Given a directed graph.
Any 2 vertices are adjacent. The edge connecting a pair of vertices may be uni-directional or bi-directional.
How do I find a Hamilton path?
Side notes:
Wikipedia says "A strongly connected simple directed graph with n vertices is Hamiltonian if every vertex has a full degree greater than or equal to n." Therefore, a solution must exist in my problem.
I understand that the general Hamilton path problem is NP-Complete. But it feels like this specific version should have a polynomial solution.
Use a variant of insertion sort to construct a path in quadratic time. Given a path
v1 v2 ... vn-1
on a subset of vertices, consider how to insert vn. If vn has an arc to v1, then prepend vn. If vn-1 has an arc to vn, then append vn. Otherwise, there exists by Sperner's lemma an index i such that vn has an arc from vi and an arc to vi+1. Insert it there.
Related
I am looking back at some of my algorithms homework(exam soon haha), and I am having troubles understanding the solution to one of the questions.
The Question
Tutor Solution
I am having difficulties visualizing the solution, I understand that if you have an odd number of cycles than your graph cannot be bipartite. But as I stated, I don't understand the shortest path from s to u and v to s.
Let's say G is bipartite. Then the vertex set can be divided into V1 and V2 s.t. every edge of G includes a vertex from each set. Then the vertices of every path in G alternate between V1 and V2, so the parity of the length of every path between every pair of vertices is the same. I.e., if G is bipartite and v,w are two vertices of G, then the length of every path connecting v & w is either even or odd.
If there is an edge (u,v) connecting two vertices in the same layer then this is violated. The BFS path to v and u have the same length, so same parity, since v & u are in the same layer. The edge between v & u gives us a path longer by 1, so of a different parity. Contradiction.
The tutor's solution isn't as clear as it could be, since it talks about cycles, and the two paths don't necessarily form a cycle since they can share vertices. And the definition of bipartite graph is slightly wrong (or uses non-standard definitions of edges, cross-product etc. where the things aren't pairs but 2-element sets).
But instead of the definition as given, I'd just say that a graph is bipartite if the vertices can be coloured black or white, such that each edge goes between different-coloured vertices. (Equivalently you can simply say "the graph is 2-colourable").
From this definition, it follows that if there's an even length path between two vertices they must be the same colour, otherwise different colours. In the BFS on a bipartite graph, a vertex u is layer i has a path of length i from s to u, so all vertices in the same layer have the same colour. Thus there can be no edge between two vertices in the same layer.
The following is the question I am working on:
Consider a directed, weighted graph
G
where all edge weights are
positive. The goal of this problem is to find the shortest path
in
G
between two pre-specified vertices
s
and
t
, but with an added twist: you are allowed to change the weight
of
exactly
one edge (of your
choosing) to zero.
In other words, you must pick an edge in
G
to set to zero that minimizes the shortest
path between
s
and
t
.
Give an efficient algorithm to achieve this goal in
O
(
E
lg
V
) time and analyze your algorithm’s running
time. Sub-optimal solutions will receive less credit.
Hint:
You may have to reverse the edges, run a
familiar algorithm a number of times, plus do some extra work
So I have tried running Dijkstra's from s to all other nodes and then I have tried reversing the edges and running it again from s to all other nodes. However, I found out that we have to run Dijskstra's from s to all other nodes and then reverse the edges and then run Dijkstra's from all other nodes to t. I am not exactly sure how this helps us to find the edge to set to zero. By my intuition I thought that we would simply set the maximum weight edge to zero. What is the point of reversing the edges?
We need to run Dijkstra's algorithm twice - once for the original graph with s as the source vertex, and once with the reversed graph and t as the source vertex. We'll denote the distance we get between vertex s and i from the first run as D(i) and the distance we get between vertex t and i second run D_rev(i).
Note that we can go follow the reversed edges backwards (i.e., follow them in the original direction), thus D_rev(i) is actually the shortest distance from vertex i to t. Similarly, D(i) is the shortest distance from vertex s to i following Dijkstra's algorithm.
We can now loop through all the edges, and for each edge e which connects v1 and v2, add up D(v1) and D_rev(v2), which corresponds to the weight of the path s -> v1 -> v2 -> t with e being the zero edge, since we can go from s to v1 with a distance of D(v1), set e to 0, go from v1 to v2, and then go from v2 to t with a distance of D_rev(v2). The minimum over these is the answer.
A rough proof sketch (and also a restatement) : if we set an edge e to 0, but don't use it in the path, we can be better off setting an edge that's in the path to 0. Thus, we need only consider paths that includes the zeroed edge. The shortest path through a zeroed edge e is to first take the shortest path from s to v1, and then take the shortest path from v2 to t, which are exactly what were computed using the Dijkstra algorithm, i.e., D and D_rev.
Hope this answer helps!
I have directed Graph G(V,E) with weight function w. so that weight of each (u,v) is a positive value. I need to find the most lightweight circle in the graph that vertex k' is part of it.
I've also given an algorithm i can use which can find the most lightweight path for a graph with positives weights ( i can use it only once).
