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pages/CSE559A/CSE559A_L1.md

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+# Lecture 1
+
+## Introducing the syllabus
+
+See the syllabus on Canvas.
+
+## Motivational introduction for computer vision
+
+Computer vision is the study of manipulating images.
+
+Automatic understanding of images and videos
+
+1. vision for measurement (measurement, segmentation)
+2. vision for perception, interpretation (labeling)
+3. search and organization (retrieval, image or video archives)
+
+### What is image
+
+A 2d array of numbers.
+
+### Vision is hard
+
+connection to graphics.
+
+computer vision need to generate the model from the image.
+
+#### Are A and B the same color?
+
+It depends on the context what you mean by "the same".
+
+todo
+
+#### Chair detector example.
+
+double for loops.
+
+#### Our visual system is not perfect.
+
+Some optical illusion images.
+
+todo, embed images here.
+
+### Ridiculously brief history of computer vision
+
+1960s: interpretation of synthetic worlds
+1970s: some progress on interpreting selected images
+1980s: ANNs come and go; shift toward geometry and increased mathematical rigor
+1990s: face recognition; statistical analysis in vogue
+2000s: becoming useful; significant use of machine learning; large annotated datasets available; video processing starts.
+2010s: Deep learning with ConvNets
+2020s: String synthesis; continued improvement across tasks, vision-language models.
+
+## How computer vision is used now
+
+### OCR, Optical Character Recognition
+
+Technology to convert scanned docs to text.
+
+

pages/CSE559A/_meta.js

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-    Math401_L1: {
-        display: 'hidden'
-    },
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-        display: 'hidden'
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-    Math401_L3: {
+    CSE559A_L1: {
         display: 'hidden'
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 }

pages/CSE559A/index.md

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+# CSE 559A: Computer Vision
+
+## Course Description
+

pages/Math401/index.md

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-# Math 401: Honors Seminar in Mathematics
-
-## Course Description
-

