Write a program or script in your language of choice, that prints the lucky numbers between 1 and n. Try to make it as fast as possible for sieving lucky numbers between 1 and a million. (Perhaps it is sensible to measure the time without printing the results.)
A lucky number is defined as a number that survives the Sieve of Josephus. I'm going to quote the original post here so you get the idea.
The 1 is lucky by definition.
The successor of 1 is 2. So, every second number gets eliminated:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ...
The next number is 3. Now, every third number gets eliminated:
1, 3, 7, 9, 13, 15, 19, 21, ...
Number 3 is lucky! It's[sic] successor is 7. Now, every seventh number gets eliminated. And so on.
See the link above for more details. I advise you to give the challenge a try if you haven't yet before you read the rest of the post since I'm about to tell you about my solutions.
I came up with two solutions, but the best solution ended up being an interesting combination of them both. You'll see what I mean.
I like the quote "Make it work, make it right, make it fast", which, as far as I can tell, this is attributed to Kent Beck. Therefore, my first solution focused on coding things exactly like the problem was provided, the easy way, whether or not this was the fastest way.
def lucky_numbers(max) possibles = (1..max).to_a 1.step do |current_index| break if current_index >= possibles.size step_size = possibles[current_index] possibles = reject_every_nth_item(possibles, step_size) # If current_index is one of the items to get wiped out # the same current index will actually point at the next # number. Reuse this index. redo if wiped_out?(ind, step_size) end possibles end def reject_every_nth_item(items, n) items.reject.each_with_index do |_, ind| ((ind + 1) % n).zero? end end def wiped_out?(ind, step_size) ((ind + 1) % step_size).zero? end
I was all proud of myself as I ran it. And then I ran it with
max = 1_000_000. And I sat. And I sat…
This is taking too long! I could do better.
As I was experimenting and testing things out, I noticed that, at the higher values of
step_size, the naive algorithm was only rejecting a number every so often. So I thought to myself, "What if we could take the step size out of the equation and make each step take roughly the same amount of time, regardless of how many elements there were in it?" So, instead of checking each item to see if it was a reject, I calculated out the size of the chunk of non-rejected numbers and copied those into a
results array, ignoring the rejects instead of explicitly rejecting them.
def lucky_numbers(max) possibles = (1..max).to_a 1.step do |current_index| break if current_index >= possibles.size step_size = possibles[current_index] possibles = grab_chunks_between_rejects(possibles, step_size) redo if wiped_out?(current_index, step_size) end possibles end def grab_chunks_between_rejects(items, n) chunk_count = (items.size / n).floor # The new array will be the same size minus one reject item # for every chunk possible result = Array.new(items.size - chunk_count) chunk_target_index = 0 # Some tricky indexing... # 1) Select chunks of size n - 1 (so as to not include the reject endpoint) # 2) Copy it into the results array # 3) Continue, each chunk after the one in front of it. (0...items.size).step(n) do |chunk_start| chunk = items[chunk_start...(chunk_start + n - 1)] result[chunk_target_index...(chunk_target_index + chunk.size)] = chunk chunk_target_index += chunk.size end result end def wiped_out?(ind, step_size) ((ind + 1) % step_size).zero? end
This one worked great! Even on my little laptop, it ran relatively quickly (don't worry, the benchmarks are coming). But something felt weird. My programmer senses were tingling.
I had no proof at this point, but it seemed like all of the setup and creation of a second array each "round" should cause this algorithm to perform poorly at small
step_sizes. It was at this point that this ceased to be a code challenge and became a science experiment.
HYPOTHESIS: There is a performance trade-off between the low and high end for my two solutions. Thus, there should be a combination of the two that is faster than either one as a stand-alone.
