Rank of a 0-1-matrix Suppose $K$ is a field of characteristic $0$. Let $M \in K^{n \times m}$ be a matrix such that every entry of $M$ is either $0$ or $1$. About this matrix, I know further that each sum over a column and a row is a constant number, i.e.
$$\sum_{i=1}^n M_{ij} = c_{\text{column}} \forall j=1,...,m$$
and
$$\sum_{j=1}^m M_{ij} = c_{\text{row}} \forall i=1,...,n$$
for some $c_{\text{column}}, c_{\text{row}}$ both lying in $\{1,2,3,...\}$.
Question: Can one say something about the rank of $M$ in this situation? If not, does somebody know a criterion that leads to some inequality like ''if condition is satisfied then 
$$\text{rank}(M) \geq f(n,m,c_{\text{column}}, c_{\text{row}})$$
for some function $f$''?
 A: There is a paper dealing with that by Odlyzko (ON THE RANKS OF SOME (0, 1)-MATRICES WITH CONSTANT ROW SUMS, Austral. Math. Soc. (Series A) 31 (1981), 193-201) stating the following: Take a matrix $M \in \{0,1\}^{a \times n}$ viewed as a matrix in $K^{a \times n}$ for any field $K$ of characteristic $0$. Assume further that the row-sum is constantly $m$, then: If rank($M$) $\leq n-1$, then there may be no more than 
$$g(n,m) := \begin{cases} 
              {n-1}\choose m & \text{if $1 \leq m < n/2$} \\
              2 \cdot {{n-2}\choose {(n-2)/2}} & \text{if $m = n/2$} \\
              {n-1}\choose {m-1} & \text{if $n/2 < m \leq n$} \\
            \end{cases}$$
Reformulated, this means that if one knows that there are strictly more than $g(n,m)$ distinct $\{0,1\}$-vectors among those $a$ rows of $M$, then $M$ has full rank.
A: I don't think you could tell a lot from this. Consider for example the matrices
$$\begin{bmatrix}1 & 1&0&0\\0 &1&1&0 \\0&0&1&1\\1&0&0&1 \end{bmatrix}$$
$$\begin{bmatrix}1 & 1&0&0\\1 &1&0&0 \\0&0&1&1\\0&0&1&1 \end{bmatrix}$$
both have all of the same parameters for $n,m,c_r,c_c$ but they have different ranks
