Find a path from s to t using as few red nodes as possible
$begingroup$
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
asked 5 hours ago
Hunter DyerHunter Dyer
284
$endgroup$
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
asked 5 hours ago
Hunter DyerHunter Dyer
284
$endgroup$
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
asked 5 hours ago
Hunter DyerHunter Dyer
284
$endgroup$
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
asked 5 hours ago
Hunter DyerHunter Dyer
284
asked 5 hours ago
Hunter DyerHunter Dyer
284
asked 5 hours ago
Hunter DyerHunter Dyer
284
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.
answered 2 hours ago
loxlox
1666
$endgroup$
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.
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
$endgroup$
$begingroup$
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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 mins ago
add a comment |
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2 Answers
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active
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2 Answers
2
active
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$begingroup$
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.
answered 2 hours ago
loxlox
1666
$endgroup$
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.
answered 2 hours ago
loxlox
1666
$endgroup$
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.
answered 2 hours ago
loxlox
1666
$endgroup$
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
answered 2 hours ago
loxlox
1666
answered 2 hours ago
loxlox
1666
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.
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
$endgroup$
$begingroup$
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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 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.
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
$endgroup$
$begingroup$
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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 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.
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
$endgroup$
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 53 mins ago
Apass.JackApass.Jack
13.7k1940
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
answered 53 mins ago
Apass.JackApass.Jack
13.7k1940
answered 53 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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 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
40 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
40 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
26 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 mins ago
$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
3 mins ago
add a comment |
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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