# 99 questions/Solutions/89

From HaskellWiki

< 99 questions | Solutions

```
import Data.List
type Node = Int
type Edge = (Node,Node)
type Graph = ([Node],[Edge])
dfsbipartite :: Graph -> [(Node, Int)] -> [Node] -> [Node] -> Bool
dfsbipartite ([],_) _ _ _ = True
dfsbipartite (_,_) [] _ _ = True
dfsbipartite (v,e) ((nv, 0):stack) odd even
| [x|x<-v,x==nv] == [] = dfsbipartite (v, e) stack odd even
| [] == intersect adjacent even = dfsbipartite (newv, e) ([(x,1)|x<-adjacent] ++ stack) odd (nv : even)
| otherwise = False
where
adjacent = [x | (x,y)<-e,y==nv] ++ [x | (y,x)<-e,y==nv]
newv = [x|x<-v,x/=nv]
dfsbipartite (v,e) ((nv, 1):stack) odd even
| [x|x<-v,x==nv] == [] = dfsbipartite (v, e) stack odd even
| [] == intersect adjacent odd = dfsbipartite (newv, e) ([(x,0)|x<-adjacent] ++ stack) (nv : odd) even
| otherwise = False
where
adjacent = [x | (x,y)<-e,y==nv] ++ [x | (y,x)<-e,y==nv]
newv = [x|x<-v,x/=nv]
bipartite :: Graph -> Bool
bipartite ([],_) = True
bipartite (top:v,e) = dfsbipartite (top:v, e) [(top,0)] [] []
```

You can call it:

```
bipartite ([1,2,3,4,5],[(1,2),(2,3),(1,4),(3,4),(5,2),(5,4)])
```