Difference between revisions of "cpp/algorithm/ranges/is permutation"

Latest revision as of 12:51, 23 April 2024

Defined in header `<algorithm>`
Call signature
template< std::forward_iterator I1, std::sentinel_for<I1> S1, std::forward_iterator I2, std::sentinel_for<I2> S2, class Proj1 = std::identity, class Proj2 = std::identity, std::indirect_equivalence_relation<std::projected<I1, Proj1>, std::projected<I2, Proj2>> Pred = ranges::equal_to > constexpr bool is_permutation( I1 first1, S1 last1, I2 first2, S2 last2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );	(1)	(since C++20)
template< ranges::forward_range R1, ranges::forward_range R2, class Proj1 = std::identity, class Proj2 = std::identity, std::indirect_equivalence_relation< std::projected<ranges::iterator_t<R1>, Proj1>, std::projected<ranges::iterator_t<R2>, Proj2>> Pred = ranges::equal_to > constexpr bool is_permutation( R1&& r1, R2&& r2, Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {} );	(2)	(since C++20)

1) Returns true if there exists a permutation of the elements in range [first1, last1) that makes the range equal to [first2, last2) (after application of corresponding projections Proj1, Proj2, and using the binary predicate Pred as a comparator). Otherwise returns false.

2) Same as (1), but uses r1 as the first source range and r2 as the second source range, as if using ranges::begin(r1) as first1, ranges::end(r1) as last1, ranges::begin(r2) as first2, and ranges::end(r2) as last2.

The function-like entities described on this page are niebloids, that is:

Explicit template argument lists cannot be specified when calling any of them.
None of them are visible to argument-dependent lookup.
When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

In practice, they may be implemented as function objects, or with special compiler extensions.

first1, last1	-	the first range of the elements
first2, last2	-	the second range of the elements
r1	-	the first range of the elements
r2	-	the second range of the elements
pred	-	predicate to apply to the projected elements
proj1	-	projection to apply to the elements in the first range
proj2	-	projection to apply to the elements in the second range

[edit] Return value

true if the range [first1, last1) is a permutation of the range [first2, last2).

[edit] Complexity

At most $\scriptsize \mathcal{O}(N^2)$ $O(N 2)$ applications of the predicate and each projection, or exactly $\scriptsize N$ $N$ if the sequences are already equal, where $\scriptsize N$ $N$ is ranges::distance(first1, last1). However if ranges::distance(first1, last1) != ranges::distance(first2, last2), no applications of the predicate and projections are made.

[edit] Notes

The permutation relation is an equivalence relation.

The ranges::is_permutation can be used in testing, e.g. to check the correctness of rearranging algorithms such as sorting, shuffling, partitioning. If p is an original sequence and q is a "mutated" sequence, then ranges::is_permutation(p, q) == true means that q consist of "the same" elements (maybe permuted) as p.

[edit] Possible implementation

struct is_permutation_fn
{
    template<std::forward_iterator I1, std::sentinel_for<I1> S1,
             std::forward_iterator I2, std::sentinel_for<I2> S2,
             class Proj1 = std::identity, class Proj2 = std::identity,
             std::indirect_equivalence_relation<std::projected<I1, Proj1>,
                                                std::projected<I2, Proj2>>
                                                    Pred = ranges::equal_to>
    constexpr bool operator()(I1 first1, S1 last1, I2 first2, S2 last2,
                              Pred pred = {}, Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        // skip common prefix
        auto ret = std::ranges::mismatch(first1, last1, first2, last2,
                                         std::ref(pred), std::ref(proj1), std::ref(proj2));
        first1 = ret.in1, first2 = ret.in2;
 
