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The question arise from my customer's real-time system (RAM model, off-course), which has very limited resources.

Given an NxM matrix of integer values, we need to verify that the number of non-zero elements in every row and every column is maximum 2. The matrix cannot be stored! It comes as a stream (M and N are known in advance but are not constants), row after row, and each element is accessed only once. The space requirement is O(1), i.e. no additional array of size N or M can be used, only a constant number of variables.

Can this be done? If not, can it be proved? Can this be done in O(log(N) + Log(M)) space?

We were taught in undergrad school that if you are unable to provide an adequate solution, prove that no one else can... So I tried to prove to my customer that this can't be done in o(M) space, but was unable to.

Thanks allot

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Checking whether the number of nonzero elements on each row is at most $2$ can be done in $O(1)$ space on a row-by-row basis by simply counting those elements.

Therefore the only interesting part is checking the condition over the columns. This can be done in $O(M)$ space by keeping a counter for each column. In details, the counter $c_i \in \{0,1,2\}$ associated with the $i$-th column will be the number of rows seen so far that contain a nonzero element in column $i$. As soon as a row with a nonzero element in column $i$ is seen and $c_i$ was already $2$, you can immediately return "false". If the whole matrix is processed without this happening you can return "true".

Unfortunately, your problem cannot be solved using $x= o(M)$ bits. Suppose this were the case and consider some matrices with $M$ of columns and $N=2M+1$ rows, of the form described in the following.

The $i$-th of the first $M$ rows will have zeros in all columns $j \neq i$, while $i$-th column will either contain a $0$ or a $1$.

The number of distinct choices for the first $M$ rows is $2^M$, yet there are at most $2^x < 2^M$ (for sufficiently large $M$) different possible states of the $x$ bits of memory. This means that there are (at least) two distinct choices $R$ and $R'$ of the first $M$ rows that leave the algorithm in the same internal state.

In other words, for any choice of the $(M+1)$-th to $(2M+1)$-th rows, the algorithm must behave in the same way, regardless of whether the set of the first $M$ rows was $R$ or $R'$.

Since $R$ and $R'$ are distinct, there must be (at least) one index $i$ such that the element $r_i$ in the $i$-th column of the $i$-th row of $R$ is different from the element $r'_i$ in the $i$-th column of the $i$-th row of $R'$. Without loss of generality assume that $r_i = 0$ and $r'_i=1$.

For $i=1,\dots,M$, let the $(M+i)$-th row be the row containing a single $1$ in the $i$-th column. Finally, choose the $(2M+1)$-th row to have exactly one $1$ in the $i$-th column.

This leads to a contradiction: On one hand, if the set of the first $M$ rows is $R$, the algorithm must return "true" since there is at most one nonzero element per row and at most two nonzero elements per column; on the other hand, if the set of the first $M$ rows is $R'$, the algorithm must return "false" since column $i$ has $3$ nonzero elements (in the $i$-th, $(M+i)$-th, and $(2M+1)$-th rows).

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