Find a path from s to t using as few red nodes as possible The Next CEO of Stack OverflowDijkstra algorithm vs breadth first search for shortest path in graphAlgorithm to find diameter of a tree using BFS/DFS. Why does it work?Finding shortest path from a node to any node of a particular typeParallel algorithm to find if a set of nodes is on an elememtry cycle in a directed/undirected graphShortest path in unweighted graph using an iterator onlyShortest Path using DFS on weighted graphsCan a 3 Color DFS be used to identify cycles (not just detect them)?Find a path that contains specific nodes without back and forward edgesChecking if there is a single path that visits all nodes in a directed graphFind shortest path that goes through at least 5 red edges
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Find a path from s to t using as few red nodes as possible
The Next CEO of Stack OverflowDijkstra algorithm vs breadth first search for shortest path in graphAlgorithm to find diameter of a tree using BFS/DFS. Why does it work?Finding shortest path from a node to any node of a particular typeParallel algorithm to find if a set of nodes is on an elememtry cycle in a directed/undirected graphShortest path in unweighted graph using an iterator onlyShortest Path using DFS on weighted graphsCan a 3 Color DFS be used to identify cycles (not just detect them)?Find a path that contains specific nodes without back and forward edgesChecking if there is a single path that visits all nodes in a directed graphFind shortest path that goes through at least 5 red edges
$begingroup$
Was doing a little interview prep. Given an undirected graph G, such that each node is colored red or blue and |E|≥|V|, find a path in O(|E|) time such that starting and ending at 2 blue nodes, s and t, that you pass through as few red nodes as possible.
Initial Impressions: Since |E|≥|V|, O(|E|) time would include O(|E|+|V|), which means the solution likely uses BFS or DFS. Modifying the graph such that causing the all red nodes must be forced down a directed path of some long length (after making the whole graph directed) in order to use out-of-the-box BFS seems not viable, as it would mean constructing a new graph would be along O(|E||V|) time.
Another method I toyed around with was propagating values to nodes based on the safest path to that node while doing a DFS search, but not all values were guaranteed to update.
I still want to try to solve this myself, but I'm really stuck right now. Was wondering if there were any hints I could get. There are much simpler ways of doing this if it weren't for the O(|E|) time. Djikstras with creating some edge weights would work, but wouldn't be within the time bound.
graphs
$endgroup$
add a comment |
$begingroup$
Was doing a little interview prep. Given an undirected graph G, such that each node is colored red or blue and |E|≥|V|, find a path in O(|E|) time such that starting and ending at 2 blue nodes, s and t, that you pass through as few red nodes as possible.
Initial Impressions: Since |E|≥|V|, O(|E|) time would include O(|E|+|V|), which means the solution likely uses BFS or DFS. Modifying the graph such that causing the all red nodes must be forced down a directed path of some long length (after making the whole graph directed) in order to use out-of-the-box BFS seems not viable, as it would mean constructing a new graph would be along O(|E||V|) time.
Another method I toyed around with was propagating values to nodes based on the safest path to that node while doing a DFS search, but not all values were guaranteed to update.
I still want to try to solve this myself, but I'm really stuck right now. Was wondering if there were any hints I could get. There are much simpler ways of doing this if it weren't for the O(|E|) time. Djikstras with creating some edge weights would work, but wouldn't be within the time bound.
graphs
$endgroup$
1
$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago
add a comment |
$begingroup$
Was doing a little interview prep. Given an undirected graph G, such that each node is colored red or blue and |E|≥|V|, find a path in O(|E|) time such that starting and ending at 2 blue nodes, s and t, that you pass through as few red nodes as possible.
Initial Impressions: Since |E|≥|V|, O(|E|) time would include O(|E|+|V|), which means the solution likely uses BFS or DFS. Modifying the graph such that causing the all red nodes must be forced down a directed path of some long length (after making the whole graph directed) in order to use out-of-the-box BFS seems not viable, as it would mean constructing a new graph would be along O(|E||V|) time.
Another method I toyed around with was propagating values to nodes based on the safest path to that node while doing a DFS search, but not all values were guaranteed to update.
