January 12, 2009
Project Euler
Yesterday I started working on Project Euler.
Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
In other words, it's geek crack. I started with the first problem at 10pm last night and before I knew it, it was 2am.
Here are my solutions for the first 11 problems:
Problem 1
x = 0
for i in range(1, 1000):
if i % 3 == 0 or i % 5 == 0:
x = x + i
print x
Problem 2
def fib():
x = 0
a,b = 0,1
while a <= 4000000:
if a % 2 == 0:
x = x + a
a,b, = b,a+b
print x
print fib()
Problem 3
from math import sqrt
def largest_prime(n):
for i in range(int(sqrt(n)) + 1, 2, -2):
if n % i == 0:
if is_prime(i):
return i
def is_prime(n):
n = abs(int(n))
if n < 2:
return False
if n == 2:
return True
if not n & 1:
return False
for i in range(3, int(sqrt(n)) + 1, 2):
if n % i == 0:
return False
return True
n = 600851475143
print largest_prime(n)
Problem 4
def is_palindrome(n):
x = str(n)
y = len(x)
if (y == 1):
return True
for j in range(0, y / 2):
if x[j] != x[y - j - 1]:
return False
return True
def largest_palindromic_product(min, max):
maxProduct = 0
for i in range(max, min, -1):
for j in range(max, min, -1):
x = i * j
if is_palindrome(x):
if x > maxProduct:
maxProduct = x
return maxProduct
print largest_palindromic_product(99, 999)
Problem 5
def is_divisible(n, x):
for i in range(n, 2, -1):
if x % i != 0:
return False
return True
x = 0
while True:
x = x + 20
if is_divisible(20, x):
print x
break
Problem 6
def sum_of_squares(n):
return reduce(lambda x, y: x + (y * y), range(1, n + 1))
def square_of_sums(n):
x = reduce(lambda x, y: x + y, range(1, n + 1))
return x * x
def diff_sums(n):
return square_of_sums(n) - sum_of_squares(n)
print diff_sums(100)
Problem 7
x = 0
y = 0
while x < 10001:
if is_prime(y):
x = x + 1
y = y + 1
print y - 1
Problem 8
def numbers_at_index(n):
return map(lambda x: int(number[x]), range(n, n + 5))
def product_at_index(n):
return reduce(lambda x, y: x * y, numbers_at_index(n))
print max(map(product_at_index, range(995)))
Problem 9
# brute force, baby!
def is_pythagorean_triplet(a, b, c):
if (a * a) + (b * b) == (c * c):
return True
return False
def find():
for i in range(1, 1001):
for j in range(i, 1001):
for k in range(j, 1001):
if (i + j + k == 1000):
if is_pythagorean_triplet(i, j, k):
return i * j * k
print find()
Problem 10
print sum(filter(is_prime, range(1, 2000000)))
Problem 11
import operator
grid = [8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8,49,49,99,40,17,81,18,57,60,87,17,40,
98,43,69,48,4,56,62,0,81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65,52,70,95,
23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91,22,31,16,71,51,67,63,89,41,92,36,54,22,40,
40,28,66,33,13,80,24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50,32,98,81,28,
64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70,67,26,20,68,2,62,12,20,95,63,94,39,63,8,
40,91,66,49,94,21,24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72,21,36,23,9,
75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95,78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,
14,9,53,56,92,16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57,86,56,0,48,35,71,
89,7,5,44,44,37,44,60,21,58,51,54,17,58,19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,
89,55,40,4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66,88,36,68,87,57,62,20,72,
3,46,33,67,46,55,12,32,63,93,53,69,4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,
36,20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16,20,73,35,29,78,31,90,1,74,
31,49,71,48,86,81,16,23,57,5,54,1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]
def right(n):
if n / 20 != n + 3 / 20: return 0
return reduce(operator.mul, grid[n : n + 4])
def down(n):
if n + 60 > 399: return 0
return reduce(operator.mul, grid[n : n + 61 : 20])
def vright(n):
if n / 20 != n + 3 / 20: return 0
if n + 63 > 399: return 0
return reduce(operator.mul, grid[n : n + 65 : 21])
def vleft(n):
if n / 20 != n - 3 / 20: return 0
if n + 57 > 399: return 0
return reduce(operator.mul, grid[n : n + 58 : 19])
print max(map(lambda i : max(right(i), down(i), vright(i), vleft(i)), range(400)))
Posted by J.P. Cummins ● Comments (1)
Comments
Brandon Hubbell Jan. 13, 2009 at 5:59 a.m.
Boo! You didn't post your run times.
You can improve your prime check (#3) by improving your even check.
Where N is >= 0,
return (N & 1) != 1;
Probably won't help much now, but you run into some more beastly prime checks later.
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