# Matrix element value counting in O(1) space

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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– Raphael
Nov 4 '19 at 19:49

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).