Update 16.3.md

docs/Chap16/16.3.md

@@ -50,7 +50,7 @@ B(T) & = f(x) d_T(x) + f(y) d_T(y) \\\\
 \end{aligned}
 $$
 
-Thus, the statement of theorem is true. Now suppose $n > 2$ and also suppose that theorem is true for trees on $n - 1$ leaves.  
+Thus, the statement of theorem is true. Now suppose $n > 2$ and also suppose that theorem is true for trees on $n - 1$ leaves.
 Let $c_1$ and $c_2$ are two sibling leaves in $T$ such that they have the same parent $p$. Letting $T'$ be the tree obtained by deleting $c_1$ and $c_2$, by induction we know that
 
 $$
@@ -88,7 +88,7 @@ $$
 \end{aligned}
 $$
 
-That is, where $i$ and $j$ are the upper and lower median,  
+That is, where $i$ and $j$ are the upper and lower median,
 respectively; and $c_i = d_T(x_i) - d_T(x_{n - i + 1})$:
 
 $$f(x_1)c_1 + \cdots + f(x_i)c_i > f(x_j)c_i + \cdots + f(x_n)c_1$$
@@ -108,7 +108,7 @@ $$
 First observe that any full binary tree has exactly $2n - 1$ nodes. We can encode the structure of our full binary tree by performing a preorder traversal of $T$.
 For each node that we record in the traversal, write a $0$ if it is an internal node and a $1$ if it is a leaf node. Since we know the tree to be full, this uniquely determines its structure.
 
-Next, note that we can encode any character of $C$ in $\lceil \lg n \rceil$ bits. Since there are $n$ characters, we can encode them in order of appearance in our preorder traversal using ndlg ne bits.
+Next, note that we can encode any character of $C$ in $\lceil \lg n \rceil$ bits. Since there are $n$ characters, we can encode them in order of appearance in our preorder traversal using $n\left\lceil \lg n \right\rceil$ bits.
 
 ## 16.3-7