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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










2












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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
    3 hours ago















2












$begingroup$


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.










share|cite|improve this question









$endgroup$







  • 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
    3 hours ago













2












2








2





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
    3 hours ago












  • 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
    3 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
3 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
3 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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:



  1. 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 vertex.



  1. 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.





share|cite|improve this answer











$endgroup$












  • $begingroup$
    the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
    $endgroup$
    – Kevin Wang
    3 mins ago


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
    $endgroup$
    – Apass.Jack
    50 mins ago











  • $begingroup$
    Although it is intuitively clear, it takes some time to explain it clearly.
    $endgroup$
    – Apass.Jack
    26 mins ago












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

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:



  1. 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 vertex.



  1. 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.





share|cite|improve this answer











$endgroup$












  • $begingroup$
    the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
    $endgroup$
    – Kevin Wang
    3 mins ago















4












$begingroup$

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:



  1. 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 vertex.



  1. 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.





share|cite|improve this answer











$endgroup$












  • $begingroup$
    the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
    $endgroup$
    – Kevin Wang
    3 mins ago













4












4








4





$begingroup$

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:



  1. 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 vertex.



  1. 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.





share|cite|improve this answer











$endgroup$



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:



  1. 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 vertex.



  1. 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.






share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 mins ago









templatetypedef

5,59911945




5,59911945










answered 2 hours ago









loxlox

1866




1866











  • $begingroup$
    the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
    $endgroup$
    – Kevin Wang
    3 mins ago
















  • $begingroup$
    the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
    $endgroup$
    – Kevin Wang
    3 mins ago















$begingroup$
the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
$endgroup$
– Kevin Wang
3 mins ago




$begingroup$
the graph is undirected, so step 1 should be to find all all-blue connected components, not SCCs, right?
$endgroup$
– Kevin Wang
3 mins ago











1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
    $endgroup$
    – Apass.Jack
    50 mins ago











  • $begingroup$
    Although it is intuitively clear, it takes some time to explain it clearly.
    $endgroup$
    – Apass.Jack
    26 mins ago
















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
    1 hour ago










  • $begingroup$
    I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
    $endgroup$
    – Apass.Jack
    50 mins ago











  • $begingroup$
    Although it is intuitively clear, it takes some time to explain it clearly.
    $endgroup$
    – Apass.Jack
    26 mins ago














1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour 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
    1 hour ago










  • $begingroup$
    I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
    $endgroup$
    – Apass.Jack
    50 mins ago











  • $begingroup$
    Although it is intuitively clear, it takes some time to explain it clearly.
    $endgroup$
    – Apass.Jack
    26 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
    1 hour ago










  • $begingroup$
    I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
    $endgroup$
    – Apass.Jack
    50 mins ago











  • $begingroup$
    Although it is intuitively clear, it takes some time to explain it clearly.
    $endgroup$
    – Apass.Jack
    26 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
1 hour 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
1 hour ago












$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
50 mins ago





$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
50 mins ago













$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
26 mins ago





$begingroup$
Although it is intuitively clear, it takes some time to explain it clearly.
$endgroup$
– Apass.Jack
26 mins ago


















draft saved

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