If there was a trade-off somewhere between the two solutions, I would be able to run a few tests and find the best possible combination. I quickly updated my
def lucky_numbers(max, strategy_switch = 10) possibles = (1..max).to_a 1.step do |current_index| break if current_index >= possibles.size step_size = possibles[current_index] # Calculate next round of numbers possibles = if step_size <= strategy_switch reject_every_nth_item(possibles, step_size) else grab_chunks_between_rejects(possibles, step_size) end redo if wiped_out?(current_index, step_size) end possibles end
And now to collect some data! For sanity's sake, I wrapped each function in a module to differentiate them.
require "benchmark" require_relative "lucky_numbers3" max_nums = 1_000_000 switch_levels = [ 2, 10, 100, 500, 1000, 5000, 10_000, 50_000, 100_000, 500_000, ] Benchmark.bm do |bm| switch_levels.each do |level| bm.report(level.to_i) do Lucky3::lucky_numbers(max_nums, level) end end end # ▶ ruby optimize_lucky3.rb # user system total real # 2 17.550718 21.655086 39.205804 ( 39.535233) # 10 18.351711 23.234818 41.586529 ( 42.615220) # 100 18.443557 22.266405 40.709962 ( 41.178138) # 500 20.278373 23.325708 43.604081 ( 44.987643) # 1000 20.725864 22.510307 43.236171 ( 43.617160) # 5000 27.245503 21.725938 48.971441 ( 49.304915) # 10000 33.828295 22.476806 56.305101 ( 56.742865) # 50000 82.069851 37.399753 119.469604 (120.174670) # 100000 120.410101 23.425739 143.835840 (144.841090) # 500000 445.835956 31.520488 477.356444 (483.688845)
And no science experiment is complete without plots! Unfortunately, the data visualization story (outside of the web) is not as cushy in Ruby as it is in Python. Fortunately, I found a wrapper for Python's
matplotlib so I could stay in Ruby land for now. Note that I used a log-based scale on the X-axis to better display my results.
# plot_lucky3.rb require "matplotlib/pyplot" plt = Matplotlib::Pyplot xs = [2, 10, 100, 500, 1000, 5000, 10_000, 50_000, 100_000, 500_000] ys = [ 39.54, 42.62, 41.18, 44.99, 43.62, 49.30, 56.74, 120.17, 144.84, 483.69, ] ax = plt.gca() ax.scatter(xs, ys) ax.set_xscale('log') plt.show()
This is the part that is confusing to me. I've run the experiment multiple times and I always end up with a dip at around 1000. We'll talk about that in a moment. Here are the benchmarks of each solution directly compared.
require "benchmark" require_relative "lucky_numbers" require_relative "lucky_numbers2" require_relative "lucky_numbers3" iterations = 1_000_000 Benchmark.bm do |bm| bm.report("Version 1") do Lucky1::lucky_numbers(iterations) end bm.report("Version 2") do Lucky2::lucky_numbers(iterations) end bm.report("Version 3") do Lucky3::lucky_numbers(iterations, 1000) end end
And the results:
▶ ruby lucky_number_timer.rb user system total real Version 1 794.304851 28.946642 823.251493 (826.377494) Version 2 23.267372 33.858966 57.126338 ( 57.238392) Version 3 19.517162 20.816684 40.333846 ( 40.470201)
CONCLUSION: My hypothesis was supported. A combination of the two solutions performs better than either one solution by itself. However, the reason behind this is probably not the one that I initially thought. Some further research is required to find out why this worked.
My wife teaches middle school science and I just got done helping her judge their science fair, so I've got experimentation on the brain. Hopefully, this little experiment makes her proud!
I'm sure my solution wasn't the best possible solution, but it gets the job done. Ruby isn't necessarily the language of choice for high-throughput calculations (although it's not necessarily a bad choice either). If anybody can come up with a convincing reason for why my data came out the way it did, including the dip at 1000, I'll personally send you a shiny new trophy emoji. 🏆 Also, if you've got a solution that you like better than these, I want to see it. Share it! (or share a link to your solution commented on Heiko's original post).
Originally posted on