I am having a discussion with a friend if the following will work:
We recently learned in a lecture about Breadth-First-Search. I know that it is a special case of Dijkstra where each edge weight is set to one. Assume now we are given a graph where the edges have integer weights of more than one. Then I would modify this graph by introducing additional vertices and connecting them by edges with weight one, e.g. assume we have an edge of weight 3 connecting the vertices u and v, then I would introduce dummy-vertices d1, d2, remove the edge connecting u and v and instead add edges {u, d1}, {d1, d2}, {d2,v} of weight one.
If I modify my whole graph this way and then apply breadth-first search starting from one of the original vertices, wouldn't this work as well?
Thank you very much in advance!
Since BFS is guaranteed to return an optimal path on unweighted graphs, and you've created the unweighted equivalent of your original graph, you'll be guaranteed to get the shortest path.
What you lose by doing this over Dijkstra's algorithm is runtime optimality. Now the runtime of your algorithm is dependent on the edge weights, whereas Dijkstra's is only dependent on the number of edges.
This sort of thought experiment is a great way to understand how Dijkstra's algorithm works (eg. how would you modify your algorithm to not require creating a new graph? Or not take 100 steps for an edge with weight 100?). In fact this is probably how Dijkstra discovered the algorithm to begin with.
Related
I am a newbie in using python and I am in need of some help.
I am trying to built a weighted and undirected k-nearest-neighbors graph for a given 13-dimensional dataset containing 200 data points.
For a start, I created an 3-dimensional embedding via PCA (preserving up to 98% of the initial data structure). I also created the embedding scatter plot using matplotlib and a similarity matrix containing each data point's distance to it's 10 nearest neighbors using sklearn.neighbors.kneighbors_graph. The resulting matrix is not a symmetric one and would lead me to a directed graph.
What I want to do is to create an undirected graph, using the distances as edge weights and each data point as a vertex. Focusing on the "undirected" part of the process, this means that:
(A) two vertices (let's say v-i and v-j) would be connected with an undirected edge if v-i is among the k-nearest-neighbors of v-j or if v-j is among the k-nearest-neighbors of v-i.
(B) two vertices would be connected with an undirected edge if v-i is and among the k-nearest_neighbors of v-j and v-j is among the k-nearest-neighbors of v-i.
The resulting similarity matrix (using either (A) or (B) would be a symmetric one).
Unfortunately, I have no idea how to do this or how to plot it. Does anyone have a clue?
Thanks in advance!!!
I tried using Networkx, but I'm afraid it doesn't work.
Given a directed weighted graph G and n, where n is the number of paths to be used to cover all the vertices in the graph G. How can I minimize the maximum cost of the longest path? (assuming that a solution always exist in this graph)
For n = 1, this obviously becomes a Travelling Salesman Problem - which is NP-hard. Thus, I wouldn't look for exact algorithms in your case.
My guess would be that a good solution for small n would be to use one of the abundant algorithms for the Travelling Salesman Problem (which usually approximate optimal solutions quite good) and then remove the (n-1) heaviest edges from the found path. That way you end with n paths.
The Wikipedia Article on TSP actually lists some pretty easy algorithmic techniques which should give you a reasonably good approximation.
How to find maximal eulerian subgraph of a given graph? By "maximal" I mean subgraph with maximal number of edges, vertices, or both. My idea is to find basis of cycle space and combine basis cycles in a proper way, but I don't know how to do it (and is it a good idea or not).
UPD. Source graph is connected.
Some thoughts. Graph is eulerian iff it is connected (with possible isolated vertices) and all vertices have even degree.
It is 'easy' to satisfy second criteria by removing (shortest) paths between pairs of odd degree vertices.
Connectivity is problematic since removing edges can produce unconnected graph.
An example which shows that 'simple' (greedy) solution is not easy to produce. Modify complete graph K5 by splitting each edge in two edges (or more). Take two these modified K5 graph and from each one take two vertices (A, B from first and C, D from second). Connect A-C and B-D. Greedy approach would remove these added edges since they are the shortest paths. With that graph becomes unconnected. Solution would be to remove paths A-B and C-D.
It seems to me that algorithm should take a care about subgraph connectivity while removing edges. For sure algorithm should preserve that each subset of odd degree vertices, of which no pair are used to remove path between them, should have connectivity larger than cardinality of subset.
