Iterate over an infinite sequence in Ruby - ruby

Question

I am trying to solve Project Euler problem #12:
The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number
would be 1 + 2 + 3 + 4 + 5 + 6 + 7 =
28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five
divisors.
What is the value of the first triangle number to have over five
hundred divisors?
Here's the solution that I came up with using Ruby:
triangle_number = 1
(2..9_999_999_999_999_999).each do |i|
triangle_number += i
num_divisors = 2 # 1 and the number divide the number always so we don't iterate over the entire sequence
(2..( i/2 + 1 )).each do |j|
num_divisors += 1 if i % j == 0
end
if num_divisors == 500 then
puts i
break
end
end
I shouldn't be using an arbitrary huge number like 9_999_999_999_999_999. It would be better if we had a Math.INFINITY sequence like some functional languages. How can I generate a lazy infinite sequence in Ruby?

Answer 1

Several answers are close but I don't actually see anyone using infinite ranges. Ruby supports them just fine.
Inf = Float::INFINITY # Ruby 1.9
Inf = 1.0/0 # Ruby before 1.9
(1..Inf).include?(2305843009213693951)
# => true
(1..Inf).step(7).take(3).inject(&:+)
# => 24.0
In your case
(2..Inf).find {|i| ((2..( i/2 + 1 )).select{|j| i % j == 0}.count+2)==42 }
=> 2880
Your brute force method is crude and can, potentially, take a very long time to finish.

Answer 2

In Ruby >= 1.9, you can create an Enumerator object that yields whatever sequence you like. Here's one that yields an infinite sequence of integers:
#!/usr/bin/ruby1.9
sequence = Enumerator.new do |yielder|
number = 0
loop do
number += 1
yielder.yield number
end
end
5.times do
puts sequence.next
end
# => 1
# => 2
# => 3
# => 4
# => 5
Or:
sequence.each do |i|
puts i
break if i >= 5
end
Or:
sequence.take(5).each { |i| puts i }
Programming Ruby 1.9 (aka "The Pickaxe Book"), 3rd. ed., p. 83, has an example of an Enumerator for triangular numbers. It should be easy to modify the Enumerator above to generate triangular numbers. I'd do it here, but that would reproduce the example verbatim, probably more than "fair use" allows.

Answer 3

Infinity is defined on Float (Ruby 1.9)
a = Float::INFINITY
puts a #=> Infinity
b = -a
puts a*b #=> -Infinity, just toying
1.upto(a) {|x| break if x >10; puts x}

Answer 4

Currrent versions of Ruby support generators heavily:
sequence = 1.step

Answer 5

In Ruby 2.6 this becomes much easier:
(1..).each {|n| ... }
Source: https://bugs.ruby-lang.org/issues/12912

Answer 6

This would be best as a simple loop.
triangle_number = 1
i = 1
while num_divisors < 500
i += 1
triangle_number += i
# ...
end
puts i

Answer 7

As Amadan mentioned you can use closures:
triangle = lambda { t = 0; n = 1; lambda{ t += n; n += 1; t } }[]
10.times { puts triangle[] }
Don't really think it is much slower than a loop. You can save state in class object too, but you will need more typing:
class Tri
def initialize
#t = 0
#n = 1
end
def next
#t += n
#n += 1
#t
end
end
t = Tri.new
10.times{ puts t.next }
Added:
For those who like longjmps:
require "generator"
tri =
Generator.new do |g|
t, n = 0, 1
loop do
t += n
n += 1
g.yield t
end
end
puts (0..19).map{ tri.next }.inspect

Answer 8

Building on Wayne's excellent answer and in the Ruby spirit of doing things with the least number of characters here is a slightly updated version:
sequence = Enumerator.new { |yielder| 1.step { |num| yielder.yield num } }
Obviously, doesn't solve the original Euler problem but is good for generating an infinite sequence of integers. Definitely works for Ruby > 2.0. Enjoy!

Answer 9

On Christmas Day 2018, Ruby introduced the endless range, providing a simple new approach to this problem.
This is implemented by ommitting the final character from the range, for example:
(1..)
(1...)
(10..)
(Time.now..)
Or to update using Jonas Elfström's solution:
(2..).find { |i| ((2..( i / 2 + 1 )).select { |j| i % j == 0 }.count + 2) == 42 }
Hope this proves useful to someone!

Answer 10

I believe that fibers (added in Ruby 1.9 I believe) may be close to what you want. See here for some information or just search for Ruby Fibers