diff --git a/pages/CSE347/CSE347_L5.md b/pages/CSE347/CSE347_L5.md index f8bc5aa..7522f95 100644 --- a/pages/CSE347/CSE347_L5.md +++ b/pages/CSE347/CSE347_L5.md @@ -6,7 +6,7 @@ - In general, we can design an algorithm to map instances of a new problem to instances of known solvable problem (e.g., max-flow) to solve this new problem! - Mapping from one problem to another which preserves solutions is called reduction. -## Reduction: Basic Idea +## Reduction: Basic Ideas Convert solutions to the known problem to the solutions to the new problem diff --git a/pages/CSE347/CSE347_L7.md b/pages/CSE347/CSE347_L7.md index e7538a4..c4781f7 100644 --- a/pages/CSE347/CSE347_L7.md +++ b/pages/CSE347/CSE347_L7.md @@ -45,7 +45,7 @@ Assumption: No clause contains both a literal and its negation. Need to: construct $S$ of positive numbers and a target $t$ -Idea of construction: +Ideas of construction: For 3-SAT instance $\Psi$: @@ -276,7 +276,7 @@ Consider an instance of SSS: $\{ a_1,a_2,\cdots,a_n\}$ and sum $b$. We can creat Then we prove that the scheduling instance is a "yes" instance if and only if the SSS instance is a "yes" instance. -Idea of proof: +Ideas of proof: If there is a subset of $\{a_1,a_2,\cdots,a_n\}$ that sums to $b$, then we can schedule the jobs in that order on one machine. diff --git a/pages/CSE347/CSE347_L9.md b/pages/CSE347/CSE347_L9.md index 3f5b47a..f643ea2 100644 --- a/pages/CSE347/CSE347_L9.md +++ b/pages/CSE347/CSE347_L9.md @@ -38,7 +38,7 @@ Answer: The adversary can make the runtime of each operation $\Theta(n)$ by simp We don't want the adversary to know the hash function based on just looking at the code. -Idea: Randomize the choice of the hash function. +Ideas: Randomize the choice of the hash function. ### Randomized Algorithm @@ -57,7 +57,7 @@ $$O(n)=E[T(n)]$$ or some other probabilistic quantity. #### Randomization can help -Idea: Randomize the choice of hash function $h$ from a family of hash functions, $H$. +Ideas: Randomize the choice of hash function $h$ from a family of hash functions, $H$. If we randomly pick a hash function from this family, then the probability that the hash function is bad on **any particular** set $S$ is small. diff --git a/pages/CSE442T/CSE442T_L12.md b/pages/CSE442T/CSE442T_L12.md index 8eaeb07..9b8f374 100644 --- a/pages/CSE442T/CSE442T_L12.md +++ b/pages/CSE442T/CSE442T_L12.md @@ -82,7 +82,7 @@ The NBT(Next bit test) is complete. If $\{X_n\}$ on $\{0,1\}^{l(n)}$ passes NBT, then it's pseudorandom. -Idea of proof: full proof is on the text. +Ideas of proof: full proof is on the text. Our idea is that we want to create $H^{l(n)}_n=\{X_n\}$ and $H^0_n=\{U_{l(n)}\}$ @@ -137,7 +137,7 @@ The other part of proof will be your homework, damn. If one-way function exists, then Pseudorandom Generator exists. -Idea of proof: +Ideas of proof: Let $f:\{0,1\}^n\to \{0,1\}^n$ be a strong one-way permutation (bijection). diff --git a/pages/CSE442T/CSE442T_L13.md b/pages/CSE442T/CSE442T_L13.md index fd3b763..1910216 100644 --- a/pages/CSE442T/CSE442T_L13.md +++ b/pages/CSE442T/CSE442T_L13.md @@ -16,7 +16,7 @@ $$ Pr[x\gets \{0,1\}^n;y=f(x);A(1^n,y)=h(x)]\leq \frac{1}{2}+\epsilon(n) $$ -Idea: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function. +Ideas: $f:\{0,1\}^n\to \{0,1\}^*$ is a one-way function. Given $y=f(x)$, it is hard to recover $x$. A cannot produce all of $x$ but can know some bits of $x$. @@ -46,7 +46,7 @@ $\langle x,1^n\rangle=x_1+x_2+\cdots +x_n\mod 2$ $\langle x,0^{n-1}1\rangle=x_ n$ -Idea of proof: +Ideas of proof: If A could reliably find $\langle x,1^n\rangle$, with $r$ being completely random, then it could find $x$ too often. diff --git a/pages/CSE442T/CSE442T_L14.md b/pages/CSE442T/CSE442T_L14.md index b3289d7..275f45a 100644 --- a/pages/CSE442T/CSE442T_L14.md +++ b/pages/CSE442T/CSE442T_L14.md @@ -123,7 +123,7 @@ $Enc_F(m):$ let $r\gets U_n$; output $(r,F(r)\oplus m)$. $Dec_F(m):$ Given $(r,c)$, output $m=F(r)\oplus c$. -Idea: Adversary sees $r$ but has no idea about $F(r)$. (we choose all outputs at random) +Ideas: Adversary sees $r$ but has no Ideas about $F(r)$. (we choose all outputs at random) If we could do this, this is MMS (multi-message secure). diff --git a/pages/CSE442T/CSE442T_L16.md b/pages/CSE442T/CSE442T_L16.md index 9191c7a..a07fc93 100644 --- a/pages/CSE442T/CSE442T_L16.md +++ b/pages/CSE442T/CSE442T_L16.md @@ -77,7 +77,7 @@ With $g^a,g^b$ no one can compute $g^{ab}$. ### Public key encryption scheme -Idea: The recipient Bob distributes opened Bob-locks +Ideas: The recipient Bob distributes opened Bob-locks - Once closed, only Bob can open it. diff --git a/pages/CSE442T/CSE442T_L17.md b/pages/CSE442T/CSE442T_L17.md index 8f5d244..21005db 100644 --- a/pages/CSE442T/CSE442T_L17.md +++ b/pages/CSE442T/CSE442T_L17.md @@ -110,7 +110,7 @@ $\{p\gets \tilde{\Pi_n};y\gets Gen_q;a,b,\bold{z}\gets \mathbb{Z}_q:(p,y,y^a,y^b So DDH assumption implies discrete logarithm assumption. -Idea: +Ideas: If one can find $a,b$ from $y^a,y^b$, then one can find $ab$ from $y^{ab}$ and compare to $\bold{z}$ to check whether $y^\bold{z}$ is a valid DDH tuple. diff --git a/pages/CSE442T/CSE442T_L19.md b/pages/CSE442T/CSE442T_L19.md index 35a24a2..61adc3c 100644 --- a/pages/CSE442T/CSE442T_L19.md +++ b/pages/CSE442T/CSE442T_L19.md @@ -28,7 +28,7 @@ This is not more than one-time secure since the adversary can ask oracle for $Si We will show it is one-time secure -Idea of proof: +Ideas of proof: Say their query is $Sign_{sk}(0^n)$ and reveals $pk_0$. diff --git a/pages/CSE442T/CSE442T_L20.md b/pages/CSE442T/CSE442T_L20.md index 06780cc..3c10868 100644 --- a/pages/CSE442T/CSE442T_L20.md +++ b/pages/CSE442T/CSE442T_L20.md @@ -104,7 +104,7 @@ One-time secure: Then ($Gen',Sign',Ver'$) is one-time secure. -Idea of Proof: +Ideas of Proof: If the digital signature scheme ($Gen',Sign',Ver'$) is not one-time secure, then there exists an adversary $\mathcal{A}$ which can ask oracle for one signature on $m_1$ and receive $\sigma_1=Sign'_{sk'}(m_1)=Sign_{sk}(h_i(m_1))$. diff --git a/pages/Math4111/Exam_reviews/Math4111_E3.md b/pages/Math4111/Exam_reviews/Math4111_E3.md index 6f27cf5..d6174dc 100644 --- a/pages/Math4111/Exam_reviews/Math4111_E3.md +++ b/pages/Math4111/Exam_reviews/Math4111_E3.md @@ -2,9 +2,9 @@ ## Relations between series and topology (compactness, closure, etc.) -Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\emptyset\}$ +Limit points $E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\backslash\{x\}\cap