Algorithm in Python - Union-Find 算法

Posted:   May 23, 2017

Edited:   September 30, 2017

Status:   Completed

Tags :   Python 算法

Categories :   Python 算法

Were equations, pictures or diagrams not properly rendered, please refresh the page. If the problem persists, you can contact me.

Union-Find in Python

Union-Find 算法(中文称并查集算法)是解决动态连通性(Dynamic Conectivity)问题的一种算法,作者以此为实例,讲述了如何分析和改进算法,本节涉及三个算法实现,分别是Quick Find, Quick Union 和 Weighted Quick Union。

动态连通性(Dynamic Connectivity)

动态连通性是计算机图论中的一种数据结构,动态维护图结构中相连接的组信息。 简单的说就是,图中各个点之间是否相连、连接后组成了多少个组等信息。我们称连接在一起就像形成了一个圈子似的,成为一个组(Component),每个组有其自己的一些特征,比如组内所有成员都有同一个标记等。

提到圈子,大家比较好理解,我们在社交网络中,彼此熟悉的人之间组成自己的圈子,”熟悉”用计算机中的语言来表示就是“Connected 连通的。圈子是会变化的,今天你又新认识了某人,明天你跟某人友尽了,这种变化是动态的,所以有了动态连通性这种数据结构和问题。

比较常见的应用有,社交网络中比如LinkedIn, 判断某个用户与其它用户是否熟悉:如果你与用户A熟悉,用户A与用户B熟悉,则认为你与用户B也是连接的,你可以看到用户B的信息。在计算机网络中,也存在类似的情况,判断网络中某两个节点是否相连。

Dynamic Connectivity 的计算机语言表述

给定一个整数对(p,q),如果p 和 q尚未连通,则使二者相连通。p 和 q 相连后,我们称 p 和 q 在同一个组内。

当p 和 q 连通时,以下关系则成立:

  • 自反性:p和p自身是相连的
  • 对称性:如果p和q相连,那么q和p也相连
  • 传递性:如果p和q相连,q和r相连,那么p和r也相连

在一个网络中,会存在很多类似的整数对(p,q),假设网络容量是 N,我们可以定义一个从 0 到 N-1的整数数组,p,q是其中的值,我们可能需要的操作有:

  • 判断 p 和 q 是否相连
  • 如果未相连,则连接 p 和 q, 如果已相连,则可以不做啥
  • 查找 p 或 q 属于哪个组中 (如圈子)

这里的一个关键是,如何确定 p 和 q 是在同一个组内。这意味着,每个组需要有一些特定的属性,我们在后面的算法中会有考虑。

Union-Find 算法描述 Dynamic Connectivity

Union-Find 算法中,提供了对应的方法来实现我们前面提到的可能的操作:

  • connected(): 判断 p 和 q 是否相连,这里要调用 find(p) 和 find(q),如果二者属于同一个组,则认为是相连的,即isConnected()返回true.
  • union(): 如果未相连,则连接 p 和 q, 如果已相连,则可以不做啥
  • find(): 查找 p 或 q 属于哪个组中 (如圈子),这里返回值是整数,作为组的标识符(component identifier)。

  • count(): 返回组的数量

算法4中的API:

class UF:
    def __init__(self,N):
    def union(self,p,q): # initialize N sites with integer names
    def find(self,p): #return component identifier for p
    def connected(self,p,q): #return true if p and q are in the same component
    def count(): #number of components

Union-Find 算法及实现

根据我们前面的描述,如果确定每个组的标识符似乎比较关键,只要确定了,就可以判断是否相连。

那用什么来作为标识符,区分各个组呢?

