CBMS-2016-09

题目来源：[[文字版题库/CBMS/2016#Problem 9|2016#Problem 9]] 日期：2024-07-31 题目主题：CS-算法-归并排序

解题思路

本题的主要任务是实现归并排序算法中的两个关键部分：合并两个有序子数组和递归地对数组进行排序。以下是详细的解题思路：

1. 合并两个有序子数组（merge_two_arrays 函数）：

   • 首先定义三个指针 i、j 和 k，分别指向两个子数组的起始位置和临时数组 y 的起始位置。
   • 使用一个 while 循环遍历两个子数组，比较每个子数组中的元素，将较小的元素放入临时数组 y 中，直到其中一个子数组的元素全部被处理完。
   • 如果其中一个子数组的元素全部被处理完，则将另一个子数组的剩余元素直接复制到临时数组 y 中。
   • 这一部分的核心是利用两个子数组的有序性，通过比较最小的元素逐一归并，确保合并后的数组依然有序。

2. 递归地对数组进行排序（merge_sort 函数）：

   • 使用递归的方法将数组不断地分成两半，直到每个子数组中只有一个元素为止。
   • 然后使用上述的 merge_two_arrays 函数将这些子数组逐步合并。
   • 递归的终止条件是子数组长度小于等于 1，此时不需要进一步分割。

3. 加速归并过程的硬件支持：

   • 提供一个名为 cmp(x1, x2, x3) 的函数，它能在一次操作中找出三个元素中最小的一个。利用这个函数，可以减少归并过程中所需的比较次数，从而提升性能。
   • 实现时，可在合并步骤中使用 cmp(x[i], x[j], INFINITY) 来比较两个有效元素和一个占位符，简化合并过程。

Solution

1. Completing the merge_two_arrays function

The task is to fill in the parts labeled as (A) and (B) to correctly merge two sorted subarrays.

void merge_two_arrays(int x[], int s, int m, int e, int y[]) {
    int i = s, j = m, k = s;
    while(i < m && j < e) {
        if(x[i] < x[j]) {
            y[k] = x[i]; // (A)
            i++; k++; // (A)
        } else {
            y[k] = x[j]; // (B)
            j++; k++; // (B)
        }
    }
    while(i < m) { y[k] = x[i]; k++; i++; }
    while(j < e) { y[k] = x[j]; k++; j++; }
}

Explanation:

• (A): When the element at x[i] is less than x[j], we copy x[i] into y[k] and increment both i and k.
• (B): When x[i] is not less than x[j], we copy x[j] into y[k] and increment both j and k.

2. Completing the merge_sort function

The task is to fill in the part labeled as (C) to correctly implement the merge sort algorithm.

void merge_sort(int x[], int s, int e, int y[]) {
    if(e - s <= 1) return;
    int m = (s + e) / 2;
    merge_sort(x, s, m, y); // (C)
    merge_sort(x, m, e, y); // (C)
    merge_two_arrays(x, s, m, e, y);
    for(int i = s; i < e; i++) { x[i] = y[i]; }
}

Explanation:

• (C): The merge sort algorithm recursively divides the array into two halves until each subarray has only one element. Then, it merges the subarrays using the merge_two_arrays function. The two recursive calls are to merge_sort(x, s, m, y) and merge_sort(x, m, e, y).

3. Worst-case time complexity of merge_sort()

The worst-case time complexity of merge sort is O(n \log n), where $n$ is the number of elements in the array. This complexity arises because the array is divided into two halves $\log n$ times, and each level of recursion requires $O(n)$ operations to merge the subarrays.

4. Accelerating merge_sort() Using the cmp() Function with Ternary Split

To further optimize the merge sort, we can leverage the hardware-supported cmp(x1, x2, x3) function by modifying the traditional two-way merge sort into a three-way (ternary) merge sort. This approach divides the array into three subarrays instead of two, which can reduce the depth of the recursive calls and potentially improve the sorting speed.