I would try (for a test) with recursive brute force solution with optimization. O is list of odd degree vertices.
def remove_edges(O, G):
if O is empty:
return solution
for f in O:
for t in O\{f}":
G2 = G without path edges between (f,t)
if G2 is unconnected:
continue
return remove_edges(O\{f,t}, G2)
Optimization can be to order sets O and O{f} by vertices that have shortest paths. That can be done by finding shortest lengths between all pairs of vertices from O before removing edges. That can be done by BFS from each O vertex.
It is proved in 1979 that determining if a given graph contains a spanning Eulerian subgraph is NP-complete.
Ref: W. R. Pulleyblank, A note on graphs spanned by
Eulerian graphs, J. Graph Theory 3, 1979, pp.
309–310,
Please refer to this
Finding the maximum size (number of edges) of spanning Eulerian subgraph of a graph (if it exists) is an active research area.
Consider the following standard definitions. Given a graph G = (V, E)
A circuit is a sequence of adjacent vertices starting and ending at
the same vertex. Circuits do not allow repeated edges but they do allow
repeated vertices.
A cycle is a special case of a circuit in which vertices also do not
repeat.
Note that circuits and Eulerian subgraphs are the same thing. This means that finding the longest circuit in G is equivalent to finding a maximum Eulerian subgraph of G. As noted above, this problem is NP-hard. So, unless P=NP, an efficient (i.e. polynomial time) algorithm for finding a maximal Eulerian subgraph in an arbitrary graph is impossible.
For undirected graphs, one way of randomly producing an Eulerian subgraph is to identify a cycle basis for G. A cycle basis is a set of cycles that, when combined using symmetric differences, can be used to form every Eulerian subgraph of the original graph G. Hence, we only need to take a random selection of cycles from this set and combine them to get our arbitrary Eulerian subgraph.
Given that an Eulerian subgraph is basically a collection of overlapping cycles, here is a greedy, polynomial-time algorithm that I'd like to suggest for finding large (but not necessarily maximum) Eulerian subgraphs. This works for both directed and undirected graphs and produces a set of edges (or arcs) E’ that define an Eulerian subgraph containing a user-defined source vertex s. The following steps are for directed graphs but can be easily modified for the undirected case.
Let U = {s} and E' = {}
while U is not empty
Let u be a random element in U
Form a cycle C from u in G
if no such cycle C exists
Remove u from U
else
Add the arcs of C to E'
Remove the arcs of C from G
Add the vertices of C to U
Here’s a few points to note about this algorithm.
Here, the set U holds the vertices that are yet to be fully considered by the algorithm.
To apply this method to undirected graphs, just replace the word
"arcs" with "edges"
This method can be seen as a generalisation of
Hierholzer's algorithm. Hence, if the input graph G is already
an Eulerian graph, then the returned set E’ will contain all of the
edges from G.
Various methods can be used to generate a cycle C from
vertex u. For directed graphs, a simple method is to create an
additional dummy vertex u' and temporarily redirect all of the incoming arcs
from u to u'. Various algorithms can then be used to determine a
u-u'-path (which represents a cycle), such as BFS, DFS, or
Wilson's algorithm.
This algorithm can be said to produce a maximal Eulerian subgraph with respect to G and s. This is because, on termination, no further cycles can be added to the solution contained in E'. Note that we should not confuse the terms maximal and maximum here: finding a maximal Eulerian subgraph is easy (using the above method); finding a maximum Eulerian subgraph is NP-hard. Similar terminology is used with matchings.
I am looking to solve a problem where I have a weighted directed graph and I must start at the origin, visit all vertices at least once and return to the origin in the shortest path possible. Essentially this would be a classic example of TSP, except I DO NOT have the constraint that each vertex can only be visited once. In my case any vertex excluding the origin can be visited any number of times along the path, if this makes the path shorter. So for example in a graph containing the vertices V1, V2, V3 a path like this would be valid, given that it is the shortest path:
ORIGIN -> V1 -> V2 -> V1 -> V3 -> V1 -> ORIGIN
As a result, I am a bit stuck on what approach to take in order to solve this, as a classic dynamic programming algorithm approach which is usually used to solve TSP problems in exponential time is not suitable.
The typical approach is to create a distance matrix that gives the shortest-path distance between any two nodes. So d(i,j) = shortest path (following the edges of the network) from i to j. This can be done using Dijkstra's algorithm.
Now just solve a classical TSP with distances d(i,j). Your TSP doesn't "know" that the actual route followed might involve visiting a node multiple times. At the same time, it will ensure that the vehicle stops at every node.
Now, as for efficiency: As #Codor points out, TSP is NP-hard and so is your variant of it, so you are not going to find a provably optimal, polynomial-time algorithm. However, there are still many, many good algorithms (both heuristic and exact) for TSP, and most of them should be suitable for your problem. (In general, DP is not the way to go for TSP.)
