Write a Nonrecursive Version of Findset() With Path Compression.
Check whether a given graph contains a cycle or not.
Example:
Input:
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Output: Graph contains Cycle. Input:
Output: Graph does not contain Cycle.
Prerequisites: Disjoint Set (Or Union-Find), Union By Rank and Path Compression
We have already discussed union-find to detect cycle. Here we discuss find by path compression, where it is slightly modified to work faster than the original method as we are skipping one level each time we are going up the graph. Implementation of find function is iterative, so there is no overhead involved.Time complexity of optimized find function is O(log*(n)), i.e iterated logarithm, which converges to O(1) for repeated calls.
Refer this link for
Proof of log*(n) complexity of Union-Find
Explanation of find function:
Take Example 1 to understand find function:
(1)call find(8) for first time and mappings will be done like this:
It took 3 mappings for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 7, Reached node 6.
From node 6, skipped node 5, Reached node 4.
From node 4, skipped node 2, Reached node 0.
(2)call find(8) for second time and mappings will be done like this:
It took 2 mappings for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 5, node 6 and node 7, Reached node 4.
From node 4, skipped node 2, Reached node 0.
(3)call find(8) for third time and mappings will be done like this:
Finally, we see it took only 1 mapping for find function to get the root of node 8. Mappings are illustrated below:
From node 8, skipped node 5, node 6, node 7, node 4, and node 2, Reached node 0.
That is how it converges path from certain mappings to single mapping.
Explanation of example 1:
Initially array size and Arr look like:
Arr[9] = {0, 1, 2, 3, 4, 5, 6, 7, 8}
size[9] = {1, 1, 1, 1, 1, 1, 1, 1, 1}
Consider the edges in the graph, and add them one by one to the disjoint-union set as follows:
Edge 1: 0-1
find(0)=>0, find(1)=>1, both have different root parent
Put these in single connected component as currently they doesn't belong to different connected components.
Arr[1]=0, size[0]=2;
Edge 2: 0-2
find(0)=>0, find(2)=>2, both have different root parent
Arr[2]=0, size[0]=3;
Edge 3: 1-3
find(1)=>0, find(3)=>3, both have different root parent
Arr[3]=0, size[0]=3;
Edge 4: 3-4
find(3)=>1, find(4)=>4, both have different root parent
Arr[4]=0, size[0]=4;
Edge 5: 2-4
find(2)=>0, find(4)=>0, both have same root parent
Hence, There is a cycle in graph.
We stop further checking for cycle in graph.
C++
#include <bits/stdc++.h>
using
namespace
std;
const
int
MAX_VERTEX = 101;
int
Arr[MAX_VERTEX];
int
size[MAX_VERTEX];
void
initialize(
int
n)
{
for
(
int
i = 0; i <= n; i++) {
Arr[i] = i;
size[i] = 1;
}
}
int
find(
int
i)
{
while
(Arr[i] != i)
{
Arr[i] = Arr[Arr[i]];
i = Arr[i];
}
return
i;
}
void
_union(
int
xr,
int
yr)
{
if
(size[xr] < size[yr])
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
int
isCycle(vector<
int
> adj[],
int
V)
{
for
(
int
i = 0; i < V; i++) {
for
(
int
j = 0; j < adj[i].size(); j++) {
int
x = find(i);
int
y = find(adj[i][j]);
if
(x == y)
return
1;
_union(x, y);
}
}
return
0;
}
int
main()
{
int
V = 3;
initialize(V);
vector<
int
> adj[V];
adj[0].push_back(1);
adj[0].push_back(2);
adj[1].push_back(2);
if
(isCycle(adj, V))
cout <<
"Gxrph contains Cycle.\n"
;
else
cout <<
"Gxrph does not contain Cycle.\n"
;
return
0;
}
Java
import
java.util.*;
class
GFG
{
static
int
MAX_VERTEX =
101
;
static
int
[]Arr =
new
int
[MAX_VERTEX];
static
int
[]size =
new
int
[MAX_VERTEX];
static
void
initialize(
int
n)
{
for
(
int
i =
0
; i <= n; i++)
{
Arr[i] = i;
size[i] =
1
;
}
}
static
int
find(
int
i)
{
while
(Arr[i] != i)
{
Arr[i] = Arr[Arr[i]];
i = Arr[i];
}
return
i;
}
static
void
_union(
int
xr,
int
yr)
{
if
(size[xr] < size[yr])
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
static
int
isCycle(Vector<Integer> adj[],
int
V)
{
for
(
int
i =
0
; i < V; i++)
{
for
(
int
j =
0
; j < adj[i].size(); j++)
{
int
x = find(i);
int
y = find(adj[i].get(j));
if
(x == y)
return
1
;
_union(x, y);
}
}
return
0
;
}
public
static
void
main(String[] args)
{
int
V =
3
;
initialize(V);
Vector<Integer> []adj =
new
Vector[V];
for
(
int
i =
0
; i < V; i++)
adj[i] =
new
Vector<Integer>();
adj[
0
].add(
1
);
adj[
0
].add(
2
);
adj[
1
].add(
2
);
if
(isCycle(adj, V) ==
1
)
System.out.print(
"Graph contains Cycle.\n"
);
else
System.out.print(
"Graph does not contain Cycle.\n"
);
}
}
Python3
def
initialize(n):
global
Arr, size
for
i
in
range
(n
+
1
):
Arr[i]
=
i
size[i]
=
1
def
find(i):
global
Arr, size
while
(Arr[i] !