Approach

1. Three-way Split:

   • Divide the array into three nearly equal parts. For an array x[s..e-1], calculate the indices m1 and m2 such that m1 = s + (e - s) / 3 and m2 = s + 2 * (e - s) / 3.
   • The subarrays are then x[s..m1-1], x[m1..m2-1], and x[m2..e-1].

2. Merge Process:

   • Utilize the cmp(x1, x2, x3) function to find the smallest element among the current elements of the three subarrays.
   • Insert the smallest element into the temporary array y, and move the corresponding pointer forward.

3. Implementation:

   • Maintain three pointers for the start of each subarray and one for the position in the merged array.
   • Continue merging until all elements are processed.

Example Code

void merge_three_arrays(int x[], int s, int m1, int m2, int e, int y[]) {
    int i = s, j = m1, k = m2, l = s;
    while (i < m1 && j < m2 && k < e) {
        int cmp_result = cmp(x[i], x[j], x[k]);
        if (cmp_result == 1) {
            y[l] = x[i];
            i++;
        } else if (cmp_result == 2) {
            y[l] = x[j];
            j++;
        } else {
            y[l] = x[k];
            k++;
        }
        l++;
    }
    // Merge remaining elements of the first two arrays
    while (i < m1 && j < m2) {
        y[l++] = (x[i] < x[j]) ? x[i++] : x[j++];
    }
    // Merge remaining elements of the first and third arrays
    while (i < m1 && k < e) {
        y[l++] = (x[i] < x[k]) ? x[i++] : x[k++];
    }
    // Merge remaining elements of the second and third arrays
    while (j < m2 && k < e) {
        y[l++] = (x[j] < x[k]) ? x[j++] : x[k++];
    }
    // Copy any remaining elements
    while (i < m1) { y[l++] = x[i++]; }
    while (j < m2) { y[l++] = x[j++]; }
    while (k < e) { y[l++] = x[k++]; }
}
 
void merge_sort_ternary(int x[], int s, int e, int y[]) {
    if(e - s <= 1) return;
    int m1 = s + (e - s) / 3;
    int m2 = s + 2 * (e - s) / 3;
    merge_sort_ternary(x, s, m1, y);
    merge_sort_ternary(x, m1, m2, y);
    merge_sort_ternary(x, m2, e, y);
    merge_three_arrays(x, s, m1, m2, e, y);
    for(int i = s; i < e; i++) { x[i] = y[i]; }
}

Conclusion

By implementing a three-way merge sort using the cmp(x1, x2, x3) function, we reduce the number of levels in the recursion tree, leading to a shallower depth. While the time complexity remains $O(n\log n)$, the base of the logarithm changes from 2 to 3, resulting in fewer recursive steps and potentially faster execution, especially with large data sets. This method efficiently utilizes the three-element comparison capability provided by the hardware, optimizing the sorting process.

知识点

归并排序复杂度分析递归排序算法

难点思路

归并排序的难点主要在于如何有效地实现合并两个有序子数组。尤其是在大数据集下，如何优化合并过程中的比较操作是一个关键问题。通过引入专用的硬件指令（如 cmp 函数），可以有效减少比较次数，提高算法效率。

解题技巧和信息

• 分治法：将一个问题分成多个子问题，递归解决每个子问题，然后合并结果。归并排序即为典型的分治算法。
• 优化比较操作：在排序过程中，比较操作的次数对整体效率影响很大。可以通过硬件支持或者改进算法来减少不必要的比较。

重点词汇

• Merge Sort: 归并排序
• Recursion: 递归
• Time Complexity: 时间复杂度
• Space Complexity: 空间复杂度
• Hardware Acceleration: 硬件加速
• Comparison Operation: 比较操作

参考资料

1. 《算法导论》第三版，第 2 章：排序与查找
2. 《计算机算法设计与分析》第 5 章：分治策略
3. 《数据结构与算法分析：C 语言描述》：归并排序章节