计算不是N周期的不同正常括号序列的数量

In this article, we're going to delve into an intriguing problem from the realm of combinatorics and string processing: "Counting distinct regular bracket sequences which are not N periodic". This problem involves generating distinct valid bracket sequences and then filtering out sequences that are N-periodic. We'll discuss the problem, provide a C++ code implementation of a brute-force approach, and explain a test case.

Understanding the Problem Statement

给定一个整数N，任务是计算长度为2N的不是N周期的不同的正则括号序列的数量。如果一个序列可以表示为字符串S重复M次，其中S的长度为N且M>1，则该序列是N周期的。

A regular bracket sequence is a string that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters. For example, the sequence "(())" is regular, while ")(" and "(()" are not.

方法

由于问题的复杂性，直接的数学解决方案可能不可行。相反，我们需要使用编程方法来生成括号序列，并计算符合我们条件的括号序列的数量。

We will use a recursive function to generate all possible bracket sequences of length 2N. While generating the sequences, we'll keep track of two important parameters: the balance of the brackets (the number of open brackets should never be less than the number of closed brackets) and the number of open brackets (should not exceed N).

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为了筛选出N周期序列，我们将检查每个生成的序列。如果该序列是长度为N的较小序列的重复，我们将从计数中排除它。

C++ Implementation

这是一个用C++解决问题的蛮力方法。该方法生成所有可能的括号序列，并检查每个序列是否是N周期的，如果不是则增加计数。这个解决方案适用于小规模的输入，因为它具有指数时间复杂度，并且对于较大的输入不具有良好的可扩展性。

Example

#include 
using namespace std;

// Function to check if string is N periodic
bool isPeriodic(string s, int N) {
   string sub = s.substr(0, N);
   for (int i = N; i < s.size(); i += N) {
      if (sub != s.substr(i, N)) return false;
   }
   return true;
}

// Function to generate all bracket sequences
void generateBrackets(string s, int open, int close, int N, int &count) {
   // If sequence is complete
   if (s.size() == 2*N) {
      if (!isPeriodic(s, N)) count++;
      return;
   }
   
   // Add an open bracket if we have some left
   if (open < N) generateBrackets(s + "(", open + 1, close, N, count);
   
   // Add a close bracket if it doesn't make the sequence invalid
   if (close < open) generateBrackets(s + ")", open, close + 1, N, count);
}

int main() {
   int N = 3;
   int count = 0;
   generateBrackets("", 0, 0, N, count);
   cout << "Count of sequences: " << count;
   return 0;
}

Output

Count of sequences: 5

Test Case

Let's consider a test case with N = 3. This code will generate all 5 distinct regular bracket sequences of length 6 that are not 3-periodic: ((())), (()()), (())(), ()(()), ()()().

Conclusion

本文介绍了一种暴力解决非N周期的不同正则括号序列计数问题的方法。虽然这种方法可以处理小规模的输入，但需要注意的是，由于需要生成和检查所有可能的括号序列，该问题具有指数复杂度，因此不适用于大规模输入。