I still want to try to solve this myself, but I'm really stuck right now. Was wondering if there were any hints I could get. There are much simpler ways of doing this if it weren't for the O(|E|) time. Djikstras with creating some edge weights would work, but wouldn't be within the time bound.
graphs
$endgroup$
Was doing a little interview prep. Given an undirected graph G, such that each node is colored red or blue and |E|≥|V|, find a path in O(|E|) time such that starting and ending at 2 blue nodes, s and t, that you pass through as few red nodes as possible.
Initial Impressions: Since |E|≥|V|, O(|E|) time would include O(|E|+|V|), which means the solution likely uses BFS or DFS. Modifying the graph such that causing the all red nodes must be forced down a directed path of some long length (after making the whole graph directed) in order to use out-of-the-box BFS seems not viable, as it would mean constructing a new graph would be along O(|E||V|) time.
Another method I toyed around with was propagating values to nodes based on the safest path to that node while doing a DFS search, but not all values were guaranteed to update.
I still want to try to solve this myself, but I'm really stuck right now. Was wondering if there were any hints I could get. There are much simpler ways of doing this if it weren't for the O(|E|) time. Djikstras with creating some edge weights would work, but wouldn't be within the time bound.
graphs
graphs
asked 5 hours ago
Hunter DyerHunter Dyer
284
284
1
$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago
add a comment |
1
$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago
1
1
$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago
$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To solve this, you need to use $BFS$. But first, manipulate $G$ so that the path will always favor blue vertices.
The solution has 2 parts:
- Use $DFS$ on blue vertices only, to find all all-blue strongly connected components, or $SCC$. Let's denote each $SCC$ by $v'$. Now, each blue $v in v'$ will be "compressed" to a single vertex $u$, and an edge $(u,x)$ will be added for every $x in N(v')$.
Note any such $x$ is necessarily red.
This step costs $O(V+E) = O(E)$, since $DFS$ is $O(V+E)$, and you have at most $V$ blue vertices, which make no more than $E$ new edges to add.
Step 1 means all paths that are blue-only will be free. On the new graph, the $BFS$ will only consider the edges which pass through a red vertice.
- Use $BFS$ from $s$. That length of the path to $t$ will essentially be the shortest path under the constraint of least red vertices in the path.
$endgroup$
add a comment |
$begingroup$
Convert $G$ to a directed graph $G'$ where we have two edges $(u,v)$ and $(v,u)$ in $G'$ for every edge $u,v$ in $G$. Let the length of $(u,v)$ be 1 if $v$ is a red node and 0 otherwise. Now run Dijkstra's algorithm on $G'$ from the starting node $s$ to the ending node $t$.
It is clear that the shortest path thus found passes as few red nodes as possible.
$endgroup$
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To solve this, you need to use $BFS$. But first, manipulate $G$ so that the path will always favor blue vertices.
The solution has 2 parts:
- Use $DFS$ on blue vertices only, to find all all-blue strongly connected components, or $SCC$. Let's denote each $SCC$ by $v'$. Now, each blue $v in v'$ will be "compressed" to a single vertex $u$, and an edge $(u,x)$ will be added for every $x in N(v')$.
Note any such $x$ is necessarily red.
This step costs $O(V+E) = O(E)$, since $DFS$ is $O(V+E)$, and you have at most $V$ blue vertices, which make no more than $E$ new edges to add.
Step 1 means all paths that are blue-only will be free. On the new graph, the $BFS$ will only consider the edges which pass through a red vertice.
- Use $BFS$ from $s$. That length of the path to $t$ will essentially be the shortest path under the constraint of least red vertices in the path.
$endgroup$
add a comment |
$begingroup$
To solve this, you need to use $BFS$. But first, manipulate $G$ so that the path will always favor blue vertices.
The solution has 2 parts:
- Use $DFS$ on blue vertices only, to find all all-blue strongly connected components, or $SCC$. Let's denote each $SCC$ by $v'$. Now, each blue $v in v'$ will be "compressed" to a single vertex $u$, and an edge $(u,x)$ will be added for every $x in N(v')$.
Note any such $x$ is necessarily red.
This step costs $O(V+E) = O(E)$, since $DFS$ is $O(V+E)$, and you have at most $V$ blue vertices, which make no more than $E$ new edges to add.
Step 1 means all paths that are blue-only will be free. On the new graph, the $BFS$ will only consider the edges which pass through a red vertice.
