[Algorithms I] Week 1-1 Union-Find

1. Dynamic Connectivity pb

pb statement

a set of N obj, indexed by 0,1,...,N-1 UNION: connect objects void union(int p, int q) FIND: is there a path connecting 2 obj? boolean connected(int p, int q)


connect components(联通分支): max set of obj that are mutually connected.


  • union(p,q): connect 2 obj
  • connected(p,q): test if p and q are connected
  • find(p): find the component id of p
  • count(): nb of components

注意: 命名不是很好, 这里的find()函数不对应FIND query, connected()函数才是真正的FIND query, find()函数是为了connected()函数而做的一个辅助函数(find(p): return the root of the node p ) 在connected里就可以调用find: return find(p)==find(q)


public interface UF{
void union(int p, int q);
boolean connected(int p, int q);
//int find(int p);
//int count();


2. Quick Find

"eager approach"

data structure

⇒ an int array id[] initialized to id[p]=p for all p interpretation: id[p] = component id of obj p ⇒ p and q are connected iff* id[p]==id[q] (ie. find very fast) 没有用find()函数

UF operations

  • FIND:

connected(p,q): very fast, just check id[p] and id[q]

  • UNION:

when merging 2 components : union(p,q): id[p]=id[q](总是让第一个参数p的id变为第二个参数q的id),
⇒ then have to modify all entries with id equal to id[p] !
⇒ too many entries to change


(class QuickFindUF implements UF)


  • FIND: cte
  • UNION: lin...

if N obj + N unions ⇒ quad time !
btw, 程序运行速度: ~10^9/s

3. Quick Union

"lazy approach"

data structure

⇒ also an int array id[] considering a set of trees, 此时每个联通分支都是一个tree
interpretation: id[p] = parent index of obj p (觉得这个数组叫做father更好....)
⇒ p is a root node

UF operations

  • FIND:

connected(p,q):check if root of p == root of q

  • UNION:

union(p,q):just set p's root to be child of q's root (把第一个参数p的那棵树放入第二个参数q的树的根节点作为子树)

  • root():

前两个的操作都需要一个函数查找一个节点的root, 需要写一个函数实现, 也很简单, 一路找parent即可:

private int root(int p){
    while (p!=id[p]) p=id[p];
    return p;



in the worst case (all elements is in a list form), root() is ~N, so:

  • FIND: lin
  • UNION: lin

quick find和quick union的问题:

4. Quick Union Improvements

improvement1: weighting

keep track of tree size ⇒ balance by taking the small tree be a child of the large tree

⇒ add an extra array: sz[] sz[i] is the size of the tree with root i


(数组int sz[] 初始全部为1)

  • 依然需要root()函数.

    private int root(int p){ while(p!=id[p]) p=id[p]; return p; }

  • FIND

    public boolean connected(int p, int q){ return root(p)==root(q); }


    public void union(int p, int q){ int rp = root(p), rq=root(q); if(rp==rq) return; // if (sz[rp]<sz[rq]){ id[rp]=rq; sz[rq]+=sz[rp]; } else{...} }


FIND: proportional to depth of p and q in their tree UNION: const if p and q are root

  • proposition

the max depth of weightedQuickUnion is lgN

[pf] considering a node x, in tree T1, dep(x) is x's depth in its tree.

dep(x) will increase by 1, iff T1 is merged into another tree T2 (and by the algo, shoud have |T1|<=|T2| )
→ x's tree's size become |T1|+|T2| >= 2|T1| ⇒ everytime dep(x) increased by 1, x's tree's size will at least double*
at first dep(x)=1, if dep(x) increases lgN times, the size of the tree will be >= N

so the root() function takes only lgN time. conclusion: both UNION and FIND will be in lgN time.

improvement 2: path compression

  • imporve the root() function:

when looking for root of a node ⇒ link all nodes in the path up to the root.
⇒ just a constant extra time compared to old implementation.

  • 2 pass implementation:

    private int root(int p){ int r = p; while(r!=id[r]) r=id[r]; while(p!=r){ int t = p; p=id[p]; id[t]=r; } return r; }

flatens the tree greatly.

  • single pass implementation:

just make all other node point to its grandparent (halving the path length) ⇒ not as flatening as before, but in practice will almost be the same. just one extra line of code: private int root(int p){ while(p!=id[p]){ id[p] = id[ id[p] ]; p=id[p];
} return p; }


(for weighet quick union with path compression — WQUPC)
very very small:

lg*() function: "iterated log function", lg(N) = the number of time to take log to get to 1 lg()几乎可以看成常数了:

ex. lg(65536) = 4* (x^16=65536)
because: lg(65536)=16 ; lg(16) = 4; lg(4)=2; lg(2)=1.
⇒ N obj, M unions will take (almost) linear time

(有人证明了不存在理论上linear的算法. )

conclusion: both UNION and FIND will be in constant time.


上面这个表格好像quick union的部分有问题? 最坏情况下应该是N+MN吧??

WQUCF reduce 30 years to 6 seconds.

5. Union Find Application

  • percolation
  • dynamic connectivity
  • Kruskal MST algo
  • Games (GO)
  • .......


  • model:

NN grid of sites
⇒ each site is open with proba=
⇒ sys percolate iff bottom and top are connected by open sites.

  • question: the percolation probability as a function of p (phase transition)

nobody knows how to get the threshold mathematically
⇒ run simulations to find out the phase transition threshold.

  • Monte Carlo simulation

→ all sites initilized to be closed
→ randomly open sites one by one
→ when the sys percolates, the vacancy percentage is an estimate of p
*(run above simulation for millions of times)


  • N^2 sites, named 0 to N^2-1

  • add 2 more vertual sites: one on top, one on bottom

  • openning a site: union to adjcent open sites (at most 4 unions)
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