191014K02 | Day 2 Lecture 2

191014K02: Day 2 Lecture 2

Work in progress.

A simple path in a graph is a sequence of distinct vertices u_1, \ldots, u_\ell such that all consecutive vertices have an edge between them. We are going to talk about the problem of finding long paths in graphs:

k-Path

Does G have a simple path on k vertices?

This problem is “of course”¹ NP-complete.

What is the probability that a random coloring with k colors makes a fixed k-path “multicolored”? It’s \frac{k!}{k^k}, and we claim that this is at least \frac{1}{e^k}.

Lower bound on the probability of the good event

Claim 1. \frac{k!}{k^k} \geqslant \frac{1}{e^k}

Proof by induction:

\begin{aligned} \frac{k^k}{k !} & = \underbrace{{\color{indianred}\frac{(k-1)^{k-1}}{(k-1) !} \cdot \frac{(k-1) !}{(k-1)^{k-1}}}}_{1} \cdot \frac{k^k}{k!} \\ & ~ \\ & = \underbrace{{\color{darkseagreen}\frac{(k-1)^{k-1}}{(k-1) !}}}_{\text{Induction Hypothesis}} \cdot \frac{(k-1) !}{(k-1)^{k-1}} \cdot \frac{k^k}{k!} \\ & \leqslant {\color{darkseagreen}e^{k-1}} \cdot \frac{(k-1)!}{(k-1)^{k-1}} \cdot \frac{k^k}{k!} \\ & ~ \\ & \leqslant e^{k-1} \cdot \frac{{\color{olivedrab}(k-1)!}}{(k-1)^{k-1}} \cdot \frac{{\color{palevioletred}k^k}}{{\color{olivedrab}k!}} \\ & \leqslant e^{k-1} \cdot {\color{olivedrab}\frac{1}{k}} \cdot \frac{{\color{palevioletred}k^{k-1}}}{(k-1)^{k-1}} \cdot {\color{palevioletred}k} \\ & \leqslant e^{k-1} \cdot \frac{1}{{\color{indianred}k}} \cdot \frac{k^{k-1}}{(k-1)^{k-1}} \cdot {\color{indianred}k} \\ & \leqslant e^{k-1} \cdot \frac{k^{k-1}}{(k-1)^{k-1}} \\ & ~ \\ & =e^{k-1} \cdot {\color{darkseagreen}\left(1+\frac{1}{k-1}\right)^{k-1}}\\ & \leqslant e^{k-1} \cdot {\color{darkseagreen}e}\\ & =e^k \end{aligned}

Chernoff Bounds

The useful way to recall what the Chernoff bound tells us is the following: for independent 0/1 random variables, the probability that X deviates from its expectation by a large amount is extremely small.

Chernoff Bound (informal)

If X is the sum of many independent random variables with “small” values, then X is very very likely to be very close to E[X].

The following is the precise statment:

Chernoff Bound: the Theorem.

Let X= X_1 + X_2 \cdots+ X_n where:

the X_i’s take values from \{0,1\}, and
the X_i’s is are independent,

then \operatorname{Pr}[{\color{indianred}|X-E[X]| \geqslant \varepsilon E[X]}] \leqslant 2e^{-\varepsilon^2 \cdot E[X]/3}

And here is a useful variation, handy for when you don’t know the expectation, but have upper and lower bounds on it.

Chernoff Bound: a Useful Variation

The: Let X= X_1 + X_2 \ldots+ X_n where:

the X_i’s take values from \{0,1\}
the X_i’s are independent

Let \mu_L \leqslant E[X] \leqslant \mu_H

Then: \begin{aligned} & \operatorname{Pr}\left[x-\mu_H \geqslant \varepsilon \mu_H\right] \leqslant e^{-\varepsilon^2 \cdot \mu_H / 3} \\ & \operatorname{Pr}\left[\mu_L-x \geqslant \varepsilon \mu_L\right] \leqslant e^{-\varepsilon^2 \cdot \mu_L / 3} \end{aligned}

Footnotes

Thanks, Hamiltonian Path.↩︎