IS CS-2018S2-03

题目来源：Problem 3 日期：2024-08-10 题目主题：CS-算法-二叉树与递归

解题思路

我们面对的是一个特殊的 AVL 树 $T_n$，这个二叉树有如下性质：

1. 每个叶子节点的高度为 $0$。
2. 每个非叶子节点 $v$ 的高度是 ${\max }_{w\in L(v)}p(v,w)$。
3. 节点 $v$ 可能有以下三种类型：
   • $v$ 是一个叶子节点。
   • $v$ 只有一个子节点，并且 $v$ 的高度为 $1$。
   • $v$ 有两个子节点，并且这两个子节点的高度相差 $1$。

题目要求我们求出 $T_n$ 的节点总数 $N_n$，以及在 $N_n$ 与黄金比例 $r=\frac{1+\sqrt{5}}{2}$ 之间建立一些不等式关系。最终，我们还需要设计一个算法来为这些节点分配整数。

Solution

Q1: Calculate $N_5$

To calculate $N_5$, we first observe the following pattern based on the problem’s constraints:

• $N_0=1$ (a single leaf node).
• $N_1=2$ (one internal node with height $1$ and one leaf node).

From the problem’s constraints, we see that:

$$N_n=1+N_{n-1}+N_{n-2}\text{ for }n\ge 2$$

Given the recursive relationship, we calculate $N_5$ as follows:

$$\begin{array}{rl}{N}_0& =1,\\ N_1& =2,\\ N_2& =1+N_1+N_0=4,\\ N_3& =1+N_2+N_1=7,\\ N_4& =1+N_3+N_2=12,\\ N_5& =1+N_4+N_3=20.\end{array}$$

Thus, $N_5=20$.

Q2: Express $N_n$ in terms of $N_{n-1}$ and $N_{n-2}$ for $n\ge 2$

The recurrence relation for $N_n$ is derived as follows:

$$N_n=1+N_{n-1}+N_{n-2}$$

This relation reflects the sum of nodes for the tree of height $n$ where the root contributes one node, and the subtrees contribute nodes according to the heights $n-1$ and $n-2$.

Q3: Prove that $N_n\ge r^n$ for every $n\ge 0$

To prove this inequality by induction:

Base Cases:

• $N_0=1\ge r^0=1$
• $N_1=2\ge r^1=r\approx 1.618$

Inductive Step:

Assume $N_k\ge r^k$ for $k\le n-1$. For $N_n$:

$$N_n=1+N_{n-1}+N_{n-2}\ge 1+r^{n-1}+r^{n-2}.$$

We need to show:

$$1+r^{n-1}+r^{n-2}\ge r^n.$$

Using the property of $r$:

$$r^n=r^{n-1}+r^{n-2}.$$

Thus:

$$1+r^{n-1}+r^{n-2}\ge r^n.$$

This completes the proof.

Q4: Prove that $N_n\le r^{n+2}$ for every $n\ge 0$

We now prove by induction that $N_n\le r^{n+2}$.

Base Cases:

• $N_0+1=2\le r^2\approx 2.618$
• $N_1+1=3\le r^3\approx 4.236$

Inductive Step:

Assume for $n\ge 2$, $N_k+1\le r^{k+2}$ holds for all $k\le n-1$. For $N_n$:

$$N_n+1=(N_{n-1}+1)+(N_{n-2}+1)\le r^{n+1}+r^n$$

Using the fact that $r^{n+2}=r^{n+1}+r^n$:

$$N_n\le r^{n+1}+r^n-1<r^{n+2}$$

Thus, $N_n\le r^{n+2}$ for all $n\ge 0$.

Q5: Show an $O(r^n)$ algorithm that computes such an assignment, with proof that it runs indeed in $O(r^n)$ time

1. Algorithm Description

We are tasked with constructing an AVL tree of height $n$ while assigning integers to each node such that the AVL properties are preserved. By precomputing subtree sizes using dynamic programming, the assignment process can be streamlined.

Algorithm Steps:

1. Precompute Subtree Sizes: Use dynamic programming to calculate the sizes of all subtrees up to height $n$. Define size(h) to store the size of a tree of height $h$:

   $$\text{size}[h]=\{\begin{array}{ll}1,& \text{if }h=0,\\ 1+\text{size}[h-1]+\text{size}[h-2],& \text{if }h>0.\end{array}$$

   This step runs in $O(n)$ time.

2. Main Function fillTree(arr, h):

   • Base case: if $h=0$, return a tree with a single node containing the first element of arr.

   • Calculate the sizes for left and right subtrees using the precomputed size[h]:

     $$\text{leftSize}=\text{size}[h-1],\text{rightSize}=\text{size}[h-2]$$

   • Use quickSelect to find the $(\text{leftSize}+1)$th smallest element in arr to be the root.

   • Recursively fill the left subtree with the smallest leftSize elements, and the right subtree with the largest rightSize elements.

   • Return the tree.

3. QuickSelect: This algorithm finds the $k$th smallest element in an array in average $O(n)$ time. It uses a pivot to partition the array and recursively selects the $k$th element from the appropriate partition.

