through two arrays simultaneously. In a more complicated problem I'd consider wrapping things up into an object, but that seems unwarranted here.
Well, that's enough for now. Something wrong or unpythonic? Hit me back.
