数学代写|CS6260 Cryptography
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CS6260 Cryptography课程简介
Cryptographic schemes are used to protect sensitive information and to ensure the integrity and authenticity of communications. Here are some of the most commonly used cryptographic schemes and their applications:
- AES (Advanced Encryption Standard) – AES is a symmetric encryption algorithm used to protect data in transit or at rest. It is widely used in applications such as file encryption, email encryption, and VPNs. AES works by transforming plaintext into ciphertext using a secret key, which is used to encrypt and decrypt the data.
- CBC (Cipher Block Chaining) – CBC is a mode of operation for block ciphers, such as AES. CBC works by breaking the plaintext into blocks and encrypting each block using the previous block’s ciphertext as an input. This ensures that the same plaintext block does not always result in the same ciphertext block, making the encryption more secure.
PREREQUISITES
No previous knowledge of cryptography is necessary. This course is about applying theory to practical problems, but it is still a theory course. The main requirement is basic “mathematical maturity”. You have to be able to read and write mathematical definitions, statements and proofs.
It is expected that you were successful in your undergraduate discrete math class and took basic algorithms and computability/complexity theory classes. In particular, you have to know how to measure the running time of an algorithm and how to do proofs by contradiction and contraposition. You also have to know the basics of probability theory and modular arithmetic. You should also have familiarity with Python for the coding portions of the course.
If you cannot recall what terms like permutation, sample space, random variable, conditional probability, big-O notation mean, you should consider taking the course in a later semester and refresh your knowledge of the above topics in the meanwhile. I recommend you review an undergraduate textbook on discrete math.
All necessary elements of number theory will be presented during the course.
CS6260 Cryptography HELP（EXAM HELP， ONLINE TUTOR）
Exercise 5.1. Let $\omega=\exp (2 \pi i / N) \in \mathbb{C}$, which is a primitive $N$ th root of unity. Let $\mathbf{V}=\left(v_{k j}\right)$ be the Vandermonde matrix with $v_{k j}=\omega^{k j}, 0 \leq k, j<N$.
(a) Show that the $\overline{\omega^j}=\omega^{N-j}$, where $\bar{z}$ denotes complex conjugation.
(b) Show that $\sum_{j=0}^{N-1}\left(\omega^k\right)^j=0$ for any $k \not 00(\bmod N)$.
(c) Show that $\mathbf{V} \mathbf{V}^=N I$, where $\mathbf{V}^$ denotes the conjugate transpose of $\mathbf{V}$.
(d) Show that when $\beta$ is the discrete Fourier transform of $\alpha$, then $\beta=$ $\frac{1}{\sqrt{N}} \mathbf{V} \boldsymbol{\alpha}$.
(a) We have $\overline{\omega^j}=\overline{\cos(2\pi j/N)+i\sin(2\pi j/N)}=\cos(2\pi j/N)-i\sin(2\pi j/N)=\omega^{N-j}$, where we used the fact that $\cos$ is an even function and $\sin$ is an odd function.
(b) When $k$ is not divisible by $N$, we have $\sum_{j=0}^{N-1}\left(\omega^k\right)^j=\frac{1-\left(\omega^k\right)^N}{1-\omega^k}=0$, where we used the fact that $\omega^N=1$ and $\omega^k\neq 1$ because $k$ is not divisible by $N$.
(c) We have $(\mathbf{V}\mathbf{V}^*){k,l}=\sum{j=0}^{N-1}v_{k,j}\overline{v_{l,j}}=\sum_{j=0}^{N-1}\omega^{kj}\omega^{-lj}=\sum_{j=0}^{N-1}\omega^{j(k-l)}$. When $k=l$, this is equal to $N$, because every term is equal to $1$. When $k\neq l$, we have $\sum_{j=0}^{N-1}\omega^{j(k-l)}=\frac{1-\omega^{N(k-l)}}{1-\omega^{k-l}}=0$, because $k-l$ is not divisible by $N$.
(d) Let $\beta$ be the discrete Fourier transform of $\alpha$, i.e., $\beta_k=\sum_{j=0}^{N-1}\alpha_j\omega^{-kj}$. Then, we have \begin{align*} \frac{1}{\sqrt{N}}\mathbf{V}\boldsymbol{\alpha}k &= \frac{1}{\sqrt{N}}\sum{j=0}^{N-1}\omega^{kj}\alpha_j \ &= \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\omega^{-jk}\alpha_j \ &= \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}\beta_j\omega^{jk} \ &= \beta_k, \end{align*} where we used the fact that $\omega^{-k}=\omega^{N-k}$ in the second step, and the fact that $\mathbf{V}$ is unitary in the third step. Therefore, we have $\beta=\frac{1}{\sqrt{N}}\mathbf{V}\boldsymbol{\alpha}$, as claimed.
Exercise 5.2. Let $\mathrm{U}$ be an $N \times N$ unitary matrix. Show that for any vector $\boldsymbol{\alpha} \in \mathbb{C}^N,|\mathbf{U} \boldsymbol{\alpha}|=|\boldsymbol{\alpha}|$
We want to study the Fourier coefficients of a very special complex vector. Let $t_0$ be an integer such that $0 \leq t_0<r$, and let $m$ be minimal such that $m r+t_0 \geq N$. Let $\alpha \in \mathbb{C}^N$ be given by
$$
\alpha_k= \begin{cases}\frac{1}{\sqrt{m}} & k=t_0+j r, \text { and } \ 0 & \text { otherwise. }\end{cases}
$$
Note that $|\boldsymbol{\alpha}|=1$ and that $1-m r / N=1-\left(N-t_0\right) / N \leq r / N \leq 1 / r$.
Let $\mathbf{U}$ be an $N \times N$ unitary matrix and $\boldsymbol{\alpha} \in \mathbb{C}^N$. We want to show that $|\mathbf{U} \boldsymbol{\alpha}|=|\boldsymbol{\alpha}|$.
Let $\boldsymbol{\beta}=\mathbf{U} \boldsymbol{\alpha}$, then we have: \begin{align*} |\boldsymbol{\beta}|^2 &=\boldsymbol{\beta}^* \boldsymbol{\beta} \ &=\left(\mathbf{U} \boldsymbol{\alpha}\right)^\left(\mathbf{U} \boldsymbol{\alpha}\right) \ &=\boldsymbol{\alpha}^ \mathbf{U}^* \mathbf{U} \boldsymbol{\alpha} \ &=\boldsymbol{\alpha}^* \boldsymbol{\alpha} && (\because \mathbf{U} \text{ is unitary, so } \mathbf{U}^* \mathbf{U} = \mathbf{I}) \ &=|\boldsymbol{\alpha}|^2. \end{align*}
Therefore, we have shown that $|\mathbf{U} \boldsymbol{\alpha}|=|\boldsymbol{\alpha}|$.
Textbooks
• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.
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