diff --git a/docs/Chap18/18.1.md b/docs/Chap18/18.1.md
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@@ -1,44 +1,101 @@
 ## 18.1-1
 
+  
+
 > Why don't we allow a minimum degree of $t = 1$?
 
+  
+
 According to the definition, minimum degree $t$ means every node other than the root must have at least $t - 1$ keys, and every internal node other than the root thus has at least $t$ children. So, when $t = 1$, it means every node other than the root must have at least $t - 1 = 0$ key, and every internal node other than the root thus has at least $t = 1$ child.
 
+  
+
 Thus, we can see that the minimum case doesn't exist, because no node exists with $0$ key, and no node exists with only $1$ child in a B-tree.
 
+  
+
 ## 18.1-2
 
+  
+
 > For what values of $t$ is the tree of Figure 18.1 a legal B-tree?
 
+  
+
 According to property 5 of B-tree, every node other than the root must have at least $t - 1$ keys and may contain at most $2t - 1$ keys. In Figure 18.1, the number of keys of each node (except the root) is either $2$ or $3$. So to make it a legal B-tree, we need to guarantee that $t - 1 \le 2 \text{ and } 2 t - 1 \ge 3$, which yields $2 \le t \le 3$. So $t$ can be $2$ or $3$.
 
+  
+
 ## 18.1-3
 
+  
+
 > Show all legal B-trees of minimum degree $2$ that represent $\\{1, 2, 3, 4, 5\\}$.
 
-We know that every node except the root must have at least $t - 1 = 1$ keys, and at most $2t - 1 = 3$ keys. Also remember that the leaves stay in the same depth. Thus, there are $2$ possible legal B-trees:
+Since $t=2$, every node (except the root) must have at least $t-1=1$ key and at most $2t-1=3$ keys.  
 
-- $$| 1, 2, 3, 4, 5 |$$
+- $$
+\begin{array}{c}
+|\,2\,| \\[6pt]
+\swarrow \quad \searrow \\[6pt]
+|\,1\,| \quad |\,3,4,5\,|
+\end{array}
+$$
 
-- $$| 3 |$$
+- $$
+\begin{array}{c}
+|\,3\,| \\[6pt]
+\swarrow \quad \searrow \\[6pt]
+|\,1,2\,| \quad |\,4,5\,|
+\end{array}
+$$
 
-    $$\swarrow \quad \searrow$$
+- $$
+\begin{array}{c}
+|\,4\,| \\[6pt]
+\swarrow \quad \searrow \\[6pt]
+|\,1,2,3\,| \quad |\,5\,|
+\end{array}
+$$
 
-    $$| 1, 2 | \qquad\qquad | 4, 5 |$$
+- $$
+\begin{array}{c}
+|\,2,4\,| \\[6pt]
+\swarrow \quad \downarrow \quad \searrow \\[6pt]
+|\,1\,| \quad |\,3\,| \quad |\,5\,|
+\end{array}
+$$
+
+  
 
 ## 18.1-4
 
+  
+
 > As a function of the minimum degree $t$, what is the maximum number of keys that can be stored in a B-tree of height $h$?
 
+  
+
 $$
+
 \begin{aligned}
+
 n & = (1 + 2t + (2t) ^ 2 + \cdots + (2t) ^ {h}) \cdot (2t - 1) \\\\
-  & = (2t)^{h + 1} - 1.
+
+& = (2t)^{h + 1} - 1.
+
 \end{aligned}
+
 $$
 
+  
+
 ## 18.1-5
 
+  
+
 > Describe the data structure that would result if each black node in a red-black tree were to absorb its red children, incorporating their children with its own.
 
-After absorbing each red node into its black parent, each black node may contain $1, 2$ ($1$ red child), or $3$ ($2$ red children) keys, and all leaves of the resulting tree have the same depth, according to property 5 of red-black tree (For each node, all paths from the node to descendant leaves contain the same number of black nodes). Therefore, a red-black tree will become a Btree with minimum degree $t = 2$, i.e., a 2-3-4 tree.
+  
+
+After absorbing each red node into its black parent, each black node may contain $1, 2$ ($1$ red child), or $3$ ($2$ red children) keys, and all leaves of the resulting tree have the same depth, according to property 5 of red-black tree (For each node, all paths from the node to descendant leaves contain the same number of black nodes). Therefore, a red-black tree will become a Btree with minimum degree $t = 2$, i.e., a 2-3-4 tree.
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