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Given integers $n,m$, I want to find a $m \times n$ binary matrix $X$ such that there does not exist any non-zero vector $y \in \{-1,0,1\}^n$ with $Xy=0$ (all operations performed over $\mathbb{Z}$). What algorithm could I use for this?

In more detail: We are given parameters $n$ and $m$. The problem is to determine if there exists $x$ such that $x_{i,j} \in \{0,1\}$, and there does not exist $y\ne (0,0,\dots,0)$ where $y_j \in \{-1,0,1\}$ for all $j$ and for all $1 \leq i \leq m$,

$$\sum_{1 \leq j \leq n} x_{i,j} y_j = 0.$$

(Notice that we require that at least one of the $y_j \ne 0$ to avoid the trivial solution.)

For example, consider $m=3,n=4$. Then, expressing $x_{i,j}$ as a matrix $X$,

$$ X=\begin{pmatrix} 0 & 1 & 1 & 0 \\ 1 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ \end{pmatrix} $$

is a valid solution for $m=3$ and $n=4$.

What algorithm can I use to solve this problem? Can I formulate this as an integer linear programming problem or maybe as a constraint programming problem?

marshall
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    What makes you think integer linear programming is a good way to solve this problem? What makes you ask for a formulation as an integer linear program? Is there some requirement to formulate it that way? Perhaps it is an exercise that requires you to formulate it in that way? To be honest, this sounds like an instance of an XY problem to me. – D.W. Dec 27 '13 at 23:49
  • @D.W. My motivation was simply to find a way to use existing software tools to solve this hard problem. – marshall Dec 28 '13 at 07:53
  • What would be a typical size for $n$ and $m$, in your application? This will affect which algorithms are efficient enough. – D.W. Dec 28 '13 at 08:11
  • @D.W. Both $n$ and $m$ will be under $20$ typically although of course it would be great to be able to solve larger instances too. – marshall Dec 28 '13 at 08:12
  • @marshall is $X$ valid if the only solution to $Xy=0$ is $y=0$ or must it have a non-trivial solution ($y\neq0$)? Also see the discussion under D.W. solution. – bcorso Dec 30 '13 at 04:38
  • @marshall, I see that you have added a bounty, asking if there is a constraint programming formulation. Sure! My SAT formulation is also a constraint programming formulation (with a particular class of constraints). If this isn't what you were looking for, can you elaborate on your requirements and how you are planning to evaluate any proposed answer? (i.e., what evaluation criteria you will use?) – D.W. Jan 04 '14 at 08:00
  • @D.W. A method that can prove if any such matrix exists for a fixed $m$ and $n$ is really my end goal. If there is more than one such suggestion, then I guess it will come down to how fast it runs :) – marshall Jan 04 '14 at 08:30

2 Answers2

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I have a method for you that will help you find valid solutions (matrices) for many possible values of $m,n$. However, it is not a complete answer to your question. It can try to find a matrix for a particular value of $m,n$, but it might fail, and if it fails, you've learned nothing; my method cannot prove that no such matrix exists.

The method is based upon the following observation:

Theorem. If we have a valid $m_1\times n_1$ matrix $X_1$ that meets all your requirements (for parameters $m_1,n_1$) and a valid $m_2\times n_2$ matrix $X_2$ that meets all your requirements (for parameters $m_2,n_2$), then we can find a valid $m\times n$ matrix that meets all your requirements (for parameters $m,n$), where $m=m_1+m_2$ and $n=n_1+n_2$.

Proof. Use the following matrix:

$$X = \begin{pmatrix} 0 &X_1 \\ X_2 &Z \end{pmatrix},$$

where $Z$ is arbitrary. Suppose $Xy=0$, where $y \in \{-1,0,1\}^n$. Then since the last $m_1$ coefficients of $Xy$ are zero, and since $X_1 y_1 =0$ implies $y_1=0$, it follows that the last $n_1$ coefficients of $y$ are zero. Thus by letting $y_2$ be the restriction of $y$ to its first $m_2$ coefficients, we find that $X_2 y_2 = 0$. But this implies $y_2 = 0$, i.e., $y=0$. In other words, if $Xy=0$, then $y=0$. This proves that $X$ is a valid matrix.

Now this lets us find many values of $m,n$ where it is possible to find a valid matrix $X$. In particular, seed things with some small matrices for various small values of $m,n$ (using any convenient method); then you can derive some larger values of $m,n$ that also have such a matrix.

Here are some observations that will help you identify seed values $m,n$ where such a matrix $X$ exists:

  • First, a trivial observation: Obviously, if $n \le m$, it is easy to find a valid solution $x$: just use the identity matrix (if $n<m$, fill in the extra rows arbitrarily). No need to use integer linear programming. So this problem is only interesting when $n>m$.

