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Project Euler: Problem 35 in Ruby

I used my prime generator from Problem 27 for this one. It would have been faster to build the rotation into my generator, but it ran fine without it.

Problem 35

How many circular primes are there below one million?

require 'prime_generator'

primer = Prime_Generator.new 1_000_000

def is_rot_prime? primer, chars
	chars.size.times do |i|
		chars = Array.new(chars.size) { |i| chars[i - 1] }
		return false if !primer.is_prime?(chars.join("").to_i)
	end
	true
end

count = 0
primer.stack.each do |n|
	count += 1 if is_rot_prime? primer, n.to_s.split("")
end

# subtract 1 because "1" doesn't count
puts count - 1

Speaking of the rotation, the ruby array initialize methods and negative indexers make it a cinch to rotate. How cool is this?

chars = Array.new(chars.size) { |i| chars[i - 1] }

Project Euler: Problem 34 in Ruby

Practically every Project Euler problem benefits from memoization, however the issues I ran into with #34 had nothing to do with my algorithm.

The first hurdle was figuring out the upper bound. After scratching around in my notebook, I figured any number I would be looking for would have 7 digits or less. That gives us an upper bound of 9! * 7.

The second hurde took me much, much longer to figure out.

Here's the secret: 0! = 1

I had taken it for granted that 0! would (of course!) be 0, and I had pre-filled my cache with the number 0. I went round, and round, and round, and round, and round, and round before figuring out (quite accidentally) my error.

It doesn't make any sense to me, but you can't argue with math!

Problem 34

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

# cache the digit factorials
factorials = [1]
(1..9).each do |i|
	factorials.push(i * factorials.last)
end

result = 0
(3..2_540_160).each do |n|
	sum = n.to_s.split("").inject(0) do |sum,n|
		sum + factorials[n.to_i]
	end
	result += n if n == sum
end
puts result

Runs in just under a minute

Project Euler : Problem 36 in Ruby

I've been ill today, but there's no better medicine than an easy Project Euler problem! I tried doing some bitwise magic, but in the end my simplest solution proved the fastest as well.

Problem 36

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

def palindrome? s
	s == s.reverse
end

sum = (1..1_000_000).inject(0) do |sum, n|
	sum += palindrome?(n.to_s) && palindrome?(n.to_s 2) ? n : 0
end

puts sum

Project Euler: Problem 33 in Ruby

It's fugly, but it works. The hardest part was understanding the question. If the longer description didn't say that there were exactly 4 fractions, I might have gone crazy.

For reals.

Problem 33

Discover all the fractions with an unorthodox cancelling method.

top, bottom = 1, 1

(10..98).each do |i|
	((i/10)..9).each do |jt|
		jt *= 10
		(1..9).each do |jo|
			j = jt + jo
			next if i >= j
			if i % 10 == j / 10 && i.to_f / j == (i / 10).to_f / (j % 10)
				top *= i
				bottom *= j
			end
		end
	end
end

puts bottom / bottom.gcd(top)