2. Complexity Analysis Using Akra-Bazzi Theorem

To rigorously analyze the time complexity of constructing the AVL tree, we use the Akra-Bazzi Theorem, which is a generalization of the Master Theorem and is more suitable for handling recurrences with multiple recursive branches and non-uniform problem sizes.

Recurrence Relation

The time complexity $T(n)$ for constructing an AVL tree of height $n$ satisfies the following recurrence relation:

$$T(n)=T\left(\frac{N_n}{r}\right)+T\left(\frac{N_n}{r^2}\right)+O(N_n),$$

where $N_n$ is the number of nodes in the tree of height $n$, and $r=\frac{1+\sqrt{5}}{2}$ is the golden ratio. Since $N_n\approx c\cdot r^n$, the recurrence can be rewritten in terms of $N$:

$$T(N)=T\left(\frac{N}{r}\right)+T\left(\frac{N}{r^2}\right)+O(N).$$

Applying the Akra-Bazzi Theorem

The Akra-Bazzi Theorem is a variation of the Master Theorem, used to solve recurrences of the form:

$$T(x)=g(x)+\sum _{i=1}^k{a}_{i}T(b_{i}x+h_i(x)),$$

where:

• $g(x)=\Theta (x^c)$,
• $a_i>0$,
• $0<b_i<1$,
• $h_i(x)=O(x/{\log }^{2}x)$.

The solution to the recurrence is given by:

$$T(x)=\Theta \left(x^p\left(1+{\int }_1^x\frac{g(u)}{u^{p+1}}\mathrm{d}u\right)\right)$$

where $p$ is the unique real solution to:

$$\sum _{i=1}^k{a}_i{b}_i^p=1.$$

Determine Parameters for the Akra-Bazzi Theorem

In our recurrence:

• $g(N)=O(N)$ (since $f(N)=O(N)$),
• $a_1=a_2=1$ (from the two recursive calls),
• $b_1=\frac{1}{r}$ and $b_2=\frac{1}{r^2}$,
• $h_i(N)=0$ (no additional terms in the subproblems).

Solving for $p$

We solve for $p$ using the equation:

$${\left(\frac{1}{r}\right)}^p+{\left(\frac{1}{r^2}\right)}^p=1.$$

Given that $r$ is the golden ratio, we can simplify the equation:

$$r^{-p}+r^{-2p}=1.$$

Multiply through by $r^{2p}$:

$$r^p+1=r^{2p}.$$

Taking the logarithm:

$$\log (r^p+1)=2p\log (r),$$

we find that $p=1$ satisfies the equation.

Final Complexity

With $p=1$, the integral term in the Akra-Bazzi Theorem solution does not introduce additional logarithmic factors, leading to:

$$T(N)=O(N).$$

Since $N_n\in O(r^n)$, the time complexity in terms of the height $n$ is:

$$T(n)=O(r^n).$$

Conclusion

By applying the Akra-Bazzi Theorem and solving for $p$, we rigorously confirm that the time complexity for constructing the AVL tree and assigning integers is indeed $O(r^n)$. This approach accurately captures the growth rate of the recursive calls and the work done at each step, leading to a precise complexity analysis.

知识点

AVL树递归关系黄金比例平衡二叉树斐波那契数列主定理

难点思路

该问题的难点在于证明节点数量 $N_n$ 与黄金比例 $r$ 之间的不等式关系。这要求对递归公式有深刻理解，并且要熟悉黄金比例的一些特殊性质。此外，理解 AVL 树的平衡条件对于推导 $N_n$ 的递归公式至关重要，这一推导不仅解释了 AVL 树的平衡性如何影响节点数量，还揭示了其与斐波那契数列之间的联系。

解题技巧和信息

• 当处理递归关系时，特别是在树形结构中，寻找类似斐波那契数列的关系是一种有效的方法。
• 使用数学归纳法来证明递归关系通常是必要的，尤其是当需要证明不等式时。
• 理解 AVL 树的平衡条件，不仅有助于推导节点数量递归公式，还帮助解释为什么 AVL 树的时间复杂度能保持在 $O(\log n)$。
• Akra-Bazzi 定理是处理这种不均等递归关系的强大工具，尤其是在分析时间复杂度时。

重点词汇

• AVL 树 (AVL Tree)
• 递归 (Recursion)
• 黄金比例 (Golden Ratio)
• 数学归纳法 (Mathematical Induction)
• 时间复杂度 (Time Complexity)
• 平衡二叉树 (Balanced Binary Tree)
• 斐波那契数列 (Fibonacci Sequence)

参考资料

1. Introduction to Algorithms, 3rd Edition, by Cormen, Leiserson, Rivest, and Stein - Chapter 12 (Binary Search Trees) and Chapter 4 (Recurrences)
2. Algorithm Design by Jon Kleinberg and Éva Tardos - Chapter 6 (Dynamic Programming)
3. Data Structures and Algorithm Analysis in C by Mark Allen Weiss - Chapter 4 (Balanced Trees)