C++ 中的排列和组合库函数
C++ 中使用最广泛的现有库是什么,可以给出 n 个元素中 k 个元素的所有组合和排列?
我不是在问算法,而是在问现有的库或方法。
谢谢。
What's the most widely used existing library in C++ to give all the combination and permutation of k elements out of n elements?
I am not asking the algorithm but the existing library or methods.
Thanks.
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我决定在这里测试 dman 和 Charles Bailey 的解决方案。我将它们分别称为解决方案 A 和 B。我的测试是访问
vector
size = 100 的每个组合,一次 5 个。以下是测试代码:测试代码
所有代码都是在 2.8 GHz Intel Core i5 上使用 clang++ -O3 编译的。
解决方案 A
解决方案 A 会导致无限循环。即使我将
n
设得很小,这个程序也永远不会完成。随后因此而被否决。解决方案 B
这是一个编辑。解决方案 B 在撰写此答案的过程中发生了变化。起初它给出了错误的答案,由于非常及时的更新,它现在给出了正确的答案。它打印出:
解决方案 C
接下来我尝试了 N2639 看起来与解决方案 A 非常相似,但工作正常。我将这个解决方案称为C,它打印出:
解决方案C比解决方案B快703倍。
解决方案D
最后找到了一个解决方案D此处。该解决方案具有不同的签名/风格,称为
for_each_combination
,其使用方式与std::for_each
非常相似。上面的驱动程序代码在计时器调用之间发生变化,如下所示:解决方案 D 打印出:
解决方案 D 比解决方案 C 快 12.9 倍,比解决方案 B 快 9000 多倍。
我认为这是一个相对较小的问题:只有 7500 万次访问。随着访问量增加到数十亿,这些算法之间的性能差异不断扩大。解决方案 B 已经很笨拙了。解决方案 C 最终变得难以处理。解决方案 D 是访问我所知道的所有组合的性能最高的算法。
显示解决方案 D 的链接还包含其他几种算法,用于枚举和访问具有各种属性的排列(循环、可逆等)。这些算法的设计都将性能作为目标之一。请注意,这些算法都不要求初始序列按排序顺序排列。这些元素甚至不需要是
LessThanComparable
。I decided to test the solutions by dman and Charles Bailey here. I'll call them solutions A and B respectively. My test is visiting each combination of of a
vector<int>
size = 100, 5 at a time. Here's the test code:Test Code
All code was compiled using clang++ -O3 on a 2.8 GHz Intel Core i5.
Solution A
Solution A results in an infinite loop. Even when I make
n
very small, this program never completes. Subsequently downvoted for this reason.Solution B
This is an edit. Solution B changed in the course of writing this answer. At first it gave incorrect answers and due to very prompt updating it now gives correct answers. It prints out:
Solution C
Next I tried the solution from N2639 which looks very similar to solution A, but works correctly. I'll call this solution C and it prints out:
Solution C is 703 times faster than solution B.
Solution D
Finally there is a solution D found here. This solution has a different signature / style and is called
for_each_combination
, and is used much likestd::for_each
. The driver code above changes between the timer calls like so:Solution D prints out:
Solution D is 12.9 times faster than solution C, and over 9000 times faster than solution B.
I consider this a relatively small problem: only 75 million visits. As the number of visits increases into the billions, the discrepancy in the performance between these algorithms continues to grow. Solution B is already unwieldy. Solution C eventually becomes unwieldy. Solution D is the highest performing algorithm to visit all combinations I'm aware of.
