How to make sure that std::random_shuffle always produces a different result?

Question

Is there some function, similar to srand(), that I need to call to make sure that std::random_shuffle() always produces different results? i.e. if I call it several times with the same data, I want the order to be different every time. How can I make sure of that?

@Blindy: It's random, but the results are not uniformly distributed. — John Bartholomew, Aug 03 '11 at 19:18
srand is quite good. If it is not good enough for your need, search for other random generators (maybe boost) — BЈовић, Aug 03 '11 at 19:18
Make sure you **NEVER** call srand() more than once in a program. — Martin York, Aug 03 '11 at 19:23
`srand()` doesn't make sure anything produces different results. @John: it's less and less random each time, because it's more and more predictable each time. — R. Martinho Fernandes, Aug 03 '11 at 19:47
@LokiAstari there are plenty of reasons to. If you care about running a MC algorithm multiple times with different parameters but the same random number stream, for example. — geometrian, Nov 03 '14 at 05:32

score 14 · Accepted Answer · answered Aug 03 '11 at 19:22

14

std::random_shuffle has two forms. One that takes 2 arguments (begin/end iterators), and one that takes 3 (begin/end iterator and a random generator).

The first form uses std::rand(), so you would use std::srand() to seed it's random number generator. You can also use the 3-argument version and provide the RNG yourself.

answered Aug 03 '11 at 19:22

Dave S

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As I understand it, how the first form implements it is its business. Some platforms may not use `std::rand`. – Tom Kerr Aug 03 '11 at 19:26
You're right. I was looking at the libstdc++ version. The original author should consult the documentation for your STL. Or if they don't want to rely on that, they should use the 3rd form and use something like `boost::random` or the C++0x `std::random`. – Dave S Aug 03 '11 at 19:30

score 6 · Answer 2 · answered Aug 03 '11 at 19:18

std::random_shuffle has a template overload for specifying the RNG.

template <class RandomAccessIterator, class RandomNumberGenerator>
  void random_shuffle ( RandomAccessIterator first, RandomAccessIterator last,
                        RandomNumberGenerator& rand );

reference

score 5 · Answer 3 · answered Feb 23 '16 at 15:20

random_shuffle is deprecated since C++14 (removed in C++17) and replaced with shuffle (exists since C++11) http://en.cppreference.com/w/cpp/algorithm/random_shuffle

possible usage:

shuffle(items.begin(), items.end(), std::default_random_engine(std::random_device()()));

score 3 · Answer 4 · edited Dec 25 '12 at 19:09

3

I think you can give a random generator functor to std::random_shuffle, so you can be able to fully control the random number generation. Looking here, this functor takes the place of the RandomNumberGenerator template argument.

edited Dec 25 '12 at 19:09

onkar

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answered Aug 03 '11 at 19:18

neodelphi

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score 3 · Answer 5 · edited Jul 21 '20 at 17:35

Generally call srand(time(NULL)) before calling std::random_shuffle() would give you what you need, it give you different result each time you call std::random_shuffle(). It's because std::random_shuffle() internally calls rand() in many popular implementations (e.g. VS-2008 and GCC).

Of course you can supple a RNG yourself if you want to call the other overloaded std::random_shuffle with a extra parameter.

score 2 · Answer 6 · answered Aug 03 '11 at 20:59

2

As a last resort, you can:

Call std::random_shuffle
Compute a hash of the sequence, store it in a std::set
Discard if the hash is already present

I fail to see how using a custom generator could guarantee that the sequence is unique.

answered Aug 03 '11 at 20:59

Alexandre C.

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How to make sure that std::random_shuffle always produces a different result?

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