Subcubic Equivalences Between Graph Centrality Problems, APSP and Diameter

Abstract

Measuring the importance of a node in a network is a major goal in the analysis of social networks, biological systems, transportation networks etc. Different centrality measures have been proposed to capture the notion of node importance. For example, the center of a graph is a node that minimizes the maximum distance to any other node (the latter distance is the radius of the graph). The median of a graph is a node that minimizes the sum of the distances to all other nodes. Informally, the betweenness centrality of a node w measures the fraction of shortest paths that have w as an intermediate node.Finally, the reach centrality of a node w is the smallest distance r such that any s-t shortest path passing through w has either s or t in the ball of radius r around w.The fastest known algorithms to compute the center and the median of a graph, and to compute the betweenness or reach centrality even of a single node take roughly cubic time in the number n of nodes in the input graph. It is open whether these problems admit truly subcubic algorithms, i.e. algorithms with running timeÕ(n 3−δ ) for some constant δ > 0 1 . We relate the complexity of the mentioned centrality problems to two classical problems for which no truly subcubic algorithm is known, namely All Pairs Shortest Paths (APSP) and Diameter. We show that Radius, Median and Betweenness Centrality are equivalent under subcubic reductions to APSP, i.e. that a truly subcubic algorithm for any of these problems implies a truly subcubic algorithm for all of them. We then show that Reach Centrality is equivalent to Diameter under subcubic reductions. The same holds for the problem of approximating Betweenness Centrality within any constant factor. Thus the latter two centrality problems could potentially be solved in truly subcubic time, even if APSP requires essentially cubic time.

Faster approximation schemes for the two-dimensional knapsack problem

“…Then, we can show that there are only O (log n) squares in total that can potentially be large. This reduces the guessing time to (log n) O (1) = O (1) · n O (1) . More generally, with our approach we can obtain such an improved bound in any setting with only few (profit) classes of input squares where for each class there are only few relevant squares if we know that the class contributes only a small number of squares to a (near-)optimal solution.…”

“…Theorem 1.2. There is an algorithm for the twodimensional knapsack problem for rectangles under (1 + )-resource augmentation with and without rotation with running time O (1) · n O (1) . This algorithm computes a solution with profit at least the profit of the optimal solution (that does not use resource augmentation).…”

“…In Section 3 we present a technique that allows us to guess the large squares in time O (1) · n O (1) . Then in Section 4 we present an EPTAS for the special case that N N , i.e., the knapsack is a rectangle with very large aspect ratio.…”

“…For each input rectangle i denote by h i ∈ N its height and by w i ∈ N its width. We reduce the running time of the known PTAS [9] of Ω n 1/ 1/ to O (1) · n O (1) and compute even a solution whose profit is at least p(OP T ), instead of a (1 + )-approximative solution. We investigate first the case without the possibility to rotate the rectangles.…”

“…This raises the question whether this highly impractical running time is necessary or whether better running times are possible of e.g., n O(1/ ) or even O (1) · n O(1) (where O (1) denotes a value that is constant for constant ). Recently, there has been a lot of research on determining the best possible running times for specific problems, see e.g., [6,1] and references therein. For some problems, we know even the best possible exponent of n under standard complexity assumptions.…”