The link showing solution D also contains several other algorithms for enumerating and visiting permutations with various properties (circular, reversible, etc.). Each of these algorithms was designed with performance as one of the goals. And note that none of these algorithms requires the initial sequence to be in sorted order. The elements need not even be
LessThanComparable
.组合:来自 Mark Nelson 的文章,我们有相同的主题
next_combination
排列:从 STL 中我们有std::next_permutation
Combinations: from Mark Nelson's article on the same topic we have
next_combination
Permutations: From STL we havestd::next_permutation
这个答案提供了一个最小的实施工作的解决方案。如果您想检索大输入范围的组合,它可能不会具有可接受的性能。
标准库具有
std::next_permutation
,您可以轻松地从中构建next_k_permutation
和next_combination
。如果您没有 tr1::bind 或 boost::bind ,您需要构建一个函数对象,将参数交换为给定的比较。当然,如果您只对
next_combination
的std::less
变体感兴趣,那么您可以直接使用std::greater
:是
next_combination
的相对安全版本。如果您可以保证范围[mid, last)
与调用next_combination
后的顺序一致,那么您可以使用更简单的方法:这也适用于 bi - 定向迭代器以及随机访问迭代器。
为了输出组合而不是 k 排列,我们必须确保每个组合仅输出一次,因此只有当它是按顺序排列的 k 排列时,我们才会返回组合。
替代方案是使用反向迭代器而不是参数交换
bind
调用,或者如果std::less
显式使用std::greater
正在使用的比较。This answer provides a minimal implementation effort solution. It may not have acceptable performance if you want to retrieve combinations for large input ranges.
The standard library has
std::next_permutation
and you can trivially build anext_k_permutation
from it and anext_combination
from that.If you don't have
tr1::bind
orboost::bind
you would need to build a function object that swaps the arguments to a given comparison. Of course, if you're only interested in astd::less
variant ofnext_combination
then you can usestd::greater
directly:This is a relatively safe version of
next_combination
. If you can guarantee that the range[mid, last)
is in order as they would be after a call tonext_combination
then you can use the simpler:This also works with bi-directional iterators as well as random access iterators.
To output combinations instead of k-permutations, we have to ensure that we output each combination only once, so we'll return a combination it only if it is a k-permutation in order.
Alternatives would be to use a reverse iterator instead of the parameter swapping
bind
call or to usestd::greater
explicitly ifstd::less
is the comparison being used.@查尔斯·贝利上面:
我可能是错的,但我认为上面的前两个算法不会删除第一个和中间之间的重复项?也许我不确定如何使用它。
4选2示例:
12 34
12 43(排序后)
13 24(next_permutation 之后)
13 42(排序后)
14 23(next_permutation 之后)
14 32(排序后)
21 34(在next_permutation之后)
所以我在返回之前添加了一个检查以查看斜体值是否按顺序排列,但绝对不会想到你写的部分(非常优雅!谢谢! )。
没有完全测试,只是粗略测试。
@ Charles Bailey above:
I could be wrong, but I think the first two algorithms above does not remove duplicates between first and mid? Maybe I am not sure how to use it.
4 choose 2 example:
12 34
12 43 (after sort)
13 24 (after next_permutation)
13 42 (after sort)
14 23 (after next_permutation)
14 32 (after sort)
21 34 (after next_permutation)
So I added a check to see if the value in italics is in order before returning, but definitely wouldn't have thought of the part you wrote though (very elegant! thanks!).
Not fully tested, just cursory tests..
也许它已经在前面的答案中说明过,但在这里我找不到关于参数类型的完整通用方法,而且除了 Boost 之外,我也没有在现有的库例程中找到它。这是我在针对各种参数变化广泛的场景构建测试用例期间所需的通用方法。也许它对你也有帮助,至少对于类似的场景。 (可用于有疑问的微小变化的排列和组合)
可能的用法:
生成:
当然,这可以进一步优化/专门化。例如,如果您想避免有效重复,您可以简单地添加哈希方案和/或避免函子。此外,由于参数作为引用保存,因此可以考虑通过删除复制/赋值构造函数和运算符来保护生成器免受可能的错误使用。
时间复杂度在理论排列复杂度范围内。
Maybe it's already stated within the previous answers, but here I cannot find a full generic way for this with respect to the parameter types and I also didn't find it within existing library routines besides Boost. This is a generic way I needed during test case construction for scenarios with a wide spread of various parameter variations. Maybe it's helpful to you too, at least for similar scenarios. (Usable for permutation and combination with minor changes in doubt)
Possible usage:
Generates:
For sure, this can be further optimized/specialized. For instance you can simply add a hashing scheme and/or an avoid functor if you want to avoid effective repetitions. Also, since the parameters are held as references, one might consider to protect the generator from possible error-prone usage via deleting copy/assignement constructors and operators.
Time complexity is within the theoretical permutation complexity range.