pages/Math4351/Math4351_L1.md

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+# Lecture 1
+
+## Toy example (RSA encryption)
+
+$Enc$
+
+1. Choose a letter, count to number of letters in the alphabet.
+2. Calculate $s^7$, then divide by 33, take the remainder.
+3. Send the remainder.
+
+$Dec: s = r^3 \mod 33$
+
+To build up such system.
+
+Step 1: Understanding the arithmetic of remainders.
+
+## Part 1: Divisibility and prime numbers
+
+### Divisibility and division algorithm
+
+Let $a, b\in \mathbb{Z}$, with $b>0$. There are unique integers, $q$ (quotient) and $r$ (remainder), such that $a = bq + r$ and $0 \leq r < b$.
+
+Example: $31=7\times 4 + 3$
+
+Proof:
+
+The quotient and remainder are unique.
+
+(1) Existence:
+
+Let $S = \{a - bk \mid k \in \mathbb{Z}, a - bk \geq 0\}$. Choose $r$ to be the smallest non-negative element of $S$. This means $r = a - bq$ for some $q \in \mathbb{Z}$. i.e. $a=bq+r$.
+
+> New notion: $\triangle FSOC$ means For the sake of contradiction.
+
+Notice that $r \geq 0$, by contradiction, suppose $r \geq b$, then $r-b \geq 0$ and $r-b \in S$, but $r-b < r$, which contradicts the minimality of $r$.
+
+Therefore, $r \geq 0$ and $r < b$.
+
+Example: $a=31, b=7$, $S = \{31-7k \mid k \in \mathbb{Z}, 31-7k \geq 0\}=\{\cdots, -32, -25, -18, -11, -4, 3, 10, 17, 24, 31, \cdots\}$, $r=3$.
+
+So We choose $q=4$ and $r=3$.
+
+(2) Uniqueness:
+
+Suppose we have two pairs $(q, r)$ and $(q', r')$ such that $a = bq + r = bq' + r'$ Suppose $q \neq q'$, without loss of generality, suppose $q > q'$, $q-q' \geq 1$. Then $b(q-q') = r'-r$.
+
+Since $r'=b(q-q')+r \geq b(q-q') \geq b$, which contradicts that $r' < b$.
+
+Therefore, $q=q'$ and $r=r'$.
+
+EOP
+
+#### Definition: Divisibility
+
+Let $a, b \in \mathbb{Z}$, we say $b$ divides $a$ and write $b \mid a$ if there exists $k\in \mathbb{Z}$ such that $a = bk$.
+
+Example: $3 \mid 12$ because $12 = 3 \times 4$.
+
+#### Properties of divisibility
+
+Let $a, b, c \in \mathbb{Z}$.
+
+(1) $b \mid a \iff r=0$ in the division algorithm.
+
+(2) If $a \mid b$ and $b \mid c$, then $a \mid c$.
+
+(3) If $a \mid b$ and $b \mid a$, then $a = \pm b$.
+
+(4) If $a \mid b$ and $a \mid c$, then $a \mid bx + cy$ for all $x, y \in \mathbb{Z}$. (We call such $bx+cy$ a linear combination of $b$ and $c$.)
+
+(5) If $c\neq 0$ and $a \mid b \iff ac \mid bc$.
+
+Some proof examples:
+
+(2) Since $a \mid b$ and $b \mid c$, there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $c = bl$. Then $c = bl = (ak)l = a(kl)$, so $a \mid c$.
+
+EOP
+
+(3) If $a \mid b$ and $b \mid a$, then there exist $k, l \in \mathbb{Z}$ such that $b = ak$ and $a = bl$. Then $a = bl = (ak)l = a(kl)$, so $a(1-kl) = 0$.
+
+Case 1: $a=0$, then $b=0$, so $a=b$.
+
+Case 2: $a \neq 0$, then $1-kl=0$, so $kl=1$. Since $k, l \in \mathbb{Z}$, $k=l=\pm 1$, so $a = \pm b$.
+
+EOP
+
+#### Definition: Divisor
+
+Let $a\in \mathbb{Z}$, we define $D(a) = \{d\in \mathbb{Z} \mid d \mid a\}$.
+
+**Note that $D(0) = \mathbb{Z}$.**
+
+Example: $D(12) = \{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\}$.
+
+#### Definition: Greatest common divisor
+
+Let $a, b \in \mathbb{Z}$, where $a,b$ not both zero, we define the greatest common divisor of $a$ and $b$ to be the largest element in $D(a) \cap D(b)$. It is denoted by $(a,b)$.
+
+> Terrible, I really hate this notation. But professor said it's unlikely to be confused with the interval $(a,b)$ since they don't show up in the same context usually.
+
+Example:
+
+$(12, 18) = 6$.
+
+**Note that $(0,0)$ is not defined. (there is no largest element in $D(0) \cap D(0)$.)**
+
+but it is okay that one of $a, b$ is zero. For example, $(0, 18) = 18$.
+
+$(n,n) = |n|$ for all $n \in \mathbb{Z}$.
+
+In general, if $(a,b)=0$ we say $a$ and $b$ are relatively prime, or coprime.
+
+$\forall a, b \in \mathbb{Z}$, $(a,b) \geq 1$.
+
+#### Theorem for calculating gcd
+
+Let $a, b \in \mathbb{Z}$, with $b\neq 0$, then for any $k\in \mathbb{Z}$, $(a,b) = (b,a-bk)$.
+
+Example: $(12, 18) = (18, 12-18) = (18, -6) = 6$.
+
+$(938,210)=(210,938-210\times 4)=(210,938-840)=(210,98)$.
+
+Proof:
+
+We will prove that $D(a) \cap D(b) = D(b) \cap D(a-bk)$.
+
+(1) $D(a) \cap D(b) \subseteq D(b) \cap D(a-bk)$:
+
+Let $d \in D(a) \cap D(b)$, then $d \mid a$ and $d \mid b$.
+
+By property of divisibility (4), If $a\mid b$ and $b\mid c$, then for all $x,y\in \mathbb{Z}$, $a\mid bx+cy$.
+
+So $d\mid a+b\cdot (-k) = a-bk$.
+
+Therefore, $d \in D(b) \cap D(a-bk)$.
+
+(2) $D(b) \cap D(a-bk) \subseteq D(a) \cap D(b)$:
+
+Let $d \in D(b) \cap D(a-bk)$, then $d \mid b$ and $d \mid a-bk$.
+
+By property of divisibility (4), $d \mid bk + (a-bk) = a$.
+
+Therefore, $d \in D(a) \cap D(b)$.
+
+EOP
+
+This theorem gives rise to the Euclidean algorithm which is a efficient way to compute the greatest common divisor of two integers. $2\Theta(\log n)+1=O(\log n)$ ([Proof in CSE 442T Lecture 7](https://notenextra.trance-0.com/CSE442T/CSE442T_L7#euclidean-algorithm)).
+
+### Euclidean algorithm
+
+We will skip this part, it's already the third time we see this algorithm in wustl.
+
+#### Theorem: Euclidean algorithm returns correct gcd
+
+Let $a>b>0$, be integers. Using the Euclidean algorithm, we can find $b>r_0>r_1>r_2>\cdots>r_n$ such that $a=bq_0+r_0, b=r_0q_1+r_1, \cdots, r_{n-1}=r_nq_{n+1}+r_{n+1}, r_n=0$. Then $(a,b)=r_n$.
+
+Proof:
+
+(a) This process terminates. $b>r_0>r_1>r_2>\cdots>r_n$ is a strictly decreasing sequence of positive integers, so it must terminate.
+

pages/Math4351/_meta.js

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+export default {
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+    "---":{
+        type: 'separator'
+    },
+    Math4351_L1: "Introduction to Number Theory and Cryptography (Lecture 1)",
+}

pages/Math4351/index.md

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+# Math 4351
+
+Number theory and cryptography.
+
+No textbook required.
+
+

pages/_meta.js

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