And algorithm with access to coin-flips
Given a graph G = (V, E) find a non-trivial S \subseteq V such that |E(S, \bar S)| is minimized. Where E(S, \bar S) is the set of edges between S and \bar S.
If the graph has two vertices, output the only possibility. Otherwise,
This returns a chosen min-cut with probability at least \frac{1}{\binom{n}{2}}. It’s runtime is O(n^4 \cdot \ln n).
If we run the algorithm T = c \cdot n^2 \cdot \ln n times and choose the best result we get a fail probability of \frac{1}{n^c}.
If the graph has two vertices, output the only possibility. Otherwise,
This returns a min-cut with probability at least \frac{1}{2\log n}. It’s runtime is O(n^2 \cdot \log n).
Given polynomials p(x) and q(x) we want to know whether they are equal for any x. In other words whether p(x) - q(x) \equiv 0. We do not know how p or q looks like, we can only evaluate their values.
Let p(x_1, \cdots, x_n) be a non-zero polynomial of degree d. Let S be a finite subset of \R with at least d elements. If x_i \sim U(S) then
P[p(x_1, \cdots, x_n) = 0] \le \frac{d}{|S|}
Distribution D, want to estimate mean of D. Draw independent samples from D and return the empirical average. How close should we expect this average to be the real mean and how often?
Given a random variable X \ge 0, for all k > 0
Pr[X \ge k] \le \frac{\mathbb E[X]}{k}
or
Pr[X \ge k \cdot \mathbb E[X]] \le \frac{1}{k}
Pr[|X - \mathbb E[X]| > \epsilon] \le \frac{Var[X]}{\epsilon^2}
or
Pr[|X - \mathbb E[X]| > k \cdot \sigma] \le \frac{1}{k^2}
where \sigma = \sqrt{Var[X]}
Let X = \sum_{i=1}^n X_i where X_i are independent and X_i = 1 with probability p_i and X_i = 0 otherwise. Let \mu = \mathbb E[X] and \delta > 0
P[X \ge (1 + \delta) \mu] \le e^{-\frac{\delta^2}{2 + \delta}\mu}
P[X \le (1 - \delta) \mu] \le e^{-\frac{\delta^2}{2}\mu}
Let X = \sum_{i=1}^n X_i where X_i are independent such that a \le X_i \le b. Let \mu = \mathbb E[X] and \delta > 0
P[X \ge (1 + \delta) \mu] \le e^{-\frac{2\delta^2\mu^2}{n(b-a)^2}}
P[X \le (1 - \delta) \mu] \le e^{-\frac{\delta^2\mu^2}{n(b-a)^2}}