fix: 修复不规范错误

.vscode/settings.json

@@ -0,0 +1,2 @@
+{
+}

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_1.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/27/KiB4xw9ncTDaLs7.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/27/KiB4xw9ncTDaLs7.png" width="640"/>
 </figure>
 
 ## **Kai**
@@ -53,6 +53,7 @@ I^2 = -E \Rightarrow I^{-1} =
     0 & 0 & -1 & 0 \\
 \end{bmatrix}
 $$
+
 $$
 Q\bar{Q} = (a^2+b^2+c^2+d^2)E \Rightarrow Q^{-1} = \frac{1}{a^2+b^2+c^2+d^2}
 \begin{bmatrix}
@@ -64,13 +65,59 @@ Q\bar{Q} = (a^2+b^2+c^2+d^2)E \Rightarrow Q^{-1} = \frac{1}{a^2+b^2+c^2+d^2}
 $$
 
 #### (3)
+
 The complete proving is to use $(a, b, c, d)$ to represent all the $Q$s below with their calculations, which is easy but tedious.
-- (a)  $$\forall Q_1, Q_2 \in H, Q_1 + Q_2 \in H$$
-- (b) $$\forall Q_1, Q_2 \in H, Q_1 + Q_2 = Q_2 + Q_1$$
-- (c) $$\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2) + Q_3 = Q_1 + (Q_2 + Q_3) $$
-- (d) $$\exist Q_0 \in H,  \forall Q \in H, Q_0 + Q = Q$$
-- (e) $$\forall Q \in H, \exist \hat{Q} \in H,  Q + Q^ = Q_0$$
-- (f) $$\forall Q_1, Q_2 \in H, Q_1Q_2 \in H$$
-- (g) $$\forall Q_1, Q_2, Q_3 \in H, (Q_1Q_2)Q_3 = Q_1(Q_2Q_3)$$
-- (h) $$\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2)Q_3 = Q_1Q_3 + Q_2Q_3$$
-- (i) $$\exist Q_1, Q_2 \in H, Q_1Q_2 \neq Q_2Q_1$$
+
+- (a)  
+
+$$
+\forall Q_1, Q_2 \in H, Q_1 + Q_2 \in H
+$$
+
+- (b)
+
+$$
+\forall Q_1, Q_2 \in H, Q_1 + Q_2 = Q_2 + Q_1
+$$
+
+- (c\) 
+
+$$
+\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2) + Q_3 = Q_1 + (Q_2 + Q_3)
+$$
+
+- (d) 
+
+$$
+\exists Q_0 \in H,  \forall Q \in H, Q_0 + Q = Q
+$$
+
+- (e) 
+
+$$
+\forall Q \in H, \exists \hat{Q} \in H,  Q + Q^ = Q_0
+$$
+
+- (f) 
+
+$$
+\forall Q_1, Q_2 \in H, Q_1Q_2 \in H
+$$
+
+- (g)
+
+$$
+\forall Q_1, Q_2, Q_3 \in H, (Q_1Q_2)Q_3 = Q_1(Q_2Q_3)
+$$
+
+- (h) 
+
+$$
+\forall Q_1, Q_2, Q_3 \in H, (Q_1 + Q_2)Q_3 = Q_1Q_3 + Q_2Q_3
+$$
+
+- (i) 
+
+$$
+\exists Q_1, Q_2 \in H, Q_1Q_2 \neq Q_2Q_1
+$$

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f1_2.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/27/fSDUupmWIkTFJr5.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/27/fSDUupmWIkTFJr5.png" width="640"/>
 </figure>
 
 ## **Kai**

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f2_1.md

@@ -13,7 +13,7 @@ tags:
 
 
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/28/gpkQ9sXUD6iECKt.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/28/gpkQ9sXUD6iECKt.png" width="640"/>
 </figure>
 
 
@@ -28,7 +28,7 @@ tags:
 #### (1)
 
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/28/XrRkah9u1AYnbU6.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/28/XrRkah9u1AYnbU6.png" width="640"/>
 </figure>
 
 #### (2)

docs/kakomonn/kyoto_university/informatics/ist_202208_kiso_f2_2.md

@@ -13,10 +13,10 @@ tags:
 
 
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/28/udKwDh7LaJgM6qs.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/28/udKwDh7LaJgM6qs.png" width="640"/>
 </figure>
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/28/YXyKcf9bNxpU47Q.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/28/YXyKcf9bNxpU47Q.png" width="640"/>
 </figure>
 
 ## **Kai**
@@ -38,35 +38,35 @@ So, there must not be a cycle in a shortest path.
 
 ### (3)
 
-1. Assume $k=1$ and insert it into $\delta(i, k, k-1)+\delta(k, j, k-1)$, we have:
+1\. Assume $k=1$ and insert it into $\delta(i, k, k-1)+\delta(k, j, k-1)$, we have:
 
-    $$
-    \delta(i, 1, 0)+\delta(1, j, 0) = l(i, 1)+l(1,j)
-    $$
+$$
+\delta(i, 1, 0)+\delta(1, j, 0) = l(i, 1)+l(1,j)
+$$
 
