diff --git a/docs/courses/软件工程学院/信息安全数学基础(二)/2024秋期末考试答案.md b/docs/courses/软件工程学院/信息安全数学基础(二)/2024秋期末考试答案.md
new file mode 100644
index 0000000..8dba8c4
--- /dev/null
+++ b/docs/courses/软件工程学院/信息安全数学基础(二)/2024秋期末考试答案.md
@@ -0,0 +1,123 @@
+---
+title: 2024Fall_期末
+category:
+ - 软件工程学院
+ - 课程资料
+tag:
+ - 试卷
+author:
+ - タクヤマ
+---
+
+## 说明
+
+本学期(2024年秋季学期)的期末试卷难度**相当**低,本人评价为有手就行,故没有特意记忆所有题目,而是挑出一些有代表性的题目以供参考
+
+我们这里再提供一道填空题,一道选择题,以显示这张卷子的**诚意**
+
+ -
+
+ 设 $G$ 是一个群,对于 $\forall a,b \in G$, $a*b=b*a$, 则 $G$ 是 ____ 群
+
+ -
+
+ 设 $R$ 是一个环,对于 $\forall a,b \in R$, $a+b=b+a$, 则 $R$ 是?
+ A. 整环 B. 交换环 C. 含幺环 D. 环
+
+
+
+很大程度上,老师的出题和给分是十分宽容的,请大家好好学习这门课程,**数学并不是妖魔鬼怪!**
+
+## 题目
+
+> **1. 设 $G=\{a,b,c,e\}$ 是一个群, $H=\{a,e\}$ 是 $G$ 的子群,写出 $H$ 的所有左陪集**
+
+
+
+解:
+
+将 $G$ 中元素 $g$ 各个代入,计算 $gH$
+
+- $g = e$:
+ $eH = H$
+
+- $g = a$:
+ $aH = \{ a \cdot a, a \cdot e \} = \{ a^2, a \}$,由于 $G$ 是群,且 $H$ 是子群,所以 $a^2$ 必须是 $G$ 中元素。
+ 故 $a^2 = e$,则: $aH = \{ e, a \} = H$
+
+- $g = b$:
+ $bH = \{ b \cdot a, b \cdot e \} = \{ ba, b \}$,
+ 假设 $ba = c$ ,则:
+ $bH = \{ c, b \}$
+
+- $g = c$:
+ $cH = \{ c \cdot a, c \cdot e \} = \{ ca, c \}$,
+ 假设 $ca = b$ ,则:
+ $cH = \{ b, c \} = bH$
+
+综上所述,H的所有子陪集是 $\{ e, a \}, \{ b, c \}$
+
+
+
+***
+
+> **2. 写出 $R=Z/6Z$ 的所有零因子**
+
+
+
+解:
+
+ $R=Z/6Z \cong Z_6$,我们只要考虑 $Z_6$ 上的性质:
+ 显然有 $2\*3 \equiv 0 \pmod{6}, 4*3 \equiv 0 \pmod{6}$,所以零因子是2,3,4
+
+
+
+***
+
+> **3. 定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$**
+
+
+
+解:
+
+ 直接计算即可,我们这里直接给出答案: $P+Q = (8, 15)$ , $2P = (14, 15)$
+
+
+
+***
+
+> **4. 求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$,并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元**
+
+
+
+解:
+
+ 这时候有同学要问了,生成元怎么找啊?其实很简单,直接验证就可以了,(出于强大的直觉和观察力)我们在这里直接验证 $x$ :
+
+ | $x^n$ | $x^n\pmod{x^4+x^3+1}$ |
+ | -------- | --------------------- |
+ | $x^0$ | $1$ |
+ | $x^1$ | $x$ |
+ | $x^2$ | $x^2$ |
+ | $x^3$ | $x^3$ |
+ | $x^4$ | $x^3 + 1$ |
+ | $x^5$ | $x^3 + x + 1$ |
+ | $x^6$ | $x^3 + x^2 + x + 1$ |
+ | $x^7$ | $1 + x^2 + x$ |
+ | $x^8$ | $x^2 + x + 1$ |
+ | $x^9$ | $x^3 + x^2 + x$ |
+ | $x^{10}$ | $x^3 + x^2 + x + 1$ |
+ | $x^{11}$ | $x^3 + x + 1$ |
+ | $x^{12}$ | $x^3 + 1$ |
+ | $x^{13}$ | $x^3$ |
+ | $x^{14}$ | $x^2$ |
+
+如上表,由 $x$ 生成的15个非零元素互不相等,所以 $x$ 确实是生成元,非零元的表示如表中所示
+
+
+
+***
+
+> **5. 证明:有限环的特征一定不为0**
+
+证明:请参考课件Chap7.pdf中27-28页
diff --git a/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试答案.md b/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试答案.md
deleted file mode 100644