  • Second, if $n$ is small enough, you can express this as a SAT instance and apply an off-the-shelf SAT solver. The SAT instance will be of exponential size: it will have more than $3^n$ constraints, so this is only helpful for very small values of $n$, but it will still help you construct some values of $m,n$ where you can find a valid matrix $X$.

  • Third, you can use bcorso's answer to handle all cases where $n=m+1$ (there is always a valid solution, for $m\ge 3$).

    In particular, you can construct a SAT instance where the $x_{i,j}$ are the variables. Now, for each possible non-zero vector $y \in \{-1,0,1\}^n$, you can add a complicated constraint enforcing the requirement that $Xy \ne 0$. (You'll need to have $m$ adders, each of which adds up to $n$ 0-or-1 values, and then a comparison to test whether the results of all of the $m$ adders are all zero or not.)

In this way, I would expect that, for each $n\le 8$ (or so), you can probably find the largest value of $m$ such that there exists a valid $m\times n$ matrix. Now once you have those seed values, you can use the Theorem above to help you find additional values of $m,n$ where such a matrix exists.

As I stated above, this is not a complete solution, but it might help you solve your problem at least some of the time.

For general $m,n$, I doubt that there's any straightforward formulation of this as a polynomial-size integer linear program (unless $\text{NP} = \text{NP}^\text{co-NP}$ or the polynomial hierarchy collapses or something like that, which is not expected to hold; or unless you use some special knowledge about the solution to this problem).

Just telling whether a candidate value of $x$ is indeed a valid solution to this problem is $\text{co-NP}$-complete. See https://cstheory.stackexchange.com/q/20277/5038. In other, recognizing a solution to this problem can't be done in polynomial time (as far as we know); just recognizing a valid solution is $\text{co-NP}$-complete. This means that the problem of finding a valid solution is in $\text{NP}^\text{co-NP}$. In contrast, integer linear programming is in $\text{NP}$. Therefore, without using some special knowledge about this problem, I don't think you can find a generic reduction from your problem to integer linear programming unless $\text{NP}^\text{co-NP} = \text{NP}$ (something that most complexity theorists believe is not likely to hold).

Don't take this too seriously. I'm not trying to prove a formal theorem or anything like that; I'm just trying to give some weak evidence why this problem does not look like it has a straightforward, generic formulation as an instance of integer linear programming.

Of course, you can probably get a formulation as an integer linear program with a number of constraints that is exponential (say, in $m$), similar to how we got a SAT instance. It'll be uglier, because expressing the constraint that some vector (namely, $Xy$) is not identically zero is ugly in ILP. But it's doable. See, e.g., Express boolean logic operations in zero-one integer linear programming (ILP). However, I'm not sure this will be any better than the SAT-based method. If I were implementing this, I would start by trying the SAT-based method, because (if you use a suitable front end, like STP) I think it will be easier to implement and might work just as well or better than an ILP-based formulation.

