Hi –
On Mon, 27 Jul 2009, Robert K. wrote:
I am unhappy with using currentNum += 2 twice. I very much want to
When you do:
statement until condition
statement isn’t executed at all unless condition is true. Since
primes[3] is false, the incrementation never takes place.
I garbled that because I was trying to answer the question and write
code that flipped the logic at the same time (see below for
clarification).
num += 2
end until primes[num]
? At least that would be a direct translation of the original
currentNum += 2
while primes[currentNum] == false do
currentNum += 2
end
This was a somewhat incomplete forward reference to my “flipped logic”
version (see below).
There’s another issue here, though. Since the default value for
non-existent array values is nil, the test for truth will fail even if
num has gone over the maximum. So you might want to flip the logic:
There’s yet another issue: the limit highestPrime is not honored during the
incrementation of current_num which likely leads to the hanging which I
assume is in fact an endless loop.
The endless loop in the original is because currentNum never gets
incremented at all. Here’s a skeletal version:
x = 3
array = [true, true, true, true]
x += 2 until array[x] == true
x will still be 3 at the end. So the outer loop executes repeatedly,
not because the limit isn’t honored but because there’s no
incrementation.
Here’s the version I cooked up the other day, which shows my flipped
logic in context:
max = 10
non_primes = Array.new(max)
num = 3
while num < max do
(num * num).step(max, num * 2) { |i| non_primes[i] = true }
begin
num += 2
end while non_primes[num]
end
Instead of setting an array to all true and then setting certain ones
to false, I prefer piggybacking on the fact that array values are nil
by default, and set them to true selectively.
David
(Though I garbled my example a bit.)