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










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
    2 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
    2 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
    2 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
    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










2 Answers
2






active

oldest

votes


















2












$begingroup$

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 vertice.



  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









$endgroup$




















    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
      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












    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    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 vertice.



    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









    $endgroup$

















      2












      $begingroup$

      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 vertice.



      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









      $endgroup$















        2












        2








        2





        $begingroup$

        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 vertice.



        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 vertice.



        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










        answered 2 hours ago









        loxlox

        1666




        1666





















            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
              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
















            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
              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














            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 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

















            • $begingroup$
              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


















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