# How to trace the path in a Breadth-First Search?

## Question:

How do you trace the path of a Breadth-First Search, such that in the following example:

If searching for key `11`

, return the **shortest** list connecting 1 to 11.

```
[1, 4, 7, 11]
```

## Answers:

You should have look at http://en.wikipedia.org/wiki/Breadth-first_search first.

Below is a quick implementation, in which I used a list of list to represent the queue of paths.

```
# graph is in adjacent list representation
graph = {
'1': ['2', '3', '4'],
'2': ['5', '6'],
'5': ['9', '10'],
'4': ['7', '8'],
'7': ['11', '12']
}
def bfs(graph, start, end):
# maintain a queue of paths
queue = []
# push the first path into the queue
queue.append([start])
while queue:
# get the first path from the queue
path = queue.pop(0)
# get the last node from the path
node = path[-1]
# path found
if node == end:
return path
# enumerate all adjacent nodes, construct a
# new path and push it into the queue
for adjacent in graph.get(node, []):
new_path = list(path)
new_path.append(adjacent)
queue.append(new_path)
print bfs(graph, '1', '11')
```

This prints: `['1', '4', '7', '11']`

Another approach would be maintaining a mapping from each node to its parent, and when inspecting the adjacent node, record its parent. When the search is done, simply backtrace according the parent mapping.

```
graph = {
'1': ['2', '3', '4'],
'2': ['5', '6'],
'5': ['9', '10'],
'4': ['7', '8'],
'7': ['11', '12']
}
def backtrace(parent, start, end):
path = [end]
while path[-1] != start:
path.append(parent[path[-1]])
path.reverse()
return path
def bfs(graph, start, end):
parent = {}
queue = []
queue.append(start)
while queue:
node = queue.pop(0)
if node == end:
return backtrace(parent, start, end)
for adjacent in graph.get(node, []):
if node not in queue :
parent[adjacent] = node # <<<<< record its parent
queue.append(adjacent)
print bfs(graph, '1', '11')
```

The above codes are based on the assumption that there’s no cycles.

I thought I’d try code this up for fun:

```
graph = {
'1': ['2', '3', '4'],
'2': ['5', '6'],
'5': ['9', '10'],
'4': ['7', '8'],
'7': ['11', '12']
}
def bfs(graph, forefront, end):
# assumes no cycles
next_forefront = [(node, path + ',' + node) for i, path in forefront if i in graph for node in graph[i]]
for node,path in next_forefront:
if node==end:
return path
else:
return bfs(graph,next_forefront,end)
print bfs(graph,[('1','1')],'11')
# >>>
# 1, 4, 7, 11
```

If you want cycles you could add this:

```
for i, j in for_front: # allow cycles, add this code
if i in graph:
del graph[i]
```

I liked qiao’s first answer very much!

The only thing missing here is to mark the vertexes as visited.

Why we need to do it?

Lets imagine that there is another node number 13 connected from node 11. Now our goal is to find node 13.

After a little bit of a run the queue will look like this:

```
[[1, 2, 6], [1, 3, 10], [1, 4, 7], [1, 4, 8], [1, 2, 5, 9], [1, 2, 5, 10]]
```

Note that there are TWO paths with node number 10 at the end.

Which means that the paths from node number 10 will be checked twice. In this case it doesn’t look so bad because node number 10 doesn’t have any children.. But it could be really bad (even here we will check that node twice for no reason..)

Node number 13 isn’t in those paths so the program won’t return before reaching to the second path with node number 10 at the end..And we will recheck it..

All we are missing is a set to mark the visited nodes and not to check them again..

This is qiao’s code after the modification:

```
graph = {
1: [2, 3, 4],
2: [5, 6],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}
def bfs(graph_to_search, start, end):
queue = [[start]]
visited = set()
while queue:
# Gets the first path in the queue
path = queue.pop(0)
# Gets the last node in the path
vertex = path[-1]
# Checks if we got to the end
if vertex == end:
return path
# We check if the current node is already in the visited nodes set in order not to recheck it
elif vertex not in visited:
# enumerate all adjacent nodes, construct a new path and push it into the queue
for current_neighbour in graph_to_search.get(vertex, []):
new_path = list(path)
new_path.append(current_neighbour)
queue.append(new_path)
# Mark the vertex as visited
visited.add(vertex)
print bfs(graph, 1, 13)
```

The output of the program will be:

```
[1, 4, 7, 11, 13]
```

Without the unneccecery rechecks..