最简单的一个办法是,所有的节点都赋予一个 ID,如果两个节点相连,则将这两个节点的 ID 设成一样的,这样,这两个节点便属于同一个组了。网络中每个组都有了一个唯一的 ID。只要节点 p 和 q 的 ID 相同,则认为节点 p 和 q 相连。我们用数组来放置节点 ID,find()方法可以快速返回 ID,所以我们的第一个算法就叫做 QuickFind。

QuickFind 算法

QuickFind 算法中,find方法比较简单,union(p,q)方法需要考虑的一点是,要将与p相连的所有节点 id 都设为q当前的 id,使p所在的组和q所在的组结合成了一个同一组。(注:也可以把与q相连的所有节点id都设为p的id)

最开始的时候,所有节点都互不相连。我们假设所有的节点由id=0到N-1的整数表示。

图片.png

代码:

# -*- coding: utf-8 -*-

class QuickFind(object):
    id=[]
    count=0

    def __init__(self,n):
        self.count = n
        i=0
        while i<n:
            self.id.append(i)
            i+=1

    def connected(self,p,q):
        return self.find(p) == self.find(q)

    def find(self,p):    
        return self.id[p]

    def union(self,p,q):
        idp = self.find(p)
        if not self.connected(p,q):
            for i in range(len(self.id)):
                if self.id[i]==idp: # 将p所在组内的所有节点的id都设为q的当前id
                    self.id[i] = self.id[q]
            self.count -= 1

我们的测试端代码如下:

# -*- coding: utf-8 -*-

import quickfind

qf = quickfind.QuickFind(10)

print "initial id list is %s" % (",").join(str(x) for x in qf.id)

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for k in list:
    p =  k[0]
    q =  k[1]
    qf.union(p,q)
    print "%d and %d is connected? %s" % (p,q,str(qf.connected(p,q)    ))

print "final id list is %s" % (",").join(str(x) for x in qf.id)
print "count of components is: %d" % qf.count

运行结果:

initial id list is 0,1,2,3,4,5,6,7,8,9
4 and 3 is connected? True
3 and 8 is connected? True
6 and 5 is connected? True
9 and 4 is connected? True
2 and 1 is connected? True
8 and 9 is connected? True
5 and 0 is connected? True
7 and 2 is connected? True
6 and 1 is connected? True
1 and 0 is connected? True
6 and 7 is connected? True
final id list is 1,1,1,8,8,1,1,1,8,8
count of components is: 2

下图是算法4中的图示,可供参考:

图片.png

QuickFind 算法分析:

find方法快速返回数组的值,但union方法最坏情况下,几乎需要遍历整个数组,如果数组很大(比如社交网络巨大) 、需要连接的节点对很多的时候,QuickFind算法的复杂度就相当大了。所以我们需要改进一下union方法。

QuickUnion 算法

前面的QuickFind算法中,union的时候可能需要遍历整个数组,导致算法性能下降。有没有什么办法可以不用遍历整个数组,又可以保证同一个组内的所有节点都有一个共同属性呢?树结构。树的所有节点都有一个共同的根节点,每个树只有一个根节点,那每个树就可以代表一个组。union(p,q)的时候,只要把p所在的树附加到q所在的树的根节点,这样,p和q就在同一树中了。

改进后的算法即是QuickUnion算法。我们同样要用到 id 数组,只是这里的 id 放的是节点所在树的根节点。

find(p): 返回的是 p 所在树的根节点 union(p,q): 将 p 所在树的根节点的 id 设为 q 所在树的根节点

图片.png

代码实现:

# -*- coding: utf-8 -*-

class QuickUnion(object):
    id=[]
    count=0

    def __init__(self,n):
        self.count = n
        i=0
        while i<n:
            self.id.append(i)
            i+=1

    def connected(self,p,q):
        if self.find(p) == self.find(q):
            return True
        else:            
            return False

    def find(self,p):   
        while (p != self.id[p]):
            p = self.id[p]
        return p

    def union(self,p,q):
        idq = self.find(q)
        idp = self.find(p)
        if not self.connected(p,q):
            self.id[idp]=idq
            self.count -=1            

类似的测试端代码:

# -*- coding: utf-8 -*-

import quickunion

qf = quickunion.QuickUnion(10)

print "initial id list is %s" % (",").join(str(x) for x in qf.id)

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for k in list:
    p =  k[0]
    q =  k[1]
    qf.union(p,q)
    print "%d and %d is connected? %s" % (p,q,str(qf.connected(p,q)    ))

print "final root list is %s" % (",").join(str(x) for x in qf.id)
print "count of components is: %d" % qf.count

