# 4.4 The recursion-tree method for solving recurrences

## 4.4-1

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\lfloor n / 2 \rfloor) + n$. Use the substitution method to verify your answer.

• The subproblem size for a node at depth $i$ is $n / 2^i$.

Thus, the tree has $\lg n + 1$ levels and $3^{\lg n} = n^{\lg 3}$ leaves.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, \lg n - 1$, is $3^i(n / 2^i) = (3 / 2)^i n$.

\begin{aligned} T(n) & = n + \frac{3}{2}n + \Big(\frac{3}{2}\Big)^2 n + \cdots + \Big(\frac{3}{2}\Big)^{\lg n - 1} n + \Theta(n^{\lg 3}) \\ & = \sum_{i = 0}^{\lg n - 1} \Big(\frac{3}{2}\Big)^i n + \Theta(n^{\lg 3}) \\ & = \frac{(3 / 2)^{\lg n} - 1}{(3 / 2) - 1}n + \Theta(n^{\lg 3}) \\ & = 2[(3 / 2)^{\lg n} - 1]n + \Theta(n^{\lg 3}) \\ & = 2[n^{\lg(3 / 2)} - 1]n + \Theta(n^{\lg 3}) \\ & = 2[n^{\lg 3 - \lg 2} - 1]n + \Theta(n^{\lg 3}) \\ & = 2[n^{\lg 3 - 1 + 1} - n] + \Theta(n^{\lg 3}) \\ & = O(n^{\lg 3}). \end{aligned}

• We guess $T(n) \le cn^{\lg 3} - dn$,

\begin{aligned} T(n) & = 3T(\lfloor n / 2 \rfloor) + n \\ & \le 3 \cdot (c(n / 2)^{\lg 3} - d(n / 2)) + n \\ & = (3 / 2^{\lg 3})cn^{\lg 3} - (3d / 2)n + n \\ & = cn^{\lg 3} + (1 - 3d / 2)n, \end{aligned}

where the last step holds for $d \ge 2$.

## 4.4-2

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n / 2) + n^2$. Use the substitution method to verify your answer.

• The subproblem size for a node at depth $i$ is $n / 2^i$.

Thus, the tree has $\lg n + 1$ levels and $1^{\lg n} = 1$ leaf.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, \lg{n - 1}$, is $1^i (n / 2^i)^2 = (1 / 4)^i n^2$.

\begin{aligned} T(n) & = \sum_{i = 0}^{\lg n - 1} \Big(\frac{1}{4}\Big)^i n^2 + \Theta(1) \\ & < \sum_{i = 0}^\infty \Big(\frac{1}{4}\Big)^i n^2 + \Theta(1) \\ & = \frac{1}{1 - (1 / 4)} n^2 + \Theta(1) \\ & = \Theta(n^2). \end{aligned}

• We guess $T(n) \le cn^2$,

\begin{aligned} T(n) & \le c(n / 2)^2 + n^2 \\ & = cn^2 / 4 + n^2 \\ & = (c / 4 + 1)n^2 \\ & \le cn^2, \end{aligned}

where the last step holds for $c \ge 4 / 3$.

## 4.4-3

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 4T(n / 2 + 2) + n$. Use the substitution method to verify your answer.

• The subproblem size for a node at depth $i$ is $n / 2^i$.

Thus, the tree has $\lg n + 1$ levels and $4^{\lg n} = n^2$ leaves.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, \lg n - 1$, is $4^i(n / 2^i + 2) = 2^i n + 2 \cdot 4^i$.

\begin{aligned} T(n) & = \sum_{i = 0}^{\lg n - 1} (2^i n + 2 \cdot 4^i) + \Theta(n^2) \\ & = \sum_{i = 0}^{\lg n - 1} 2^i n + \sum_{i = 0}^{\lg n - 1} 2 \cdot 4^i + \Theta(n^2) \\ & = \frac{2^{\lg n} - 1}{2 - 1}n + 2 \cdot \frac{4^{\lg n} - 1}{4 - 1} + \Theta(n^2) \\ & = (2^{\lg n} - 1)n + \frac{2}{3} (4^{\lg n} - 1) + \Theta(n^2) \\ & = (n - 1)n + \frac{2}{3}(n^2 - 1) + \Theta(n^2) \\ & = \Theta(n^2). \end{aligned}

• We guess $T(n) \le c(n^2 - dn)$,

\begin{aligned} T(n) & = 4T(n / 2 + 2) + n \\ & \le 4c[(n / 2 + 2)^2 - d(n / 2 + 2)] + n \\ & = 4c(n^2 / 4 + 2n + 4 - dn / 2 - 2d) + n \\ & = cn^2 + 8cn + 16c - 2cdn - 8cd + n \\ & = cn^2 - cdn + 8cn + 16c - cdn - 8cd + n \\ & = c(n^2 - dn) - (cd - 8c - 1)n - (d - 2) \cdot 8c \\ & \le c(n^2 - dn), \end{aligned}

where the last step holds for $cd - 8c - 1 \ge 0$.

