Product of lengths of all cycles in an undirected graph

Given an undirected and unweighted graph. The task is to find the product of the lengths of all cycles formed in it.

Example 1:

The above graph has two cycles of length 4 and 3, the product of cycle lengths is 12.

Example 2:

The above graph has two cycles of length 4 and 3, the product of cycle lengths is 12.



Approach: Using the graph coloring method, mark all the vertex of the different cycles with unique numbers. Once the graph traversal is completed, push all the similar marked numbers to an adjacency list and print the adjacency list accordingly. Given below is the algorithm:

  • Insert the edges into an adjacency list.
  • Call the DFS function which uses the coloring method to mark the vertex.
  • Whenever there is a partially visited vertex, backtrack till the current vertex is reached and mark all of them with cycle numbers. Once all the vertexes are marked, increase the cycle number.
  • Once Dfs is completed, iterate for the edges and push the same marked number edges to another adjacency list.
  • Iterate in the another adjacency list, and keep the count of number of vertex in a cycle using map and cycle numbers
  • Iterate for cycle numbers, and multiply the lengths to get the final product which will be the answer.

Below is the implementation of the above approach:

C++

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// C++ program to find the
// product of lengths of cycle
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
  
// variables to be used
// in both functions
vector<int> graph[N];
  
// Function to mark the vertex with
// different colors for different cycles
void dfs_cycle(int u, int p, int color[],
            int mark[], int par[], int& cyclenumber)
{
  
    // already (completely) visited vertex.
    if (color[u] == 2) {
        return;
    }
  
    // seen vertex, but was not completely 
    // visited -> cycle detected.
    // backtrack based on parents to find
    // the complete cycle.
    if (color[u] == 1) {
  
        cyclenumber++;
        int cur = p;
        mark[cur] = cyclenumber;
  
        // backtrack the vertex which are
        // in the current cycle thats found
        while (cur != u) {
            cur = par[cur];
            mark[cur] = cyclenumber;
        }
        return;
    }
    par[u] = p;
  
    // partially visited.
    color[u] = 1;
  
    // simple dfs on graph
    for (int v : graph[u]) {
  
        // if it has not been visited previously
        if (v == par[u]) {
            continue;
        }
        dfs_cycle(v, u, color, mark, par, cyclenumber);
    }
  
    // completely visited.
    color[u] = 2;
}
  
// add the edges to the graph
void addEdge(int u, int v)
{
    graph[u].push_back(v);
    graph[v].push_back(u);
}
  
// Function to print the cycles
int productLength(int edges, int mark[], int& cyclenumber)
{
    unordered_map<int, int> mp;
  
    // push the edges that into the
    // cycle adjacency list
    for (int i = 1; i <= edges; i++) {
        if (mark[i] != 0)
            mp[mark[i]]++;
    }
    int cnt = 1;
  
    // prodcut all the length of cycles
    for (int i = 1; i <= cyclenumber; i++) {
        cnt = cnt * mp[i];
    }
    if (cyclenumber == 0)
        cnt = 0;
  
    return cnt;
}
  
// Driver Code
int main()
{
  
    // add edges
    addEdge(1, 2);
    addEdge(2, 3);
    addEdge(3, 4);
    addEdge(4, 6);
    addEdge(4, 7);
    addEdge(5, 6);
    addEdge(3, 5);
    addEdge(7, 8);
    addEdge(6, 10);
    addEdge(5, 9);
    addEdge(10, 11);
    addEdge(11, 12);
    addEdge(11, 13);
    addEdge(12, 13);
  
    // arrays required to color the
    // graph, store the parent of node
    int color[N];
    int par[N];
  
    // mark with unique numbers
    int mark[N];
  
    // store the numbers of cycle
    int cyclenumber = 0;
    int edges = 13;
  
    // call DFS to mark the cycles
    dfs_cycle(1, 0, color, mark, par, cyclenumber);
  
    // function to print the cycles
    cout << productLength(edges, mark, cyclenumber);
  
    return 0;
}

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Python3

# Python3 program to find the
# product of lengths of cycle
from collections import defaultdict

# Function to mark the vertex with
# different colors for different cycles
def dfs_cycle(u, p, color, mark, par):

global cyclenumber

# already (completely) visited vertex.
if color[u] == 2:
return

# seen vertex, but was not completely
# visited -> cycle detected.
# backtrack based on parents to find
# the complete cycle.
if color[u] == 1:

cyclenumber += 1
cur = p
mark[cur] = cyclenumber

# backtrack the vertex which are
# in the current cycle thats found
while cur != u:
cur = par[cur]
mark[cur] = cyclenumber

return

par[u] = p

# partially visited.
color[u] = 1

# simple dfs on graph
for v in graph[u]:

# if it has not been visited previously
if v == par[u]:
continue

dfs_cycle(v, u, color, mark, par)

# completely visited.
color[u] = 2

# add the edges to the graph
def addEdge(u, v):

graph[u].append(v)
graph[v].append(u)

# Function to print the cycles
def productLength(edges, mark, cyclenumber):

mp = defaultdict(lambda:0)

# push the edges that into the
# cycle adjacency list
for i in range(1, edges+1):
if mark[i] != 0:
mp[mark[i]] += 1

cnt = 1

# prodcut all the length of cycles
for i in range(1, cyclenumber + 1):
cnt = cnt * mp[i]

if cyclenumber == 0:
cnt = 0

return cnt

# Driver Code
if __name__ == “__main__”:

N = 100000
graph = [[] for i in range(N)]

# add edges
addEdge(1, 2)
addEdge(2, 3)
addEdge(3, 4)
addEdge(4, 6)
addEdge(4, 7)
addEdge(5, 6)
addEdge(3, 5)
addEdge(7, 8)
addEdge(6, 10)
addEdge(5, 9)
addEdge(10, 11)
addEdge(11, 12)
addEdge(11, 13)
addEdge(12, 13)

# arrays required to color the
# graph, store the parent of node
color, par = [None] * N, [None] * N

# mark with unique numbers
mark = [None] * N

# store the numbers of cycle
cyclenumber, edges = 0, 13

# call DFS to mark the cycles
dfs_cycle(1, 0, color, mark, par)

# function to print the cycles
print(productLength(edges, mark,
cyclenumber))

# This code is contributed by Rituraj Jain

Output:

12

Time Complexity: O(N), where N is the number of nodes in the graph.



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