# Does a subcubic algorithm exist for the following problem?

Given a symmetric real $n \times n$ matrix $A=(a_{ij})$, is there an algorithm which computes the sum $$\sum_{i,j,k}\max(a_{ij},a_{ik},a_{jk})$$ over all $1\leq i<j<k\leq n$ with time-complexity better than $O(n^3)$?

• Note that this is at least as hard as counting the number of triangles in a given graph. If your input matrix encodes a graph such that "0" indicates an edge and "1" indicates a missing edge, then $\max(a_{ij}, a_{ik}, a_{jk}) = 0$ if and only if there is a triangle formed by nodes $i$, $j$, and $k$, and otherwise it is $1$. – Jukka Suomela Jun 4 '18 at 11:24
• I think the only known significantly subcubic algorithms for triangle counting are based on fast matrix multiplication? It might be tricky to apply those techniques here in this problem. Also, if you are looking for something practical, anything based on fast matrix multiplication isn't going to be helpful. – Jukka Suomela Jun 4 '18 at 11:25

There exists a quite practical approach that works in $$O(n^3 / w)$$ time, where $$w$$ is the number of bits in the processor word. The main idea is that you iterate over the elements of the matrix one by one in the increasing order (break ties arbitrarily) and "switch them on". Consider the moment when the largest element of some triple $$a_{ij}, a_{ik}, a_{jk}$$ is switched on. For simplicity, let's assume that the said element is $$a_{ij}$$. It is natural to add the value of the triple to the answer now, when the last element is switched on. So we have to count the number of possible $$k$$'s such that $$a_{ik}$$ and $$a_{jk}$$ are already switched on (that would be the number of triples, here $$a_{ij}$$ is the largest element, so they were completely switched on just now). Here we can speed up the naive $$O(n)$$ implementation by using bit optimisation.

For details you can refer to the following implementation in C++11 that should work for $$n \leqslant 5000$$, $$|a_{ij}| \leqslant 10^9$$ (it is not very optimized; however, it still beats naive summation for $$n = 5000$$ by large margin at least on my machine).

// code is not very elegant,
// but should be understandable
// here the matrix a has dimensions n x n
// a has to be symmetric!
int64_t solve (int n, const vector<vector<int32_t>> &a)
{
std::vector<boost::dynamic_bitset<int64_t>> mat
(n, boost::dynamic_bitset<int64_t>(n));

vector<pair<int, int>> order;
for (int j = 1; j < n; j++)
for (int i = 0; i < j; i++)
order.emplace_back(i, j);
sort(order.begin(), order.end(),
[&] (const pair<int, int> &l, const pair<int, int> &r)
{return a[l.first][l.second] < a[r.first][r.second];});

int64_t ans = 0;
for (const auto &position : order)
{
int i, j;
tie (i, j) = position;
mat[i][j] = mat[j][i] = 1;
// here it is important that conditions
// mat[i][i] = 0 and mat[j][j] = 0 always hold
ans += (mat[i] & mat[j]).count() * int64_t(a[i][j]);
}

return ans;
}


If you consider using bit optimisations cheating, you can use four russians method to the same result here, yielding a $$O(n^3 / \log n)$$ algorithm, which should be less practical (because $$w$$ is pretty large on most modern hardware) but is theoretically better. Indeed, let's choose $$b \approx \log_2 n$$ and keep each row of the matrix as an array of $$\lceil \frac{n}{b} \rceil$$ integers from $$0$$ to $$2^b - 1$$, where the $$i$$-th number in the array corresponds to the bits of the row ranging from $$ib$$ inclusive to $$\min(n, (i + 1)b)$$ exclusive in $$0$$-indexation. We can precalculate the scalar products of every two such blocks in $$O(2^{2b} b)$$ time. Updating a position in the matrix is fast because we are changing only one integer. To find the scalar product of rows $$i$$ and $$j$$ just iterate over arrays corresponding to that rows, look up scalar products of the corresponding blocks in the table and sum up the obtained products.

The above paragraph assumes that operations with integers $$\leqslant n$$ take $$O(1)$$ time. It is quite common assumption, because it usually does not actually change the comparative speed of the algorithms (for example, if we don't use that assumption, brute force method actually works in $$O(n^3 \log n)$$ time (here we measure time in bit operations) if $$a_{ij}$$ take integer values with absolute values at least up to $$n^{\varepsilon}$$ for some constant $$\varepsilon > 0$$ (and otherwise we can solve the problem with $$O(n^{\varepsilon})$$ matrix multiplications anyway); however the four russians method suggested above uses $$O(n^3 / \log n)$$ operations with numbers of size $$O(\log n)$$ in that case; so it makes $$O(n^3)$$ bit operations, which is still better than brute force despite the change of the model).

The question about existence of $$O(n^{3 - \varepsilon})$$ approach is still interesting, though.

Techniques (bit optimisations and four russians method) presented in this answer are by no means original and are presented here for completeness of the exposition. However, finding a way to apply them was not trivial.

• Firstly, your suggestion indeed appears to be helpful in practical terms, I might just try it out in my use case. Thanks! Secondly, your algorithms computational complexity is still $O(n^3)$ for any fixed width numerical type. Could you elaborate about the $O(n^3 / \log n)$ approach? I do not get how we could find the scalar product of mat[i] and mat[j] faster than $O(n)$ (which would be required if we access all their elements). – user89217 Nov 15 '18 at 20:41
• Also, your code does not define mat which seems to be important. I understand how it could be defined, but I wonder if (mat[i] & mat[j]).count() would work as desired with any STL container. – user89217 Nov 15 '18 at 20:44
• Regarding mat - I guess we must use std::vector<boost::dynamic_bitset<int64_t>>. – user89217 Nov 15 '18 at 21:27
• Regarding mat: yes, I had standard bitset in mind actually, but boost::dynamic_bitset is even better in this case, because its size does not have to be compile-time constant. Will edit the answer to add this detail and to clarify the four russians approach. – Kaban-5 Nov 15 '18 at 22:04
• Great, this looks solid to me. A minor point: since the transdichotomous model assumes we can perform operations on machine words in $O(1)$, there is no need to precalculate any scalar products. In fact, the model assumes that $w\ge\log_2 n$ , so $O(n^3/w)$ is at least as good as $O(n^3/\log n)$. And, as you say, precalculating scalar products makes no practical sense (an array look up will be slower than the binary op). – user89217 Nov 16 '18 at 14:25