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在《数据结构与算法分析（Java语言描述）》（第三版）这本教材中，我们看到涉及了关于数据结构、算法以及程序设计的基础概念。在给出的文档中，部分题目和答案涵盖了递归、数学推理、文件处理以及计算理论等多个方面的知识。 1. 文件处理：在问题1.4中提到了一个通用的处理文件的方法`void processFile(String fileName)`。这个方法用于打开文件，执行所需处理，然后关闭文件。如果文件中包含`#include SomeFile`这样的语句，程序会递归调用`processFile( SomeFile )`来处理被包含的文件。这种方法体现了编程中的文件I/O操作和递归思想，用于处理嵌套包含的文件。 2. 递归算法：问题1.5展示了求整数n的个位数之和的递归算法`ones(int n)`。这个函数通过将n不断除以2并取余，递归地计算个位数之和，直至n小于2时返回n，体现了递归算法的基本构造。 3. 数学归纳法证明：问题1.7(a)中，使用了数学归纳法证明了一个关于对数的性质。首先验证基础情况，然后假设性质对于某个区间内的所有数都成立，并推导出下一个区间的性质，以此类推，证明了对数不等式。 4. 数列求和：问题1.8涉及数列求和的计算。例如在(b)中，通过等比数列的性质，我们能够计算出一个特殊的求和公式，这里使用了差分技巧，将两个等式相减，简化求和过程。 5. 模运算与指数定律：在问题1.10中，利用模运算和指数定律，证明了2^100模5的结果，展示了指数运算在模算术中的应用。 6. 大O符号和对数估计：问题1.9涉及到大O符号的使用，估计了1到N的和与对数关系，这在算法分析中是非常重要的，因为它帮助我们理解算法的时间复杂度。这些题目和解答展示了《数据结构与算法分析》课程中的核心概念，包括递归、数学推理、文件处理、计算复杂度分析等。通过解决这些问题，学生可以加深对这些关键概念的理解，并锻炼他们在实际编程场景中应用这些知识的能力。

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. 4S  1  2  3 .

. 4S  1  4  9  16 .

4243

CHAPTER 1

Introduction

1.4 The general way to do this is to write a procedure with heading

void processFile( String fileName );

which opens fileName, does whatever processing is needed, and then closes it. If a line of the form

#include SomeFile

is detected, then the call

processFile( SomeFile );

is made recursively. Self-referential includes can be detected by keeping a list of files for which a call to

processFile has not yet terminated, and checking this list before making a new call to processFile.

1.5 public static int ones( int n )

{

if( n < 2 )

return n;

return n % 2 + ones( n / 2 );

}

1.7 (a) The proof is by induction. The theorem is clearly true for 0 < X  1, since it is true for X = 1, and for X <

1, log X is negative. It is also easy to see that the theorem holds for 1 < X  2, since it is true for X = 2, and

for X < 2, log X is at most 1. Suppose the theorem is true for p < X  2p (where p is a positive integer), and

consider any 2p < Y  4p (p  1). Then log Y = 1 + log(Y/2)< 1 + Y/2 < Y/2 + Y/2  Y, where the first

inequality follows by the inductive hypothesis.

(b) Let 2

= A. Then A

= (2

)

= 2

. Thus log A

= XB. Since X = log A, the theorem is proved.

1.8 (a) The sum is 4/3 and follows directly from the formula.

(b)

S 



Subtracting the first equation from the second gives

3S  1 



. By part (a), 3S = 4/3 so S = 4/9.

(c)

S 



Subtracting the first equation from the second gives

(III) The running time is O(N

(IV) The running time is O(N

(V) j can be as large as i

, which could be as large as N

. k can be as large as j, which is N

. The running time

is thus proportional to NN



, which is O(N

(VI) The if statement is executed at most N

times, by previous arguments, but it is true only O(N

)

times (because it is true exactly i times for each i). Thus the innermost loop is only executed O(N

) times.

Each time through, it takes O(j

) = O(N

) time, for a total of O(N

). This is an example where multiplying

loop sizes can occasionally give an overestimate.

2.8 (a) It should be clear that all algorithms generate only legal permutations. The first two algorithms have tests

to guarantee no duplicates; the third algorithm works by shuffling an array that initially has no duplicates, so

none can occur. It is also clear that the first two algorithms are completely random, and that each permutation

is equally likely. The third algorithm, due to R. Floyd, is not as obvious; the correctness can be proved by

induction. See J. Bentley, “Programming Pearls,” Communications of the ACM 30 (1987), 754–757. Note

that if the second line of algorithm 3 is replaced with the statement

swap References( a[i], a[ ran dint( 0, n–1 ) ] );

then not all permutations are equally likely. To see this, notice that for N = 3, there are 27 equally likely ways

of performing the three swaps, depending on the three random integers. Since there are only 6 permutations,

and 6 does not evenly divide 27, each permutation cannot possibly be equally represented.

(b) For the first algorithm, the time to decide if a random number to be placed in a[i] has not been used

earlier is O(i). The expected number of random numbers that need to be tried is N/(N – i). This is obtained

as follows: i of the N numbers would be duplicates. Thus the probability of success is (N – i)/N. Thus the

expected number of independent trials is N/(N – i). The time bound is thus

N 1



N 1



i



i





i0





O(N

j 1

log N )

The second algorithm saves a factor of i for each random number, and thus reduces the time bound to

O(N log N) on average. The third algorithm is clearly linear.

(c,d) The running times should agree with the preceding analysis if the machine has enough memory. If not,

the third algorithm will not seem linear because of a drastic increase for large N.

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