I have a question regarding the maximum flow in a network. I was trying to find a cut set in a graph that could disconnect the source and the destination. I explored all the edge independent paths in the graph from a source to the destination. Next I picked an edge from each of these paths and grouped them together. So basically I enumerated all the possible combinations of taking one edge from each path.
So I have a set of such groups. Does this mean that I have eventually found the cut sets of the network for that particular source and destination? Is this an efficient method?
This sounds like it has exponential complexity. I can't say exactly, because I don't know what "all the edge independent paths" means. For example:
A
|
B
/ \
C D
\ /
E
There are two paths from A to E, but they're not edge independent.
A maximum flow on a graph is dual to a minimum cut, and there are plenty of standard algorithms that can find one in (small) polynomial time. If you're satisfied with any cut at all, just remove all the edges -- this runs in time O(E).
What are your constraints?
Related
I've been trying to find a way to replicate the following network graph format in R using DiagrammeR/GraphViz, but without success (ignore the thick black arrow on N1): https://i.stack.imgur.com/oHpQz.png
The graph is a directed graph and each edge in a certain direction either ends with an arrowhead (-->) if the edge value is positive, or a circle/odot (--o) if the edge value is negative. Between a pair of nodes (ex. N1 -- A1), there can be an edge N1 --> A1 and an edge A1 --o N1, and these need to be concentrated so that the two edges look like one line with an arrowhead on one end and a circlehead on the opposite end (like this: o--->). These cannot be parallel or look like two edges ideally.
Another requirement is that the nodes have to be in very specific positions and remain there throughout model simulations where edges might change. From what I have tried and the documentation I have read, this is not possible to do in DOT format, but is possible in neato format.
This is where I get a problem. In neato, I can align the nodes exactly where I want them by defining their x,y positions. However, when I use concentrate = true to create the o---> edge from two otherwise parallel edges, only one type of arrowhead remains. So an edge that's supposed to look like o---> ends up looking like ---> or o---.
This is not a problem in DOT format as concentrate = true does what I want it to do, but in DOT I cannot assign exact node positions. I have tried getting around this using node ranks but without much luck. It seems to stack nodes I want in different ranks within the same rank. As well, concentrate = true doesn't seem to work for edges between nodes within the same rank, as it leaves them as two separate curved edges ---> and o--- without concentrating them.
The reason why I need this to work is because I'm running model simulations where the edges change, and I need to generate hundreds of such graphs. For easy comparison, the nodes need to stay in the same place for consistency.
This is the closest I could come up with using neato format (nodes are positioned the way I want but it's not showing the proper o---> for all the black edges minus self-edges; red edges are true one-way links): https://i.stack.imgur.com/YJBY7.jpg
If only the edges showed up as the proper o---> format, this would be perfect for my needs. If you know of any way to fix this issue using DiagrammeR/GraphViz, or even another program, I would be so grateful. Thanks!
You probably don't need concentrate. Look at arrowtail and dir (https://www.graphviz.org/doc/info/attrs.html#d:arrowtail and https://www.graphviz.org/doc/info/attrs.html#d:dir) and neato -n
digraph c {
graph[label="can neato do the work?"]
node[shape=circle]
a [pos="100,100"]
b [pos="200,100"]
c [pos="300,100"]
a->b [dir=both arrowtail=odot]
c->c [dir=both arrowtail=odot arrowhead=none]
}
Giving:
Given (N+1) points. All the N points lie on x- axis. Remaining one point (HEAD point) lies anywhere in the coordinate plane.
Given a START point on x- axis.
Find the shortest distance to cover all the points starting from START point.We can traverse a point multiple times.
Example N+1=4
points on x axis
(0,1),(0,2),(0,3)
HEAD Point
(1,1) //only head point can lie anywhere //Rest all on x axis
START Point
(0,1)
I am looking for a method as of how to approach this problem.
Whether we should visit HEAD point first or HEAD point in between.
I tried to find a way using Graph Theory to simplify this problem and reduce the paths that need to be considered. If there is an elegant way to represent this problem using graphs to identify a solution, I was not able to find it. This approach becomes very inefficient as the n increases - the time and memory is O(2^n).
Looking at this as a tree graph, the root node would be the START point, then each of its child nodes would be the points it is connected to.
Since the START point and the rest of the points aside from the HEAD all lay on the x-axis, all non-HEAD points only need to be connected to adjacent points on the x-axis. This is because the distance of the path between any two points is the sum of the distances between any adjacent points along the path between those two points (the subset of nodes representing points on the x-axis does not need to form a complete graph). This reduces the brute force approach some.
