Consider the following optimization problem:

Given $n\leq 10^3$ vectors $v_i\in\mathbb{R}^2$, all of which are small, i.e., $\|v_i\| \leq 1$, find a subset $S$ of them that minimizes $ \| w + \sum_{i\in S} v_i \|^2$, where $w$ is a fixed known vector.

I know that this can be reduced to an unconstrained 0-1 quadratic program (with unknowns $x_i\in\{0,1\}$, $x_i=1$ whenever $i\in S$), but that is a hard problem in general.

However, since the constraints are so specific, is there a way to get a reasonably simple efficient approximation algorithm for it? One that doesn't involve integer programming? It's not even clear to me how to efficiently implement a greedy algorithm for it.

Are there special cases of this problem that are easier? For example, would it help if all the vectors were unit length, $\|v_i\|=1$?

  • $\begingroup$ Related: cs.stackexchange.com/q/45718/755, cs.stackexchange.com/q/33788/755. And maybe: cs.stackexchange.com/q/56158/755 $\endgroup$
    – D.W.
    Commented May 24, 2017 at 23:33
  • $\begingroup$ Your problem is the Closest Vector Problem (CVP) in an integer lattice. In general, the CVP is fairly hard. However, you are working in only 2 dimensions, so the problem might become tractable. It's known that the 2D SVP can be solved in polynomial time (algorithm due to Gauss) if you start with only 2 vectors. I don't know if it is solvable if you start with $n>2$ vectors, or if it extends to the CVP. $\endgroup$
    – D.W.
    Commented May 24, 2017 at 23:50
  • $\begingroup$ @D.W. I don't follow you: in a lattice, the coefficients are just arbitrary integers. How do you reduce the 0-1 problem to a lattice problem? $\endgroup$
    – Kirill
    Commented May 25, 2017 at 0:22
  • $\begingroup$ You are absolutely right; my mistake. I overlooked that we need the coefficients to be not just integers, but either zero or one. To try to save my idea: I think we still might be able to formulate as a lattice problem by replacing the vector $(x_i,y_i)$ with a vector of the form $(x_i,y_i,0,0,\dots,0,K,0,\dots,0)$ where $K$ is a sufficiently large constant, but then we're not in a lattice of dimension two. That might lead to an approximation algorithm using LLL lattice reduction, but I haven't thought about it carefully. $\endgroup$
    – D.W.
    Commented May 25, 2017 at 16:42
  • $\begingroup$ Why isn't the objective function to be minimized the following? $$ \left\| \left( \sum_{i\in S} v_i \right) - w \right\|_2^2$$ $\endgroup$ Commented Jun 4, 2017 at 2:45

2 Answers 2


Let $\rm V$ be the $2 \times n$ matrix whose $i$-th column is vector $\rm v_i$. We have the Boolean optimization problem in $\mathrm z \in \{0,1\}^n$

$$\min_{\mathrm z \in \{0,1\}^n} \| \mathrm V \mathrm z + \mathrm w \|_2^2$$


$$\rm z = \frac 12 (x + 1_n)$$

where $\mathrm x \in \{\pm 1\}^n$. Hence, we have an instance of the Boolean least-squares (BLS) problem

$$\min_{\mathrm x \in \{\pm 1\}^n} \| \mathrm A \mathrm x - \mathrm b \|_2^2$$


$$\rm A = \frac 12 V \qquad \qquad \qquad b = - \left( \frac 12 V 1_n + w \right)$$

Since $\{\pm 1\}$ is the solution set of the quadratic equation $x_i^2 = 1$, we have the following (non-convex) quadratically constrained quadratic program (QCQP) in $\mathrm x \in \mathbb R^n$

$$\begin{array}{ll} \text{minimize} & \| \mathrm A \mathrm x - \mathrm b \|_2^2\\ \text{subject to} & x_i^2 = 1 \quad \forall i \in \{1,2,\dots,n\}\end{array}$$

Exhaustive evaluation of the objective function at all $2^n$ points is only feasible for small $n$.

SDP relaxation

Note that

$$\begin{array}{rl} \| \mathrm A \mathrm x - \mathrm b \|_2^2 &= (\mathrm A \mathrm x - \mathrm b)^{\top} (\mathrm A \mathrm x - \mathrm b)\\ &= \mbox{tr} \left( (\mathrm A \mathrm x - \mathrm b) (\mathrm A \mathrm x - \mathrm b)^{\top} \right)\\ &= \mbox{tr} \left( \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix}^{\top} \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix}^{\top} \right)\\ &= \mbox{tr} \left( \begin{bmatrix} \mathrm A^{\top}\\ -\mathrm b^{\top}\end{bmatrix} \begin{bmatrix} \mathrm A & -\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix} \begin{bmatrix} \mathrm x^{\top} & 1\end{bmatrix} \right)\\ &= \mbox{tr} \left( \begin{bmatrix} \,\,\,\, \mathrm A^{\top} \mathrm A & -\mathrm A^{\top}\mathrm b\\ -\mathrm b^{\top}\mathrm A & \,\,\,\, \mathrm b^{\top}\mathrm b\end{bmatrix} \begin{bmatrix} \mathrm x \mathrm x^{\top} & \mathrm x\\ \mathrm x^{\top} & 1\end{bmatrix} \right)\end{array}$$