I like both @Qiao first answer and @Or’s addition. For a sake of a little less processing I would like to add to Or’s answer.

In @Or’s answer keeping track of visited node is great. We can also allow the program to exit sooner that it currently is. At some point in the for loop the `current_neighbour`

will have to be the `end`

, and once that happens the shortest path is found and program can return.

I would modify the the method as follow, pay close attention to the for loop

```
graph = {
1: [2, 3, 4],
2: [5, 6],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}
def bfs(graph_to_search, start, end):
queue = [[start]]
visited = set()
while queue:
# Gets the first path in the queue
path = queue.pop(0)
# Gets the last node in the path
vertex = path[-1]
# Checks if we got to the end
if vertex == end:
return path
# We check if the current node is already in the visited nodes set in order not to recheck it
elif vertex not in visited:
# enumerate all adjacent nodes, construct a new path and push it into the queue
for current_neighbour in graph_to_search.get(vertex, []):
new_path = list(path)
new_path.append(current_neighbour)
queue.append(new_path)
#No need to visit other neighbour. Return at once
if current_neighbour == end
return new_path;
# Mark the vertex as visited
visited.add(vertex)
print bfs(graph, 1, 13)
```

The output and everything else will be the same. However, the code will take less time to process. This is especially useful on larger graphs. I hope this helps someone in the future.

Very easy code. You keep appending the path each time you discover a node.

```
graph = {
'A': set(['B', 'C']),
'B': set(['A', 'D', 'E']),
'C': set(['A', 'F']),
'D': set(['B']),
'E': set(['B', 'F']),
'F': set(['C', 'E'])
}
def retunShortestPath(graph, start, end):
queue = [(start,[start])]
visited = set()
while queue:
vertex, path = queue.pop(0)
visited.add(vertex)
for node in graph[vertex]:
if node == end:
return path + [end]
else:
if node not in visited:
visited.add(node)
queue.append((node, path + [node]))
```

With cycles included in the graph, would not something like this work better?

```
from collections import deque
graph = {
1: [2, 3, 4],
2: [5, 6, 3],
3: [10],
4: [7, 8],
5: [9, 10],
7: [11, 12],
11: [13]
}
def bfs1(graph_to_search, start, end):
queue = deque([start])
visited = {start}
trace = {}
while queue:
# Gets the first path in the queue
vertex = queue.popleft()
# Checks if we got to the end
if vertex == end:
break
for neighbour in graph_to_search.get(vertex, []):
# We check if the current neighbour is already in the visited nodes set in order not to re-add it
if neighbour not in visited:
# Mark the vertex as visited
visited.add(neighbour)
trace[neighbour] = vertex
queue.append(neighbour)
path = [end]
while path[-1] != start:
last_node = path[-1]
next_node = trace[last_node]
path.append(next_node)
return path[::-1]
print(bfs1(graph,1, 13))
```

This way only new nodes will be visited and moreover, avoid cycles.

# Javascript version and search first/all paths …

P.S, Graph with cylces works well.

Your can `convert`

it to `python`

, it’s easy

```
function search_path(graph, start, end, exhausted=true, method='bfs') {
// note. Javascript Set is ordered set...
const queue = [[start, new Set([start])]]
const visited = new Set()
const allpaths = []
const hashPath = (path) => [...path].join(',') // any path hashing method
while (queue.length) {
const [node, path] = queue.shift()
// visited node and its path instant. do not modify it others place
visited.add(node)
visited.add(hashPath(path))
for (let _node of graph.get(node) || []) {
// the paths already has the node, loops mean nothing though.
if (path.has(_node))
continue;
// now new path has no repeated nodes.
let newpath = new Set([...path, _node])
if (_node == end){
allpaths.push(newpath)
if(!exhausted) return allpaths; // found and return
}
else {
if (!visited.has(_node) || // new node till now
// note: search all possible including the longest path
visited.has(_node) && !visited.has(hashPath(newpath))
) {
if(method == 'bfs')
queue.push([_node, newpath])
else{
queue.unshift([_node, newpath])
}
}
}
}
}
return allpaths
}
```

output like this..

```
[
[ 'A', 'C' ],
[ 'A', 'E', 'C'],
[ 'A', 'E', 'F', 'C' ] // including F in `A -> C`
]
```