I've taken ill for the last two days, so I've been working on a couple Project Euler problems in between trips to the bathroom.
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.
Since you start with n = 0, you know that b always has to be prime in order to satisfy n = 0.
Next, if b must be prime, and all primes greater than 2 are odd, and we don't care about expressions resulting in less than 3 consecutive primes (example expression has 40), then we know that all values of a must be odd in order to satisfy n = 1!
I also suspect that for the number of consecutive primes we need to 'win', that we only need to look at negative values of a, but I'm having a heck of a time trying to prove it.
require 'prime_generator'
# pre-calculate primes
MAX = 1_000
primer = Prime_Generator.new MAX
primes = primer.stack
product = 0
highest = 0
# a must be odd
(0..MAX).each do |i|
next if i & 1 == 0
# b must be prime
primes.each do |b|
# a can be positive or negative
[i,-i].each do |a|
n = 0
while n += 1 do
break unless primer.is_prime?(n ** 2 + a * n + b)
end
if highest < n
highest = n
product = a * b
end
end
end
end
puts product
And here's the prime generator I'm using:
class Prime_Generator
attr_reader :stack
def initialize max = 3
@stack = [1,2,3]
fill_to max
end
def fill_to max
n = 1
while true do
n += 4
return @stack if n > max
@stack << n if is_prime? n
n += 2
return @stack if n > max
@stack << n if is_prime? n
end
end
def is_prime? n
return false if n <= 0
max = Math.sqrt(n).floor
fill_to(max + 1) if max > @stack.last
@stack.each do |i|
next if i == 1
return true if i > max
return false if n % i == 0
end
true
end
end
You can find more Project Euler solutions here: https://svn2.assembla.com/svn/joe-zack-personal/projects/euler/ruby/
Tags: project euler, ruby
I realize this isn't the fast solution, but the more I optimized, the uglier it got so I'm done playing with it. The hardest part was figuring out what the upper bound limit was.
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
power, total = 5, 0
(power * 9**power).times do |i|
total += i if i == i.to_s.split('').inject(0) {
|sum, n|
sum + n.to_i**power
}
end
puts total - 1
Tags: project euler, ruby
I knew I'd be implementing my own division algorithm for this problem, but I had a hard time figuring out a good way to detect the repeating sequence.
That's all I have to say about that.
Find the value of d 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
def divide n, d, repo = []
return repo.size - repo.index(n) if repo.include? n
divide 10 * (n - (n / d) * d), d, repo << n
end
highest = {"d" => 1, "count" => 1}
(1..499).each do |i|
x = i * 2 + 1
count = divide 1, x
if count > highest['count']
highest = {"d" => x, "count" => count}
end
end
puts highest["d"]
Tags: project euler, ruby
I could see that this problem was solvable by hand, but that's no fun.
Instead I decided to implement it with a tree. Not the most efficient solution to be certain, especially the way I rolled it, but it was a fun exercise. I don't get to play with trees often enough.
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
def nth_lexical_permutation values, key = "", chain = ""
chain = chain + key.to_s
children = values.select { |i| i != key }
$count -= 1 if children.size == 0
return chain if $count == 0
children.each do |i|
result = nth_lexical_permutation children, i, chain
return result if result.size > 0
end
""
end
$count = 1_000_000
set = (0..9).to_a
puts nth_lexical_permutation(set)
Yep, there's a global in there. It just made things easier. There are actually quite a few things I don't like about this program. But it's late. And I'm tired.
Tags: project euler, ruby
Looks like I took a pretty common approach on this one. First I calculated all the abundant numbers and dropped the sums in a hash. Then I just summed up to the given upper bound of 28,124 excluding those abundant sums.
Ruby1.9 solves it on my computer in about 15 seconds. After reading through the forums, some people were using a much lower upper limit of 20,200 which brought my run time down to a much respectable 6 seconds.
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
def sum_proper_divisors n
sum = 1
(2..Math.sqrt(n)).each do |i|
if n % i == 0
result = n / i
sum += result unless result == i
sum += i
end
end
sum
end
def next_abundant repo
i = repo.last + 1
while( sum_proper_divisors(i) <= i)
i += 1
end
i
end
max = 20_200
repo = [12]
sums = {24 => nil}
total = 0
while repo.last < max do
repo << next_abundant(repo)
repo.each do |i|
sum = repo.last + i
if sum > max
break
end
sums.store sum, nil
end
end
max.times do |i|
total += i unless sums.include? i
end
puts total
Tags: project euler, ruby