I'm trying to use graph-tool to quickly calculate the distance from a source vertex to all vertices within a maximum distance, using a cost property that I have available for each edge.
I suppose I have to use the dijkstra_search function, but how do I specify the stop-criterium ? I have a working example, but I think it traverses the entire graph (taking several seconds as it's the entire road network of Holland).
Second, what's the fastest way of generating a list of: (vertex-id, distance) once the dijkstra_search function finishes ?
Turns out it was pretty easy:
import graph_tool.all as gt
dm = G.new_vp("double", np.inf)
gt.shortest_distance(G, source=sourcenode, weights=EdgePropertyMap, dist_map = dm, max_dist=0.1)
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I am trying to find a max value of a curve fitted plot for a certain region in this plot. I have a 4th order fit, and when i use max(x), the ans for this is an extrapolated value, while I am actually looking fot the max value of the 'bump' in my data.
So question, how do I select the max for only a certain region in the data while using a cfit? Or how do I exclude a part of the fit?
LF = pol4Fit(L,F);
Coefs= coeffvalues(LF);
This code does only give the optimum (the max value) of the real points:
L_opt = feval(LF,L);
[F_opt,Num_Length]= max (L_opt);
Opt_Length= L(Num_Length);
So now I was trying something like: y=max(LF(F)), but this is not specific to select a region.
Try to only evaluate the region you are interested in.
For instance, let's say the specific region is a vector named S.
You can simply rewrite your code like below:
L_opt = feval(LF,S);
Use the specific domain region S instead of the whole domain L and it only evaluates the region you are concerned with. Then using max function should work properly for you.
I've been struggling with a problem for a while:) in Matlab.
I have an image (A.tif) in which I would like to find maxima (with defined threshold) but more specific coordinates of these maxima. My goal is to create short profiles on the image crossing these maxima (let say +- 20 pixels on both sides of the maximum)
I tried this:
[r c]=find(A==max(max(A)));
I suppose that r and c are coordinates of maximum (only one/first or every maximum?)
How can I implement these coordinates into ,for example improfile function?
I think it should be done using nested loops?
Thanks for every suggestion
Your code is working but it finds only global maximum coordinates.I would like to find multiple maxima (with defined threshold) and properly address its coordinates to create multiple profiles crossing every maximum found. I have little problem with improfile function :
improfile(IMAGE,[starting point],[ending point]) .
Lets say that I get [rows, columns] matrix with coordinates of each maximum and I'm trying to create one direction profile which starts in the same row where maximum is (about 20 pixels before max) and of course ends in the same row (also about 20 pixels from max) .
is this correct expression :improfile(IMAGE,[rows columns-20],[rows columns+20]); It plots something but it seems to only joins maxima rather than making intensity profiles
You're not giving enough information so I had to guess a few things. You should apply the max() to the vectorized image and store the index:
[~,idx] = max(I(:))
Then transform this into x and y coordinates:
[ix,iy] = ind2sub(size(I),idx)
This is your x and y of the maximum of the image. It really depends what profile section you want. Something like this is working:
I = imread('peppers.png');
Ir = I(:,:,1);
[~,idx]=max(Ir(:))
[ix,iy]=ind2sub(size(Ir),idx)
improfile(Ir,[0 ix],[iy iy])
EDIT:
If you want to instead find the k largest values and not just the maximum you can do an easy sort:
[~,idx] = sort(I(:),'descend');
idxk = idx(1:k);
[ix,iy] = ind2sub(size(I),idxk)
Please delete your "reply" and instead edit your original post where you define your problem better
I'm working with Mean shift, this procedure calculates where every point in the data set converges. I can also calculate the euclidean distance between the coordinates where 2 distinct points converged but I have to give a threshold, to say, if (distance < threshold) then this points belong to the same cluster and I can merge them.
How can I find the correct value to use as threshold??