I thought about creating a sub graph G' where all vertices and edges that are strongly connected components. find the graph which k' is part of it. then find for the most lightweight adjacent edge from k' to some v of vertices. from that v i can run the algorithm given and find the lightweight path then add the weight of the vertex missing ( (k',v) ).
is that seems correct ? I'm in the beginning of this course and I feel i'm not there yet.
It is a single-source shortest-path problem, where you exclude k->k self-loop as a solution, and find a longer path from k to k. The trick is always expand the shortest path thread.
Given this definition, you can start Googling...
I can't imagine why you called your source vertex k'. Anyway...
Add a new vetrex w that has the same outgoing edges as k'.
Then use Dijkstra's algorithm to find the shortest path from w to k'.
Substitute k' for w in the path, and you have the smallest cycle including k'.
Very interesting problem. I am assuming that there are no negative values in the graph, or otherwise the following solutions requires first normalizing the vertices such that the negative values become at least 0. First method (trivial) is to detect all cycles starting from the target vertex (k). Then compute the weight of all those cycles and take the minimum. The second method is to run Dijkstra algorithm (again watch out negative weights) from the target node (k). Then iterate over all incoming edges of (k), and select the source node that has the minimum Dijkstra value. Now the lightest cycle includes the single path (formed by Dijkstra traversal) from (k) to the chosen node + the bridge to come back to (k). I hope that helps :)
I have a graph with about 6500 vertices, some of which have the label 'c'. I need to devise an algorithm that finds the shortest path between any two of the vertices that includes AT LEAST ONE of these 'c' vertices. This is simple enough, but the problem is that the required complexity is O(ElogV) where E is the number of edges, and V is the number of vertices. I have already implemented Dijkstra's Algorithm using a min-heap, so I can find the general shortest path in O(ElogV), but I am having trouble extending the problem. Any suggestions?
Note that calling Dijkstra's from source to c + c to destination iteratively does not fall within the complexity restraints, as it ends up being O(CElogV)
Let G be your graph, with vertices v1...vn.
Make a new graph that consists of two copies of your original vertices: v1..vn, v1'..vn'. In this new graph, let there be an edge between vi and vj or vi' and vj' if there's an edge between vi and vj in your original graph. Also let there be an edge between vi and vj' if there's an edge between vi and vj in your original graph, and vj' is labelled c.
Then, given two vertices vi, vj, the shortest path between vi and vj' in the new graph is the shortest path between vi and vj in the original graph that passes through at least one vertex labelled c.
Because the number of vertices in the new graph is doubled, and the number of edges at most tripled, the complexity doesn't change from O(VlogE) (where V and E are the number of vertices/edges in the original graph).
If you have an undirected graph:
Say you're searching for a shortest path between S and T.
find all the shortest paths from S to any node using Dijkstra's algorithm
find all the shortest paths from T to any node (same algorithm)
iterate over all marked nodes c and find the shortest path S to c combined with c to T using previously calculated shortest paths.
Is there a graph algorithm that given a start(v) and an end(u) will find a shortest path through the given set of edges, but if u is a disconnected vertex, it will also determine the shortest path to add missing edges until u is no longer disconnected?
I have a pixel matrix where lines are made of 255's(black) and 0's(white). lines(255) can have breaks or spurs and I must get rid of both. I could have a pixel matrix forest with say 7 or so trees of black pixels. I need to find the true end points of each tree, find the single shortest path of each tree, then union all skew trees together to form 1 single line(ie, a single shortest path from the furthest 2 end points in the original matrix). all edge weights could be considered 1.
Thanks
How about running Dijkstra's algorithm and if disconnected, connect v and u? What's your criteria for "best place to add a missing edge?" Do edges have weights (like distance)?
Edit:
For one idea of "the best place," you could try the path that has minimal sum of shortest paths between all connected pairs. Floyd–Warshall algorithm can be used to find shortest paths between all pairs. So, run Floyd-Warshall for each node in v's tree and u.
Your problem isn't well defined for disconnected graphs. I can always add and edge between v and u.
If you meant that given an acyclic undirected disconnected graph, actually known as a forest, and given a subset of edges as a subgraph, can you find the shortest path between vertices, than this problem is trivial since if there is a path in the full graph, there is one path only.
If this is a general graph G, and you are talking about a forest subgraph G', than we need more info. Is this weighted? Is it only positive weights? If it is unweighted, do some variant of Dijksta. Define the diameter of a tree to be the length of the longest path between two leaves (this is well defined in a tree, since ther is only one such path). Let S be the sum of the diameters of all the trees in G', then set the weight all edges out of G' to 2S, then Dijkstra's algorithm will automatically prefer to step through G', stepping outside G' only when there is no choice, since walking through G' would always be cheaper.
Here you have following scenarios:
case:1. (all edges>0)
Negate all edges
find the longest path in graph using this:
https://www.geeksforgeeks.org/longest-path-undirected-tree/
negate the final result
case2: edges might be or might not be negative
Read Floyd–Warshall algorithm for this