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170 lines
4.5 KiB
Markdown
170 lines
4.5 KiB
Markdown
---
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title: Union Find-Disjoint set
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sr-due: 2022-04-17
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sr-interval: 30
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sr-ease: 270
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---
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tags: #review
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---
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# 1 Union Find/Disjoint set
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## 1 Example
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- We have 12 'objects'
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- *Some* pairs have been connected
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- Nodes with a sequence of edges between them form a group
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- e.g., 0 5, 2, 1 4 6 9, 3 8 10 11, 7
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- Groups with no connecting edges are *disjoint* sets
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## 2 Requirements
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- Make(n) - make a set of n vertices with no edges between them
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- Union(x, y) - connect x and y by an edge (merge their two groups)
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- y becomes the representative node for the whole group
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- e,g,. Union(2, 1)
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- now : 0 5 2 1 4 6 9 3 8 10 11 7
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- the representative node of the new group is 1
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- the number of groups is always : n - number of union operations between elements of different groups
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- Find(x) Find and return a representative of the group the x belongs to.
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- If x and y are in the same group then Find(x) == Find(y)
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-
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## 3 Implementation
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### 3.1 UF 1
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```java
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int[] reps;
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public void make(int n){
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reps = new int[n];
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for(int = 0; i < n; i++) reps[i] = i;
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}
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public int find(int x){
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return reps[x];
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}
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public void union(int x, int y){
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rx = reps[x]
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ry = reps[y]
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for i = 0 to n-1
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if reps[i] = rx then
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reps[i] = ry
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end if
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end for
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}
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``````
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Operation | Cost | reason
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-------------|------| --
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make | $\Theta(n)$ | filling n place of an array
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find(x) | $\Theta(1)$ | find value in array is constant
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union(x, y) | $\Theta(n)$ | When x and y's rep are different, the whole array must be examined
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Total possible number of union calls where x and y's rep are different is n-1
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So the Total possible cost of all union calls is $\theta(n^2)$
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### 3.2 UF 2
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``` java
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int[] reps;
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public void make(int n){
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reps = new int[n];
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for(int = 0; i < n; i++) reps[i] = i;
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}
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public int find(int x){
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if(reps[x]==x) return x;
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return find(reps[x]);
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}
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public void union(int x, int y){
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reps[find(x)] = find(y);
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}
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```
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Operation | Cost | reason
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-------------|------| --
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make | $\Theta(n)$ | filling n place of an array
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find(x) | $\Theta(n)$ | need to look through chain nodes for representative
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union(x, y) | $\Theta(n)$ | bounded by two calls to find
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Total possible number of union calls where x and y's rep are different is n-1
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So the Total possible cost of all union calls is $\theta(n^2)$
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### 3.3 UF 3
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For each rep, let its rank be the length of the longest chain of local reps that reaches it
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When union(x,y) make the rep with the larger rank the rep of the other
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If equal ranks -> make the second the rep of the first
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``` java
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int[] reps;
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int[] rank;
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public void make(int n){
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reps = new int[n];
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rank = new int[n];
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for(int = 0; i < n; i++) reps[i] = i;
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}
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public int find(int x){
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if(reps[x]==x) return x;
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return find(reps[x]);
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}
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public void union(int x, int y){
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rootUnion(find(x), find(y))
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}
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//x and y are known to be representatives
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private void rootUnion(x, y){
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if(rank[x] > rank[y]){
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reps[y] = x;
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} else if (rank[y] > rank[x]){
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reps[x] = y;
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} else { //rank[x] == rank[y]
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reps[x] = y;
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rank[y] ++;
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}
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}
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```
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Operation | Cost | reason
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------------|------| --
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make | $\Theta(n)$ | filling n place of an array
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find(x) | $\Theta(lg\ n)$ | rank is bounded by $lg\ n$
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union(x, y) | $\Theta(lg\ n)$ | bounded by two calls to find
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Total possible number of union calls where x and y's rep are different is n-1
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So the Total possible cost of all union calls is $\theta(n^2)$
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trade off means this requires an extra $\theta(n)$ space
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#### 3.3.1 Min size of set of rank k
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for k = 0 -> size must be at least 1
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for k = 1 -> size must be at least 2
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for larger k -> the set must have been formed by the union of two sets of rank k-1. So its size must be at least twice the min size of a set of rank k-1
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--> min size of set of rank k is $2^k$
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#theorem
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>a set of rank k must contain at least $2^k$ elements
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$\therefore$ The maximum rank of an element is $\log_2(n)$ -> $lg(n)$
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since the time for $Find$ is big-$\theta$ of the rank of the representative found we get $O(lg n)$ bounds for both find and union
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^we used $O$ not $\theta$ because we dont know that the worst case will always occur.
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If could happen that the sequence of Union operations does not create a rank that is as big as i could be
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^this is an example of a semi-formal proof by [Induction](content/notes/induction.md)
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### 3.4 UF 4
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Change find so it implements [path compression](None) to "flatten" the chains
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```java
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if (x != reps[x]) {
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reps[x] = find(reps[x]);
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}
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return reps[x];
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```
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