The Combination Function and Iterator using Depth First Search A

  • 时间:2020-09-13 14:33:25
  • 分类:网络文摘
  • 阅读:157 次

The combination algorithm returns the sequence for a list or string without considering the order. For example, the Combination of string “abc” with two characters would be “ab”, “ac” and “bc”.

The number of combination for n-size elements using m elements can be computed as:

combination-formula The Combination Function and Iterator using Depth First Search Algorithm algorithms c / c++ DFS python

combination-formula

For example, Picking 2 elements out of 3 has three solutions: 3!/(2!*(3-2)!) = 3

Solving Combination via DFS Algorithm

We can use the DFS (Depth First Search) algorithm to start picking elements until we have picked the desired number of items. For each iteration, we can choose to pick or not pick the current element. The DFS can be easily implemented via Recursion.

In the following C++ function, we define a recursive combination function that will push all the results into a vector (pass by reference). We pass a position variable and we will only pick in current iteration the elements to the right of this position.

1
2
3
4
5
6
7
8
9
10
11
12
void combination(string alphabeta, int toPick, vector<string> &result, string s = "", int pos = 0) {
    if (s.size() == toPick) { // found a solution
        result.push_back(s);            
    }
    if (pos == alphabeta.size()) { // reach the end of items
        return;
    }
    for (int i = pos; i < alphabeta.size(); ++ i) { 
        // picking this element, then start picking elements to the right
        combination(alphabeta, toPick, result, s + alphabeta[i], i + 1);
    }
}
void combination(string alphabeta, int toPick, vector<string> &result, string s = "", int pos = 0) {
    if (s.size() == toPick) { // found a solution
        result.push_back(s);            
    }
    if (pos == alphabeta.size()) { // reach the end of items
        return;
    }
    for (int i = pos; i < alphabeta.size(); ++ i) { 
        // picking this element, then start picking elements to the right
        combination(alphabeta, toPick, result, s + alphabeta[i], i + 1);
    }
}

However, the DFS may not be optimal, the complexity is N elements first iteration, N-1 second iteration etc thus N*(N-1)*(N-2)*…(N-M), roughly to O(N^M).

Python Combination Function

We can implement in Python the following combination function that works for both List and String:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
def combinations(data, length):
    r = []
    def dfs(cur, pos):
        if len(cur) == length:
            r.append(cur)
        if pos == len(data):
            return
        if type(data) is list:
            for i in range(pos, len(data)):
                dfs(cur + [data[i]], i + 1)
        else:
            for i in range(pos, len(data)):
                dfs(cur + data[i], i + 1)
    if type(data) is list:        
        dfs([], 0)
    else:        
        dfs("", 0)
    return r
def combinations(data, length):
    r = []
    def dfs(cur, pos):
        if len(cur) == length:
            r.append(cur)
        if pos == len(data):
            return
        if type(data) is list:
            for i in range(pos, len(data)):
                dfs(cur + [data[i]], i + 1)
        else:
            for i in range(pos, len(data)):
                dfs(cur + data[i], i + 1)
    if type(data) is list:        
        dfs([], 0)
    else:        
        dfs("", 0)
    return r

A local recursive DFS function is defined, and we use type() to deal with the list (array) or the string. Example usage:

1
2
3
4
5
6
print(combinations([1, 2, 3, 4], 2))
>>>  [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
print(combinations(['a', 'b', 'c', 'd'], 2))
>>>  [['a', 'b'], ['a', 'c'], ['a', 'd'], ['b', 'c'], ['b', 'd'], ['c', 'd']]
print(combinations('abcd', 2))
>>>  ['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
print(combinations([1, 2, 3, 4], 2))
>>>  [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
print(combinations(['a', 'b', 'c', 'd'], 2))
>>>  [['a', 'b'], ['a', 'c'], ['a', 'd'], ['b', 'c'], ['b', 'd'], ['c', 'd']]
print(combinations('abcd', 2))
>>>  ['ab', 'ac', 'ad', 'bc', 'bd', 'cd']

Python Combination Iterator

If we are not using all the combination results at once, we can modify the above Python algorithm to return a iterator – which avoids sucking up memory at once. Each result is yield-ed.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
def combinations(data, length):
    def dfs(cur, pos):
        if len(cur) == length:
            yield cur
        if pos == len(data):
            return
        if type(data) is list:
            for i in range(pos, len(data)):
                for j in dfs(cur + [data[i]], i + 1):
                    yield j
        else:
            for i in range(pos, len(data)):
                for j in dfs(cur + data[i], i + 1):
                    yield j
    if type(data) is list:        
        return dfs([], 0)
    else:        
        return dfs("", 0)
def combinations(data, length):
    def dfs(cur, pos):
        if len(cur) == length:
            yield cur
        if pos == len(data):
            return
        if type(data) is list:
            for i in range(pos, len(data)):
                for j in dfs(cur + [data[i]], i + 1):
                    yield j
        else:
            for i in range(pos, len(data)):
                for j in dfs(cur + data[i], i + 1):
                    yield j
    if type(data) is list:        
        return dfs([], 0)
    else:        
        return dfs("", 0)

As the combinations function now returns an iterator, we can use it like this:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
for i in combinations_iterator([1, 2, 3, 4], 2):
    print(i)
# [1, 2]
# [1, 3]
# [1, 4]
# [2, 3]
# [2, 4]
# [3, 4]
for i in combinations_iterator(['a', 'b', 'c', 'd'], 2):
    print(i)
# ['a', 'b']
# ['a', 'c']
# ['a', 'd']
# ['b', 'c']
# ['b', 'd']
# ['c', 'd']
for i in combinations_iterator('abcd', 2):
    print(i)
# ab
# ac
# ad
# bc
# bd
# cd
for i in combinations_iterator([1, 2, 3, 4], 2):
    print(i)
# [1, 2]
# [1, 3]
# [1, 4]
# [2, 3]
# [2, 4]
# [3, 4]
for i in combinations_iterator(['a', 'b', 'c', 'd'], 2):
    print(i)
# ['a', 'b']
# ['a', 'c']
# ['a', 'd']
# ['b', 'c']
# ['b', 'd']
# ['c', 'd']
for i in combinations_iterator('abcd', 2):
    print(i)
# ab
# ac
# ad
# bc
# bd
# cd
python The Combination Function and Iterator using Depth First Search Algorithm algorithms c / c++ DFS python

python

Another interesting read to implement the combination using bitmasking algorithm: Using Bitmasking Algorithm to Compute the Combinations of an Array

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读:
近期百度SEO优化的规则有哪些变化?  影响中小企业SEO排名的五个地方  权重的高低与网站收录有何关系?两者有什么关联性!  Digital Business Cards App Will Change The Way You Network Forev  3 Things You Need To Know When Launching Your Startup’s Blog  Instagram Influencer Marketing Is A Billion Dollar Industry [Inf  5 Social Adverts for Driving Stellar Webinar Attendance (Infogra  5 Ways to Serve Up a Tastier Food Blog to Your Audience  Meet These Single Moms That Created Successful Blogs  Boost Your SERP Rankings with Better Marketing Automation 
评论列表
添加评论