index 5641ec1..0000000
--- a/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试答案.md
+++ /dev/null
@@ -1,67 +0,0 @@
-# 答案
-
-##
-
-### 1.设 $G=\{a,b,c,e\}$ 是一个群, $H=\{a,e\}$ 是 $G$ 的子群,写出 $H$ 的所有左陪集
-
-解:
- 将 $G$ 中元素 $g$ 各个代入,计算 $gH$
-
-- $g = e$:
- $eH = H$
-
-- $g = a$:
- $aH = \{ a \cdot a, a \cdot e \} = \{ a^2, a \}$,由于 $G$ 是群,且 $H$ 是子群,所以 $a^2$ 必须是 $G$ 中元素。
- 故 $a^2 = e$,则: $aH = \{ e, a \} = H$
-
-- $g = b$:
- $bH = \{ b \cdot a, b \cdot e \} = \{ ba, b \}$,
- 假设 $ba = c$ ,则:
- $bH = \{ c, b \}$
-
-- $g = c$:
- $cH = \{ c \cdot a, c \cdot e \} = \{ ca, c \}$,
- 假设 $ca = b$ ,则:
- $cH = \{ b, c \} = bH$
-
-综上所述,H的所有子陪集是 $\{ e, a \}, \{ b, c \}$
-
-**2.写出 $R=Z/6Z$ 的所有零因子**
-
-解:
- $R=Z/6Z \cong Z_6$,我们只要考虑 $Z_6$ 上的性质:
- 显然有 $2\*3 \equiv 0 \pmod{6}, 4*3 \equiv 0 \pmod{6}$,所以零因子是2,3,4
-
-**3.定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$**
-
-解:
- 直接计算即可,我们这里直接给出答案: $P+Q = (8, 15)$ , $2P = (14, 15)$
-
-**4.求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$,并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元**
-
-解:
- 这时候有同学要问了,生成元怎么找啊?其实很简单,直接验证就可以了,(出于强大的直觉和观察力)我们在这里直接验证 $x$ :
-
- | $x^n$ | $x^n\pmod{x^4+x^3+1}$ |
- |------|----------|
- | $x^0$ | $1$ |
- | $x^1$ | $x$ |
- | $x^2$ | $x^2$ |
- | $x^3$ | $x^3$ |
- | $x^4$ | $x^3 + 1$ |
- | $x^5$ | $x^3 + x + 1$ |
- | $x^6$ | $x^3 + x^2 + x + 1$ |
- | $x^7$ | $1 + x^2 + x$ |
- | $x^8$ | $x^2 + x + 1$ |
- | $x^9$ | $x^3 + x^2 + x$ |
- | $x^{10}$ | $x^3 + x^2 + x + 1$ |
- | $x^{11}$ | $x^3 + x + 1$ |
- | $x^{12}$ | $x^3 + 1$ |
- | $x^{13}$ | $x^3$ |
- | $x^{14}$ | $x^2$ |
-
-如上表,由 $x$ 生成的15个非零元素互不相等,所以 $x$ 确实是生成元,非零元的表示如表中所示
-
-### 5.证明:有限环的特征一定不为0
-
-证明:请参考课件Chap7.pdf中27-28页
diff --git a/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试(部分)题目.md b/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试(部分)题目.md
deleted file mode 100644
index 4499a66..0000000
--- a/docs/courses/软件工程学院/信息安全数学基础(二)/24FallFinalExam/2024秋期末考试(部分)题目.md
+++ /dev/null
@@ -1,19 +0,0 @@
-## 说明:
-本学期(2024年秋季学期)的期末试卷难度**相当**低,本人评价为有手就行,故没有特意记忆所有题目,而是挑出一些有代表性的题目以供参考
-
-
- - 设 $G=\{a,b,c,e\}$ 是一个群, $H=\{a,e\}$ 是 $G$ 的子群,写出 $H$ 的所有左陪集
- - 写出 $R=Z/6Z$ 的所有零因子
- - 定义在有限域 $F_{17}$ 上的椭圆曲线 $E: x^3 + 2x + 3 = y^2$ 上有点 $P(2, 7), Q(11, 8)$ , 计算 $P+Q$ , $2P$
- - 求域 $F_{16}=F_2[x]/(x^4+x^3+1)$的一个生成元 $g(x)$,并用 $g(x)$ 的幂表示 $F_{16}$ 中的所有非零元
- - 证明:有限环的特征一定不为0
-
-
-我们这里再提供一道填空题,一道选择题,以显示这张卷子的**诚意**
-
- - 设 $G$ 是一个群,对于 $\forall a,b \in G$, $a*b=b*a$, 则 $G$ 是 ____ 群
- - 设 $R$ 是一个环,对于 $\forall a,b \in R$, $a+b=b+a$, 则 $R$ 是?