- Use $BFS$ from $s$. That length of the path to $t$ will essentially be the shortest path under the constraint of least red vertices in the path.
$endgroup$
add a comment |
$begingroup$
To solve this, you need to use $BFS$. But first, manipulate $G$ so that the path will always favor blue vertices.
The solution has 2 parts:
- Use $DFS$ on blue vertices only, to find all all-blue strongly connected components, or $SCC$. Let's denote each $SCC$ by $v'$. Now, each blue $v in v'$ will be "compressed" to a single vertex $u$, and an edge $(u,x)$ will be added for every $x in N(v')$.
Note any such $x$ is necessarily red.
This step costs $O(V+E) = O(E)$, since $DFS$ is $O(V+E)$, and you have at most $V$ blue vertices, which make no more than $E$ new edges to add.
Step 1 means all paths that are blue-only will be free. On the new graph, the $BFS$ will only consider the edges which pass through a red vertice.
- Use $BFS$ from $s$. That length of the path to $t$ will essentially be the shortest path under the constraint of least red vertices in the path.
$endgroup$
To solve this, you need to use $BFS$. But first, manipulate $G$ so that the path will always favor blue vertices.
The solution has 2 parts:
- Use $DFS$ on blue vertices only, to find all all-blue strongly connected components, or $SCC$. Let's denote each $SCC$ by $v'$. Now, each blue $v in v'$ will be "compressed" to a single vertex $u$, and an edge $(u,x)$ will be added for every $x in N(v')$.
Note any such $x$ is necessarily red.
This step costs $O(V+E) = O(E)$, since $DFS$ is $O(V+E)$, and you have at most $V$ blue vertices, which make no more than $E$ new edges to add.
Step 1 means all paths that are blue-only will be free. On the new graph, the $BFS$ will only consider the edges which pass through a red vertice.
- Use $BFS$ from $s$. That length of the path to $t$ will essentially be the shortest path under the constraint of least red vertices in the path.
answered 2 hours ago
loxlox
1666
1666
add a comment |
add a comment |
$begingroup$
Convert $G$ to a directed graph $G'$ where we have two edges $(u,v)$ and $(v,u)$ in $G'$ for every edge $u,v$ in $G$. Let the length of $(u,v)$ be 1 if $v$ is a red node and 0 otherwise. Now run Dijkstra's algorithm on $G'$ from the starting node $s$ to the ending node $t$.
It is clear that the shortest path thus found passes as few red nodes as possible.
$endgroup$
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
add a comment |
$begingroup$
Convert $G$ to a directed graph $G'$ where we have two edges $(u,v)$ and $(v,u)$ in $G'$ for every edge $u,v$ in $G$. Let the length of $(u,v)$ be 1 if $v$ is a red node and 0 otherwise. Now run Dijkstra's algorithm on $G'$ from the starting node $s$ to the ending node $t$.
It is clear that the shortest path thus found passes as few red nodes as possible.
$endgroup$
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
add a comment |
$begingroup$
Convert $G$ to a directed graph $G'$ where we have two edges $(u,v)$ and $(v,u)$ in $G'$ for every edge $u,v$ in $G$. Let the length of $(u,v)$ be 1 if $v$ is a red node and 0 otherwise. Now run Dijkstra's algorithm on $G'$ from the starting node $s$ to the ending node $t$.
It is clear that the shortest path thus found passes as few red nodes as possible.
$endgroup$
Convert $G$ to a directed graph $G'$ where we have two edges $(u,v)$ and $(v,u)$ in $G'$ for every edge $u,v$ in $G$. Let the length of $(u,v)$ be 1 if $v$ is a red node and 0 otherwise. Now run Dijkstra's algorithm on $G'$ from the starting node $s$ to the ending node $t$.
It is clear that the shortest path thus found passes as few red nodes as possible.
answered 42 mins ago
Apass.JackApass.Jack
13.7k1940
13.7k1940
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
add a comment |
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
Yeah that definitely works, but the runtime of Dijkstra's is O(|E| + |V|log|V|) which is more than O(|E|).
$endgroup$
– Hunter Dyer
29 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
15 mins ago
add a comment |
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$begingroup$
Try a variant of BFS in which you first find all red nodes reachable only via blue nodes, and so on.
$endgroup$
– Yuval Filmus
2 hours ago