《算法导论》第四章-思考题（参考答案）

算法导论（第三版）参考答案：思考题4.1，思考题4.2，思考题4.3，思考题4.4，思考题4.5，思考题4.6

Problem 4.1 (Recurrence examples)

Give asymptotic upper and lower bound for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for $n≤2$. Make your bounds as tight as possible, and justify your answers.

1. $T(n) = 2T(n/2) + n^4$
2. $T(n) = T(7n/10) + n$
3. $T(n) = 16T(n/4) + n^2$
4. $T(n) = 7T(n/3) + n^2$
5. $T(n) = 7T(n/2) + n^2$
6. $T(n) = 2T(n/4) + \sqrt{n}$
7. $T(n) = T(n - 2) + n^2$

1. 主方法情况三，$T(n)=\Theta(n^4)$
2. 主方法情况三，$T(n)=\Theta(n)$
3. 主方法情况二，$T(n)=\Theta(n^2\lg{n})$
4. 主方法情况一，$T(n)=\Theta(n^2)$
5. 主方法情况一，$T(n)=\Theta(n^{\lg_27})$
6. 主方法情况二，$T(n)=\Theta(\sqrt{n}\lg_{n})$
7. 递归树法，假设n为偶数，$T(n)=n^2+(n-2)^2+(n-4)^2+\cdots+T(2)= \sum_{i=0}^{\frac{n}{2}-1}(n -2i)^2 + T(2)= \Theta(n^3)$

Problem 4.2 (Parameter-passing costs)

Throughout this book, we assume that parameter passing during procedure calls takes constant time, even if an N-element array is being passed. This assumption is valid in most systems because a pointer to the array is passed, not the array itself. This problem examines the implications of three parameter-passing strategies:

1. An array is passed by pointer. Time $=Θ(1)$
2. An array is passed by copying. Time $=Θ(N)$, where $N$ is the size of the array.
3. An array is passed by copying only the subrage that might be accessed by the called procedure. Time $=Θ(q−p+1)$ if the subarray $A[p…q]$ is passed.

So:

1. Consider the recursive binary search algorithm for finding a number in a sorted array (see Exercise 2.3-5). Give recurrences for the worst-case running times of binary search when arrays are passed using each of the three methods above, and give good upper bounds on the solutions of the recurrences. Let $N$ be the size of the original problems and n be the size of a subproblem.
2. Redo part (a) for the MERGE-SORT algorithm from Section 2.3.1.

1. 二分查找

   a.

   $T(n) = T(n/2) + c = \Theta(\lg{n})$

   b.

   $T(n) = T(n/2) + cN = 2cN + T(n/4) = 3cN + T(n/8) = \sum_{i=0}^{\lg{n}-1}(2^icN/2^i) = cN\lg{n} = \Theta(n\lg{n})$ c.

   $T(n) = T(n/2) + cn = \Theta(n)$

2. 归并排序

   a.

   $T(n) = 2T(n/2) + cn = \Theta(n\lg{n})$

   b.
   

   $$\begin{align} T(n) &= 2T(n/2) + cn + 2N = \sum_{i=0}^{\lg{n}-1}(cn + 2^iN) = \sum_{i=0}^{\lg{n}-1}cn + N\sum_{i=0}^{\lg{n}-1}2^i \\ &= cn\lg{n} + N\frac{2^{\lg{n}} - 1}{2-1} = cn\lg{n} + nN - N = \Theta(nN) = \Theta(n^2) \end{align}$$
   c.

   $T(n) = 2T(n/2) + cn + 2n/2 = 2T(n/2) + (c+1)n = \Theta(n\lg{n})$

Problem 4.3 (More recurrence examples)

Give asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for sufficiently small n. Make your bounds as tight as possible, and justify your answers.

1. $T(n) = 4T(n/3) + n\lg{n}$
2. $T(n) = 3T(n/3) + n/\lg{n}$
3. $T(n) = 4T(n/2) + n^2\sqrt{n}$
4. $T(n) = 3T(n/3 - 2) + n/2$
5. $T(n) = 2T(n/2) + n/\lg{n}$
6. $T(n) = T(n/2) + T(n/4) + T(n/8) + n$
7. T(n)=T