通过重复生成所有排列

Posted 2023-02-22

【中文标题】通过重复生成所有排列

【英文标题】：Generating all permutations with repetition

【发布时间】：2013-05-08 17:06:58

【问题描述】：

我们如何在任何项目可以重复任意次数的时候生成 n（给定）不同项目的所有可能排列 r？

Combinatorics 告诉我会有 n^r 个，只是想知道如何用 C++/python 生成它们？

【参考方案1】：

这是一个可能的 C++ 实现，与标准库函数 std::next_permutation 类似

//---------------------------------------------------------------------------
// Variations with repetition in lexicographic order
// k: length of alphabet (available symbols)
// n: number of places
// The number of possible variations (cardinality) is k^n (it's like counting)
// Sequence elements must be comparable and increaseable (operator<, operator++)
// The elements are associated to values 0÷(k-1), max=k-1
// The iterators are at least bidirectional and point to the type of 'max'
template <class Iter>
bool next_variation(Iter first, Iter last, const typename std::iterator_traits<Iter>::value_type max)

    if(first == last) return false; // empty sequence (n==0)

    Iter i(last); --i; // Point to the rightmost element
    // Check if I can just increase it
    if(*i < max)  ++(*i); return true;  // Increase this element and return

    // Find the rightmost element to increase
    while( i != first )
       
        *i = 0; // reset the right-hand element
        --i; // point to the left adjacent
        if(*i < max)  ++(*i); return true;  // Increase this element and return
       

    // If here all elements are the maximum symbol (max=k-1), so there are no more variations
    //for(i=first; i!=last; ++i) *i = 0; // Should reset to the lowest sequence (0)?
    return false;
 // 'next_variation'

这就是用法：

std::vector<int> b(4,0); // four places initialized to symbol 0
do
   for(std::vector<int>::const_iterator ib=b.begin(); ib!=b.end(); ++ib)
      
       std::cout << std::to_string(*ib);
      
   std::cout << '\n';
  
while( next_variation(b.begin(), b.end(), 2) ); // use just 0-1-2 symbols

【参考方案2】：

将您的排列视为基于 n 的数字系统中的 r 位数字。从 000...0 开始，将“数字”加一：0000, 0001, 0002, 000(r-1), 0010, 0011, ...

代码很简单。

【讨论】：

能详细说明一下吗？

你要求 Inspired 离开这个国家吗？

@Quixotic：嗯，一定是谷歌搜索时打错了，因为我得到的只是“你是说外籍人士吗？”我现在正在寻找它。谢谢你的新词:)

好吧，如果我们不考虑离开我的国家的计划:) 原始问题的解决方案如下：int P[n] = 0; int k; dO /* do whatever you like with a permutation in p[] */ k = n-1; while (k >= 0) P[k]++; if (P[k] == r) P[k] = 0; k--; else break; while (k >= 0);

【参考方案3】：

这是@Inspired 方法的示例，其中 n 为字母表的前三个字母，r = 3：

alphabet = [ 'a', 'b', 'c' ]

def symbolic_increment( symbol, alphabet ):
    ## increment our "symbolic" number by 1
    symbol = list(symbol)
    ## we reverse the symbol to maintain the convention of having the LSD on the "right"
    symbol.reverse()
    place = 0;
    while place < len(symbol):
        if (alphabet.index(symbol[place])+1) < len(alphabet):
            symbol[place] = alphabet[alphabet.index(symbol[place])+1]
            break
        else:
            symbol[place] = alphabet[0];
            place+=1
    symbol.reverse()
    return ''.join(symbol)

permutations=[]
r=3
start_symbol = alphabet[0] * (r)
temp_symbol = alphabet[0] * (r)
while 1:
    ## keep incrementing the "symbolic number" until we get back to where we started
    permutations.append(temp_symbol)
    temp_symbol = symbolic_increment( temp_symbol, alphabet)
    if( temp_symbol == start_symbol ): break

你也可以用 itertools 来做：

from itertools import product

r=3
for i in xrange(r-1):
    if (i==0):
        permutations = list(product(alphabet, alphabet))
    else:
        permutations = list(product(permutations, alphabet))
    permutations = [ ''.join(item) for item in permutations ]