- A. 整环 B.交换环 C.含幺环 D.环
-
-
-很大程度上,老师的出题和给分是十分宽容的,请大家好好学习这门课程,**数学并不是妖魔鬼怪!**
diff --git a/docs/courses/软件工程学院/形式语言与自动机理论/index.md b/docs/courses/软件工程学院/形式语言与自动机理论/index.md
index e248e6d..ccbb1c4 100644
--- a/docs/courses/软件工程学院/形式语言与自动机理论/index.md
+++ b/docs/courses/软件工程学院/形式语言与自动机理论/index.md
@@ -1,5 +1,5 @@
---
-title: 信息安全数学基础(二)
+title: 形式语言与自动机理论
category:
- 软件工程学院
- 课程评价
@@ -9,5 +9,4 @@ tag:
- 抽象代数
---
-### 这门课是抽象代数,但是并没有涉及到抽象代数真正复杂的部分,较为浅尝辄止,所以课程本身的学习难度并不高。
-### 同样地,这门课实际考试不难,老师会捞,请大家不要害怕。
+
diff --git a/package-lock.json b/package-lock.json
index b42a27d..bb8546b 100644
--- a/package-lock.json
+++ b/package-lock.json
@@ -1846,76 +1846,76 @@
]
},
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- "version": "3.12.0",
- "resolved": "https://registry.npmjs.org/@shikijs/core/-/core-3.12.0.tgz",
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"dev": true,
"license": "MIT",
"dependencies": {
- "@shikijs/types": "3.12.0",
+ "@shikijs/types": "3.12.1",
"@shikijs/vscode-textmate": "^10.0.2",
"@types/hast": "^3.0.4",
"hast-util-to-html": "^9.0.5"
}
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"dev": true,
"license": "MIT",
"dependencies": {
- "@shikijs/types": "3.12.0",
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}
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+ "version": "3.12.1",
+ "resolved": "https://registry.npmjs.org/@shikijs/themes/-/themes-3.12.1.tgz",
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"dev": true,
"license": "MIT",
"dependencies": {
- "@shikijs/types": "3.12.0"
+ "@shikijs/types": "3.12.1"
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- "resolved": "https://registry.npmjs.org/@shikijs/transformers/-/transformers-3.12.0.tgz",
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+ "version": "3.12.1",
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+ "integrity": "sha512-crGh3cSZf6mwg3K2W8i79Ja+q4tVClRHdHLnUGi5arS58+cqdzsbkrEZBDMyevf9ehmjFUWDTEwCMEyp9I3z0g==",
"dev": true,
"license": "MIT",
"dependencies": {
- "@shikijs/core": "3.12.0",
- "@shikijs/types": "3.12.0"
+ "@shikijs/core": "3.12.1",
+ "@shikijs/types": "3.12.1"
}
},
"node_modules/@shikijs/types": {
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- "resolved": "https://registry.npmjs.org/@shikijs/types/-/types-3.12.0.tgz",
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+ "version": "3.12.1",
+ "resolved": "https://registry.npmjs.org/@shikijs/types/-/types-3.12.1.tgz",
+ "integrity": "sha512-Is/p+1vTss22LIsGCJTmGrxu7ZC1iBL9doJFYLaZ4aI8d0VDXb7Mn0kBzhkc7pdsRpmUbQLQ5HXwNpa3H6F8og==",
"dev": true,
"license": "MIT",
"dependencies": {
@@ -2912,6 +2912,16 @@
"vuepress": "2.0.0-rc.24"
}
},
+ "node_modules/@vuepress/plugin-redirect/node_modules/commander": {
+ "version": "14.0.0",
+ "resolved": "https://registry.npmjs.org/commander/-/commander-14.0.0.tgz",
+ "integrity": "sha512-2uM9rYjPvyq39NwLRqaiLtWHyDC1FvryJDa2ATTVims5YAS4PupsEQsDvP14FqhFr0P49CYDugi59xaxJlTXRA==",
+ "dev": true,
+ "license": "MIT",
+ "engines": {
+ "node": ">=20"
+ }
+ },
