Suppose we have a network and in some cases there is the same weight between two servers
how the algorithm is gonna work on that case ? ...for example:
C---> F weight is 2
C--->B weight is 2
Here's the problem...find the shortest way out from A to K.While applying the standard algorithm I get stuck, facing two egdes with the same weight, for instance from C---> F weight is 2 and C--->B weight is 2.
Related
I am designing a fuzzy controller and for that, I have to define 3 triangular function sets. They are:
1 large
2 medium
3 small
But my problem is I have following data only:
Maximum input = 3 Minimum input= 0.1
Maximum output = 5.5 Minimum output= 0.8
How to define 3 triangular set range based on only this given information?
Here is the formula for a triangular membership function
f=0 if x<=a
f=(x-a)/(b-a) if a<=x<=b
f=(c-x)/(c-b) if b<=x<=c
f=0 if x>c
where a is the min, c is the max and b is the midpoint.
In your case, take the top situation where the max is 3 and the min is 0.1. The midpoint is (3+0.1)/2=1.55, so you have
f=0 if x<=0.1
f=(x-0)/(1.55-1) if 0.1<=x<=1.55
f=(3-x)/(3-1.55) if 1.55<=x<=3
f=0 if x>3
You should be able to take the 2nd example from here, but if not let me know. Something worth pointing out is that the midpoint may not be the ideal b in your situation. Any point between a and c could serve as your b, just know that it is the point where the membership function equals 1.
It is difficult to tell, but it looks like maybe you just have given parameters for two of the functions, perhaps for small and large or medium and large. You may need to use some judgement for the 3rd membership function.
For an experiment I need to pseudo randomize a vector of 100 trials of stimulus categories, 80% of which are category A, 10% B, and 10% C. The B trials have at least two non-B trials between each other, and the C trials must come after two A trials and have two A trials following them.
At first I tried building a script that randomized a vector and sort of "popped" out the trials that were not where they should be, and put them in a space in the vector where there was a long series of A trials. I'm worried though that this is overcomplicated and will create an endless series of unforeseen errors that will need to be debugged, as well as it not being random enough.
After that I tried building a script which simply shuffles the vector until it reaches the criteria, which seems to require less code. However now that I have spent several hours on it, I am wondering if these criteria aren't too strict for this to make sense, meaning that it would take forever for the vector to shuffle before it actually met the criteria.
What do you think is the simplest way to handle this problem? Additionally, which would be the best shuffle function to use, since Shuffle in psychtoolbox seems to not be working correctly?
The scope of this question moves much beyond language-specific constructs, and involves a good understanding of probability and permutation/combinations.
An approach to solving this question is:
Create blocks of vectors, such that each block is independent to be placed anywhere.
Randomly allocate these blocks to get a final random vector satisfying all constraints.
Part 0: Category A
Since category A has no constraints imposed on it, we will go to the next category.
Part 1: Make category C independent
The only constraint on category C is that it must have two A's before and after. Hence, we first create random groups of 5 vectors, of the pattern A A C A A.
At this point, we have an array of A vectors (excluding blocks), blocks of A A C A A vectors, and B vectors.
Part 2: Resolving placement of B
The constraint on B is that two consecutive Bs must have at-least 2 non-B vectors between them.
Visualize as follows: Let's pool A and A A C A A in one array, X. Let's place all Bs in a row (suppose there are 3 Bs):
s0 B s1 B s2 B s3
Where s is the number of vectors between each B. Hence, we require that s1, s2 be at least 2, and overall s0 + s1 + s2 + s3 equal to number of vectors in X.
The task is then to choose random vectors from X and assign them to each s. At the end, we finally have a random vector with all categories shuffled, satisfying the constraints.
P.S. This can be mapped to the classic problem of finding a set of random numbers that add up to a certain sum, with constraints.
It is easier to reduce the constrained sum problem to one with no constraints. This can be done as:
s0 B s1 t1 B s2 t2 B s3
Where t1 and t2 are chosen from X just enough to satisfy constraints on B, and s0 + s1 + s2 + s3 equal to number of vectors in X not in t.
Implementation
Implementing the same in MATLAB could benefit from using cell arrays, and this algorithm for the random numbers of constant sum.
You would also need to maintain separate pools for each category, and keep building blocks and piece them together.
Really, this is not trivial but also not impossible. This is the approach you could try, if you want to step aside from brute-force search like you have tried before.
I have training set of size 54 * 65536 and a testing set of 18 * 65536.
I want to use a knn classifier, but I have some questions:
1) How should I define trainlabel?
Class = knnclassify(TestVec,TrainVec, TrainLabel,k);
Is it a vector of size 54 * 1 that defines to which group each row in training set belongs? Here the group is numbered as 1 ,2,..
2) To find the accuracy I used this:
cp = classperf(TrainLabel);
Class = knnclassify(TestVec,TrainVec, TrainLabel);
cp = classperf(TestLabel,Class);
cp.CorrectRate*100
Is this right? Is there another method to calculate it?
3) How can I enhance the accuracy?
4) How do I choose the best value of k?
I do not know matlab nor the implementation of the knn you are providing, so I can answer only a few of your questions.
1) You assumption is correct. trainlabel is a 54*1 vector or an array of size 54 or something equivalent that defines which group each datapoint (row) in training set belongs to.
2) ... MATLAB / implementation related, sorry.
3) That is a very big discussion. Possible ways are:
Choose a better value of K.
Preprocess the data (or make preprocessing better if already applied).
Get a better / bigger trainset.
to name a few...
4) You can use different values while measuring the accuracy for each one and keep the best. (Note: If you do that, make sure you do not measure the accuracy of the classifier per value of k only once, but rather you use some technique like 10-Folding or something).
There is more than a fair chance that the library you are using for the K-NNclassifier provides such utilities.
I am trying to train a random forest classificator on a very imbalanced dataset with 2 classes (benign-malign).
I have seen and followed the code from a previous question (How to set up and use sample weight in the Orange python package?) and tried to set various higher weights to the minority class data instances, but the classificators that I get work exactly the same.
My code:
data = Orange.data.Table(filename)
st = Orange.classification.tree.SimpleTreeLearner(min_instances=3)
forest = Orange.ensemble.forest.RandomForestLearner(learner=st, trees=40, name="forest")
weight = Orange.feature.Continuous("weight")
weight_id = -10
data.domain.add_meta(weight_id, weight)
data.add_meta_attribute(weight, 1.0)
for inst in data:
if inst[data.domain.class_var]=='malign':
inst[weight]=100
classifier = forest(data, weight_id)
Am I missing something?
Simple tree learner is simple: it's optimized for speed and does not support weights. I guess learning algorithms in Orange that do not support weight should raise an exception if the weight argument is specified.
If you need them just to change the class distribution, multiply data instances instead. Create a new data table and add 100 copies of each instance of malignant tumor.
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))