        // iterate over the rest, counting how many times each element
        // from [first1, last1) appears in [first2, last2)
        for (auto i {first1}; i != last1; ++i)
        {
            const auto i_proj {std::invoke(proj1, *i)};
            auto i_cmp = [&]<typename T>(T&& t)
            { 
                return std::invoke(pred, i_proj, std::forward<T>(t));
            };
 
            if (i != ranges::find_if(first1, i, i_cmp, proj1))
                continue; // this *i has been checked
 
            if (const auto m {ranges::count_if(first2, last2, i_cmp, proj2)};
                m == 0 or m != ranges::count_if(i, last1, i_cmp, proj1))
                return false;
        }
        return true;
    }
 
    template<ranges::forward_range R1, ranges::forward_range R2,
             class Proj1 = std::identity, class Proj2 = std::identity,
             std::indirect_equivalence_relation<
                 std::projected<ranges::iterator_t<R1>, Proj1>,
                 std::projected<ranges::iterator_t<R2>, Proj2>>
                     Pred = ranges::equal_to>
    constexpr bool operator()(R1&& r1, R2&& r2, Pred pred = {},
                              Proj1 proj1 = {}, Proj2 proj2 = {}) const
    {
        return (*this)(ranges::begin(r1), ranges::end(r1),
                       ranges::begin(r2), ranges::end(r2),
                       std::move(pred), std::move(proj1), std::move(proj2));
    }
};
 
inline constexpr is_permutation_fn is_permutation {};

[edit] Example

Run this code

#include <algorithm>
#include <array>
#include <cmath>
#include <iostream>
#include <ranges>
 
auto& operator<<(auto& os, std::ranges::forward_range auto const& v)
{
    os << "{ ";
    for (const auto& e : v)
        os << e << ' ';
    return os << "}";
}
 
int main()
{
    static constexpr auto r1 = {1, 2, 3, 4, 5};
    static constexpr auto r2 = {3, 5, 4, 1, 2};
    static constexpr auto r3 = {3, 5, 4, 1, 1};
 
    static_assert(
        std::ranges::is_permutation(r1, r1) &&
        std::ranges::is_permutation(r1, r2) &&
        std::ranges::is_permutation(r2, r1) &&
        std::ranges::is_permutation(r1.begin(), r1.end(), r2.begin(), r2.end()));
 
    std::cout
        << std::boolalpha
        << "is_permutation(" << r1 << ", " << r2 << "): "
        << std::ranges::is_permutation(r1, r2) << '\n'
        << "is_permutation(" << r1 << ", " << r3 << "): "
        << std::ranges::is_permutation(r1, r3) << '\n'
 
        << "is_permutation with custom predicate and projections: "
        << std::ranges::is_permutation(
            std::array {-14, -11, -13, -15, -12},  // 1st range
            std::array {'F', 'E', 'C', 'B', 'D'},  // 2nd range
            [](int x, int y) { return abs(x) == abs(y); }, // predicate
            [](int x) { return x + 10; },          // projection for 1st range
            [](char y) { return int(y - 'A'); })   // projection for 2nd range
        << '\n';
}

Output:

is_permutation({ 1 2 3 4 5 }, { 3 5 4 1 2 }): true
is_permutation({ 1 2 3 4 5 }, { 3 5 4 1 1 }): false
is_permutation with custom predicate and projections: true

[edit] See also

ranges::next_permutation (C++20)	generates the next greater lexicographic permutation of a range of elements (niebloid)[edit]
ranges::prev_permutation (C++20)	generates the next smaller lexicographic permutation of a range of elements (niebloid)[edit]
is_permutation (C++11)	determines if a sequence is a permutation of another sequence (function template) [edit]
next_permutation	generates the next greater lexicographic permutation of a range of elements (function template) [edit]
prev_permutation	generates the next smaller lexicographic permutation of a range of elements (function template) [edit]
equivalence_relation (C++20)	specifies that a `relation` imposes an equivalence relation (concept) [edit]

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Difference between revisions of "cpp/algorithm/ranges/is permutation"

Latest revision as of 12:51, 23 April 2024

Contents

[edit] Parameters