#1478 B-3c

appendixb/problems/3/a.tex

@@ -1,15 +1,17 @@
 \subproblem
 The fact trivially holds for binary trees with at most three nodes, so in the rest of the proof we will assume that $n\ge4$.
-For a given binary tree consider a sequence of nodes $\langle x_0$, $x_1$, \dots, $x_k\rangle$, where $x_0$ is the root, $x_k$ is a leaf, and for $i=1$, 2, \dots, $k$, $x_i$ is the root of the subtree of $x_{i-1}$ with more nodes.
+We define a \concept{branch} of a binary tree as a path $\langle x_0$, $x_1$, \dots, $x_k\rangle$, where $x_0$ is the root, $x_k$ is a leaf, and for $i=1$, 2, \dots, $k$, $x_i$ is the root of the subtree of $x_{i-1}$ with more nodes.
+Note that a tree may have more than one branch.
 Let's denote by $s(x)$ the number of nodes in the subtree rooted at $x$.
 For every $i=0$, 1, \dots, $k-1$,
-\[
+\begin{equation} \label{eq:subtree-sizes-on-branch}
     s(x_i) \le 2s(x_{i+1})+1.
-\]
+\end{equation}
 Of course, the sequence $\langle s(x_0)$, $s(x_1)$, \dots, $s(x_k)\rangle$ decreases from $n$ to 1, so there must be $j$, such that $s(x_j)>n/4$ and $s(x_{j+1})\le n/4$.
 Thus, by removing edge $(x_{j-1},x_j)$, we partition the nodes of the tree into two sets of sizes
 \begin{align*}
-    s(x_j) &\le 2n/4+1 \\
+    s(x_j) &\le 2s(x_{j+1})+1 \tag{by inequality \eqref{eq:subtree-sizes-on-branch}} \\
+    &= 2n/4+1 \\
     &\le 3n/4
 \end{align*}
 and

appendixb/problems/3/c.tex

@@ -0,0 +1,18 @@
+\subproblem
+Assume that $n\ge6$.
+We'll remove one edge from the original tree to cut off a subtree of at most $\lfloor n/2\rfloor$ vertices.
+Consider any of the tree's branches $\langle x_0$, $x_1$, \dots, $x_k\rangle$, as well as the function $s$, both defined in the solution to part (a).
+There is an edge $(x_j,x_{j+1})$, such that $s(x_j)>\lfloor n/2\rfloor$ and $s(x_{j+1})\le\lfloor n/2\rfloor$.
+From inequality \eqref{eq:subtree-sizes-on-branch} we have
+\begin{align*}
+    s(x_{j+1}) &\ge (s(x_j)-1)/2 \\
+    &> (\lfloor n/2\rfloor-1)/2 \\
+    &\ge \lfloor n/2\rfloor/3.
+\end{align*}
+We cut off the subtree rooted at $x_{j+1}$ and continue partitioning the remaining tree, with at most
+\[
+    \lfloor n/2\rfloor-s(x_{j+1}) < (2/3)\lfloor n/2\rfloor
+\]
+vertices still to remove from it.
+In each subsequent step the number of nodes remaining to be cut is reduced by a factor of $2/3$.
+Therefore, we'll need to remove $O(\log_{3/2}n)=O(\lg n)$ edges.

appendixb/problems/3/main.tex

@@ -1,3 +1,4 @@
 \problem{Bisecting trees}
 \subimport{./}{a}
 \subimport{./}{b}
+\subimport{./}{c}

clrs4e-solutions.tex

@@ -81,6 +81,8 @@
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+
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