运行结果:

initial id list is 0,1,2,3,4,5,6,7,8,9
4 and 3 is connected? True
3 and 8 is connected? True
6 and 5 is connected? True
9 and 4 is connected? True
2 and 1 is connected? True
8 and 9 is connected? True
5 and 0 is connected? True
7 and 2 is connected? True
6 and 1 is connected? True
1 and 0 is connected? True
6 and 7 is connected? True
final root list is 1,1,1,8,3,0,5,1,8,8
count of components is: 2

算法4中的图示供参考理解:

图片.png

QuickUnion 算法分析:

union方法已经很快速了现在,find方法比QuickFind慢了,其最坏的情况下,如下图,一次find需要访问1+..+N次数组,union方法中需要调用两次find方法,即复杂度变成2(1+…+N)=(N+1)N,接近N的平方了。

图片.png

Weighted Quick Union 算法

前面的QuickUnion算法中,union的时候只是简单的将两个树合并起来,并没有考虑两个树的大小,所以导致最坏情况的发生。改进的方法可以是,在union之前,先判断两个树的大小(节点数量),将小点的树附加到大点的树上,这样,合并后的树的深度不会变得非常大。

示例如下:

图片.png

要判断树的大小,需要引进一个新的数组,size 数组,存放树的大小。初始化的时候 size 各元素都设为 1。

代码:

# -*- coding: utf-8 -*-

class WeightedQuickUnion(object):
    id=[]
    count=0
    sz=[]

    def __init__(self,n):
        self.count = n
        i=0
        while i<n:
            self.id.append(i)
            self.sz.append(1) # inital size of each tree is 1
            i+=1

    def connected(self,p,q):
        if self.find(p) == self.find(q):
            return True
        else:            
            return False

    def find(self,p):   
        while (p != self.id[p]):
            p = self.id[p]
        return p

    def union(self,p,q):
        idp = self.find(p)
        print "id of %d is: %d" % (p,idp)
        idq = self.find(q)
        print "id of %d is: %d" % (q,idq)
        if not self.connected(p,q):            
            print "Before Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
            print "Before Connected: tree size of %d's id is: %d" % (q,self.sz[idq])
            if (self.sz[idp] < self.sz[idq]):
                print "tree size of %d's id is smaller than %d's id" %(p,q)
                print "id of %d's id (%d) is set to %d" % (p,idp,idq)
                self.id[idp] = idq

                print "tree size of %d's id is incremented by tree size of %d's id" %(q,p)
                self.sz[idq] += self.sz[idp]    
                print "After Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
                print "After Connected: tree size of %d's id is: %d" % (q,self.sz[idq])         
            else:                  
                print "tree size of %d's id is larger than or equal with %d's id" %(p,q)
                print "id of %d's id (%d) is set to %d" % (q,idq,idp)
                self.id[idq] = idp
                print "tree size of %d's id is incremented by tree size of %d's id" %(p,q)
                self.sz[idp] += self.sz[idq]   
                print "After Connected: tree size of %d's id is: %d" % (p,self.sz[idp])
                print "After Connected: tree size of %d's id is: %d" % (q,self.sz[idq])         

            self.count -=1    

测试端代码:

# -*- coding: utf-8 -*-

import weightedquickunion

qf = weightedquickunion.WeightedQuickUnion(10)

print "initial id list is %s" % (",").join(str(x) for x in qf.id)

list = [
        (4,3),
        (3,8),
        (6,5),
        (9,4),
        (2,1),
        (8,9),
        (5,0),
        (7,2),
        (6,1),
        (1,0),
        (6,7)
        ]

for k in list:
    p =  k[0]
    q =  k[1]
    print "." * 10 + "unioning %d and %d"  % (p,q)  + "." * 10
    qf.union(p,q)
    print "%d and %d is connected? %s" % (p,q,str(qf.connected(p,q)    ))

print "final id list is %s" % (",").join(str(x) for x in qf.id)
print "count of components is: %d" % qf.count