Since $\mathrm x \in \{\pm 1\}^n$, all $n$ entries on the main diagonal of $\mathrm x \mathrm x^{\top}$ are equal to $1$. Thus,

$$\begin{bmatrix} \mathrm x \mathrm x^{\top} & \mathrm x\\ \mathrm x^{\top} & 1\end{bmatrix}$$

is symmetric, positive semidefinite and has only ones on its main diagonal, i.e., it is a correlation matrix. Its rank is $1$. Let

$$\mathrm C := \begin{bmatrix} \,\,\,\, \mathrm A^{\top} \mathrm A & -\mathrm A^{\top}\mathrm b\\ -\mathrm b^{\top}\mathrm A & \,\,\,\, \mathrm b^{\top}\mathrm b\end{bmatrix}$$

Hence, the BLS problem can be written as the following rank-constrained optimization problem in $(n+1) \times (n+1)$ symmetric matrix $\rm Y$

$$\begin{array}{ll} \text{minimize} & \langle \mathrm C , \mathrm Y \rangle\\ \text{subject to} & y_{ii} = 1, \quad \forall i \in \{1,2,\dots,n+1\}\\ & \mathrm Y \succeq \mathrm O_{n+1}\\ & \mbox{rank} (\mathrm Y) = 1\end{array}$$

which is a hard problem due to the rank constraint. Relaxing the optimization problem above by discarding the rank constraint, we then obtain the following semidefinite program (SDP) in $\rm Y$

$$\boxed{\begin{array}{ll} \text{minimize} & \langle \mathrm C , \mathrm Y \rangle\\ \text{subject to} & y_{ii} = 1, \quad \forall i \in \{1,2,\dots,n+1\}\\ & \mathrm Y \succeq \mathrm O_{n+1}\end{array}}$$

which is convex and computationally tractable. This SDP relaxation provides a lower bound on the minimum of the original BLS problem. In the very, very fortunate case where the optimal solution of the SDP happens to be rank-$1$, we have solved the BLS problem.


  • $\begingroup$ Thank you for the answer. Do you know if size-$n=10^3$ SDP problems should be considered feasible? It seems like it might be too big. $\endgroup$
    – Kirill
    Commented May 26, 2017 at 20:41
  • $\begingroup$ @Kirill That should be feasible. Take a look at Do they use semidefinite programming in industry?. $\endgroup$ Commented May 27, 2017 at 15:58

Consider the following approach to approximate this problem.

First, construct an efficient data structure using your vectors that can perform Nearest Neighbor related searching, such as a KD Tree. Then, formulate your problem as an optimization that's a function of $N$, the number of nearest neighbors you will compute. To get there, we can do the following with the cost function:

\begin{align} J &= \left\lVert w + \sum_{i \in S} v_i\right\Vert^2 \\ &= \frac{N^2}{N^2}\left\lVert w + \sum_{i \in S_N} v_i\right\Vert^2 \\ &= N^2\left\lVert \frac{w}{N} + \frac{1}{N}\sum_{i \in S_N} v_i\right\Vert^2 \end{align}

Note that $S_N$ is then an approximate solution for $S$. We can then approximate this problem by choosing $N$ such that we generate a set of vectors, which can be represented by the index set $S_N$, that are the $N$ nearest neighbors to $\frac{-w}{N}$. The logic is that selecting $N$ nearest neighbors from $\lbrace v_i \rbrace$ about $\frac{-w}{N}$ will produce a value for $\frac{1}{N}\sum_{i \in S_N} v_i$ close to $\frac{-w}{N}$ and in turn help minimize the norm.

The resulting optimization then ends up being:

\begin{align} N^{*} = \arg \min_{N} N^2 \left\lVert \frac{w}{N} + \frac{1}{N}\sum_{i \in S_N} v_i\right\Vert^2 \end{align}

How you go about tackling the above optimization is then up to you.

  • $\begingroup$ Thank you for the answer. Does this assume that the vectors are uniformly distributed? I don't want to assume that, so I'm not sure about the step where the average of $N$ nearest neighbours should be close to $\frac{-w}{N}$. $\endgroup$
    – Kirill
    Commented May 26, 2017 at 20:46
  • $\begingroup$ @Kirill This approximation scheme does not depend on that, though I expect it would perform better if a subset of the vectors were at least uniformly distributed near $\frac{-w}{N}$. The level of accuracy this might achieve certainly is dependent on the distribution of the vectors. $\endgroup$
    – spektr
    Commented May 26, 2017 at 21:05

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