Hykilpikonna
281 lines
11 KiB
TeX
281 lines
11 KiB
TeX
\documentclass[fontsize=11pt]{article}
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\usepackage{amsfonts}
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\usepackage{amsmath}
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\usepackage{amsthm}
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\usepackage[utf8]{inputenc}
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\usepackage[margin=0.75in]{geometry}
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\title{CSC110 Assignment 4: Number Theory, Cryptography, and Algorithm Running Time}
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\author{Azalea Gui \& Peter Lin}
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\date{\today}
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% Some useful LaTeX commands. You are free to use these or not, and also add your own.
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\newcommand{\N}{\mathbb{N}}
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\newcommand{\Z}{\mathbb{Z}}
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\cO}{\mathcal{O}}
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\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}
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\newcommand{\code}[1]{\texttt{#1}}
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\begin{document}
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\maketitle
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\section*{Part 1: Practice with Proofs}
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\begin{enumerate}
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\item[1.] Statement to prove:
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$$\forall a, k, n \in \Z,~ \gcd(a, n) = 1 \Rightarrow \gcd(a + kn, n) = 1$$
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\begin{proof} : \\
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$d = \gcd(a,n)$ is defined as $(a=0 \land n=0 \implies d=0) \land \\ (a \neq 0 \lor n \neq 0 \implies d | a \land d | n \land (\forall e \in \N, e | a \land e | n \implies e \le d))$ \\
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\\
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Let $a,k,n \in \Z$ \\
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Assume that $\gcd(a,n) = 1$ \\
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Since $d = 1 \neq 0$, this implies that $a \neq 0 \lor n \neq 0$ \\
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Since $a \neq 0 \lor n \neq 0$, we know $1 | a \land 1 | n \land (\forall e_2 \in \N, e_2 | a \land e_2 | n \implies e_2 \le 1)$ is also true. \\
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We need to prove: $(a + kn) \neq 0 \lor n \neq 0 \implies 1 | (a + kn) \land 1 | n \land (\forall e \in \N, e | (a + kn) \land e | n \implies e \le 1)$ \\
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\\
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Suppose $(a + kn) \neq 0 \lor n \neq 0$ \\
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We need to prove: $1 | (a + kn) \land 1 | n \land (\forall e \in \N, e | (a + kn) \land e | n \implies e \le 1)$
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\begin{enumerate}
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\item[1.] Proving for: $1 | (a + kn)$ \\
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That is: $\exists c \in \Z$ s.t. $(a + kn) = 1 \cdot c$ \\
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Take $c = (a + kn)$ \\
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$(a + kn) = 1 \cdot (a + kn)$ is true.
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\item[2.] $1 | n$ is given to be true.
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\item[3.] Proving for $\forall e \in \N, e | (a + kn) \land e | n \implies e \le 1$ \\
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Let $e \in \N$ \\
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Suppose $e | (a + kn) \land e | n$ \\
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$a + kn = ex \land n = ey$ for some $x,y \in \Z$ \\
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$a + key = ex$ for some $x,y \in \Z$ \\
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$a = e(x - ky)$ for some $x,y \in \Z$ \\
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Let $c = (x - ky)$ \\
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By substitution, we now have: $a = ec$ \\
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Therefore, $\exists c \in \Z$ s.t. $a = ec$ is true. \\
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Which means $e | a$ is true. \\
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Since we are given $\forall e_2 \in \N, e_2 | a \land e_2 | n \implies e_2 \le 1$ \\
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Take $e_2 = e$, we now have $e | a \land e | n \implies e \le 1$ \\
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And since we now know $e | a$ and $e | n$, we can conclude $e \le 1$. \\
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Which is what we want to show.
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\end{enumerate}
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\end{proof}
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\item[2.] Statement to prove (we've expanded the definition of Omega for you!):
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$$\exists c, n_0 \in \R^+,~ \forall n \in \N,~ n \geq n_0 \Rightarrow \log_{3} n - \log_{11} n \geq c \cdot \log_{14} n$$
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\begin{proof}
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TODO: Your proof goes here.
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\end{proof}
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\item[3.] Statement to prove (we haven't expanded the definition of Big-O for you, but we encourage you to do so yourself):
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$$\forall f, g: \N \to \R^{\geq 0},~ g \in \cO(f) \land \big(\forall m \in \N,~ f(m) \geq 1 \big) \Rightarrow g \in \cO(\floor{f})$$
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\begin{proof}
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TODO: Your proof goes here.
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\end{proof}
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\end{enumerate}
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\newpage
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\section*{Part 2: Generating Coprime Numbers}
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\begin{enumerate}
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\item[1.]
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Not to be handed in.
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\item[2.]
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Complete this part in the provided \texttt{a4\_part2.py} starter file.
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Do \textbf{not} include your solution in this file.
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\item[3.]
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Prove that each loop invariant holds.
