nikhil - 1 year ago 68

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 Source

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
```

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.