"node_modules/@vuepress/plugin-rtl": {
"version": "2.0.0-rc.112",
"resolved": "https://registry.npmjs.org/@vuepress/plugin-rtl/-/plugin-rtl-2.0.0-rc.112.tgz",
@@ -3070,15 +3080,15 @@
}
},
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- "resolved": "https://registry.npmjs.org/@vueuse/core/-/core-13.8.0.tgz",
- "integrity": "sha512-rmBcgpEpxY0ZmyQQR94q1qkUcHREiLxQwNyWrtjMDipD0WTH/JBcAt0gdcn2PsH0SA76ec291cHFngmyaBhlxA==",
+ "version": "13.9.0",
+ "resolved": "https://registry.npmjs.org/@vueuse/core/-/core-13.9.0.tgz",
+ "integrity": "sha512-ts3regBQyURfCE2BcytLqzm8+MmLlo5Ln/KLoxDVcsZ2gzIwVNnQpQOL/UKV8alUqjSZOlpFZcRNsLRqj+OzyA==",
"dev": true,
"license": "MIT",
"dependencies": {
"@types/web-bluetooth": "^0.0.21",
- "@vueuse/metadata": "13.8.0",
- "@vueuse/shared": "13.8.0"
+ "@vueuse/metadata": "13.9.0",
+ "@vueuse/shared": "13.9.0"
},
"funding": {
"url": "https://github.com/sponsors/antfu"
@@ -3088,9 +3098,9 @@
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},
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+ "version": "13.9.0",
+ "resolved": "https://registry.npmjs.org/@vueuse/metadata/-/metadata-13.9.0.tgz",
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"dev": true,
"license": "MIT",
"funding": {
@@ -3098,9 +3108,9 @@
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},
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- "resolved": "https://registry.npmjs.org/@vueuse/shared/-/shared-13.8.0.tgz",
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+ "version": "13.9.0",
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"license": "MIT",
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@@ -3591,13 +3601,13 @@
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- "resolved": "https://registry.npmjs.org/commander/-/commander-14.0.0.tgz",
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+ "resolved": "https://registry.npmjs.org/commander/-/commander-8.3.0.tgz",
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"dev": true,
"license": "MIT",
"engines": {
- "node": ">=20"
+ "node": ">= 12"
}
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@@ -4533,16 +4543,6 @@
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- "resolved": "https://registry.npmjs.org/commander/-/commander-8.3.0.tgz",
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- "license": "MIT",
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"resolved": "https://registry.npmjs.org/kind-of/-/kind-of-6.0.3.tgz",
@@ -5929,18 +5929,18 @@
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+ "resolved": "https://registry.npmjs.org/shiki/-/shiki-3.12.1.tgz",
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- "@shikijs/langs": "3.12.0",
- "@shikijs/themes": "3.12.0",
- "@shikijs/types": "3.12.0",
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+ "@shikijs/engine-oniguruma": "3.12.1",
+ "@shikijs/langs": "3.12.1",
+ "@shikijs/themes": "3.12.1",
+ "@shikijs/types": "3.12.1",
"@shikijs/vscode-textmate": "^10.0.2",
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}