E\neq\phi\}$ -Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\emptyset\}$ +Closure $\overline{E}=E\cup E'=\{x\in\mathbb{R}:\forall r>0, B_r(x)\cap E\neq\phi\}$ $p_n\to p\implies \forall \epsilon>0, \exists N$ such that $\forall n\geq N, p_n\in B_\epsilon(p)$ @@ -24,7 +24,7 @@ Rudin Proof: Rudin's proof uses a fact from Chapter 2. -If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\emptyset$). +If $E$ is compact, and $S\subseteq E$ is infinite, then $S$ has a limit point in $E$ ($S'\cap E\neq\phi$). ## Examples of Cauchy sequence that does not converge diff --git a/pages/Math4111/Math4111_L12.md b/pages/Math4111/Math4111_L12.md index 31bca58..37dfda6 100644 --- a/pages/Math4111/Math4111_L12.md +++ b/pages/Math4111/Math4111_L12.md @@ -30,7 +30,7 @@ A 2-cell is a set of the form $[a_1,b_1]\times[a_2,b_2]$ Theorem 2.38 replace with "closed and bounded intervals" to "k-cells". -Idea of Proof: +Ideas of Proof: Apply the Theorem to each dimension separately. diff --git a/pages/Math4111/Math4111_L16.md b/pages/Math4111/Math4111_L16.md index 654251d..0629f80 100644 --- a/pages/Math4111/Math4111_L16.md +++ b/pages/Math4111/Math4111_L16.md @@ -146,7 +146,7 @@ This proves the claim. By definition of supremum, the claim implies that $\forall \epsilon>0$, $diam(\overline{E})\leq 2\epsilon+diam E$. So $diam(\overline{E})\leq diam E$. -(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \emptyset$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0. +(b) By **Theorem 2.36**, $\bigcap_{n=1}^{\infty}K_n\neq \phi$. Suppose for contradiction that there are at least two distinct points $p,q\in \bigcap_{n=1}^{\infty}K_n$. Then for all $n\in \mathbb{N}$, $x,y\in K_n$ so $diam K_n\geq d(p,q)>0$. Then diameter of $K_n$ does not converge to 0. EOP diff --git a/pages/Math4111/Math4111_L20.md b/pages/Math4111/Math4111_L20.md index 9f7fb30..6a9ebfa 100644 --- a/pages/Math4111/Math4111_L20.md +++ b/pages/Math4111/Math4111_L20.md @@ -78,7 +78,7 @@ So if $\limsup_{n\to\infty} t_n \leq \limsup_{n\to\infty} s_n$, then $\lim_{n\to Now we will show $\limsup_{n\to\infty} t_n \geq e$. -Idea: (special case of the argument) +Ideas: (special case of the argument) If $n\geq 2$, then diff --git a/pages/Math4111/Math4111_L21.md b/pages/Math4111/Math4111_L21.md index ca5b396..25f2f3e 100644 --- a/pages/Math4111/Math4111_L21.md +++ b/pages/Math4111/Math4111_L21.md @@ -230,7 +230,7 @@ $$ \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty a_{f(n)} $$ -Idea of proof: +Ideas of proof: Let $f:\mathbb{N}\to \mathbb{N}$ be a bijection. diff --git a/pages/Math4111/Math4111_L24.md b/pages/Math4111/Math4111_L24.md index f92a2b5..3f66ca7 100644 --- a/pages/Math4111/Math4111_L24.md +++ b/pages/Math4111/Math4111_L24.md @@ -1 +1,188 @@ -# Lecture 24 \ No newline at end of file +# Lecture 24 + +## Reviews + +Let $f: X\to Y$. Consider the following statement: + +"$f$ is continuous $\iff$ for every open set $V\in Y$, $f^{-1}(V)$ is open in $X$." + +1. To give a direct proof of the $\implies$ direction, what must be the first few steps be? + +2. To give a direct proof of the $\impliedby$ direction, what must be the first few steps be? +3. Try to complete the proofs of both directions. + +> A function $f:X\to Y$ is continuous if $\forall p\in X$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))$. (_For every point in a ball of $B_\delta(p)$, there is a ball of $B_\epsilon(f(p))$ that contains the image of the point._) +> +> A set $V\subset Y$ is open if $\forall q\in V$, $\exists r>0$ such that $B_r(q)\subset V$. + +## New materials + +### Continuity and open sets + +#### Theorem 4.8 + +A function $f:X\to Y$ is continuous if and only if for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$. + +Proof: + +$\implies$: Suppose $f$ is continuous. Let $V\subset Y$ be open. Let $p\in f^{-1}(V)$. Since $f(p)\in V$, $\exists \epsilon > 0$ such that $B_\epsilon(f(p))\subset V$. + +Since $f$ is continuous, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))\subset V$. Therefore, $B_\delta(p)\subset f^{-1}(V)$. This shows that $f^{-1}(V)$ is open. + +$\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$. Let $p\in X$ and $\epsilon > 0$. Let $B_\epsilon(f(p))\in V$. Then $f^{-1}(B_\epsilon(f(p)))$ is open in $X$. + +Since $p\in f^{-1}(B_\epsilon(f(p)))$ and $f^{-1}(B_\epsilon(f(p)))$ is open, $\exists \delta > 0$ such that $B_\delta(p)\subset f^{-1}(B_\epsilon(f(p)))$. Therefore, $f(B_\delta(p))\subset B_\epsilon(f(p))$. This shows that $f$ is continuous. + +EOP + +#### Corollary 4.8 + +$f$ is continuous if and only if for every closed set $C\subset Y$, $f^{-1}(C)$ is closed in $X$. + +Ideas of proof: + +- $C$ closed in $Y\iff Y\backslash C$ open in $Y$ +- $f^{-1}(C)$ closed in $X\iff f^{-1}(Y\backslash C)$ open in $X$ +- $f^{-1}(Y\backslash C) = X\backslash f^{-1}(C)$ + +Continue this proof by yourself. + +#### Theorem 4.7 + +Composition of continuous functions is continuous. + +Suppose $X,Y,Z$ are metric spaces, $E\subset X$, $f:E\to Y$ is continuous, and $g:Y\to Z$ is continuous. Then $g\circ f:E\to Z$ is continuous. + +Ideas of proof: + +- Let $B_\epsilon(g(f(p)))\subset Z$ +- $g(f(B_\delta(p)))\subset B_\epsilon(g(f(p)))$ +- $f(B_\delta(p))$ is open in $Y$ +- $g^{-1}(B_\epsilon(g(f(p)))$ is open in $Y$ +- $(g\circ f)^{-1}(B_\epsilon(g(f(p)))) = f^{-1}(g^{-1}(B_\epsilon(g(f(p)))))$ +- $f^{-1}(g^{-1}(B_\epsilon(g(f(p))))) = (g\circ f)^{-1}(B_\epsilon(g(f(p))))$ + +Apply Theorem 4.8 to complete the proof. + +#### Theorem 4.9 + +For $f:X\to \mathbb{C},g:X\to \mathbb{C}$ are continuous, then, $f+g,f/g$ are continuous. + +Ideas of proof: + +We can reduce this theorem to Theorem about limits and apply what you learned in chapter 3. + +#### Examples of continuous functions 4.11 + +> $\forall p\in \mathbb{R}$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in \mathbb{R}$, $|x-p|<\delta\implies |f(x)-f(p)|<\epsilon$. + +(a). $f(x) = \mathbb{R}\to \mathbb{R},f(x) = x$ is continuous. boring. + +Proof: + +Let $p\in \mathbb{R}$ and $\epsilon > 0$. Let $\delta = \epsilon$. Then, $\forall x\in \mathbb{R}$, if $|x-p|<\delta$, then $|f(x)-f(p)| = |x-p| < \delta = \epsilon$. + +EOP + +Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous. + +(b). $f:\mathbb{R}^k\to \mathbb{R},f(x)=|x|$ is continuous. + +Ideas of proof: + +- $|f(x)-f(p)| = ||x|-|p||\leq |x-p|$ By reverse triangle inequality. +- Let $\epsilon > 0$. Let $\delta = \epsilon$. + +### Continuity and compactness + +#### Theorem 4.13 + +A mapping of $f$ of a