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










3












$begingroup$


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















3












$begingroup$


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













3












3








3





$begingroup$


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 6 hours ago









Hunter DyerHunter Dyer

334




334







  • 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
    19 mins ago










  • $begingroup$
    I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
    $endgroup$
    – Apass.Jack
    11 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.



While we are running Dijkstra's algorithm, we are in one of two kinds of stages alternatively. One kind of stage is when we are exploring towards red nodes. The other kind of stage is when we are exploring towards blue nodes. Since each edge is checked/visited as most twice, the running time is $O(|E|)$.






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
    1 hour 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









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
    19 mins ago










  • $begingroup$
    I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
    $endgroup$
    – Apass.Jack
    11 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
    19 mins ago










  • $begingroup$
    I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
    $endgroup$
    – Apass.Jack
    11 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 28 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
    19 mins ago










  • $begingroup$
    I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
    $endgroup$
    – Apass.Jack
    11 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
    19 mins ago










  • $begingroup$
    I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
    $endgroup$
    – Apass.Jack
    11 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
19 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
19 mins ago












$begingroup$
I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
$endgroup$
– Apass.Jack
11 mins ago




$begingroup$
I believe that is a typo by lox. Yes, it should be connected components instead of SCC.
$endgroup$
– Apass.Jack
11 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.



While we are running Dijkstra's algorithm, we are in one of two kinds of stages alternatively. One kind of stage is when we are exploring towards red nodes. The other kind of stage is when we are exploring towards blue nodes. Since each edge is checked/visited as most twice, the running time is $O(|E|)$.






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
    1 hour 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.



While we are running Dijkstra's algorithm, we are in one of two kinds of stages alternatively. One kind of stage is when we are exploring towards red nodes. The other kind of stage is when we are exploring towards blue nodes. Since each edge is checked/visited as most twice, the running time is $O(|E|)$.






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
    1 hour 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.



While we are running Dijkstra's algorithm, we are in one of two kinds of stages alternatively. One kind of stage is when we are exploring towards red nodes. The other kind of stage is when we are exploring towards blue nodes. Since each edge is checked/visited as most twice, the running time is $O(|E|)$.






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.



While we are running Dijkstra's algorithm, we are in one of two kinds of stages alternatively. One kind of stage is when we are exploring towards red nodes. The other kind of stage is when we are exploring towards blue nodes. Since each edge is checked/visited as most twice, the running time is $O(|E|)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 14 mins ago

























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
    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
    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




$begingroup$
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
1 hour ago





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


















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Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e