(I can use every value and from it depends the result, but I need the optimal value)
I've implemented mean-shift clustering several times and have run into this same issue. Depending on how many iterations you're willing to shift each point for, or what your termination criteria is, there is usually some post-processing step where you have to group the shifted points into clusters. Points that theoretically shift to the same mode need not practically end up on directly top of each other.
I think the best and most general way to do this is to use a threshold based on the kernel bandwidth, as suggested in the comments. In the past my code to do this post processing has usually looked something like this:
threshold = 0.5 * kernel_bandwidth
clusters = []
for p in shifted_points:
cluster = findExistingClusterWithinThresholdOfPoint(p, clusters, threshold)
if cluster == null:
// create new cluster with p as its first point
newCluster = [p]
clusters.add(newCluster)
else:
// add p to cluster
cluster.add(p)
For the findExistingClusterWithinThresholdOfPoint function I usually use the minimum distance of p to each currently defined cluster.
This seems to work pretty well. Hope this helps.
I am trying to update a MST by adding a new vertex in the MST. For this, I have been following "Updating Spanning Tree" by Chin and Houck. http://www.computingscience.nl/docs/vakken/al/WerkC/UpdatingSpanningTrees.pdf
A step in the paper requires me to find the largest edge in the path/paths between two given vertices. My idea is to find all the possible paths between the vertices and then, subsequently find the largest edge from the paths. I have been trying to implement this in MATLAB. However, so far, I have been unsuccessful. Any lead / clear algorithm to find all paths between two vertices or even the largest edge in the path between two given nodes/ vertices would be really welcome.
For reference, I would like to put forward an example. If the graph has following edges 1-2, 1-3, 2-4 and 3-4, the paths between 4 and 4 are:
1) 4-2-1-3-4
2) 4-3-1-2-4
Thank you
The algorithm works by lowering the t value to exclude large edges from the new MST. When the algorithm completes, t will be the lowest edge that remains to be inserted to complete the MST.
The m value represents the largest edge on a path from r to z, local to each run of INSERT. m is lowered at each iteration of the loop if possible, thereby removing the previous m edge as a possible candidate for t.
It's not easy to explain in words, I recommend doing a run of the algorithm on paper until the steps are clear.
I made a quick attempt to sketch the steps here: http://jacob.midtgaard-olesen.dk/?p=140
But basically, the algorithm adds edges from the old MST unless it finds a smaller edge to add between the new node z and another node in the old MST. In the example, the edge (A,B) is not in the new tree, since a better connection to B was found by the algorithm.
Note that on selecting h and k, if t and (w,r) have equal edge value, I believe you should choose (w,r)
Finally you should probably go trough the proof following the algorithm to understand why the algorithm works. (I didn't read it all :) )
Can we use Dijkstra's algorithm with negative weights?
STOP! Before you think "lol nub you can just endlessly hop between two points and get an infinitely cheap path", I'm more thinking of one-way paths.
An application for this would be a mountainous terrain with points on it. Obviously going from high to low doesn't take energy, in fact, it generates energy (thus a negative path weight)! But going back again just wouldn't work that way, unless you are Chuck Norris.
I was thinking of incrementing the weight of all points until they are non-negative, but I'm not sure whether that will work.
As long as the graph does not contain a negative cycle (a directed cycle whose edge weights have a negative sum), it will have a shortest path between any two points, but Dijkstra's algorithm is not designed to find them. The best-known algorithm for finding single-source shortest paths in a directed graph with negative edge weights is the Bellman-Ford algorithm. This comes at a cost, however: Bellman-Ford requires O(|V|·|E|) time, while Dijkstra's requires O(|E| + |V|log|V|) time, which is asymptotically faster for both sparse graphs (where E is O(|V|)) and dense graphs (where E is O(|V|^2)).
In your example of a mountainous terrain (necessarily a directed graph, since going up and down an incline have different weights) there is no possibility of a negative cycle, since this would imply leaving a point and then returning to it with a net energy gain - which could be used to create a perpetual motion machine.
