## Posts Tagged ‘**simple**’

## Number of Cycles in a Graph

**Question : Find the number of simple cycles in a simple graph.**

Simple Graph – An undirected graph that has no loops and no more than one edge between any two different vertices.

Simple Cycle – A closed (simple) path, with no other repeated vertices or edges other than the starting and ending vertices.

Given a graph of N vertices and M edges, we will look at an algorithm with time complexity O(*2*^{N}*N*^{2}). We will use dynamic programming to do so. Let there be a matrix map, such that map[i][j] is equal to 1 if there is a edge between i and j and 0 otherwise. Let there be another array f[1<<N][N] which denotes the number of simple paths.

Let, i denote a subset S of our vertices k be the smallest set bit of i then f[i][j] is the number of simple paths from j to k that contains vertices only from the set S.

In our algorithm first we will find f[i][j] and then check if there is a edge between k and j, if yes, we can complete every simple path from j to k into a simple cycle and hence we add f[i][j] to our result of total number of simple cycles.Now how to find f[i][j].

For very subset i we iterate through all edges j. Once we have set k, we look for all vertices 'l' that can be neighbors of j in our subset S. So if l is a vertex in subset S and there is edge from j to l then f[i][j] = f[i][j] + the number of simple paths from l to i in the subset {S – j}. Since a simple graph is undirected or bidirectional, we have counted every cycle twice and so we divide our result by 2. Here's a sample C++ code which takes N, M and the edges as input.

#include<iostream> using namespace std; #define SIZE 20 bool map[SIZE][SIZE],F; long long f[1<<SIZE][SIZE],res=0; int main() { int n,m,i,j,k,l,x,y; scanf("%d%d",&n,&m); for (i=0;i<m;i++) { scanf("%d%d",&x,&y); x--;y--; if (x>y) swap(x,y); map[x][y]=map[y][x]=1; f[(1<<x)+(1<<y)][y]=1; } for (i=7;i<(1<<n);i++) { F=1; for (j=0;j<n;j++) if (i&(1<<j) && f[i][j]==0) { if (F) { F=0; k=j; continue; } for (l=k+1;l<n;l++) { if (i&(1<<l) && map[j][l]) f[i][j]+=f[i-(1<<j)][l]; } if (map[k][j]) res+=f[i][j]; } } printf("%lld\n",res/2); }

NJOY!

-fR0DDY