diff --git a/chapter4/sections/4/4.tex b/chapter4/sections/4/4.tex
index a0deb8f..6470427 100644
--- a/chapter4/sections/4/4.tex
+++ b/chapter4/sections/4/4.tex
@@ -1 +1,36 @@
-\workinprogress % TODO
+We'll use the recursion-tree method to guess an upper bound on $T(n)$.
+Let's assume that $0<\alpha\le1/2$, because the case when $1/2\le\alpha<1$ is symmetric, and as we'll see, the choice of $\alpha$ doesn't affect the order of growth of an upper bound on $T(n)$.
+\refFigure{4.4-4} shows such a tree.
+\begin{figure}[htb]
+    \input{4.tikz}
+    \caption{A recursion tree for the recurrence $T(n)=T(\alpha n)+T((1-\alpha)n)+cn$, where $0<\alpha\le1/2$.} \label{fig:4.4-4}
+\end{figure}
+
+Let $n_0>0$ be the implicit threshold constant such that $T(n)=\Theta(1)$ for $0<n<n_0$, and let $c$ represent the upper-bound constant hidden by the $\Theta(n)$ term for $n\ge n_0$.
+The root-to-leaf path formed by right-going edges is the longest, where the node at depth $i$ represents a subproblem of size $(1-\alpha)^in$.
+Let $h$ be the height of the tree.
+At depth $h$ the rightmost node is a leaf, so $(1-\alpha)^hn<n_0\le(1-\alpha)^{h-1}n$, which gives $h=\lfloor\log_{1/(1-\alpha)}(n/n_0)\rfloor+1$, and thus $h=\Theta(\lg n)$.
+The recursion divides the problem into two subproblems, whose sizes sum up to the size of the problem.
+Since the cost incurred by a problem of size $m$ is $cm$, the sum of costs incurred by the subproblems is also $cm$.
+Therefore, the total cost of all nodes at any depth is bound by $cn$.
+Summing the costs of internal nodes across each level, we have at most $cn$ per level times $h=\Theta(\lg n)$ for a total cost of $O(n\lg n)$ for all internal nodes.
+
+The number of leaves of the tree is described by the following recurrence:
+\[
+    L(n) =
+    \ccases{
+        1 & \text{if $n<n_0$}, \\
+        L(\alpha n)+L((1-\alpha)n) & \text{if $n\ge n_0$}.
+    }
+\]
+We can apply the substitution method to show it has solution $L(n)=O(n)$ for any $0<\alpha<1$.
+Using the inductive hypothesis $L(n)\le dn$ for some constant $d>0$, and assuming that the inductive hypothesis holds for all values less than $n$, we have
+\begin{align*}
+    L(n) &= L(\alpha n)+L((1-\alpha)n) \\
+    &\le d\alpha n+d(1-\alpha)n \\
+    &= dn,
+\end{align*}
+which holds for any $d>0$ and any $0<\alpha<1$.
+Picking $d=1$ suffices to handle the base case $L(1)=1$ for $0<n<n_0$.
+
+Returning to recurrence $T(n)$, we obtain its upper bound of $O(n\lg n)+O(n)=O(n\lg n)$.
diff --git a/chapter4/sections/4/4.tikz b/chapter4/sections/4/4.tikz
new file mode 100644
index 0000000..2ef7405
--- /dev/null
+++ b/chapter4/sections/4/4.tikz
@@ -0,0 +1,64 @@
+\begin{tikzpicture}[
+	level/.style = {level distance=3\vertexsize, sibling distance=16\vertexsize/2^#1},
+	level 3/.style = {edge from parent/.style = {transition edge={solid}{very densely dashed}{0.7}}},
+	level arrow/.style = {->, thick, blue},
+	total node/.style = {align=right, text width=10mm},
+]
+
+\node (root) {$cn$}
+	child {node (0) {$c\alpha n$}
+		child {node[text depth=0.5ex, text height=1.5ex] (00) {$c\alpha^2n$}
+			child {node (000) {}}
+			child {node (001) {}}
+		}
+		child {node[text depth=0.5ex, text height=1.5ex] (01) {$c\alpha(1-\alpha)n$}
+			child {node (010) {}}
+			child {node (011) {}}
+		}
+	}
+	child {node (1) {$c(1-\alpha)n$}
+		child {node[text depth=0.5ex, text height=1.5ex] (10) {$c\alpha(1-\alpha)n$}
+			child {node (100) {}}
+			child {node (101) {}}
+		}
+		child {node[text depth=0.5ex, text height=1.5ex] (11) {$c(1-\alpha)^2n$}
+			child {node (110) {}}
+			child {node (111) {}}
+		}
+	};
+\node[below=\vertexsize of 000] (leaf 0) {$\Theta(1)$};
+\node[below=2\vertexsize of 001] (leaf 1) {$\Theta(1)$};
+\node[below=1.75\vertexsize of 010] (leaf 2) {$\Theta(1)$};
+\node[below=2.75\vertexsize of 011] (leaf 3) {$\Theta(1)$};
+\node[below=2.5\vertexsize of 100] (leaf 4) {$\Theta(1)$};
+\node[below=3.5\vertexsize of 101] (leaf 5) {$\Theta(1)$};
+\node[below=3.25\vertexsize of 110] (leaf 6) {$\Theta(1)$};
+\node[below=4.25\vertexsize of 111] (leaf 7) {$\Theta(1)$};
+\foreach \x in {0, ..., 7} {
+	\draw[very densely dashed] (leaf \x.north) -- +(0mm, 2.5mm);
+}
+
+\node[total node, right=of leaf 7] (level h total) {$\Theta(1)$};
+\node[total node] at (level h total |- 11) (level 2 total) {$cn$};
+\node[total node] at (level h total |- 1) (level 1 total) {$cn$};
+\node[total node] at (level h total |- root) (level 0 total) {$cn$};
+
+\path (level 2 total.east) -- (level h total.east) node[midway, xshift=-5mm, font=\bfseries] (total ellipsis) {$\vdots$};
+\node[font=\bfseries] at (total ellipsis -| root) {$\vdots$};
+
+\draw[level arrow, shorten <=2mm] (root) -- (level 0 total);
+\draw[level arrow, shorten <=2mm] (1) -- (level 1 total);
+\draw[level arrow, shorten <=2mm] (11) -- (level 2 total);
+\draw[level arrow, shorten <=2mm] (leaf 7) -- (level h total);
+
+\draw[orange, decorate, decoration={brace, amplitude=8pt, mirror}]
+	(leaf 0.south west |- leaf 7.south east) -- (leaf 7.south east) node[midway, yshift=-5mm, text=black] (leaves total) {$O(n)$};
+
+\coordinate (total line end) at (leaves total -| level h total.east);
+\draw (1 |- total line end) -- (total line end) node[at end, anchor=east, yshift=-4mm] {Total: $O(n\lg n)$};
+
+\coordinate (bottom) at (leaf 0.west |- level h total);
+\coordinate[left=14mm of bottom] (bottom shifted);
+\draw[<->, orange] (bottom shifted) -- (bottom shifted |- root) node[pos=0.5, fill=white, text=black] {$\lfloor\log_{1/(1-\alpha)}(n/n_0)\rfloor+1$};
+
+\end{tikzpicture}