25 votes
Accepted

Why can't we find shortest paths with negative weights by just adding a constant so that all weights are positive?

Adding a weight to every edge adds more weight to long paths than short paths. (Long in the sense of having many edges.) For example, suppose the lowest-cost edge has weight $-2$ and there are ...
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23 votes

What is the meaning of 'breadth' in breadth first search?

Consider the data structure used to represent the search. In a BFS, you use a queue. If you come across an unseen node, you add it to the queue. The “frontier” is the set of all nodes in the search ...
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16 votes

Minimum spanning tree vs Shortest path

Though Minimum Spanning Tree and Shortest Path algorithms computation looks similar they focus on 2 different requirements. In MST, requirement is to reach each vertex once (create graph tree) and ...
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  • 271
14 votes
Accepted

A* graph search time-complexity

These are basically two different perspectives or two different ways of viewing the running time. Both are valid (neither is incorrect), but $O(b^d)$ is arguably more useful in the settings that ...
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  • 141k
13 votes
Accepted

Example of graph with exponential many s-t minpaths and min cuts

I can offer an example for super-exponentially many shortest paths and super-polynomially many minimum cuts. An example for many shortest s-t-paths you probably came up with is the layer graph, ...
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  • 70.9k
12 votes

Why does Dijkstra's algorithm fail on a negative weighted graphs?

Adding a constant amount to each edge length can change the shortest path for the simple reason that it increases the length of a path with many edges by more than it increases the length of a path ...
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12 votes

How does consistency imply that a heuristic is also admissible?

To proof the statement in your question, let us proof that consistency implies admissibility whereas the opposite is not necessarily true. This would make consistency a stronger condition than the ...
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11 votes
Accepted

Shortest path between two points with n hops

If vertices can be visited more than once, then yes: you can create $n+1$ copies of the graph, with each vertex $v$ in the original graph becoming the $n+1$ vertices $v_1, \dots, v_{n+1}$ and each ...
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11 votes
Accepted

Find shortest path between two vertices that uses at most one negative edge

You can use Dijkstra twice to find in your $G'$ the cost for each vertex $v \in V$, the cost of the optimal $s$-$v$-path and the cost of the optimal $v$-$t$-path. Store this in a table creatively ...
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  • 13.3k
11 votes

Find shortest path between two vertices that uses at most one negative edge

Another approach is to create a single graph $H$ as follows: each vertex in $G$ has two counterparts in $H$: vertex $s$ becomes $s_A$ and $s_B$, vertex $t$ becomes $t_A$ and $t_B$, and so on. each ...
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  • 521
10 votes

Minimum spanning tree vs Shortest path

I think an example will make it clearer.. The spanning tree looks like below. This is because if we add up the edges in this configuration, we get the least total cost possible: 2+5+14+4=25. ...
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  • 363
10 votes
Accepted

Finding all edges on any shortest path between two nodes

Off the top of my head, you could do this. (Let's say you want to find all edges on a shortest path from $s$ to $t$.) Run an All-Points Shortest Path (APSP) algorithm to store the shortest-path ...
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  • 6,940
9 votes
Accepted

What role is the set, S playing in Dijkstra's algorithm given in the book CLRS?

No, you are not missing anything if you remove $S$ completely. You could implement and run Dijkstra's algorithm correctly still. Set $S$ is used later in the book to help explain the algorithm and ...
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  • 34.1k
8 votes

Minimum spanning tree vs Shortest path

The difference lies in what is the ultimate goal of this algorithms- Dijkstra's - Here the goal is to reach from start to end. You are concerned about only this 2 points, and optimize your path ...
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8 votes

What is the fastest algorithm for finding all shortest paths in a sparse graph?

You can try to make a version of Floyd-Warshall that is faster on sparse matrices. First, let us recall what this algorithm does: Let $M$ be a matrix of distances. At the beginning of the algorithm $...
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8 votes

Why can't DFS be used to find shortest paths in unweighted graphs?

