# 数学代写|密码学代写cryptography theory代考|CS127

## 数学代写|密码学代写cryptography theory代考|FACTORING USING A QUANTUM COMPUTER

Our goal is to set up a quantum system with an amplitude distribution like (5.2) and apply the quantum Fourier transform, which will result in a quantum system where we know that the likelihood of measuring a $k$ satisfying (5.1) is at least $4 / \pi^2$. We then measure such a $k$ and compute a rational approximation which gives us the period of the function $f$, which allows us to factor $n$.
We begin with a two entangled quantum registers, one with $\log 2 N$ qubits and the other with $\left\lceil\log _2 n\right\rceil$ qubits. The first register should contain a superposition of the integers $0,1,2, \ldots, N-1$, all with the same amplitude, while the second register should contain 0 . We have the state $$\sum{j=0}^{N-1} \frac{1}{\sqrt{N}}|j\rangle|0\rangle .$$

Using a quantum circuit, we compute $f$ on the first register, storing the result in the second register. Since any classical computation can be done on a quantum computer, computing $f$ is easy. Our two quantum registers now contain a superposition of $(k, f(k))$ for $k=0,1, \ldots, N-1$, all being equally likely, and we have the new state
$$\sum_{j=0}^{N-1} \frac{1}{\sqrt{N}}|j\rangle|f(j)\rangle .$$
Next, we measure the second register. Suppose we measure the value $s$. Since we have not yet measured the first register, we do not know what it contains, but since it was entangled with the second register, every value we could possible measure must be consistent with $s$. Suppose $t_0$ is the smallest non-negative integer such that $f\left(t_0\right)=s$. Then the only values we can measure in the first register are the integers $t_0+j r$.

Since we had a uniform probability before we measured the second register, we must have a uniform probability after measuring. We now have $m$ possible states, so we have the state
$$\sum_{j=0}^{m-1} \frac{1}{\sqrt{m}}\left|t_0+j r\right\rangle|s\rangle .$$
In other words, if we ignore the second register which is constant, our first register now has exactly the amplitudes given by (5.2).

We then use a second quantum circuit to compute the quantum Fourier transform on the first register. And then we measure the first register. As discussed, we will with significant probability measure a $k$ satisfying (5.1), which will allow us to factor $n$.

## 数学代写|密码学代写cryptography theory代考|STATISTICAL DISTANCE

We shall sometimes need to consider sampling algorithms that are slightly different. The answer to the Decision Diffie-Hellman problem is not unique, and sometimes it would be convenient if it was.

Example 6.7. The $\mathrm{DDH}{G, g}^{\prime}$ is identical to the $\mathrm{DDH}{G, g}$ problem, except that the sampling algorithm samples $z_1 \longleftarrow G \backslash\left{z_0\right}$, not $z_1 \longleftarrow r G$.
We need tools to reason about such small differences.
Definition 6.1. Let $\mathcal{X}0, \mathcal{X}_1$ be probability spaces on a finite set $S$. The statistical distance between $\mathcal{X}_0$ and $\mathcal{X}_1$ is $$\Delta\left(\mathcal{X}_0, \mathcal{X}_1\right)=\frac{1}{2} \sum{s \in S}\left|\operatorname{Pr}\left[\mathcal{X}0=s\right]-\operatorname{Pr}\left[\mathcal{X}_1=s\right]\right| .$$ Two probability spaces are $\epsilon$-close if the statistical distance is at most $\epsilon$. Exercise 6.8. Compute the statistical distance between the instance sampling algorithm for $\mathrm{DDH}{G, g}$ and $\mathrm{DDH}_{G, g}^{\prime}$.

Exercise 6.9. Fix a finite set $S$. Show that the statistical distance $\Delta$ is a metric on the set of probability spaces on $S$.

If we consider a fixed algorithm and give it input sampled from two different probability spaces, the statistical distance between the outputs will be bounded by the statistical distance of the inputs.

Exercise 6.10. Let $\mathcal{A}$ be an algorithm that takes as input elements of a finite set $S$ and outputs elements of a finite set $T$, and let $\mathcal{X}_0, \mathcal{X}_1$ be probability spaces on $S$. Define $\mathcal{Y}_i$ to be the probability space on $T$ induced by sampling $s \leftarrow \mathcal{X}_i$ and computing $u \leftarrow \mathcal{A}(s)$. Show that
$$\Delta\left(\mathcal{Y}_0, \mathcal{Y}_1\right) \leq \Delta\left(\mathcal{X}_0, \mathcal{X}_1\right)$$
Two probability spaces are statistically indistinguishable if their statistical distance is small. The origin of the phrase lies in the following result.

Proposition 6.1. Let $\mathcal{X}0, \mathcal{X}_1$ be probability spaces on a finite set $S$, and let $\mathcal{A}$ be a distinguisher for $\mathcal{X}_0$ and $\mathcal{X}_1$. Then $$\operatorname{Adv}{\mathcal{X}_0, \mathcal{X}_1}^{\text {dist }}(\mathcal{A}) \leq 2 \Delta\left(\mathcal{X}_0, \mathcal{X}_1\right)$$

# 密码学代考

## 数学代写|密码学代写cryptography theory代考|FACTORING USING A QUANTUM COMPUTER

$$\sum_{j=0}^{m-1} \frac{1}{\sqrt{m}}\left|t_0+j r\right\rangle|s\rangle .$$

## 数学代写|密码学代写cryptography theory代考|STATISTICAL DISTANCE

$$\Delta\left(\mathcal{Y}_0, \mathcal{Y}_1\right) \leq \Delta\left(\mathcal{X}_0, \mathcal{X}_1\right)$$

$$\operatorname{Adv} \mathcal{X}_0, \mathcal{X}_1{ }^{\text {dist }}(\mathcal{A}) \leq 2 \Delta\left(\mathcal{X}_0, \mathcal{X}_1\right)$$

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