-    Obviously, compare $l(i, 1) + l(1,j)$ with $\delta(i, j, 0) = l(i, j)$, and we immediately have the shortest path from $i$ to $j$ through $k=1$, which means:
+Obviously, compare $l(i, 1) + l(1,j)$ with $\delta(i, j, 0) = l(i, j)$, and we immediately have the shortest path from $i$ to $j$ through $k=1$, which means:
 
-    $$
-    \delta(i, j, 1)=min\{\delta(i,j,0), \delta(i, 1, 0)+\delta(1, j, 0)\}
-    $$
+$$
+\delta(i, j, 1)=min\{\delta(i,j,0), \delta(i, 1, 0)+\delta(1, j, 0)\}
+$$
 
-2. Assume that when $k=k_0$, we have: 
+2\. Assume that when $k=k_0$, we have: 
 
-   $$
-   \delta(i, j, k_0)=min\{\delta(i,j,k_0-1), \delta(i, k_0, k_0-1)+\delta(k_0, j, k_0-1)\}
-   $$
+$$
+\delta(i, j, k_0)=min\{\delta(i,j,k_0-1), \delta(i, k_0, k_0-1)+\delta(k_0, j, k_0-1)\}
+$$
 
-   And for $k=k_0+1$, consider
+And for $k=k_0+1$, consider
 
-   $$
-   \delta(i, k_0+1, k_0)+\delta(k_0+1, j, k_0)
-   $$
+$$
+\delta(i, k_0+1, k_0)+\delta(k_0+1, j, k_0)
+$$
 
-   which introduces a new vertice $k_0+1$ and gives **the shortest path that must go through** $k_0+1$. Thus, compare it to $\delta(i, j, k_0)$, which is the shortest path from $i$ to $j$ that must not considering $k_0+1$, we immediately have the shortest path from $i$ to $j$ that considers $k_0 + 1$. That being said,
+which introduces a new vertice $k_0+1$ and gives **the shortest path that must go through** $k_0+1$. Thus, compare it to $\delta(i, j, k_0)$, which is the shortest path from $i$ to $j$ that must not considering $k_0+1$, we immediately have the shortest path from $i$ to $j$ that considers $k_0 + 1$. That being said,
 
-   $$
-   \delta(i,j,k_0+1)=min\{\delta(i,j,k_0), \delta(i, k_0+1, k_0)+\delta(k_0+1, j, k_0)\}
-   $$
+$$
+\delta(i,j,k_0+1)=min\{\delta(i,j,k_0), \delta(i, k_0+1, k_0)+\delta(k_0+1, j, k_0)\}
+$$
 
 Combine *1* and *2*, we can draw the conlusion that the asked formula is proved true.
 

docs/kakomonn/kyoto_university/informatics/ist_202208_senmon_s_2.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/07/01/gXipzcs9QeadjOU.png" width="480"/>
+  <img src="https://s2.loli.net/2024/07/01/gXipzcs9QeadjOU.png" width="640"/>
 </figure>
 
 
@@ -46,6 +46,7 @@ By the Binomial Theorem, we can insert $\sum^{z}_{i=0}\frac{z!}{(z-i)!i!}\lambda
 $$
 f_{Z}(z) = \frac{e^{-(\lambda_1 + \lambda_2)}}{z!}(\lambda_1 + \lambda_2)^z
 $$
+
 So is the PMF for a Poisson Distribution with the parameter $(\lambda_1 + \lambda_2)$
 
 **PS**: A easier solution is to use Moments Generating Function.
@@ -130,7 +131,8 @@ $$
 \end{align}
 $$
 
-Insert the values, the answer is 
+Insert the values, the answer is
+
 $$
 Pr[T|S^2] = \frac{128}{133}
 $$

docs/kakomonn/kyoto_university/informatics/ist_202208_senmon_s_4.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/07/01/7wrBFAyl6jYgT8k.png" width="480"/>
+  <img src="https://s2.loli.net/2024/07/01/7wrBFAyl6jYgT8k.png" width="640"/>
 </figure>
 
 
@@ -35,6 +35,7 @@ For the same reason, we propose the solution
 $$
 S_2 = \{000, 001, 010, 101, 100\}
 $$
+
 while the transition diagram is 
 
 <figure style="text-align:center;">

docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_2.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/26/atJ2ghTnPMlXucW.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/26/atJ2ghTnPMlXucW.png" width="640"/>
 </figure>
 
 

docs/kakomonn/kyoto_university/informatics/ist_202308_senmon_s_4.md

@@ -11,7 +11,7 @@ tags:
 
 ## **Description**
 <figure style="text-align:center;">
-  <img src="https://s2.loli.net/2024/06/26/gWhJczvi4enyYFx.png" width="480"/>
+  <img src="https://s2.loli.net/2024/06/26/gWhJczvi4enyYFx.png" width="640"/>
 </figure>