Search Results

Search found 1313 results on 53 pages for 'levenstein distance'.

Page 1/53 | 1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >

  • Find distance between two points using MKMapKit

    - by mag725
    Hi, I'm attempting to find the euclidean distance in meters between two points on an MKMapView using iPhone OS 3.2. The problem is that I have these coordinates in terms of latitude and longitude, which, mathematically provides me enough data to find the distance, but it's going to take some tricky trigonometry. Is there any simpler solution? Thanks!

    Read the article

  • How can I generate signed distance fields (2D) in real time, fast?

    - by heishe
    In a previous question, it was suggested that signed distance fields can be precomputed, loaded at runtime and then used from there. For reasons I will explain at the end of this question (for people interested), I need to create the distance fields in real time. There are some papers out there for different methods which are supposed to be viable in real-time environments, such as methods for Chamfer distance transforms and Voronoi diagram-approximation based transforms (as suggested in this presentation by the Pixeljunk Shooter dev guy), but I (and thus can be assumed a lot of other people) have a very hard time actually putting them to use, since they're usually long, largely bloated with math and not very algorithmic in their explanation. What algorithm would you suggest for creating the distance fields in real-time (favourably on the GPU) especially considering the resulting quality of the distance fields? Since I'm looking for an actual explanation/tutorial as opposed to a link to just another paper or slide, this question will receive a bounty once it's eligible for one :-). Here's why I need to do it in real time: There's something else:

    Read the article

  • How can I generate signed distance fields in real time, fast?

    - by heishe
    In a previous question, it was suggested that signed distance fields can be precomputed, loaded at runtime and then used from there. For reasons I will explain at the end of this question (for people interested), I need to create the distance fields in real time. There are some papers out there for different methods which are supposed to be viable in real-time environments, such as methods for Chamfer distance transforms and Voronoi diagram-approximation based transforms (as suggested in this presentation by the Pixeljunk Shooter dev guy), but I (and thus can be assumed a lot of other people) have a very hard time actually putting them to use, since they're usually long, largely bloated with math and not very algorithmic in their explanation. What algorithm would you suggest for creating the distance fields in real-time (favourably on the GPU) especially considering the resulting quality of the distance fields? Since I'm looking for an actual explanation/tutorial as opposed to a link to just another paper or slide, this question will receive a bounty once it's eligible for one :-). Here's why I need to do it in real time:

    Read the article

  • How can I optimize retrieving lowest edit distance from a large table in SQL?

    - by Matt
    Hey, I'm having troubles optimizing this Levenshtein Distance calculation I'm doing. I need to do the following: Get the record with the minimum distance for the source string as well as a trimmed version of the source string Pick the record with the minimum distance If the min distances are equal (original vs trimmed), choose the trimmed one with the lowest distance If there are still multiple records that fall under the above two categories, pick the one with the highest frequency Here's my working version: DECLARE @Results TABLE ( ID int, [Name] nvarchar(200), Distance int, Frequency int, Trimmed bit ) INSERT INTO @Results SELECT ID, [Name], (dbo.Levenshtein(@Source, [Name])) As Distance, Frequency, 'False' As Trimmed FROM MyTable INSERT INTO @Results SELECT ID, [Name], (dbo.Levenshtein(@SourceTrimmed, [Name])) As Distance, Frequency, 'True' As Trimmed FROM MyTable SET @ResultID = (SELECT TOP 1 ID FROM @Results ORDER BY Distance, Trimmed, Frequency) SET @Result = (SELECT TOP 1 [Name] FROM @Results ORDER BY Distance, Trimmed, Frequency) SET @ResultDist = (SELECT TOP 1 Distance FROM @Results ORDER BY Distance, Trimmed, Frequency) SET @ResultTrimmed = (SELECT TOP 1 Trimmed FROM @Results ORDER BY Distance, Trimmed, Frequency) I believe what I need to do here is to.. Not dumb the results to a temporary table Do only 1 select from `MyTable` Setting the results right in the select from the initial select statement. (Since select will set variables and you can set multiple variables in one select statement) I know there has to be a good implementation to this but I can't figure it out... this is as far as I got: SELECT top 1 @ResultID = ID, @Result = [Name], (dbo.Levenshtein(@Source, [Name])) As distOrig, (dbo.Levenshtein(@SourceTrimmed, [Name])) As distTrimmed, Frequency FROM MyTable WHERE /* ... yeah I'm lost */ ORDER BY distOrig, distTrimmed, Frequency Any ideas?

    Read the article

  • How to calculate short & long distance via Haversine?

    - by Jeroen
    Hi, I am looking for a way to calculate the distance between 2 points on the globe. We've been told to use Haversine, which works fine to calculate the shortest distance between the 2 points. Now, I'd like to calculate the "long distance" between to points. So suppose you have 2 cities, A in the west and B in the east. I want to know the distance from B to A if I would travel eastwards around the globe and then reach A coming from the west. I've tried changing a couple of things in the haversine function, but doesn't seem to work. Anyone know how I can simply do this using small adjustments to the haversine function? This is what I'm using now: lat1, lat2, lng1, lng2 are in radians part1 = sin(lat2) * sin(lat1); part2 = cos(lat2) * cos(lat1) * cos(lng1 - lng2); distance = EARTH_RADIUS * acos(part1 + part2); Tnx Jeroen

