Metric Violation Distance: Hardness and Approximation

Abstract

Metric data plays an important role in various settings, for example, in metric-based indexing, clustering, classification, and approximation algorithms in general. Due to measurement error, noise, or an inability to completely gather all the data, a collection of distances may not satisfy the basic metric requirements, most notably the triangle inequality. In this paper we initiate the study of the metric violation distance problem: given a set of pairwise distances, modify the minimum number of distances such that the resulting set forms a metric. Three variants of the problem are considered, based on whether distances are allowed to only decrease, only increase, or the general case which allows both decreases and increases. We show that while the decrease only variant is polynomial time solvable, the increase only and general variants are NP-Complete, and moreover cannot in polynomial time be approximated to any ratio better than the minimum vertex cover problem. We then provide approximation algorithms for the increase only and general variants of the problem, by proving interesting necessary and sufficient conditions on the optimal solution, which are used to approximately reduce to a purely combinatorial problem for which we provide matching asymptotic upper and lower bounds.

A Matching Lower Bound

The following is a matching lower bound to the combinatorial problem from Lemma 4.7.

Lemma A.1

Let $G=(V,E)$ be a graph whose edge set E is the union of the edges of a collection of 4-cycles, ${\mathbb {C}}$, such that no two 4-cycles in ${\mathbb {C}}$ can share a chord. (Note this applies to all chords in the complete graph on V, i.e. regardless of whether they appear in E.) Then in the worst case $|{\mathbb {C}}| = \Omega (m^{4/3})$, where $m=|E|$.

Proof

Construct a graph $G=(V,E)$, where V is the disjoint union of four sets of vertices $X_1$, $X_2$, $X_3$, and $X_4$, each containing exactly t vertices, where t is a value to be determined shortly. The edge set E is sampled as follows. For $1\le i\le 4$, for each pair $(u,v) \in X_i \times X_{i+1}$ (where $X_5 = X_1$), the edge (u, v) is sampled into E independently with probability

$$\begin{aligned} p = \frac{m}{8t^2} \ll 1. \end{aligned}$$

Let ${\mathbb {C}}$ be the set of 4-cycles defined by E. Any cycle $C\in {\mathbb {C}}$ must contain exactly one vertex from each of $X_1$, $X_2$, $X_3$, and $X_4$. The probability that any quadruple of vertices $(i_1, i_2, i_3,i_4)\in X_1\times X_2\times X_3\times X_4$ defines a cycle in ${\mathbb {C}}$ is $p^4$. As such, the expected size of ${\mathbb {C}}$ is

$$\begin{aligned} \alpha = p^4t^4 = \!\left( {\frac{m}{8t^2}}\right) ^4t^4 = \!\left( {\frac{m}{8t}}\right) ^4. \end{aligned}$$

Consider such a cycle $C = (i_1,i_2,i_3,i_4)$ that we know exists in the graph. Any cycle which shares the chord $\{i_1,i_3\}$ with C clearly shares the vertices $i_1$ and $i_3$. Now such a cycle either shares a third vertex or not. The expected number of cycles which share the chord $\{i_1,i_3\}$ and no other vertex is at most $t^2p^4$ The expected number of cycles which share the chord $\{i_1,i_3\}$ and one other vertex is at most $2tp^2$. Let $X_C$ be a random variable denoting the number of cycles sharing either chord (i.e., $\{i_1,i_3\}$ or $\{i_2,i_4\}$) with C. Assuming $tp^2 \le 1$ we have,

$$\begin{aligned} \mathop {\mathbf {E}}\!\left[ {X_C | \text { { C} exists} } \right] \le 2(2tp^2 + t^2 p^4) \le 2(2 + t p^2)tp^2 \le 6tp^2. \end{aligned}$$

Assume further that $6 t p^2\le 1/10$, then by Markov’s inequality we have

$$\begin{aligned} \beta (C)&= \mathop {\mathbf {Pr}}\!\left[ { \text {no cycle shares a chord with } C \mid C \text { exists}} \right] \\&= 1 - \mathop {\mathbf {Pr}}\!\left[ {X_C \ge 1 | C \text {exists}} \right] \ge \frac{9}{10}. \end{aligned}$$

Let Y be a random variable denoting the number of cycles that exists in the graph and don’t share a chord with any other cycle that exists in the graph. We have that

$$\begin{aligned} \delta&= \mathop {\mathbf {E}}\!\left[ {Y} \right] \\&=\sum _{C} \mathop {\mathbf {Pr}}\!\left[ { (\text {no cycle shares chord with } C) \cap \!\left( { C \text { exists}}\right) } \right] \\&=\sum _{C} \beta (C) \cdot \mathop {\mathbf {Pr}}\!\left[ { C \text { exists}} \right] \ge \frac{9}{10} \alpha . \end{aligned}$$

Note that as $\alpha$ was the expected number of cycles overall, this implies $\delta = \mathop {\mathbf {E}}\!\left[ {Y} \right] = \Theta (\alpha )$.

Recall that we assumed $6tp^2 \le 1/10$, which plugging in for p becomes,

$$\begin{aligned} \frac{1}{10}\ge 6t \!\left( {\frac{m}{8t^2}}\right) ^2 \ge \frac{3}{32} \frac{m^2}{t^3} \quad \implies \quad t \ge (30/32)^{1/3} m^{2/3}. \end{aligned}$$

Thus setting $t= m^{2/3}$ (which up to constants minimizes $\alpha$) implies the expected number of cycles that do not share a chord is

$$\begin{aligned} \delta = \Theta (p^4t^4)&= \Theta \!\left( { \!\left( {\frac{m}{t^2}}\right) ^4 t^4 }\right) = \Theta \!\left( { \frac{m^4}{t^4} }\right) \\&= \Theta \!\left( { \frac{m^4}{m^{8/3}} }\right) = \Theta \!\left( { m^{4/3}}\right) . \end{aligned}$$

On the other hand, the expected number of edges is $4t^2 p = m/2$, and moreover by the Chernoff bound with high probability is at most m. Thus by the probabilistic method there exists a graph where $|E| \le m$ and the number of 4-cycles which don’t share a chord is $\Omega (m^{4/3})$. (Note to match the lemma statement, in the above construction one should only keep edges which were in cycles that did not share a chord with any other cycle.) $\square$