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D.W.
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  • You may be right that integer programming is the wrong tool but that would be a shame. I want to avoid naive enumeration of all possibilities. – marshall Dec 28 '13 at 07:51
  • @marshall, thank you! I had $n$ and $m$ backwards; thank you for the correction. I've corrected my answer accordingly. Incidentally, do look at the CSTheory question I looked at; there's a (remote) chance that the reductions considered there might be helpful to you. – D.W. Dec 28 '13 at 08:01
  • Not sure what you mean by, "if $n≤m$, it is easy to find a valid solution $X$: just use the identity matrix." A simple example of this failing is if $n=m$, then $X=I$ would give $y=0$, so it not a valid solution to the problem. – bcorso Dec 30 '13 at 01:50
  • @bcorso, thanks for the comment. I'm sorry, but I don't understand your comment. What do you mean by "$X=I$ would give $y=0$"? The definition of what it means for $X$ to be valid is that (a) all entries of $X$ are $0$ or $1$, and (b) if $Xy=0$ and all entries of $y$ are $-1$, $0$, or $1$, then $y=0$ (then all entries of $y$ are $0$). The identity matrix satisfies both of these requirements, so it is a valid solution to the problem. – D.W. Dec 30 '13 at 01:55
  • @D.W. the OP specifies in his question, "we require that at least one of the $y_j≠0$ to avoid the trivial solution." – bcorso Dec 30 '13 at 02:00
  • @bcorso, now you're repeating what I said. As the problem statement says and as my comment says, the requirement is that the all-zeros vector $y=0$ is the only solution to $Xy=0$ with $y \in {-1,0,1}^n$. The identity matrix meets this requirement. I confess I'm not able to understand why you think the identity matrix is not a valid solution to the problem; it seems to me like it clearly qualifies. Would you like to elaborate? Is it possible you have misunderstood the problem statement? – D.W. Dec 30 '13 at 02:08
  • @D.W. the OP says, "Not all the $y_j$s can equal zero." Perhaps I am misunderstanding, but I take that to mean a non-trivial solution to $y$ must exist, such that $y\notin{−1,0,1}^n$. For example, If you take a look at my answer I first show that a non-trivial solution exists for $y$, and then prove that $y\notin{−1,0,1}^n$. – bcorso Dec 30 '13 at 02:20
  • @bcorso, yup, I think you're probably misunderstanding, as I don't see that anywhere in the question. I saw the comment you are quoting, but I suspect the OP was speaking informally in the comment; the OP also says "I edited the question", and the question is more precise. In particular, if we have $X$ for which there is no solution to $Xy=0$ other than $y=0$, then I believe that $X$ counts as valid. It certainly meets all the requirements in the question as it is currently stated. Feel free to ask the OP in a comment underneath the question if you'd like to ask the OP to clarify. – D.W. Dec 30 '13 at 02:35
  • @D.W. well I'm also going off the OP's example. In his example, there is a non-trivial solution, but it's of the form $y=y_2(2,1,-1,-1)$ so it is valid since $y\notin−1,0,1^n$. From what you are saying, the OP's example would not be valid... – bcorso Dec 30 '13 at 02:45
  • @bcorso, no, that's not an accurate characterization of what I'm saying. The OP's example is also valid. Nothing I've said contradicts that. (Perhaps you're reading my comments a bit too quickly. I did say that if $X$ is such that the only $y$ with $Xy=0$ is $y=0$, then $X$ is valid, but I never suggested this the only way that $X$ can be valid, and if you read the rest of my comments I think it's clear that that is not what I'm saying.) – D.W. Dec 30 '13 at 02:47
  • Haha, okay... well then I don't get what you are saying in your above comment, "As the problem statement says and as my comment says the requirement is that the all-zeros vector $y=0$ is the only solution to $Xy=0$ with $y\in{−1,0,1}^n$". – bcorso Dec 30 '13 at 02:49
  • I think I get what you are saying now... and it seems we are at an impasse until the OP comes to correct us. Anyway, thanks for talking this out with me. – bcorso Dec 30 '13 at 02:53
  • @bcorso The problem is trivial with $n=m$ as you can just use the identity matrix. The first non-trivial case is when $n=m+1$. – marshall Dec 30 '13 at 07:36
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Below is an exact solutions for the case of $3 \leq n-1 \leq m$ . Thus, you would only need to manually check cases where $m < n-1$, ($n>3$).

$\mathbf{Theorem:}$ for $3 \leq n-1 \leq m$ there always exists a binary matrix $X$ such that no (non-trivial) solution exists to the equation $Xy=0$. Furthermore, $X$ has the form:

$$ X_{(m\ \times\ n)}=\begin{pmatrix} M_{(n-1\ \times\ n)} \\ 0_{(m-n-1\ \times\ n)} \\ \end{pmatrix} $$

Where $0_{(m-n-1\ \times\ n)}$ is a matrix of zeros and:

$$ M_{(n-1 \ \times\ n)}=\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 1 & 0 & 0 & \cdots & 0 \\ 1 & 0 & 0 & 1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \cdots & \vdots \\ 1 & 0 & 0 & 0 & \cdots & 1 & 0 \\ 0 & 1 & 1 & 0 & \cdots & 0 & 1 \\ \end{pmatrix}, $$

$\mathbf{Proof:}$ Using the above definitions of $X$, $Xy=0$ reduces to $My=0$, which produces the following system of equation:

$$ \begin{pmatrix} y_1 = t \\ y_1 + y_2 = 0 \\ y_1 + y_3 = 0 \\ y_1 + y_4 = 0 \\ \vdots \\ y_1 + y_{n-1} = 0 \\ y_2 + y_3 + y_n = 0 \\ \end{pmatrix} \Rightarrow y=t\ \begin{pmatrix} 1 \\ -1 \\ -1 \\ -1 \\ \vdots \\ -1 \\ 2 \\ \end{pmatrix} $$ Which will never have a solution for all $y_i \in \{-1,0,1\}$ for any choice of $t$ except the trivial solution $t=0$. $\Box$

bcorso
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  • The case $n \le m$ is trivial and already covered in my answer. It is only $n>m$ that is interesting. So this only provides a characterization for $n=m+1$. That's incrementally helpful, but it leaves all the rest ($n>m+1$) uncharacterized. Basically, your solution is primarily the easy cases; it leaves most of the space of non-trivial cases unanswered. – D.W. Dec 30 '13 at 00:14
  • @D.W. saying it is trivial is not a proof. I applaud you for taking a stab at the harder case, but I still believe this case will be useful for others so I'll keep it up, Thanks! – bcorso Dec 30 '13 at 01:46