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\begin{enumerate}
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\item[a.] Loop Invariant 1
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\begin{proof} : \\
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Variables: In this proof, $N$ is the abbreviation for the list \texttt{nums\_so\_far}. \\
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\\
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Assumption 1: The loop invariant 1 is true for the previous iteration. \\
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That is $\forall k_2 \in N, gcd(k_2, 2) = 1 \land gcd(k_2, 3) = 1$ \\
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\\
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Assumption 2: The statement proven in Part 1.1: \\
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$\forall a,k,n \in \Z, gcd(a,n) = 1 \implies gcd(a + kn, n) = 1$ \\
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\\
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We need to prove: $\forall k \in N \cup \{ N[-2] + 6 \}, gcd(k, 2) = 1 \land gcd(k, 3) = 1$ \\
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Which is equivalent to: $\forall k \in N, gcd(k, 2) = 1 \land gcd(k, 3) = 1$ and \\
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$gcd(N[-2] + 6, 2) = 1 \land gcd(N[-2] + 6, 3) = 1$ \\
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\\
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Since the first part is the same as the previous iteration, it is true. \\
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What we need to prove becomes:
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$gcd(N[-2] + 6, 2) = 1 \land gcd(N[-2] + 6, 3) = 1$ \\
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\\
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Pick $k_2 = N[-2] \in N$ \\
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By assumption 1, we know that $gcd(N[-2], 2) = 1$ and $gcd(N[-2], 3) = 1$
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\begin{enumerate}
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\item[1.] Proving for $gcd(N[-2] + 6, 2) = 1$ \\
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Pick $a = N[-2], k = 3, n = 2$ \\
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Since we know $gcd(N[-2], 2) = 1$, $gcd(a, n) = 1$ is true. \\
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Therefore, by assumption 2, we know that $gcd(a + kn, n) = 1$ is also true. \\
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Substituting the varaibles back, we know $gcd(N[-2] + 6, 2) = 1$.
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\item[2.] Proving for $gcd(N[-2] + 6, 3) = 1$ \\
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Pick $a = N[-2], k = 2, n = 3$ \\
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Since we know $gcd(N[-2], 3) = 1$, $gcd(a, n) = 1$ is true. \\
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Therefore, by assumption 2, we know that $gcd(a + kn, n) = 1$ is also true. \\
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Substituting the varaibles back, we know $gcd(N[-2] + 6, 3) = 1$.
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\end{enumerate}
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\end{proof}
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\item[b.] Loop Invariant 2
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\begin{proof} : \\
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Variables: In this proof, $N$ is the abbreviation for the list \texttt{nums\_so\_far}, and $|N|$ represents the size of $N$. \\
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\\
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Assumption 1: The loop invariant 2 is true for the previous iteration. \\
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That is $\forall i_2 \in \{ 0, \dots, |N| - 3\}, N[i_2] + 6 = N[i_2 + 2]$ \\
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\\
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Let $M = N \cup \{ N[-2] + 6 \}$ be the list of the current iteration. \\
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We need to prove: $\forall i \in \{0, \dots, |M| - 3\}, M[i] + 6 = M[i + 2] $
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\begin{enumerate}
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\item[1.] Let $i < |M| - 3$ \\
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Since the new entry added to $M$ is not included in $i$, this case is equivalent to the previous iteration, and we know that is true by assumption 1.
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\item[2.] Let $i = |M| - 3$ \\
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We need to prove: $M[|M| - 3] + 6 = M[|M| - 3 + 2]$ \\
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That is $M[-3] + 6 = M[-1]$ \\
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\\
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Let's start with a true statement: \\
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$N[-2] = N[-2]$ \\
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Since $N[-2] = (N \cup \{ N[-2] + 6 \})[-3] = M[-3]$, \\
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$M[-3] = N[-2]$ \\
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Add 6 to both sides: \\
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$M[-3] + 6 = N[-2] + 6$ \\
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Since $M[-1] = (N \cup \{ N[-2] + 6 \})[-1] = N[-2] + 6$ is it's last entry, \\
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$M[-3] + 6 = M[-1]$ \\
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Which is what we want to show.