To answer the question in part, the problem described in the question does not admit a polynomial-time algorithm unless P=NP by the following argument. Clearly, the proposed problem includes instances which are Euclidean. However, no optimal solution to a Euclidean instance has repeated nodes, as such a solution can be improved by deleting additional nodes, using the triangle inequality. However, according to the Wikipedia article on TSP, Euclidean TSP is still NP-hard. This means that any polynomial-time algorithm for the problem in the question would be able to solve the Euclidean TSP to optimality on polynomial time, which is impossible unless P=NP.
This is a supervised learning problem.
I have a directed acyclic graph (DAG). Each edge has a vector of features X, and each node (vertex) has a label 0 or 1. The task is to find a cost function w(X), so that the shortest path between any pair of nodes has the highest ratio of 1s to 0s (minimum classification error).
The solution must generalize well. I tried logistic regression, and the learned logistic function predicts fairly well the label of a node giving the features of a incoming edge. However, the graph's topology is not taken into account by that approach, so the solution in the whole graph is non-optimal. In other words, the logistic function is not a good weight function given the problem setup above.
Although my problem setup is not the typical binary classification problem setup, here is a good intro to it:
http://en.wikipedia.org/wiki/Supervised_learning#How_supervised_learning_algorithms_work
Here are some more details:
Each feature vector X is a d-dimensional list of real numbers.
Each edge has a vector of features. That is, given the set of edges E = {e1, e2, .. en} and set of feature vectors F = {X1, X2 ... Xn}, then edge ei is associated to vector Xi.
It is possible to come up with a function f(X), so that f(Xi)
gives the likelihood that edge ei points to a node labeled with a 1.
An example of such function is the one I mentioned above, found through logistic
regression. However, as I mentioned above, such function is non-optimal.
SO THE QUESTION IS:
Given the graph, a starting node and an finish node, how do I learn the optimal cost function w(X), so that the ratio of nodes 1s to 0s is maximized (minimum classification error)?
This is not really an answer, but we need to clarify the question. I might come back later for a possible answer though.
Below is an example DAG.
Suppose the red node is the starting node, and the yellow one is the end node. How do you define the shortest path in terms of
the highest ratio of 1s to 0s (minimum classification error) ?
Edit: I add names for each node and two example names for the top two edges.
It seems to me you cannot learn such a cost function that takes feature vectors as inputs and whose output (edge weights? or whatever) can guide you to take a shortest path toward any node considering the graph topology. The reason is stated below:
Let's assume you don't have the feature vectors you stated. Given a graph as above, if you want to find all-pair-shortest-path with respective to the ratio of 1s to 0s, it's perfect to use Bellman equation or more specifically Dijkastra plus a proper heuristic function (e.g., percentage of 1s in the path). Another possible model-free approach is to use q-learning in which we get reward +1 for visiting a 1 node and -1 for visiting a 0 node. We learn a lookup q-table for each target node one at a time. Finally we have the all-pair-shortest-path when all nodes are treated as target nodes.
Now suppose, you magically obtained the feature vectors. Since you are able to find the optimal solution without those vectors, how come they will help when they exist?
There is one possible condition that you can use the feature vector to learn a cost function which optimize edge weights, that is, the feature vectors are dependent on the graph topology (the links between nodes and the position of 1s and 0s). But I did not see this dependency in your description at all. So I guess it does not exist.
This looks like a problem where a genetic algorithm has excellent potential. If you define the desired function as e.g. (but not limited to) a linear combination of the features (you could add quadratic terms, then cubic, ad inifititum), then the gene is the vector of coefficients. The mutator can be just a random offset of one or more coefficients within a reasonable range. The evaluation function is just the average ratio of 1's to 0's along shortest paths for all pairs according to the current mutation. At each generation, pick the best few genes as ancestors and mutate to form the next generation. Repeat until the ueber gene is at hand.
I believe your question is very close to the field of Inverse Reinforcement Learning, where you take certain "expert demonstrations" of optimal paths and try to learn a cost function such that your planner (A* or some reinforcement learning agent) outputs the same path as the expert demonstration. This training is done in an iterative way. I think that in your case, the expert demonstrations could be created by you to be paths that go through maximum number of 1 labelled edges. Here is a link to a good paper on the same: Learning to Search: Functional Gradient Techniques for Imitation Learning. It is from the robotics community where motion planning is usually setup as a graph-search problem and learning cost functions is essential for demonstrating desired behavior.