# Check if there is a cycle with odd weight sum in an undirected graph

Given a weighted and undirected graph, we need to find if a cycle exist in this graph such that the sum of weights of all the edges in that cycle comes out to be odd.

Examples:

```Input : Number of vertices, n = 4,
Number of edges, m = 4
Weighted Edges =
1 2 12
2 3 1
4 3 1
4 1 20
Output : No! There is no odd weight
cycle in the given graph Input : Number of vertices, n = 5,
Number of edges, m = 3
Weighted Edges =
1 2 1
3 2 1
3 1 1
Output : Yes! There is an odd weight
cycle in the given graph
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The solution is based on the fact that “If a graph has no odd length cycle then it must be Bipartite, i.e., it can be colored with two colors

The idea is to covert given problem to a simpler problem where we have to just check if there is cycle of odd length or not. To convert, we do following

1. Convert all even weight edges into two edges of unit weight.
2. Convert all odd weight edges to a single edge of unit weight.

Let’s make an another graph for graph shown above (in example 1) Here, edges [1 — 2] have be broken in two parts such that [1-pseudo1-2] a pseudo node has been introduced. We are doing this so that each of our even weighted edge is taken into consideration twice while the edge with odd weight is counted only once. Doing this would help us further when we color our cycle. We assign all the edges with weight 1 and then by using 2 color method traverse the whole graph. Now we start coloring our modified graph using two colors only. In a cycle with even number of nodes, when we color it using two colors only, none of the two adjacent edges have the same color. While if we try coloring a cycle having odd number of edges, surely a situation arises where two adjacent edges have the same color. This is our pick! Thus, if we are able to color the modified graph completely using 2 colors only in a way no two adjacent edges get the same color assigned to them then there must be either no cycle in the graph or a cycle with even number of nodes. If any conflict arises while coloring a cycle with 2 colors only, then we have an odd cycle in our graph.

 `// C++ program to check if there is a cycle of  ` `// total odd weight ` `#include ` `using` `namespace` `std; ` ` `  `// This function returns true if the current subpart ` `// of the forest is two colorable, else false. ` `bool` `twoColorUtil(vector<``int``>G[], ``int` `src, ``int` `N,  ` `                                  ``int` `colorArr[]) { ` `     `  `    ``// Assign first color to source ` `    ``colorArr[src] = 1; ` `     `  `    ``// Create a queue (FIFO) of vertex numbers and ` `    ``// enqueue source vertex for BFS traversal ` `    ``queue <``int``> q; ` `    ``q.push(src); ` `     `  `    ``// Run while there are vertices in queue  ` `    ``// (Similar to BFS) ` `    ``while` `(!q.empty()){ ` `         `  `        ``int` `u = q.front(); ` `        ``q.pop(); ` `         `  `        ``// Find all non-colored adjacent vertices ` `        ``for` `(``int` `v = 0; v < G[u].size(); ++v){ ` `      `  `            ``// An edge from u to v exists and ` `            ``// destination v is not colored ` `            ``if` `(colorArr[G[u][v]] == -1){ ` `             `  `                ``// Assign alternate color to this  ` `                ``// adjacent v of u ` `                ``colorArr[G[u][v]] = 1 - colorArr[u]; ` `                ``q.push(G[u][v]); ` `            ``} ` ` `  `            ``//  An edge from u to v exists and destination ` `            ``// v is colored with same color as u ` `            ``else` `if` `(colorArr[G[u][v]] == colorArr[u])            ` `                ``return` `false``; ` `        ``} ` `    ``} ` `    ``return` `true``; ` `} ` ` `  `// This function returns true if graph G[V][V] is two ` `// colorable, else false        ` `bool` `twoColor(vector<``int``>G[], ``int` `N){ ` `     `  `     `  `    ``// Create a color array to store colors assigned  ` `    ``// to all veritces. Vertex number is used as index ` `    ``// in this array. The value '-1' of  colorArr[i]  ` `    ``// is used to indicate that no color is assigned ` `    ``// to vertex 'i'.  The value 1 is used to indicate  ` `    ``// first color is assigned and value 0 indicates  ` `    ``// second color is assigned. ` `    ``int` `colorArr[N]; ` `    ``for` `(``int` `i = 1; i <= N; ++i) ` `        ``colorArr[i] = -1; ` `     `  `    ``// As we are dealing with graph, the input might ` `    ``// come as a forest, thus start coloring from a  ` `    ``// node and if true is returned we'll know that ` `    ``// we successfully colored the subpart of our  ` `    ``// forest and we start coloring again from a new  ` `    ``// uncolored node. This way we cover the entire forest. ` `    ``for` `(``int` `i = 1; i <= N; i++) ` `        ``if` `(colorArr[i] == -1) ` `           ``if` `(twoColorUtil(G, i, N, colorArr) == ``false``) ` `             ``return` `false``; ` `          `  `        ``return` `true``; ` `} ` ` `  `// Returns false if an odd cycle is present else true ` `// int info[][] is the information about our graph ` `// int n is the number of nodes ` `// int m is the number of informations given to us ` `bool` `isOddSum(``int` `info[],``int` `n,``int` `m){ ` `     `  `    ``// Declaring adjacency list of a graph ` `    ``// Here at max, we can encounter all the edges with  ` `    ``// even weight thus there will be 1 pseudo node ` `    ``// for each edge ` `    ``vector<``int``> G[2*n]; ` `     `  `    ``int` `pseudo = n+1; ` `    ``int` `pseudo_count = 0; ` `    ``for` `(``int` `i=0; i

Output:

```No
```

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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