Here's a simple example:
The upper left shows the original problem: points on the x-axis along with the START and HEAD points.
In the upper right, this has been transformed into a graph with each node representing a point from the original problem. The edges represent the paths that can be taken between points. This assumes that the START point only represents the first point in the path. Unlike the other nodes, it is only included in the path once. If that is not the case and the path can return to the START point, this would approximately double the possible paths, but the same approach can be followed.
In the bottom left, the START point, a, is the root of a tree graph, and each node connected to the START point is a child node. This process is repeated for each child node until either:
A path that is obviously not optimal is identified, in which case that node can just be excluded from the graph. See the nodes in red boxes; going back and forth between the same nodes is unnecessary.
All points are included when traversing the tree from the root to that node, producing a potential solution.
Note that when creating the tree graph, each time a node is repeated, its "potential" child nodes are the same as the first time the node was included. By "potential", I mean cases above still need to be checked, because the result might include a nonsensical path, in which case that node would not be included. It is also possible a potential solution results from the path after its child nodes are included.
The last step is to add up the distances for each of the potential solutions to determine which path is shortest.
This requires a careful examination of the different cases.
Assume for now START (S) is on the far left, and HEAD (H) is somewhere in the middle the path maybe something like
H
/ \
S ---- * ----*----* * --- * ----*
Or it might be shorter to from H to and from the one of the other node
H
//
S ---- * --- * -- *----------*---*
If S is not at one end you might have something like
H
/ \
* ---- * ----*----* * --- * ----*
--------S
Or even going direct from S to H on the first step
H
/ |
* ---- * ----*----* |
S
A full analysis of cases would be quite extensive.
Actually solving the problem, might depend on the number of nodes you have. If the number is small < 10, then compete enumeration might be possible. Just work out every possible path, eliminate the ones which are illegal, and choose the smallest. The number of paths is I think in the order of n!, so its computable for small n.
For large n you can break the problem into small segments. I think its enough just to consider a small patch with nodes either side of H and a small patch with nodes either side of S.
This is not really a solution, but a possible way to think about tackling the problem.
(To be pedantic stackoverflow.com is not the right site for this question in the stack exchange network. Computational Science : algorithms might be a better place.
This is a fun problem. First, lets try to find a brute force solution, as Poosh did.
Observations about the Shortest Path
No repeated points
You are in an Euclidean geometry, thus the triangle inequality holds: For all points a,b,c, the distance d(a,b) + d(b,c) <= d(a,c). Thus, whenever you have an optimal path that contains a point that occurs more than once, you can remove one of them, which means it is not an optimal path, which leads to a contradiction and proves that your optimal path contains each point exactly once.
Permutations
Our problem is thus to find the permutation, lets call it M_i, of the numbers 1...n for points P1...Pn (where P0 is the fixed start point and Pn the head point, P1...Pn-1 are ordered by increasing x value) that minimizes the sum of |(P_M_i)-(P_M_(i-1))| for i from 1 to n, || being the vector length sqrt(v_x²+v_y²).
The number of permutations of a set of size n is n!. In this case we have n+1 points, so a brute force approach testing all permutations would have complexity (n+1)!, which is higher than even 2^n and definitely not practical, so we need further observations to improve this.
Next Steps
My next step would now be to see if there are any other sequences that can be proven to be not optimal, leading to a reduction in the number of candidates to be tested.
Paths of non-head points
Lets look at all paths (sequences of indices of points that don't contain a head point and that are parts of the optimal path. If we don't change the start and end point of a path, then any other transpositions have no effect on the outside environment and we can perform purely local optimizations. We can prove that those sequences must have monotonic (increasing or decreasing) x coordinate values and thus monotonic indices (as they are ordered by ascending x coordinate between indices 0 and n-1):
We are in a purely one dimensional subspace and the total distance of the path is thus equal to the sum of the absolute values of the differences in x coordinates between one such point and the next. It is clear that this sum is minimized by ordering by x coordinate in either ascending or descending order and thus ordering the indices in the same way. Note that this is true for maximal such paths as well as for all continuous "subpaths" of them.
Wrapping it up
The only choices we have left are:
where do we place the head node in the optimal path?
which way do we order the two paths to the left and right?
This means we have n values for the index of the head node (1...n, 0 is fixed as the start node) and 2x2 values for sort order. So we have 4n choices which we can all calculate and pick the shortest one. One of the sort orders probably determines the other but I leave that to you.
