Divide and Conquer algorithm for counting inversions

Question

I've written a program that counts the number of inversions using a Divide and Conquer algorithm, written in C++11.

#include <iostream>
#include <tuple>
#include <vector>
using IntVec = std::vector<int>;
std::tuple<IntVec, long long> count_inv(IntVec xs);
std::tuple<IntVec, long long> count_split_inv(IntVec xs, IntVec ys);
std::tuple<IntVec, long long> count_inv(IntVec xs) {
 int n = xs.size();
 if (n < 2) return std::make_tuple(xs, 0);
 IntVec::iterator middle = xs.begin() + (n / 2);
 IntVec left(xs.begin(), middle);
 IntVec right(middle, xs.end());
 long long left_inv, right_inv, split_inv;
 IntVec as, bs, cs;
 std::tie(as, left_inv) = count_inv(left);
 std::tie(bs, right_inv) = count_inv(right);
 std::tie(cs, split_inv) = count_split_inv(as, bs);
 return std::make_tuple(cs, left_inv + right_inv + split_inv);
}
std::tuple<IntVec, long long> count_split_inv(IntVec xs, IntVec ys) {
 int n = xs.size() + ys.size();
 IntVec zs(n);
 long long split_inv = 0;
 int i = 0, j = 0, k = 0;
 for (; k < n && i < xs.size() && j < ys.size(); ++k) {
 if (xs[i] <= ys[j]) {
 zs[k] = xs[i];
 ++i;
 } else {
 zs[k] = ys[j];
 ++j;
 split_inv += xs.size() - i;
 }
 }
 for (; i < xs.size(); ++k, ++i)
 zs[k] = xs[i];
 for (; j < ys.size(); ++k, ++j)
 zs[k] = ys[j];
 return std::make_tuple(zs, split_inv);
}
int main() {
 std::vector<int> xs;
 int x;
 while (std::cin >> x) {
 xs.push_back(x);
 }
 IntVec ys;
 long long inv;
 std::tie(ys, inv) = count_inv(xs);
 std::cout << inv << std::endl;
 return 0;
}

I find that the code is very verbose in some areas, such as specifying the type of the tuple. Also, parts of it are confusing. How could I rewrite this code such that it is more concise and clearer?

• Deduplicator: Number of inversions of what? And according to which rules? Also, it's "divide and conquer", not the other way around...
• wei2912: @Deduplicator Whoops, no idea why I wrote it that way. It should print out the number of inversions of the array of numbers, assuming that the numbers are to be sorted in ascending order.
• Deduplicator: Do you mean number of swaps, instead of inversions? I'm whistling in the dark here, as I still have no idea what your program does (should do), or why...
• JS1: @Deduplicator I believe that an inversion is a pair of array entries that are in the wrong order. In other words every pair (i,j) where i < j && array[i] > array[j]. At least, this is what would make sense, because this program is using the "mergesort method" of counting the number of inversions in an array. See this link, for example.
• wei2912: What JS1 wrote is what I'm looking for. I do not need to find the inversions themselves, just the number of inversions.

Answer 1

More clear by using extra argument?

If you passed in a count by reference to each function, and each function incremented this count instead of returning a new count in a tuple, the code might look cleaner. For example, this:

IntVec as, bs, cs;
std::tie(as, left_inv) = count_inv(left);
std::tie(bs, right_inv) = count_inv(right);
std::tie(cs, split_inv) = count_split_inv(as, bs);
return std::make_tuple(cs, left_inv + right_inv + split_inv);

would become this:

IntVec as = count_inv(left, count);
IntVec bs = count_inv(right, count);
return count_split_inv(as, bs, count);

And in main(), this:

long long inv;
std::tie(ys, inv) = count_inv(xs);

would become this:

long long inv = 0;
count_inv(xs, inv);

• wei2912: I'm not certain if it's good practice to use pointers in this case.
• JS1: @wei2912 As opposed to creating tuples and returning them? Why are you opposed to using pointers?
• Not a real meerkat: It certainly looks cleaner, but in these calls to count_inv(xs, inv), it's not entirely clear that the inv/count variables will be changed as a side-effect. I also agree that returning a tuple is much worse, so I don't have an imediate solution for this.