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\end{enumerate}
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\end{proof}
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\item[c.] Loop Invariant 3
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\begin{proof} : \\
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Variables: In this proof, $N$ is the abbreviation for the list \texttt{nums\_so\_far}, and $|N|$ represents the size of $N$. \\
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\\
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Assumption 1: Loop invariant 3 is true for the previous iteration. \\
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That is $\forall i_2 \in \{0, \dots, |N| - 2\}, N[i_2] < N[i_2 + 1]$ \\
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\\
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Assumption 2: Loop invariant 2 is true for the previous iteration. \\
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That is $\forall i_3 \in \{0, \dots, |N| - 3\}, N[i_3] + 6 = N[i_3 + 2]$ \\
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\\
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Let $M = N \cup \{ N[-2] + 6 \}$ be the list of the current iteration. \\
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We need to prove: $\forall i \in \{0, \dots, |M| - 2\}, M[i] < M[i + 1] $ \\
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\\
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Assumption 3: Loop invariant 2 is true for the current iteration. \\
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That is $\forall i_4 \in \{0, \dots, |M| - 3\}, M[i_4] + 6 = M[i_4 + 2]$ \\
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\\
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Let's first prove an intermediate statment, statment 4: \\
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$\forall i_5 \in \{0, \dots, |N| - 2\}, 0 < N[i_5 + 1] - N[i_5] < 6$ \\
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Let $i_5 \in \{0, \dots, |N| - 2\}$, \\
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We want to show $0 < N[i_5 + 1] - N[i_5] < 6$ \\
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\\
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Pick $i_2 = i_5$, \\
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We know that $N[i_5] < N[i_5 + 1]$ by assumption 1. \\
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Which means $0 < N[i_5 + 1] - N[i_5]$ \\
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\\
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Then, we need to prove $N[i_5 + 1] - N[i_5] < 6$ \\
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Since $N = \text{list}[1, 5]$ before the first iteration, the base case $N[1] < N[0] + 6$ is true. \\
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For the inductive step, let's look at the true statement again: \\
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$N[i_5] < N[i_5 + 1]$ \\
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Add 6 to both sides: \\
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$N[i_5] + 6 < N[i_5 + 1] + 6$ \\
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Pick $i_3 = i_5$ \\
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We know that $N[i_5] + 6 = N[i_5 + 2]$ by asssumption 2. \\
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By substitution, our true statement becomes: \\
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$N[i_5 + 2] < N[i_5 + 1] + 6$ \\
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$N[(i_5 + 1) + 1] < N[i_5 + 1] + 6$ \\
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Which is the end of our induction. \\
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\\
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We have proven the intermediate statement 4. \\
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We now need to prove: $\forall i \in \{0, \dots, |M| - 2\}, M[i] < M[i + 1] $
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\begin{enumerate}
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\item[1.] Let $i < |M| - 2$ \\
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Since the new entry added to $M$ is not included in $i$, this case is equivalent to the previous iteration, and we know that is true by assumption 1.
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\item[2.] Let $i = |M| - 2$ \\
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We need to prove: $M[|M| - 2] < M[|M| - 2 + 1]$ \\
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That is $M[-2] < M[-1]$ \\
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\\
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Pick $i_5 = |N| - 2$ \\
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We know that $0 < N[-1] - N[-2] < 6$ by statement 4 \\
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Since $N[-1] = (N \cup \{ N[-2] + 6 \})[-2] = M[-2]$, \\
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Since $N[-2] = (N \cup \{ N[-2] + 6 \})[-3] = M[-3]$, \\
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$0 < M[-2] - M[-3] < 6$ \\
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$M[-2] < M[-3] + 6$ \\
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Pick $i_4 = |M| - 3$ \\
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We know that $M[-3] + 6 = M[-1]$ by assumption 3 \\
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By substitution, we now have: \\
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$M[-2] < M[-1]$
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Which is what we want to show.
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\end{enumerate}
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\end{proof}
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\item[d.] Loop Invariant 4
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\begin{proof}
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TODO: Your proof goes here.
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\end{proof}
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\end{enumerate}
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\item[4.]
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Complete this part in the provided \texttt{a4\_part2.py} starter file.
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Do \textbf{not} include your solution in this file.
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\item[5.]
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Complete this part in the provided \texttt{a4\_part2.py} starter file.
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Do \textbf{not} include your solution in this file.
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\end{enumerate}
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\newpage
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\section*{Part 3: Running-Time Analysis}
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\begin{enumerate}
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\item[1.]
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Running-time analysis of \texttt{coprime\_to\_2\_and\_3}.
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Let $N$ be the list \code{nums\_so\_far}, and let $n$ be the input integer.
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The loop in the function runs until $N[-2] + 6 < n$. Since $N = \text{list}[1, 5]$ before the first iteration, the loop will run $\text{(amount of coprimes below n)} - 2$ times. Since the loop appends $N[-2] + 6$ to $N$, $N[-2]$ will increase for the next iteration by either $5 - 1 = 4$ or $1 + 6 - 5 = 2$ in an alternating way, which is an increase of $3$ on average. This loop will run $(n - 6) / 3 = n / 3 - 2$ times on average. However, since $\cO(n / 3 - 2) = \cO(n)$, we can be conservative here and say that the loop will run at most $n$ times.
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Since the loop contained only constant-time operations, the runtime of the loop is $\cO(n)$.
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Since there are only constant-time operations outside the loop, the runtime of the entire function is $\cO(n)$.
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\item[2.]
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Running-time analysis of \texttt{starting\_coprime\_numbers}.
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\item[3.]
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TODO: Running-time analysis of \texttt{coprime\_to\_all}.
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\end{enumerate}
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\section*{Part 4: Two New Cryptosystems}
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Complete this part in the provided \texttt{a4\_part4.py} starter file.
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Do \textbf{not} include your solution in this file.
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\end{document}
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