# 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++

`// 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.

## Recommended Posts:

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- Eulerian Path in undirected graph
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