Consider the following variant of the balanced allocation...

Consider the following variant of the balanced allocation paradigm: $n$ balls are placed sequentially in $n$ bins, with the bins labeled from 0 to $n-1$. Each ball chooses a bin $i$ uniformly at random, and the ball is placed in the least loaded of bins $i, i+1 \bmod n, i+2 \bmod n, \ldots, i+d-1 \bmod n$. Argue that, when $d$ is a constant, the maximum load grows as $\Theta(\ln n / \ln \ln n)$. That is, the balanced allocation paradigm does not yield an $O(\ln \ln n)$ result in this case.

Key Concepts

Balls-into-Bins Model

This foundational probabilistic model studies how items (balls) are distributed among containers (bins) under random assignments. It helps in understanding the typical and extreme behaviors—like the maximum load—of systems subject to random allocation.

Balanced Allocation Paradigm

This concept involves assigning items to bins while using information from multiple choices to reduce load imbalances. By choosing the least loaded option among several candidate bins, the allocation scheme aims to achieve a more uniform load distribution across bins.

Multiple Choices vs. Constant Choices

The idea of using multiple choices generally leads to significant improvements in maximum load, as seen in the ‘power of two choices’ scenario. However, when the number of choices is kept constant (d constant), the improvement is limited, resulting in maximum loads that grow as Theta(ln n/ln ln n) rather than as low as O(ln ln n).

Asymptotic Analysis and Theta Notation

Asymptotic analysis provides a high-level understanding of the behavior of algorithms for large input sizes using notations like Theta. In this context, Theta(ln n/ln ln n) describes the tight bound on the maximum load in the system, capturing the precise growth rate as the number of bins n increases.

Transcript

00:01 Given a sequence of xi, xi are sequences of iid distribution, we're required to show the expectation of s, sn is defined as the summation of the first n terms for this sequence.

00:21 For m less or equal to n, we're required to compute the expectation of some fraction, i mean expectation of xm over sn, we're going to show it is equal to m over n.

00:50 Okay, first notice the expectation of sn over sn is just equal to 1.

00:59 Okay, then separate this summation, i mean separate this fraction, we know sn is equal to x1 plus x2 plus blah blah blah plus x.

01:19 And in addition, all of those xi's are, i mean all of those xi's have the same distribution, so the first step we can separate it as summation, from the linearity of the expectation we have xi over sn, i goes from 1 to n is just equal to 1.

01:44 And as all xi's have the same distribution, we know the expectation of xi over sn is just equal to the expectation of xj over sn, for any ij.

01:59 So we can assume it is just equal to some, let's say, suppose it is equal to some constant c, so we know the summation over some constant c is equal to 1.

02:12 This means c is equal to 1 over n, because we have a summation over n terms, that means the expectation of x1 over sn, and all of them are the same, so we can use the single term to represent all of those values, is equal to 1 over n.

02:39 Now for any n less or equal to n, sn is equal to x1 plus x2 plus xn, and that means the expectation of sn over sn is equal to the expectation of x1 over sn plus x2 over sn plus blah blah blah, to get the expectation of xm over sn.

03:13 And again, by our assumption, all of those terms are equal to 1 over n.

03:18 So we have n terms, that means this result is equal to m over n...

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