# Random Tools

This page regroup the tools in domain of probability of statistic.

## 1. Source Point distribution

To choice a random point, we use uniform distribution on cirumscribed circle.

Thus circle $\mathcal{C}$ (we suppose th center is the origin) of ray $R$.

We want to find a method to create a random point with thise uniform distribution.

\begin{eqnarray} P = \begin{pmatrix} R \sqrt{\xi_2} cos 2\pi \xi_1 \\ R \sqrt{\xi_2} sin 2\pi \xi_1 \end{pmatrix} \end{eqnarray}

With $\xi_1$ ans $\xi_2$ random point on uniform distribution on $[0,1]$.

## 2. Source Point Complexity

We want to know the complexity of algorithm Source Point (with triangle $\mathcal{T}$). This complexity is regit by a geometrical distribution.

### 2.1. Geometrical Distribution

Rappel of geometrical distribution :

The probability distribution of the number X of Bernoulli trials needed to get one success.

If $p$ is the probability of success :

$\mathbb{P} \left( X = k \right) = \left( 1-p \right)^{k-1} p$

We have some reults :

• esperance : $\mathbb{E}(X) = \frac{1}{p}$

• variance : $Var(X) = \frac{1-p}{p^2}$

• variance : $\sigma(X) = \sqrt{\frac{1-p}{p^2}} = \frac{\sqrt{1-p}}{p}$

The choise of random point is regit by distribuni. The complexity of algorithm Source Point is regit by geometrical distribution of parameter :

$p = \frac{\int_{O \in \mathcal{T}}{dO}}{\int_{O \in \mathcal{C}}{dO}} = \frac{Area(\mathcal{T})}{Area(\mathcal{C})}$

With $\mathcal{C}$ the circumscribed circle of center $C$ and ray $R$.

### 2.3. Complexity

To calculus of complexity of Source Point (with triangle $\mathcal{T}$), we take :

$\text{complexity} = \mathbb{E}(X) = \frac{1}{p}$

With the ratio $p = \frac{Area(\mathcal{T})}{Area(\mathcal{C})}$

### 2.4. Example

#### 2.4.1. Equilateral triangle

We take the equilateral triangle $\mathcal{T}$ (of side $1$). We calculate its areas and areas of its circumcribled circle.

• $Area(\mathcal{T}) = \frac{\sqrt{3}}{2}$

• $Area(\mathcal{C}) = \frac{2\pi}{3}$

• $p = \frac{3\sqrt{3}}{4\pi} \thickapprox 0.413$

• Complexity $born \thickapprox 2.41 \leq 3$

We can suppose the number of loop of algorithm is a $3$ for the case of equilateral triangle.

Remarks : the case of equilateral triangle maximizes $p$ (I think).

#### 2.4.2. Right triangle

We take the right triangle $\mathcal{T}$ (of side $1$). We calculate its areas and areas of its circumcribled circle.

• $Area(\mathcal{T}) = 1$

• $Area(\mathcal{C}) = 2\pi$

• $p = \frac{1}{2\pi} \thickapprox 0.159$

• Complexity $born \thickapprox 6.2$

We can suppose the number of loop of algorithm is a $6$ for the case of right triangle.