=
i):
Arr[i]
=
Arr[Arr[i]]
i
=
Arr[i]
return
i
def
_union(xr, yr):
global
Arr, size
if
(size[xr] < size[yr]):
Arr[xr]
=
Arr[yr]
size[yr]
+
=
size[xr]
else
:
Arr[yr]
=
Arr[xr]
size[xr]
+
=
size[yr]
def
isCycle(adj, V):
global
Arr, size
for
i
in
range
(V):
for
j
in
range
(
len
(adj[i])):
x
=
find(i)
y
=
find(adj[i][j])
if
(x
=
=
y):
return
1
_union(x, y)
return
0
MAX_VERTEX
=
101
Arr
=
[
None
]
*
MAX_VERTEX
size
=
[
None
]
*
MAX_VERTEX
V
=
3
initialize(V)
adj
=
[[]
for
i
in
range
(V)]
adj[
0
].append(
1
)
adj[
0
].append(
2
)
adj[
1
].append(
2
)
if
(isCycle(adj, V)):
print
(
"Graph contains Cycle."
)
else
:
print
(
"Graph does not contain Cycle."
)
C#
using
System;
using
System.Collections.Generic;
class
GFG
{
static
int
MAX_VERTEX = 101;
static
int
[]Arr =
new
int
[MAX_VERTEX];
static
int
[]size =
new
int
[MAX_VERTEX];
static
void
initialize(
int
n)
{
for
(
int
i = 0; i <= n; i++)
{
Arr[i] = i;
size[i] = 1;
}
}
static
int
find(
int
i)
{
while
(Arr[i] != i)
{
Arr[i] = Arr[Arr[i]];
i = Arr[i];
}
return
i;
}
static
void
_union(
int
xr,
int
yr)
{
if
(size[xr] < size[yr])
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
static
int
isCycle(List<
int
> []adj,
int
V)
{
for
(
int
i = 0; i < V; i++)
{
for
(
int
j = 0; j < adj[i].Count; j++)
{
int
x = find(i);
int
y = find(adj[i][j]);
if
(x == y)
return
1;
_union(x, y);
}
}
return
0;
}
public
static
void
Main(String[] args)
{
int
V = 3;
initialize(V);
List<
int
> []adj =
new
List<
int
>[V];
for
(
int
i = 0; i < V; i++)
adj[i] =
new
List<
int
>();
adj[0].Add(1);
adj[0].Add(2);
adj[1].Add(2);
if
(isCycle(adj, V) == 1)
Console.Write(
"Graph contains Cycle.\n"
);
else
Console.Write(
"Graph does not contain Cycle.\n"
);
}
}
Javascript
<script>
var
MAX_VERTEX = 101;
var
Arr = Array(MAX_VERTEX).fill(0);
var
size = Array(MAX_VERTEX).fill(0);
function
initialize(n)
{
for
(
var
i = 0; i <= n; i++) {
Arr[i] = i;
size[i] = 1;
}
}
function
find(i)
{
while
(Arr[i] != i)
{
Arr[i] = Arr[Arr[i]];
i = Arr[i];
}
return
i;
}
function
_union(xr, yr)
{
if
(size[xr] < size[yr])
{
Arr[xr] = Arr[yr];
size[yr] += size[xr];
}
else
{
Arr[yr] = Arr[xr];
size[xr] += size[yr];
}
}
function
isCycle(adj, V)
{
for
(
var
i = 0; i < V; i++) {
for
(
var
j = 0; j < adj[i].length; j++) {
var
x = find(i);
var
y = find(adj[i][j]);
if
(x == y)
return
1;
_union(x, y);
}
}
return
0;
}
var
V = 3;
initialize(V);
var
adj = Array.from(Array(V), ()=>Array());
adj[0].push(1);
adj[0].push(2);
adj[1].push(2);
if
(isCycle(adj, V))
document.write(
"Graph contains Cycle.<br>"
);
else
document.write(
"Graph does not contain Cycle.<br>"
);
</script>
Output:
Graph contains Cycle.
Time Complexity(Find) : O(log*(n))
Time Complexity(union) : O(1)
Write a Nonrecursive Version of Findset() With Path Compression.
Source: https://www.geeksforgeeks.org/union-find-algorithm-union-rank-find-optimized-path-compression/
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