Anyways, the complexity of this algorithm is O(4n) = O(n). Because reading in the input of the problems is in O(n) and writing the output is as well, I believe that is an algorithm of optimal complexity. However, if we could reformulate the problem somewhat, so that we could read and write the input and output in some compressed form, as in only the parameters that we actually need to solve the problem, then it is possible that we could do better.
P.S.: I'm not a mathematician so I probably used wrong words for some concepts and missed the usual notation for the variables and functions. I would be glad for some expert to check this for any obvious errors.
I have about 100 atoms in a 3-D space. Each atom is a node. Edges are added between two nodes when they are closer than 0.32 nm with weight equals distance. I want to find a path from source node to destination node. Since the 100 atoms are not fully connected, sometimes I can't find a path.
What I want to do is to add one or more edges to make source and destination connected. Meanwhile, I also want to minimize the total weights of the new added edges. Again, weight is calculated from the two nodes' distance.
It is kind of a reverse problem of minimum cut. Is there any algorithm helps to do this?
Thanks a lot!
It seems like one way would be to make use of a graph search algorithm for finding the shortest path, like Dijkstra's algorithm, and perhaps work from both ends (source and destination).
The only difference is that you can't know if any edge actually exists or not, and so you are creating the graph as you go. So if you start at A, and the nodes of the graph are A, B, C, D, E. Then you need to check if A-B, A-C, A-D, and A-E exists. If only A-B exists, then you check B-C, B-D, and B-E.
This will be O(|V|^2), but will really depend on how many edges get explored.
The same idea applies if you are only interested in adding edges that are longer than 0.32 nm. It's just that the path length calculation changes. Any edges that are less than .32 nm are zero-length, or simply a lot shorter (as they become less important). If that last bit doesn't work, then it gets a bit trickier.
A similar question is posted here.
I have an undirected graph with Vertex V and Edge E. I am looking for an algorithm to identify all the cycle bases in that graph. An example of such a graph is shown below:
Now, all the vertex coordinates are known ( unlike previous question, and contrary to the explanation in the above diagram), therefore it is possible to find the smallest cycles that encompass the whole graph.
In this graph, it is possible that there are edges that don't form any cycles.
What is the best algorithm to do this?
Here's another example that you can take a look at:
Assuming that e1 is the edge that gets picked first, and the arrow shows the direction of the edge.
I haven't tried this and it is rather greedy but should work:
Pick one node
Go to one it's neighbors's
Keep on going until you get back to your starting node, but you're not allowed to visit an old node.
If you get a cycle save it if it doesn't already exist or a subset of those node make up a cycle. If the node in the cycle is a subset of the nodes in another cycle remove the larger cycle (or maybe split it in two?)
Start over at 2 with a new neighbor.
Start over at 1 with a new node.
Comments: At 3 you should of course do the same thing as for step 2, so take all possible paths.
Maybe that's a start? As I said, I haven't tried it so it is not optimized.
EDIT: An undocumented and not optimized version of one implementation of the algorithm can be found here: https://gist.github.com/750015. But, it doesn't solve the solution completely since it can only recognize "true" subsets.
I'm looking to build an algorithm (or reuse one) that organizes nodes and edges on a 2 dimensional canvas where edges can have corresponding weights.
Any starting material and info would be helpful.
What would the weights do to affect their placement on your canvas?
That being said, you might want to look into graphviz and, more specifically, the DOT language, which organizes nodes on a canvas.
Many graph visualization frameworks use a force-based simulation, in which all nodes exert a repulsive force against each other (with their mass being their size), and edges exert tension on the nodes they connect. This creates aesthetically-arranged graph visualizations.
Although again, I'm not sure where you want node "weights" to come into play. Do you want weighted nodes to be more in the center? To be larger? More further apart?
Many graph/network layout algorithms are implicitly capable of handling weighted networks, but you may need to do some pre-processing and tweaks to the implementation to get it to work. Usually the first step is to determine if your weights represent "similarities" (usually interpreted to mean that stronger weights should place nodes closer togeter) or "dissimilarities" (stronger weights = father apart). The most common case is the former, so you will need to translate them to dissimilarities, often done by subtracting each edge value from the maximum observed edge value in the network. The matrix of dissimilarity values for each edge can then be fed to the algorithm and interpreted as desired distances in the layout space for each edge (i.e. "spring lengths")--usually after multiplying by some constant to transform to display units (pixels).
If you tell me what language you are using, I may be able to point you to some code examples.