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We consider a matrix $beginbmatrixa_i,jendbmatrix$ with $r$ rows and $c$ columns. We fill this matrix only with zeros and ones.

How many ones (maximally!) we can write into the matrix $rtimes c$ to not have any rectangle with ones at its vertices? Formally:
$$
forall_i_1,i_2in1,ldots,r forall_j_1,j_2in1,ldots,c big( (i_1neq i_2 wedge j_1neq j_2) Rightarrow 0ina_i_1,j_1,a_i_1,j_2, a_i_2,j_1, a_i_2,j_2 big).
$$

For example, for $3times 3$ matrix we can fill it with 6 ones:
$$
beginbmatrix
1 & 1 & 0 \
1 & 0 & 1 \
0 & 1 & 1
endbmatrix.
$$
But when we use 7 ones, then there always is a rectangle with ones at its vertices. For example:
$$beginbmatrix
mathbf 1 & mathbf 1 & 0 \
1 & 0 & 1 \
mathbf 1 & mathbf 1 & 1
endbmatrix.
$$
But what is the maximal number of ones for matrix $rtimes c$?

It seems you just want to maximise the number of ones so that there is no
$$left[matrix1&1\1&1right]$$
as a not-necessarily-contiguous $2times 2$ submatrix.

Equivalently, how many edges can a bipartite graph with sides of size $r$ and $s$ have without having any 4-cycles?

This is the first non-trivial case of the Zarankiewicz problem.

The square case is A001197(n)-1.

There is a table of values and bounds in this paper. I have more exact values near the diagonal.

For an $n times n$ matrix, you can get $3 (n-1)$ ones by putting them in all entries except $(1,1)$ of the first row, first column and main diagonal. That's optimal at least for $n le 5$.

But for $6 times 6$ the optimal solution has $16$ ones, e.g.

$$ left[ begin arraycccccc 0&1&1&1&0&0\ 0&1&0&0
&1&1\ 1&1&0&0&0&0\ 1&0&1&0&1&0
\ 1&0&0&1&0&1\ 0&0&0&1&1&0
end array right]
$$