CBMS-2020-07

题目来源：[[做题/文字版题库/CBMS/2020#Question 7|2020#Question 7]] 日期：2024-07-23 题目主题：CS-算法-动态规划

解题思路

我们将从 5 个子问题开始解决，并逐步分析其结果和相关性质。最后，将这些性质结合起来，用于设计计算 $S(A)$ 的算法，并分析其时间复杂度。

Solution

Question 1

Given $A=\{4,1,10,3,2\}$:

For $W=11$

To find $S(A)$ for $W=11$, we need to consider the operation condition $a_i+a_{i+1}<W$. We start with the initial array and systematically apply all valid swaps.

Initial array: $\{4,1,10,3,2\}$

We need to check all possible adjacent swaps under the condition $a_i+a_{i+1}<11$:

1. Swap $4$ and $1$ (since $4+1=5<11$):

$$\{1,4,10,3,2\}$$

1. Swap $1$ and $4$ back (since $1+4=5<11$):

$$\{4,1,10,3,2\}$$

1. Swap $10$ and $3$ (since $10+3=13\nless 11$):

$$\text{No swap}$$

1. Swap $3$ and $2$ (since $3+2=5<11$):

$$\{4,1,10,2,3\}$$

All valid permutations for $W=11$ are:

1. $\{4,1,10,3,2\}$
2. $\{1,4,10,3,2\}$
3. $\{4,1,10,2,3\}$
4. $\{1,4,10,2,3\}$

Therefore, for $W=11$:

$$S(A)=4$$

For $W=12$

For $W=12$, we need to explore more possible swaps as the condition is more lenient.

Initial array: $\{4,1,10,3,2\}$

We start with the initial array and apply all valid swaps:

1. Swap $4$ and $1$ (since $4+1=5<12$):

$$\{1,4,10,3,2\}$$

1. Swap $1$ and $10$ (since $1+10=11<12$):

$$\{4,10,1,3,2\}$$

1. Swap $3$ and $2$ (since $3+2=5<12$)

$$\{4,1,10,2,3\}$$

Now explore permutations of $\{1,4,10,3,2\}$:

1. Swap $3$ and $2$ (since $3+2=5<12$):

$$\{1,4,10,2,3\}$$

Now explore permutations of $\{4,10,1,3,2\}$:

1. Swap $1$ and $3$ (since $1+3=4<12$):

$$\{4,10,3,1,2\}$$

1. Swap $3$ and $2$ (since $3+2=5<12$):

$$\{4,10,1,2,3\}$$

Now explore permutations of $\{4,10,3,1,2\}$:

1. Swap $1$ and $2$ (since $1+2=3<12$):

$$\{4,10,3,2,1\}$$

Now explore permutations of $\{4,10,1,3,2\}$:

Similarly, we get:

$$\{4,10,2,3,1\},\{4,10,2,1,3\}$$

Thus, the valid permutations for $W=12$ are:

1. $\{4,1,10,3,2\}$
2. $\{1,4,10,3,2\}$
3. $\{4,1,10,2,3\}$
4. $\{1,4,10,2,3\}$
5. $\{4,10,1,3,2\}$
6. $\{4,10,1,2,3\}$
7. $\{4,10,2,3,1\}$
8. $\{4,10,2,1,3\}$
9. $\{4,10,3,1,2\}$
10. $\{4,10,3,2,1\}$

Therefore, for $W=12$:

$$S(A)=10$$

Question 2

When $a_h+a_l<W$, prove $S(A)=N\cdot S(A-a_l)$.

Given $a_h+a_l<W$, the smallest number $a_l$ can be swapped with any adjacent element and can move to any position in the array.

Therefore, the number of permutations of $A$ is equal to the number of permutations of $A-a_l$, multiplied by the number of possible positions for $a_l$, which is $N$. Hence, we have:

$$S(A)=N\cdot S(A-a_l)$$

P.S: 排列组合计数问题中的插板法

Question 3

When $a_h+a_l\ge W$, prove $S(A)=S(\{a_1,a_2,\dots ,a_{h-1}\})\cdot S(\{a_{h+1},a_{h+2},\dots ,a_N\})$.

If $a_h+a_l\ge W$, then $a_h$ and $a_l$ , or $a_h$ and any other element of the array, cannot be swapped, effectively dividing $A$ into two independent subarrays. Thus, the total number of distinct arrays is the product of the number of distinct arrays of each subarray:

$$S(A)=S(\{a_1,a_2,\dots ,a_{h-1}\})\cdot S(\{a_{h+1},a_{h+2},\dots ,a_N\})$$

P.S: 排列组合计数问题中的乘法原理

Question 4

Algorithm to compute $S(A)$:

def compute_S(A, W):
    if len(A) == 0:
        return 1
	
	# O(N)
    a_h = max(A)
    a_l = min(A)
    
    if a_h + a_l < W:
	    # O(N)
        A_minus_a_l = [x for x in A if x != a_l]
        return len(A) * compute_S(A_minus_a_l, W)
        # T(N-1)
    else:
        index_h = A.index(a_h)
        left_part = A[:index_h]
        right_part = A[index_h+1:]
        return compute_S(left_part, W) * compute_S(right_part, W)
		# 2*T(N/2)
 
# Example
A = [4, 1, 10, 3, 2]
W = 11
print(compute_S(A, W))  # Output: 4
W = 12
print(compute_S(A, W))  # Output: 10

Question 5

Given that the worst-case scenario involves removing the smallest element each time, the time complexity can be analyzed as follows:

1. Initial size: $N$
2. Operation: Finding the smallest element (which takes $O(N)$ time) and removing it, reducing the problem size by 1.
3. Recurrence relation: The total time complexity is the sum of the times taken for each step as we reduce the size of the array from $N$ to 0.

The time complexity can be expressed as:

$$T(N)=N+(N-1)+(N-2)+\dots +1$$

This is the sum of the first $N$ natural numbers:

$$T(N)=\sum _{i=1}^{N}i=\frac{N(N+1)}{2}$$

Therefore, the worst-case time complexity is:

$$T(N)=O(N^2)$$

Question 6

Assuming that each of the $N!$ permutations of the array is equally likely, and given that the average scenario does not always involve removing the smallest element each time, the time complexity will be different.

In the average case, the algorithm will involve both scenarios of removing the smallest element and splitting the array. However, the average number of operations will not always hit the worst-case scenario.

Considering the balanced approach where the split operation happens frequently, the recurrence relation can be described more favorably compared to the worst-case:

$$T_{\text{avg}}(N)=2T(N/2)+O(N)$$

Using the Master Theorem for $a=2$, $b=2$, and $f(N)=O(N)$:

$$T_{\text{avg}}(N)=O(N\log N)$$

Thus, the average time complexity is:

$$T_{\text{avg}}(N)=O(N\log N)$$

知识点

组合计数递归复杂度分析主定理

解题技巧和信息

1. Recursive Reduction: Understand how reducing a problem by removing an element or splitting the array affects complexity.
2. Complexity Analysis: Sum of first $N$ numbers is $\frac{N(N+1)}{2}$.
3. Master Theorem: Useful for solving recurrence relations in divide-and-conquer algorithms.

重点词汇

• worst-case 最坏情况
• average-case 平均情况
• time complexity 时间复杂度
• recurrence relation 递推关系

参考资料

1. Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein, Chapter 4.