set $E$ into a metric space $Y$ is said to be **bounded** if there is a real number $M$ such that $|f(x)|\leq M$ for all $x\in E$. + +#### Theorem 4.14 + +$f:X\to Y$ is continuous. If $X$ is compact, then $f(X)$ is compact. + +Proof strategy: + +For every open cover $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$, there exists a corresponding open cover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$. + +Since $X$ is compact, there exists a finite subcover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$. Let the finite subcover be $\{f^{-1}(V_\alpha)\}_{i=1}^n$. + +Then, $\{V_\alpha\}_{i=1}^n$ is a finite subcover of $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$. + +See the detailed proof in the textbook. + +#### Theorem 4.16 (Extreme Value Theorem) + +Suppose $X$ is a compact metric space and $f:X\to \mathbb{R}$ is continuous. Then $f$ has a maximum and a minimum on $X$. + +i.e. + +$$ +\exists p_0,q_0\in X\text{ such that }f(p_0) = \sup f(X)\text{ and }f(q_0) = \inf f(X). +$$ + +Proof: + +By Theorem 4.14, $f(X)$ is compact. + +By Theorem 2.41, $f(X)$ is closed and bounded. + +By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\in X$ such that $f(p_0) = \sup f(X)$. Let $q_0\in X$ such that $f(q_0) = \inf f(X)$. + +EOP + +### Continuity and connectedness + +> **Definition 2.45**: Let $X$ be a metric space. $A,B\subset X$ are **separated** if $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$. +> +> $E\subset X$ is **disconnected** if there exist two separated sets $A$ and $B$ such that $E = A\cup B$. +> +> $E\subset X$ is **connected** if $E$ is not disconnected. + +#### Theorem 4.22 + +$f:X\to Y$ is continuous, $E\subset X$. If $E$ is connected, then $f(E)$ is connected. + +Proof: + +We will prove the contrapositive statement: if $f(E)$ is disconnected, then $E$ is disconnected. + +Suppose $f(E)$ is disconnected. Then there exist two separated sets $A$ and $B\in Y$ such that $f(E) = A\cup B$. + +Let $G = f^{-1}(A)\cap E$ and $H = f^{-1}(B)\cap E$. + +We have: + +$f(E)=A\cup B\implies E = G\cup H$ + +Since $A$ and $B$ are nonempty, $A,B\subset f(E)$, this implies that $G$ and $H$ are nonempty. + +To complete the proof, we need to show $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$. + +We have $G\subset f^{-1}(A)\cap E\subset f^{-1}(A)\subset f^{-1}(\overline{A})$ Since $\overline{A}$ is closed, $f^{-1}(\overline{A})$ is closed. This implies that $\overline{G}\subset f^{-1}(\overline{A})$. + +So $\overline{G}\subset f^{-1}(\overline{A})$ and $\overline{H}\subset f^{-1}(\overline{B})$. + +Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$. + +Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$. + +EOP + +#### Theorem 4.23 (Intermediate Value Theorem) + +Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c$ is a real number between $f(a)$ and $f(b)$, then there exists a point $x\in [a,b]$ such that $f(x) = c$. + +Ideas of proof: + +Use Theorem 2.47. A subset $E$ of $\mathbb{R}$ is connected if and only if it has the following property: if $x,y\in E$ and $xdim(W)$, then there are no inj Suppose $V,W$ are finite dimensional with $dim(V)0$ +Ideas of Proof: relies on **Theorem 3.21** $dim(null(T))>0$ ### Linear Maps and Linear Systems 3EX-1