Find a path from s to t using as few red nodes as possible The Next CEO of Stack...
Is it possible to create a QR code using text?
What does it mean 'exit 1' for a job status after rclone sync
Simplify trigonometric expression using trigonometric identities
How should I connect my cat5 cable to connectors having an orange-green line?
Could you use a laser beam as a modulated carrier wave for radio signal?
Does Germany produce more waste than the US?
Find the majority element, which appears more than half the time
My boss doesn't want me to have a side project
Compilation of a 2d array and a 1d array
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
Is it a bad idea to plug the other end of ESD strap to wall ground?
Can I cast Thunderwave and be at the center of its bottom face, but not be affected by it?
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
How dangerous is XSS
How exploitable/balanced is this homebrew spell: Spell Permanency?
What happens if you break a law in another country outside of that country?
Car headlights in a world without electricity
That's an odd coin - I wonder why
Calculate the Mean mean of two numbers
How seriously should I take size and weight limits of hand luggage?
Why does freezing point matter when picking cooler ice packs?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Planeswalker Ability and Death Timing
Could a dragon use its wings to swim?
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
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 40 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 mins ago
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "419"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106337%2ffind-a-path-from-s-to-t-using-as-few-red-nodes-as-possible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
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.
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 40 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 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 40 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 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 40 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 40 mins ago
Apass.JackApass.Jack
13.7k1940
answered 40 mins ago
Apass.JackApass.Jack
13.7k1940
answered 40 mins ago
Apass.JackApass.Jack
13.7k1940
answered 40 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 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
27 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
27 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 mins ago
$begingroup$
I will show shortly that this particular run of Dijkstra's algorithm actually takes $O(|E|)$ time.
$endgroup$
– Apass.Jack
13 mins ago
add a comment |
Thanks for contributing an answer to Computer Science Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106337%2ffind-a-path-from-s-to-t-using-as-few-red-nodes-as-possible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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