You can!!! Mark the nodes as visited while you are going depth and unmark while you return, while returning as you find another branch(es) repeat same. Save cost/path for all possible search where ...
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  • 181
8 votes
Accepted

Minimum distance between start and end by going through must visit points in a maze

Most real mazes are trees and there is a polynomial time algorithm for this problem on trees. For ease of description, orient all edges of the tree to point away from the start vertex. Now, for each ...
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8 votes

Is Dijkstras algorithm used in modern route-finding systems?

Google Maps in 2009 used Contraction Hierarchies - see this tech talk. Since then, some mind-blowing methods have been discovered, capable of doing cross-country routing in fractional milliseconds - ...
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  • 2,209
8 votes
Accepted

Why do we need to run the bellman-ford algorithm for n-1 times?

Answering your first question is simple. Try out a few graphs and see for yourself why it doesn't work. As to your second question, $n-1$ is the maximal length of a shortest path in the graph. After $...
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7 votes

Example of graph with exponential many s-t minpaths and min cuts

Take any graph $G$ on $n$ vertices which has $2^{\Omega(n)}$ minimal $s$-$t$ paths. Add to $G$ an independent set of size $n$. Now it has at least $2^n$ minimum cuts.
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7 votes
Accepted

How to find the shortest path from some vertex in set $S$ to set $S'$

If all edge lengths are non-negative, then this can be solved in $O(|E| \lg |V|)$ time using a single invocation of Dijkstra's algorithm. We're going to modify the graph slightly by adding a new ...
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  • 141k
7 votes
Accepted

Linear time algorithm for finding $k$ shortest paths from $s$ to $t$

First of all, the answer that applies here was already given by Raphael in the comments to the question: "Given that we don't even know how to find one simple shortest path in linear time, I doubt it."...
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7 votes
Accepted

Can we find k shortest paths between all pairs faster than solving the pairwise problem repeatedly?

First of all, a crucial difference in computing $k$-shortest paths is if the paths need to be simple or not. A path is called simple, if it does not contain nodes repeatedly. A path with a loop, for ...
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  • 86
7 votes

Can I run Dijkstra's algorithm using priority queue?

Yes, you can use priority queues to improve the complexity of the algorithm from $O(V^2)$ to $O(|E| + |V| \log|V|)$ where $E$ is the number of edges and $V$ is the number of nodes. You ...
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  • 171
7 votes
Accepted

Shortest walk through a given subset of edges

This is NP-hard, so it's very unlikely that a polynomial-time algorithm exists. Given any instance $G=(V, E)$ of Hamiltonian Path, create a new graph $G'=(V', E')$ in which every vertex $v \in V$ ...
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7 votes
Accepted

Finding shortest paths in undirected graphs with possibly negative edge weights

I contacted one of the authors (Kevin Wayne; thanks) of the textbook "Algorithms, 4th Edition" and got a hint: Try adding "t-joins" or "perfect matching" to your web searches. Following this, I ...
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  • 9,209
7 votes
Accepted

Dijkstra with max instead of sum

Yes, it is true. Let $w: E(G) \to \mathbb{R}$ be a weight function on the edges of $G$, $s \in V(G)$ be the start vertex. Let $p(v) = \min\{\max\{w(e_1), \ldots, w(e_k)\} \mid e_1, \ldots, e_k \text{...
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7 votes
Accepted

Shortest paths between given red vertices and arbitrary blue vertices

I assume no negative weights in this solution (since you stated Dijkstra). My solution also uses Dijkstra. Let $G'$ be the graph resulting from $G$ by contracting all the blue vertices into one vertex ...
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6 votes
Accepted

Finding all vertices on negative cycles

If you don't constrain yourself to simple cycles, you can actually use the Bellman-Ford algorithm to find all the relevant vertices. Start by running a DFS on the graph to find its strongly connected ...
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  • 553
6 votes
Accepted

Proof of Dijkstra Algorithm Optimality

There is no non-trivial complexity lower bound on any interesting problem. In particular, any algorithm not running in $O(|V|+|E|)$ is not known to be tight. That said, you might be able to show a ...
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