## 4.4-4

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 2T(n - 1) + 1$. Use the substitution method to verify your answer.

• The subproblem size for a node at depth $i$ is $n - i$.

Thus, the tree has $n$ levels and $2^{n - 1}$ leaves.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, n - 1$, is $2^i$.

\begin{aligned} T(n) & = \sum_{i = 0}^{n - 1} 2^i \\ & = \frac{2^n - 1}{2 - 1} \\ & = 2^n - 1 \\ & = \Theta(2^n). \end{aligned}

• We guess $T(n) \le c2^n + n$,

\begin{aligned} T(n) & \le 2 \cdot c2^{n - 1} + (n - 1) + 1 \\ & = c2^n + n \\ & = O(2^n). \end{aligned}

## 4.4-5

Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n - 1) + T(n / 2) + n$. Use the substitution method to verify your answer.

This is a curious one. The tree makes it look like it is exponential in the worst case. The tree is not full (not a complete binary tree of height $n$), but it is not polynomial either. It's easy to show $O(2^n)$ and $\Omega(n^2)$.

To justify that this is a pretty tight upper bound, we'll show that we can't have any other choice. If we have that $T(n) \le cn^k$, when we substitue into the recurrence, the new coefficient for $n^k$ can be as high as $c(1 + \frac{1}{2^k})$ which is bigger than $c$ regardless of how we choose the value $c$.

• We guess $T(n) \le c2^n - 4n$,

\begin{aligned} T(n) & \le c2^{n - 1} - 4(n - 1) + c2^{n / 2} - 4n / 2 + n \\ & = c(2^{n - 1} + 2^{n / 2}) - 5n + 4 & (n \ge 1 / 4) \\ & \le c(2^{n - 1} + 2^{n / 2}) - 4n & (n \ge 2)\\ & = c(2^{n - 1} + 2^{n - 1}) - 4n \\ & \le c2^n - 4n \\ & = O(2^n). \end{aligned}

• We guess $T(n) \ge cn^2$,

\begin{aligned} T(n) & \ge c(n - 1)^2 + c(n / 2)^2 + n \\ & = cn^2 - 2cn + c + cn^2 / 4 + n \\ & = (5 / 4)cn^2 + (1 - 2c)n + c \\ & \ge cn^2 + (1 - 2c)n + c & (c \le 1 / 2) \\ & \ge cn^2 \\ & = \Omega(n^2). \end{aligned}

## 4.4-6

Argue that the solution to the recurrence $T(n) = T(n / 3) + T(2n / 3) + cn$, where $c$ is a constant, is $\Omega(n\lg n)$ by appealing to the recursion tree.

We know that the cost at each level of the tree is $cn$ by examining the tree in figure 4.6. To find a lower bound on the cost of the algorithm, we need a lower bound on the height of the tree.

The shortest simple path from root to leaf is found by following the leftest child at each node. Since we divide by $3$ at each step, we see that this path has length $\log_3 n$. Therefore, the cost of the algorithm is

$$cn(\log_3 n + 1) \ge cn\log_3 n = \frac{c}{\log_3} n\log n = \Omega(n\log n).$$

## 4.4-7

Draw the recursion tree for $T(n) = 4T(\lfloor n / 2 \rfloor) + cn$, where $c$ is a constant, and provide a tight asymptotic bound on its solution. Verify your answer with the substitution method.

• The subproblem size for a node at depth $i$ is $n / 2^i$.

Thus, the tree has $\lg n + 1$ levels and $4^{\lg n} = n^{\lg 4} = n^2$ leaves.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, \lg n - 1$, is $4^i(cn / 2^i) = 2^icn$.

\begin{aligned} T(n) & = \sum_{i = 0}^{\lg n - 1} 2^icn + \Theta(n^2) \\ & = \frac{2^{\lg n} - 1}{2 - 1}cn + \Theta(n^2) \\ & = \Theta(n^2). \end{aligned}

• For $O(n^2)$, we guess $T(n) \le dn^2 - cn$,

\begin{aligned} T(n) & \le 4d(n / 2)^2 - 4c(n / 2) + cn \\ & = dn^2 - cn. \end{aligned}

• For $\Omega(n^2)$, we guess $T(n) \ge dn^2 - cn$,

\begin{aligned} T(n) & \ge 4d(n / 2)^2 - 4c(n / 2) + cn \\ & = dn^2 - cn. \end{aligned}

## 4.4-8

Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(n - a) + T(a) + cn$, where $a \ge 1$ and $c > 0$ are constants.

