Upcoming (next week!) is the First Anniversary Session of Caltech's online Machine Learning course (http://work.caltech.edu/telecourse[^]) . It is also the final session. There will be no future sessions.
The course has attracted more than 200,000 participants since its launch last year, and has gained wide acclaim. This is the last chance for anyone who wishes to take the course.
I took this course last fall and I highly recommend this!
It is well done, with video lectures on YouTube (so you don't need to "attend" at a specific time).
There is an on-line forum for discussion with Teaching Assistant(s) and other participants.
The text is an excellent book (a good value on Amazon).
Lobster Thermidor aux crevettes with a Mornay sauce, served in a Provençale manner with shallots and aubergines, garnished with truffle pate, brandy and a fried egg on top and Spam - Monty Python Spam Sketch
I was rather thinking of going commercial, I made the videos because people wanted to see what I was working on.Right now I'am making android apps using the same technology, they will be on Google Play in about two months time from now.
“Everything is simple when you take your time to analyze it.”
I know that the Traveling Salesman Problem has a solution space of (n-1)! / 2 for n = nodes, when the symetrick TSP is considered.
But, is there any mathematical way to show how big a solution space a Arc Routing Problem, like the Capacitated Arc Routing Problem have? Or any other way to show (describe) why ít is a complex problem?
In all the literature it is only described as NP-hard, and thats it.
The standard way to show a problem is NP is to show that it reduces to a known NP problem. In other words, you show transformations (which must run in P (polynomial) time) that transform any instance of the problem to an instance of a known NP problem.
Hello out there, I have been working on a project to design a new computer vision system and I have succeeded, the only stage remaining is the homography estimation stage.But to estimate the homography matrix the algorithm has to compute the singular value decomposition or svd of the measurement matrix. I can easily use openCV or LAPACK for that but I want to implement my own Jacobi SVD method and the openCV and Lapack are too big, I don't want my computer vision system library to be dependent on them. Does anyone have a copyright free example code of a fast jacobi svd implementation that I can modify without using the big free libraries? I want a simplified implementation, even pseudocode would suffice.
I will create a tech demo video for this computer vision system. It out performs state of the art systems in object detection/recognition. I'am surely going to acknowledge any help I can get.
Thanks in advance.
“Everything is simple when you take your time to analyze it.”
My problem goes something like this. I have a variable of 5 arrays a for storing the 5nodes of a network. The network is randomly formed by the random number generator. I have the time of travel for each paths of the network. All the nodes may not be connected to eachother. The network and the time of travel of each path is as per the user has assigned. Now I have to check the elements of each array and assign the penalty as per the conditions. My main task is to find a network path traveling each node of the network only once such that the node visited is not repeated. Its similar to TSP but the difference is that in TSP has the condition that it is possible to travel from each node to every other node. But for my case the network is pre defined and nodes are not connected to every other nodes in the network.
a) Like if the network stored is 2-2-3-4-5 then I find that a=a so I will have to assign the penalty (type 1 )as the network has a path 2-2 which is not possible as the starting and end of the path is same. At the same time for rest of the paths from node 2 to 3, node 3 to 4, node 4 to 5 I will have to calculate the sum of total time of travel when I finally reach the node 5.
b) Also there is another penalty condition if the path is not a possible path in the network given by user then I will have to assign penaly( type 2) for such kind of condition like if 1-2-4-3-5 is a network to be checked. Here I am supposing that all the paths are possible but the path 4-3 doesn’t exist in the real network provided by the user, So for such cases I will have to calculate the sum of the time of travel for the paths which are possible and also assign the penalty (type 2)
c) There is also a condition that I need the end point of the network as node 5.So in the network generated randomly if the end node is not the node 5 then I will have to assign the penalty (type 3) and at the same time calculated the sum of the rest possible paths. For example the node 3-5-2-1-4 here the end node is 4 so I will have to assign the penalty(type 4)
d) Since I am not allowed to visit the same node twice I will have to assign the (penalty type 4) if the network has repeatition of nodes like if the network is 2-4-3-1-4. Here the node 4 is occurring twice so I will have to assign the penalty(type 4) and also calculate the sum of the time of travel of the other paths.
Considering all the above penalty conditions I will have to check the network s. I am really not being able to use any logic on how do I start. Need some hint on how do I do it.
I'll give you a starting point. This puzzle can be solved efficiently with three algorithms:
1. The simple case: For numbers not in the rightmost column or the last two rows, move tiles to create a blank space between the next tile to place, and the correct location for this tile. Then move the tile into the blank space, bringing you one step closer to correct placement for this tile. Repeat until the tile is in the correct location.
2. For a position in the rightmost column (but not the last two rows), the problem is getting the wrong tile out and the right tile in. Get the right tile under the correct location (in the rightmost column). Then move another tile in the same row as the correct location down one (temporarily moving it out of its correct location), slide the rest of the row to the left, move the right tile up (to its correct position), get a blank spot under the wrong tile, move the wrong tile down, move the rest of the row right, then move the tile you moved out of its correct position back, thus completing the row.
3. For the last two rows, move them in a circular pattern until you get the opportunity to swap two tiles that are in the wrong order. Repeat until the last two rows are correct.
I posted this on C# and realized later this might be a more appropriate location for it.
I have been searching around for a single elimination algorithm / pseudo code for tournament brackets and I haven't found a single site on it. The best I found was Wikipedia explaining single elimination. I have been reading code project articles for a while now and I know there are many experience and very knowledgeable developers here. Since this is fairly simple to code I was hoping if somebody could help me out with the code/pseudo code or an algorithm for it. I would appreciate it a lot. Looking forward to your replies.
You create a binary tree where the leaves represent the tournament's contestants. Each pair of contestants under an interior node play a match, and the winner then occupies that interior node.
Likewise, an interior node that has two other interior nodes under it represents the winner of the match between the winners corresponding to those two interior nodes.
The root of the binary tree represents the winner of the tournament.
Tree construction: One approach uses the Composite design pattern, an abstract Node class with two derived classes: Contestant and InteriorNode. Start with a list of Contestants. While that list has more than one element, combine two elements under a new InteriorNode, remove those elements from the list, and insert the new InteriorNode in the list. The list should be ordered by increasing depth, so that, for example, all Contestants are paired before any InteriorNodes are paired.
When the list is down to one element, that's the root of the tree.