4 ms·

Of course if you write Python, Ruby, Perl and PHP code in the style of Java, and benchmark it against idiomatic Java, then Java will come out on top. This doesn

Of course if you write Python, Ruby, Perl and PHP code in the style of Java, and benchmark it against idiomatic Java, then Java will come out on top. This doesn't prove anything, other than you have a very flawed benchmark. Language implementations are optimised for the most common operations in that language. Allowing only one implementation to take advantage of this renders the benchmark useless.

deleted 17y ago
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devinj 17y ago
Well, Java would still come out on top, in all likelihood. I don't think even the idiomatic solutions in Python/Ruby/etc. would be faster unless they have some really nice JIT compilers I'm not aware of, or they changed the algorithm. They'd sure as hell be a lot smaller, though. The real Python code I'd use to implement the defined algorithm would be a fifth of the size and running time (7 lines, sans whitespace, runs in 16.4 µs versus 92.7 µs under his timing scheme (it should really use min/timeit, not average/custom-timing-solution)). import collections def kill(size, nth): chain = collections.deque(reversed(xrange(size))) while len(chain) > 1: chain.rotate(nth) chain.popleft() return chain.pop()
- keefe 17y ago
  Thank you for this addition - this is a great article, exactly why I visit this site. I'm fairly surprised about the groovy execution time, I wonder if the ruby etc. enthusiasts could generate another set of benchmarks that support their particular bias.
- algorias 17y ago
  How about: def kill(size, nth): lst = range(size) offset = nth-1 len_ = len(lst) while len_ > 1: del lst[offset::nth] offset = (offset - len_)% nth len_ = len(lst) return lst[0] It will reduce the list by 1/3rd each pass, so it's significantly faster at larger sizes.
  - devinj 17y ago
    Looks great, except for the corner case of nth = 1. I was too lazy to figure out the slicing solution myself. :)

deleted 17y ago

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devinj 17y ago

Well, Java would still come out on top, in all likelihood. I don't think even the idiomatic solutions in Python/Ruby/etc. would be faster unless they have some really nice JIT compilers I'm not aware of, or they changed the algorithm. They'd sure as hell be a lot smaller, though. The real Python code I'd use to implement the defined algorithm would be a fifth of the size and running time (7 lines, sans whitespace, runs in 16.4 µs versus 92.7 µs under his timing scheme (it should really use min/timeit, not average/custom-timing-solution)). import collections def kill(size, nth): chain = collections.deque(reversed(xrange(size))) while len(chain) > 1: chain.rotate(nth) chain.popleft() return chain.pop()

keefe 17y ago
Thank you for this addition - this is a great article, exactly why I visit this site. I'm fairly surprised about the groovy execution time, I wonder if the ruby etc. enthusiasts could generate another set of benchmarks that support their particular bias.
algorias 17y ago
How about: def kill(size, nth): lst = range(size) offset = nth-1 len_ = len(lst) while len_ > 1: del lst[offset::nth] offset = (offset - len_)% nth len_ = len(lst) return lst[0] It will reduce the list by 1/3rd each pass, so it's significantly faster at larger sizes.
- devinj 17y ago
  Looks great, except for the corner case of nth = 1. I was too lazy to figure out the slicing solution myself. :)