Power of dd Choices with Simple Tabulation

Suppose that we are to place $m$ balls into $n$ bins sequentially using the $d$-choice paradigm: For each ball we are given a choice of $d$ bins, according to $d$ hash functions $h_1,\dots,h_d$ and we place the ball in the least loaded of these bins breaking ties arbitrarily. Our interest is in the number of balls in the fullest bin after all $m$ balls have been placed. Azar et al. [STOC'94] proved that when $m=O(n)$ and when the hash functions are fully random the maximum load is at most $\frac{\lg \lg n }{\lg d}+O(1)$ whp (i.e. with probability $1-O(n^{-\gamma})$ for any choice of $\gamma$). In this paper we suppose that the $h_1,\dots,h_d$ are simple tabulation hash functions. Generalising a result by Dahlgaard et al [SODA'16] we show that for an arbitrary constant $d\geq 2$ the maximum load is $O(\lg \lg n)$ whp, and that expected maximum load is at most $\frac{\lg \lg n}{\lg d}+O(1)$. We further show that by using a simple tie-breaking algorithm introduced by V\"ocking [J.ACM'03] the expected maximum load drops to $\frac{\lg \lg n}{d\lg \varphi_d}+O(1)$ where $\varphi_d$ is the rate of growth of the $d$-ary Fibonacci numbers. Both of these expected bounds match those of the fully random setting. The analysis by Dahlgaard et al. relies on a proof by P\u{a}tra\c{s}cu and Thorup [J.ACM'11] concerning the use of simple tabulation for cuckoo hashing. We need here a generalisation to $d>2$ hash functions, but the original proof is an 8-page tour de force of ad-hoc arguments that do not appear to generalise. Our main technical contribution is a shorter, simpler and more accessible proof of the result by P\u{a}tra\c{s}cu and Thorup, where the relevant parts generalise nicely to the analysis of $d$ choices.Comment: Accepted at ICALP 201

Load Balancing with Dynamic Set of Balls and Bins

In dynamic load balancing, we wish to distribute balls into bins in an environment where both balls and bins can be added and removed. We want to minimize the maximum load of any bin but we also want to minimize the number of balls and bins affected when adding or removing a ball or a bin. We want a hashing-style solution where we given the ID of a ball can find its bin efficiently. We are given a balancing parameter $c=1+\epsilon$, where $\epsilon\in (0,1)$. With $n$ and $m$ the current numbers of balls and bins, we want no bin with load above $C=\lceil c n/m\rceil$, referred to as the capacity of the bins. We present a scheme where we can locate a ball checking $1+O(\log 1/\epsilon)$ bins in expectation. When inserting or deleting a ball, we expect to move $O(1/\epsilon)$ balls, and when inserting or deleting a bin, we expect to move $O(C/\epsilon)$ balls. Previous bounds were off by a factor $1/\epsilon$. These bounds are best possible when $C=O(1)$ but for larger $C$, we can do much better: Let $f=\epsilon C$ if $C\leq \log 1/\epsilon$, $f=\epsilon\sqrt{C}\cdot \sqrt{\log(1/(\epsilon\sqrt{C}))}$ if $\log 1/\epsilon\leq C<\tfrac{1}{2\epsilon^2}$, and $C=1$ if $C\geq \tfrac{1}{2\epsilon^2}$. We show that we expect to move $O(1/f)$ balls when inserting or deleting a ball, and $O(C/f)$ balls when inserting or deleting a bin. For the bounds with larger $C$, we first have to resolve a much simpler probabilistic problem. Place $n$ balls in $m$ bins of capacity $C$, one ball at the time. Each ball picks a uniformly random non-full bin. We show that in expectation and with high probability, the fraction of non-full bins is $\Theta(f)$. Then the expected number of bins that a new ball would have to visit to find one that is not full is $\Theta(1/f)$. As it turns out, we obtain the same complexity in our more complicated scheme where both balls and bins can be added and removed.Comment: Accepted at STOC'2

One-Way Trail Orientations

Given a graph, does there exist an orientation of the edges such that the resulting directed graph is strongly connected? Robbins\u27 theorem [Robbins, Am. Math. Monthly, 1939] asserts that such an orientation exists if and only if the graph is 2-edge connected. A natural extension of this problem is the following: Suppose that the edges of the graph are partitioned into trails. Can the trails be oriented consistently such that the resulting directed graph is strongly connected? We show that 2-edge connectivity is again a sufficient condition and we provide a linear time algorithm for finding such an orientation. The generalised Robbins\u27 theorem [Boesch, Am. Math. Monthly, 1980] for mixed multigraphs asserts that the undirected edges of a mixed multigraph can be oriented to make the resulting directed graph strongly connected exactly when the mixed graph is strongly connected and the underlying graph is bridgeless. We consider the natural extension where the undirected edges of a mixed multigraph are partitioned into trails. It turns out that in this case the condition of the generalised Robbin\u27s Theorem is not sufficient. However, we show that as long as each cut either contains at least 2 undirected edges or directed edges in both directions, there exists an orientation of the trails such that the resulting directed graph is strongly connected. Moreover, if the condition is satisfied, we may start by orienting an arbitrary trail in an arbitrary direction. Using this result one obtains a very simple polynomial time algorithm for finding a strong trail orientation if it exists, both in the undirected and the mixed setting