Increasing all the weights by a constant value so that they are non-negative will not work. To see this, consider the graph where there are two paths from A to B, one traversing a single edge of length 2, and one traversing edges of length 1, 1, and -2. The second path is shorter, but if you increase all edge weights by 2, the first path now has length 4, and the second path has length 6, reversing the shortest paths. This tactic will only work if all possible paths between the two points use the same number of edges.
If you read the proof of optimality, one of the assumptions made is that all the weights are non-negative. So, no. As Bart recommends, use Bellman-Ford if there are no negative cycles in your graph.
You have to understand that a negative edge isn't just a negative number --- it implies a reduction in the cost of the path. If you add a negative edge to your path, you have reduced the cost of the path --- if you increment the weights so that this edge is now non-negative, it does not have that reducing property anymore and thus this is a different graph.
I encourage you to read the proof of optimality --- there you will see that the assumption that adding an edge to an existing path can only increase (or not affect) the cost of the path is critical.
You can use Dijkstra's on a negative weighted graph but you first have to find the proper offset for each Vertex. That is essentially what Johnson's algorithm does. But that would be overkill since Johnson's uses Bellman-Ford to find the weight offset(s). Johnson's is designed to all shortest paths between pairs of Vertices.
http://en.wikipedia.org/wiki/Johnson%27s_algorithm
There is actually an algorithm which uses Dijkstra's algorithm in a negative path environment; it does so by removing all the negative edges and rebalancing the graph first. This algorithm is called 'Johnson's Algorithm'.
The way it works is by adding a new node (lets say Q) which has 0 cost to traverse to every other node in the graph. It then runs Bellman-Ford on the graph from point Q, getting a cost for each node with respect to Q which we will call q[x], which will either be 0 or a negative number (as it used one of the negative paths).
E.g. a -> -3 -> b, therefore if we add a node Q which has 0 cost to all of these nodes, then q[a] = 0, q[b] = -3.
We then rebalance out the edges using the formula: weight + q[source] - q[destination], so the new weight of a->b is -3 + 0 - (-3) = 0. We do this for all other edges in the graph, then remove Q and its outgoing edges and voila! We now have a rebalanced graph with no negative edges to which we can run dijkstra's on!
The running time is O(nm) [bellman-ford] + n x O(m log n) [n Dijkstra's] + O(n^2) [weight computation] = O (nm log n) time
More info: http://joonki-jeong.blogspot.co.uk/2013/01/johnsons-algorithm.html
Actually I think it'll work to modify the edge weights. Not with an offset but with a factor. Assume instead of measuring the distance you are measuring the time required from point A to B.
weight = time = distance / velocity
You could even adapt velocity depending on the slope to use the physical one if your task is for real mountains and car/bike.
Yes, you could do that with adding one step at the end i.e.
If v ∈ Q, Then Decrease-Key(Q, v, v.d)
Else Insert(Q, v) and S = S \ {v}.
An expression tree is a binary tree in which all leaves are operands (constants or variables), and the non-leaf nodes are binary operators (+, -, /, *, ^). Implement this tree to model polynomials with the basic methods of the tree including the following:
A function that calculates the first derivative of a polynomial.
Evaluate a polynomial for a given value of x.
[20] Use the following rules for the derivative: Derivative(constant) = 0 Derivative(x) = 1 Derivative(P(x) + Q(y)) = Derivative(P(x)) + Derivative(Q(y)) Derivative(P(x) - Q(y)) = Derivative(P(x)) - Derivative(Q(y)) Derivative(P(x) * Q(y)) = P(x)*Derivative(Q(y)) + Q(x)*Derivative(P(x)) Derivative(P(x) / Q(y)) = P(x)*Derivative(Q(y)) - Q(x)*Derivative(P(x)) Derivative(P(x) ^ Q(y)) = Q(y) * (P(x) ^(Q(y) - 1)) * Derivative(Q(y))