I recently discovered a website called Project Euler and took a stab at
their questions. The first 2 were fine but then when I got to the 3rd
question, all hell broke loose and I started really questioning my
progress with Ruby.

The problem asks to find the largest prime factor of a number. I’ve
peeked at other peoples’ solutions online and noticed all of them used
recursion. However, I want to solve it without recursion. Without
further ado:

Assuming that num is an odd number, find the largest prime factor. So:
prime(99) => 11 prime(13195) => 29

Here is my code. The output keeps coming out as [3] if num = 99 and I
can’t figure out why.

def prime(num)
arr = []
p = []
not_p = []
# first I find all the numbers that num is divisible by
for i in (2..num/2)
if num % i == 0
arr << i
end
end # this should output [3, 9, 11, 33]
arr.each do |x| # I loop through each element in the above

array
for i in (2…(x/2)) # I divide each element - x - by 2 b/c
it cannot be divisble by anything greater than its half
if x % i == 0 # if x is divisble by i
not_p << i # I push the i into array#not_p
end # keep looping until i reaches x/2
end
if not_p.length == 0 # if there are no values in
array#not_p, then I know x is a prime factor
p << x # so I push x into array#p
end
end
return p[-1] # returns the last element of the array, which is
the largest
end

I recently discovered a website called Project Euler and took a stab at
their questions. The first 2 were fine but then when I got to the 3rd
question, all hell broke loose and I started really questioning my
progress with Ruby.

The problem asks to find the largest prime factor of a number. I’ve
peeked at other peoples’ solutions online and noticed all of them used
recursion. However, I want to solve it without recursion. Without
further ado:

Assuming that num is an odd number, find the largest prime factor. So:
prime(99) => 11 prime(13195) => 29

Here is my code. The output keeps coming out as [3] if num = 99 and I
can’t figure out why.

def prime(num)
arr = []
p = []
not_p = []
# first I find all the numbers that num is divisible by
for i in (2..num/2)
if num % i == 0
arr << i
end
end # this should output [3, 9, 11, 33]
arr.each do |x| # I loop through each element in the above

array
for i in (2…(x/2)) # I divide each element - x - by 2 b/c
it cannot be divisble by anything greater than its half
if x % i == 0 # if x is divisble by i
not_p << i # I push the i into array#not_p
end # keep looping until i reaches x/2
end
if not_p.length == 0 # if there are no values in
array#not_p, then I know x is a prime factor
p << x # so I push x into array#p
end
end
return p[-1] # returns the last element of the array, which is
the largest
en

not_p initialise should be in the ‘for i in (2…(x/2))’ loop

Here is a same algo, code more rubysh :

def bprime(num)
divs=(2…num/2).select {|i| num%i==0}
primes= divs.reject {|x| (2…(x/2)).any? {|i| x%i==0}}
return primes.last || num
end
p bprime 4
p bprime 5
p bprime 9
p bprime 12
p bprime 131711737
p bprime 99