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# 数学代写|数论代写Number Theory代考|STAT7604 Flipping a coin until a head appears

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## 数学代写|数论代写Number Theory代考|Flipping a coin until a head appears

In this and subsequent sections of this chapter, we discuss a number of specific probabilistic algorithms.

Let us begin with the following simple algorithm (which was already presented in Example 7.1) that essentially flips a coin until a head appears:
\begin{aligned} \text { repeat } & \ b & \leftarrow R{0,1} \ \text { until } b & =1 \end{aligned}
Let $X$ be a random variable that represents the number of loop iterations made by the algorithm. It should be fairly clear that $X$ has a geometric distribution, where the associated probability of success is $1 / 2$ (see Example 6.30). However, let us derive this fact from more basic principles. Define random variables $B_1, B_2, \ldots$, where $B_i$ represents the value of the bit assigned to $b$ in the $i$ th loop iteration, if $X \geq i$, and $\star$ otherwise. Clearly, exactly one $B_i$ will take the value 1 , in which case $X$ takes the value $i$.
Evidently, for each $i \geq 1$, if the algorithm actually enters the $i$ th loop iteration, then $B_i$ is uniformly distributed over ${0,1}$, and otherwise, $B_i=\star$. That is:
$$\begin{gathered} \mathrm{P}\left[B_i=0 \mid X \geq i\right]=1 / 2, \quad \mathrm{P}\left[B_i=1 \mid X \geq i\right]=1 / 2 \ \mathrm{P}\left[B_i=\star \mid X<i\right]=1 \end{gathered}$$

## 数学代写|数论代写Number Theory代考|Generating a random number from a given interval

Suppose we want to generate a number $n$ uniformly at random from the interval ${0, \ldots, M-1}$, for a given integer $M \geq 1$.

If $M$ is a power of 2 , say $M=2^k$, then we can do this directly as follows: generate a random $k$-bit string $s$, and convert $s$ to the integer $I(s)$ whose base-2 representation is $s$; that is, if $s=b_{k-1} b_{k-2} \cdots b_0$, where the $b_i$ are bits, then
$$I(s):=\sum_{i=0}^{k-1} b_i 2^i .$$
In the general case, we do not have a direct way to do this, since we can only directly generate random bits. However, suppose that $M$ is a $k$-bit number, so that $2^{k-1} \leq M<2^k$. Then the following algorithm does the job:

Algorithm RN:
repeat
\begin{aligned} & s \leftarrow R{0,1}^{\times k} \ & n \leftarrow I(s) \ & \text { until } n<M \ & \text { output } n\end{aligned}
Let $X$ denote the number of loop iterations of this algorithm, $Y$ its running time, and $N$ its output.

In every loop iteration, $n$ is uniformly distributed over $\left{0, \ldots, 2^k-1\right}$, and the event $n<M$ occurs with probability $M / 2^k$; moreover, conditioning on the latter event, $n$ is uniformly distributed over ${0, \ldots, M-1}$. It follows that $X$ has a geometric distribution with an associated success probability $p:=M / 2^k \geq 1 / 2$, and that $N$ is uniformly distributed over ${0, \ldots, M-1}$. We have $\mathrm{E}[X]=1 / p \leq 2$ (see Example 6.35) and $Y \leq c k X$ for some implementation-dependent constant $c$, from which it follows that
$$\mathrm{E}[Y] \leq c k \mathrm{E}[X] \leq 2 c k$$

## 数学代写|数论代写数论代考|投掷硬币直至出现头像

$$\纹理 { repeat } b {quad {leftarrow R 0,1 {text { until } b=1$$

$s\leftarrow R 0,1^{times k}。\夸父 n /leftarrow I(s)$ 直到 $n<M /quad$ 输出 $n$

$$y]\mathrm{E}[Y]\leq c k\mathrm{E}[X]\leq 2 c k$$

## MATLAB代写

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