Given an undirected weighted graph G = (V,E). Each vertex represents a city and the weight of an edge connected a and b is the number of years that it will take to finish building a high speed route between city a and city b. Describe an algorithm that will find the least number of years before one can travel between any two cities in the graph.
The routes are being built simultaneously, so that if we have three cities a, b and c and an edge between a and b with weight 1, another edge between b and c with weight 2, then the output should be 2.
The comment above is pointing you the right answer, in my opinion this sounds like a classic Prim's algorithm problem. http://en.wikipedia.org/wiki/Prim's_algorithm
Related
Here's the problem:
Consider three mutually independent classifiers, A, B, C, with equal error probabilities:
Pr(errA) = Pr(errB) = Pr(errC) = t
Let D be another classifier that takes the majority vote of A, B, and C.
• What is Pr(errD)?
• Plot Pr(errD) as a function of t.
• For what values of t, the performance of D is better than any of the other three classifiers?
My questions are:
(1) I couldn't figure out the error probability of D. I thought it would be 1 minus alpha (1 - α), but I am not sure.
(2) How to plot t(Pr(errD))? I assume without finding Pr(errD) then I can plot it.
(3) Here as well, I couldn't figure it out. Comparatively, how should I determine the performance of D?
If I understand well, your problem can be formulated with simple terms without any ensemble learning.
Given that D is the result of a vote by 3 classifiers, D is wrong if and only if at most one of the estimators is right.
A,B,C are independent, so:
the probability of none being right is t^3
the probability of one being right while the other two are wrong is 3(1-t)t^2 (the factor 3 is because there are three ways to achieve this)
So P(errD) = t^3 + 3(1-t)t^2 = -2t^3 + 3t^2
You should be able to plot this as a function of t in the interval [0:1] without too many difficulties.
As for your third question, just solve P(errA) - P(errD) >0 (this means that the error probability of D is smaller than for A and so that its performance is better). If you solve this, you should find that the condition is t<0.5.
To come back to ensemble learning, note that the assumption of independence between your estimators is usually not verified in practice.
I am working on solving the following problem and implement the solution in C++.
Let us assume that we have an oriented weighted graph G = (V, A, w) and P a set of persons.
We receive a number of queries such that every query gives a person p and two vertices s and d and asks to compute the minimum weighted path between s and d for the person p. One person can have multiple paths.
After the end of all queries I have a number k <= |A| and I should give k arcs such that the number of persons using at least one of the k arcs is maximal (this is a maximum coverage problem).
To solve the first part I implemented the Djikistra algorithm using priority_queue and I compute the minimal weight between s and d. (Is this a good way to do ?)
To solve the second part I store for every arc the set of persons that use this arc and I use a greedy algorithm to compute the set of arcs (at each stage, I choose an arc used by the largest number of uncovered persons). (Is this a good way to do it ?)
Finally, if my algorithms are goods how can I implement them efficiently in C++?
I trying to find an algorithm to the following question with one different :
the edge are not distinct.
Give an efficient algorithm to test if T remains the minimum-cost spanning tree with the new edge added to G.
in this link- there is a solution but it is not for the different I wrote up:
the edges are not nessecerliy distinct.
Updating a Minimum spanning tree when a new edge is inserted
someone has an idea?
Well, the naive approach of just using Prim or Kruskal to find the min cost spanning tree of the new graph and then see which one has a lower total cost isn't too bad at O(|E|log|E|).
But we don't need to look at the whole graph.
Suppose your new edge connects vertices A and B. Let C be the parent of A. If B is not a descendent of A, then if A-B is lower cost than A-C, then T is no longer the MST and B should be the new parent of the subtree rooted at A.
If B is a descendant of A, then if A-B is shorter than any of the branches in T along the path from A to B, then T is no longer the MST, and the highest cost edge along that path should be removed, B is the root of the newly disconnected component, and should be added as a child of A.
I believe you may need to check these things a second time, reversing which vertices are A and B. The complexity of this is log|V| where the base of the log is the average number of children per node of T. In the case of T being a straight line, it's O(|V|), but otherwise, I think you could say it is O(log|V|).
First find an MST using one of the existing efficient algorithms.
Now adding an edge (v,w) creates a cycle in the MST. If the newly added edge has the maximum cost among the edges on the cycle then the MST remains as it is. If some other edge on the cycle has the maximum cost, then that's the edge to be removed to get a tree with lower cost.
So we need an efficient way to find the edge with the maximum value on the cycle. You can climb from v and w until you reach LCA(v, w) (the least common ancestor of v and w) to get the edge with the max cost. This takes linear time in the worst case.
If you are going to answer multiple such queries then pre-processing the MST is probably better. You can pre-process the MST to get a sparse table data structure in O(N lg N) time and then use this data structure to answer max queries in O(lg N) time in the worst case.
I am going to find a appropriate function in order to obtain accurate similarity between two persons according to their favourites.
for instance persons are connected to tags and their desire to each tags will be kept on the edge of tag nodes as a numeric values. I want to recommend similar persons to each persons.
I have found two solutions:
Cosine Similarity
There is Cosine function in Neo4j that just accept one input while in above function I need to pass vectores to this formula. Such as:
for "a": a=[10, 20, 45] each number indicates person`s desire to each tag.
for "b": b=[20, 50, 70]
Pearson Correlation
When I was surfing on the net and your documentation I found:
http://neo4j.com/docs/stable/cypher-cookbook-similarity-calc.html#cookbook-calculate-similarities-by-complex-calculations
My question is what is your logic behind this formula?
What is difference between r and H?
Because at the first glance I think H1 or H2 are always equals one. Unless I should consider the rest of the graph.
Thank you in advanced for any helps.
I think the purpose of H1 and H2 are to normalize the results of the times property (the number of times the user ate the food) across food types. You can experiment with this example in this Neo4j console
Since you mention other similarity measures you might be interested in this GraphGist, Similarity Measures For Collaborative Filtering With Cypher. It has some simple examples of calculating Pearson correlation and Jaccard similarity using Cypher.
This example makes it a little bit hard to understand what is going on. In this example, H1 and H2 are both 1. a better example would show each person eating different types of food, so you'd be able to see the value of H changing. If "me" also ate "vegetables", "pizza", and "hotdogs", their H would be 4.
Can't help you with Neo4J, just want to point out that Cosine Similarity and Pearsons' correlation coefficient are essentially the same thing. If you decode the different notations, you'll find that the only difference is that Pearsons zero-centers the vectors first. So you can define Pearsons as follows:
Pearsons(a, b) = Cosine(a - mean(a), b - mean(b))
I have this question related to "Maximum Likelihood Estimation"...I've tried to solve it but I couldn't ... would you please help!
Suppose for an event X, there are three possible
values, A, B and C. Now we repeat X for N times. The number of times that we observe A or B
is N1, the number of times that we observe A or C is N2. Let pA be the unknown frequency of
value A. Please give the maximum likelihood estimation of pA
This question doesn't really belong in Stackoverflow, but I will answer it anyway.
You can look at the number of observations of A (I am calling this N_A) as a hidden variable and maximize with respect to the marginal distribution (the sum over all possible values of N_A).
There is no closed form solution, in general, for the MLE parameters - solutions are the zeros of a polynomial constrained to the simplex. Below, I have derived an Expectation Maximization updates.