• The tree has $n / a + 1$ levels.

The total cost over all nodes at depth $i$, for $i = 0, 1, 2, \ldots, n / a - 1$, is $c(n - ia)$.

\begin{aligned} T(n) & = \sum_{i = 0}^{n / a} c(n - ia) + (n / a)ca \\ & = \sum_{i = 0}^{n / a} cn - \sum_{i = 0}^{n / a} cia + (n / a)ca \\ & = cn^2/a - \Theta(n) + \Theta(n) \\ & = \Theta(n^2). \end{aligned}

• For $O(n^2)$, we guess $T(n) \le cn^2$,

\begin{aligned} T(n) & \le c(n - a)^2 + ca + cn \\ & \le cn^2 - 2can + ca + cn \\ & \le cn^2 - c(2an - a - n) & (a > 1 / 2, n > 2a) \\ & \le cn^2 - cn \\ & \le cn^2 \\ & = \Theta(n^2). \end{aligned}

• For $\Omega(n^2)$, we guess $T(n) \ge cn^2$,

\begin{aligned} T(n) & \ge c(n - a)^2 + ca + cn \\ & \ge cn^2 - 2acn + ca + cn \\ & \ge cn^2 - c(2an - a - n) & (a < 1 / 2, n > 2a) \\ & \ge cn^2 + cn \\ & \ge cn^2 \\ & = \Theta(n^2). \end{aligned}

## 4.4-9

Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(\alpha n) + T((1 - \alpha)n) + cn$, where $\alpha$ is a constant in the range $0 < \alpha < 1$, and $c > 0$ is also a constant.

We can assume that $0 < \alpha \le 1 / 2$, since otherwise we can let $\beta = 1 − \alpha$ and solve it for $\beta$.

Thus, the depth of the tree is $\log_{1 / \alpha} n$ and each level costs $cn$. And let's guess that the leaves are $\Theta(n)$,

\begin{aligned} T(n) & = \sum_{i = 0}^{\log_{1 / \alpha} n} cn + \Theta(n) \\ & = cn\log_{1 / \alpha} n + \Theta(n) \\ & = \Theta(n\lg n). \end{aligned}

We can also show $T(n) = \Theta(n\lg n)$ by substitution.

To prove the upper bound, we guess that $T(n) \le dn\lg n$ for a constant $d > 0$,

\begin{aligned} T(n) & = T(\alpha n) + T((1 - \alpha)n) + cn \\ & \le d\alpha n\lg(\alpha n) + d(1 - \alpha)n\lg((1 - \alpha)n) + cn \\ & = d\alpha n\lg\alpha + d\alpha n\lg n + d(1 - \alpha)n\lg(1 - \alpha) + d(1 - \alpha)n\lg n + cn \\ & = dn\lg n + dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn \\ & \le dn\lg n, \end{aligned}

where the last step holds when $d \ge \frac{-c}{\alpha\lg\alpha + (1 - \alpha)\lg(1 - \alpha)}$.

We can achieve this result by solving the inequality

\begin{aligned} dn\lg n + dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn & \le dn\lg n \\ \implies dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn & \le 0 \\ \implies d(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) & \le -c \\ \implies d & \ge \frac{-c}{\alpha\lg\alpha + (1 - \alpha)\lg(1 - \alpha)}, \end{aligned}

To prove the lower bound, we guess that $T(n) \ge dn\lg n$ for a constant $d > 0$,

\begin{aligned} T(n) & = T(\alpha n) + T((1 - \alpha)n) + cn \\ & \ge d\alpha n\lg(\alpha n) + d(1 - \alpha)n\lg((1 - \alpha)n) + cn \\ & = d\alpha n\lg\alpha + d\alpha n\lg n + d(1 - \alpha)n\lg(1 - \alpha) + d(1 - \alpha)n\lg n + cn \\ & = dn\lg n + dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn \\ & \ge dn\lg n, \end{aligned}

where the last step holds when $0 < d \le \frac{-c}{\alpha\lg\alpha + (1 - \alpha)\lg(1 - \alpha)}$.

We can achieve this result by solving the inequality

\begin{aligned} dn\lg n + dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn & \ge dn\lg n \\ \implies dn(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) + cn & \ge 0 \\ \implies d(\alpha \lg\alpha + (1 - \alpha) \lg(1 - \alpha)) & \ge -c \\ \implies 0 < d & \le \frac{-c}{\alpha\lg\alpha + (1 - \alpha)\lg(1 - \alpha)}, \end{aligned}

Therefore, $T(n) = \Theta(n\lg n)$.