代码运行结果:

initial id list is 0,1,2,3,4,5,6,7,8,9
..........unioning 4 and 3..........
id of 4 is: 4
id of 3 is: 3
Before Connected: tree size of 4's id is: 1
Before Connected: tree size of 3's id is: 1
tree size of 4's id is larger than or equal with 3's id
id of 3's id (3) is set to 4
tree size of 4's id is incremented by tree size of 3's id
After Connected: tree size of 4's id is: 2
After Connected: tree size of 3's id is: 1
4 and 3 is connected? True
..........unioning 3 and 8..........
id of 3 is: 4
id of 8 is: 8
Before Connected: tree size of 3's id is: 2
Before Connected: tree size of 8's id is: 1
tree size of 3's id is larger than or equal with 8's id
id of 8's id (8) is set to 4
tree size of 3's id is incremented by tree size of 8's id
After Connected: tree size of 3's id is: 3
After Connected: tree size of 8's id is: 1
3 and 8 is connected? True
..........unioning 6 and 5..........
id of 6 is: 6
id of 5 is: 5
Before Connected: tree size of 6's id is: 1
Before Connected: tree size of 5's id is: 1
tree size of 6's id is larger than or equal with 5's id
id of 5's id (5) is set to 6
tree size of 6's id is incremented by tree size of 5's id
After Connected: tree size of 6's id is: 2
After Connected: tree size of 5's id is: 1
6 and 5 is connected? True
..........unioning 9 and 4..........
id of 9 is: 9
id of 4 is: 4
Before Connected: tree size of 9's id is: 1
Before Connected: tree size of 4's id is: 3
tree size of 9's id is smaller than 4's id
id of 9's id (9) is set to 4
tree size of 4's id is incremented by tree size of 9's id
After Connected: tree size of 9's id is: 1
After Connected: tree size of 4's id is: 4
9 and 4 is connected? True
..........unioning 2 and 1..........
id of 2 is: 2
id of 1 is: 1
Before Connected: tree size of 2's id is: 1
Before Connected: tree size of 1's id is: 1
tree size of 2's id is larger than or equal with 1's id
id of 1's id (1) is set to 2
tree size of 2's id is incremented by tree size of 1's id
After Connected: tree size of 2's id is: 2
After Connected: tree size of 1's id is: 1
2 and 1 is connected? True
..........unioning 8 and 9..........
id of 8 is: 4
id of 9 is: 4
8 and 9 is connected? True
..........unioning 5 and 0..........
id of 5 is: 6
id of 0 is: 0
Before Connected: tree size of 5's id is: 2
Before Connected: tree size of 0's id is: 1
tree size of 5's id is larger than or equal with 0's id
id of 0's id (0) is set to 6
tree size of 5's id is incremented by tree size of 0's id
After Connected: tree size of 5's id is: 3
After Connected: tree size of 0's id is: 1
5 and 0 is connected? True
..........unioning 7 and 2..........
id of 7 is: 7
id of 2 is: 2
Before Connected: tree size of 7's id is: 1
Before Connected: tree size of 2's id is: 2
tree size of 7's id is smaller than 2's id
id of 7's id (7) is set to 2
tree size of 2's id is incremented by tree size of 7's id
After Connected: tree size of 7's id is: 1
After Connected: tree size of 2's id is: 3
7 and 2 is connected? True
..........unioning 6 and 1..........
id of 6 is: 6
id of 1 is: 2
Before Connected: tree size of 6's id is: 3
Before Connected: tree size of 1's id is: 3
tree size of 6's id is larger than or equal with 1's id
id of 1's id (2) is set to 6
tree size of 6's id is incremented by tree size of 1's id
After Connected: tree size of 6's id is: 6
After Connected: tree size of 1's id is: 3
6 and 1 is connected? True
..........unioning 1 and 0..........
id of 1 is: 6
id of 0 is: 6
1 and 0 is connected? True
..........unioning 6 and 7..........
id of 6 is: 6
id of 7 is: 6
6 and 7 is connected? True
final id list is 6,2,6,4,4,6,6,2,4,4
count of components is: 2

算法4中的图示:

图片.png

Comments


Be the first one to comment on this page!
You can use extended GitHub flavored markdown in your comment. Commenting FAQs & Guidelines