    Read the article

  • SQLAuthority News – Advantages of Distance Learning

    - by Pinal Dave
    Distance education is extremely popular – almost overnight, it seems.  Almost everyone has taken an online course, or knows someone who has, or is considering joining an online school.  There are many advantages and disadvantages to attending an online school – but the same can be said of attending a physical school!  Let’s take a look at the top reasons to use distance education. 1) Flexibility.  Physical universities are usually willing to make some concessions to student – like night classes, study hours, and online networks.  However, nothing is going to beat the flexibility of distance education.  You can attend classes and take notes anytime, anywhere, wearing anything you’d like! 2) Affordability.  We don’t need to get into hard numbers to understand how an expensive university can be.  Students are taking on more and more debt just to get an education.  Many of these fees pay for room, board, and facilities.   Distance education cuts out all these costs, and makes attending school much more affordable for the average student. 3) Try before you buy.  Did you know that the average college student changes his or her major 10 times before they graduate?  You can imagine that this kind of indecision plays a huge part in WHEN you graduate – not being able to make up your mind can cost you big bucks if you have to stay in school for extra years!  Distance education allows you to take different classes from a wide range of disciplines.  Do you want to study forensic science or English literature?  Now you don’t have to pay for classes you can’t afford just to find out. 4) Pace yourself.  Some students struggle in a traditional classroom setting – classes can be taught too fast, too slow, or there are too many distractions.  Distance education allows mature students to set the pace themselves.  They can rewatch lectures they didn’t catch the first time, or go through classes quickly if they are already familiar with the material – cutting out the chance of burning out or getting bored. 5) Lifelong learning.  Maybe you already have a degree, but would like to learn more about your field, or a related field, or maybe even about something completely unrelated – just because you are curious!  Distance education allows you to learn whatever you want ,whenever you want (and yes, wearing anything you’d like!). 6) Attend whatever college you want.  Because of the popularity of distance education, physical campuses are getting in on the game by offering online courses – often just uploaded versions of classes already taught at their campus.  Ever wanted to attend Harvard, but knew you couldn’t get in?  Take a class online!  Of course, you probably should not attempt to lie and say you have a Harvard degree, but Ivy League colleges are prestigious because they are the best in their field – take advantage of the best by taking an online course! I am a big believer in continuing education, whether it is online courses, returning to school, or even take informal classes online.  Distance education can be a great way to accomplish these goals and become a lifelong learner. My friends at provides training through virtual classrooms for students who want to avoid travelling. Distance learning course allows IT aspirants to connect with trainers using the internet.  I encourage everyone to check it out! Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, SQL Training, T SQL, Technology

    Read the article

  • Distance between two 3D objects' faces

    - by Arthur Gibraltar
    I'm really newbie on programming and I'm making some tests. I couldn't find nowhere on Internet how could I calculate the distance between two 3D objects' faces. Is there anyway? Detailing, as an example, I have two 3D cubes. Each one has a vector3 position designating it's center on the 3D space and an orientation matrix. And each cube has a size (float width, float height and float length). I could get a simple distance between them by calling Vector3.Distance(), but it doesn't consider its sizes, just the position. Then the distance would be between its centers. Is there any way to calculate the distance between the faces? Thanks for any reply.

    Read the article

  • Why distance field text rendering have clear outline?

    - by jinhwan
    http://www.valvesoftware.com/publications/2007/SIGGRAPH2007_AlphaTestedMagnification.pdf All the process for doing distance rendering is clear, but 'how does it work' is not clear for me. It looks like that distance field pixels which are created around original pixel may affect 2d texture sampling interpolation process. But I can't understand the interpolation process. I've read that the distance field rendering is processed under nearest-neighbour interpolation. If it is true, shouldn't the distance field redering creates non interpolated result? In my thought, they should looks liked retro style pixel art. Where do i misunderstand in this process? So far, It is no difference with alpha test for me. Both of them throw away all pixcel which are not in. How does extra distance field pixel affect rendering under nearest-neighbour interpolation?

    Read the article

  • Edit Distance in Python

    - by Alice
    I'm programming a spellcheck program in Python. I have a list of valid words (the dictionary) and I need to output a list of words from this dictionary that have an edit distance of 2 from a given invalid word. I know I need to start by generating a list with an edit distance of one from the invalid word(and then run that again on all the generated words). I have three methods, inserts(...), deletions(...) and changes(...) that should output a list of words with an edit distance of 1, where inserts outputs all valid words with one more letter than the given word, deletions outputs all valid words with one less letter, and changes outputs all valid words with one different letter. I've checked a bunch of places but I can't seem to find an algorithm that describes this process. All the ideas I've come up with involve looping through the dictionary list multiple times, which would be extremely time consuming. If anyone could offer some insight, I'd be extremely grateful. Thanks!

    Read the article

  • How to get Distance Kilometer in android?

    - by user1787493
    i am very new to Google maps i want calculate the distance between two places in android .for that i get the two places lat and lag positions for that i write the following code: private double getDistance(double lat1, double lat2, double lon1, double lon2) { double dLat = Math.toRadians(lat2 - lat1); double dLon = Math.toRadians(lon2 - lon1); double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2)) * Math.sin(dLon / 2) * Math.sin(dLon / 2); double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a)); double temp = 6371 * c; temp=temp*0.621; return temp; } the above code cant give the accurate distance between two places .what is the another way to find distance please give me any suggestions thanks in advance....

    Read the article

  • google maps v3 distance

    - by Shane
    Trying to create a new version of the map functions seen here: http://www.daftlogic.com/projects-google-maps-distance-calculator.htm but using the v3 api. So far I am able to set markers on click and can draw the geodesic polyline. The issues I am currently running into are: Updating the poly-line on marker drag I'm pretty sure I have to put each marker in an array and do a for loop so that I can keep clicking and adding points that will add to the total distance. Properly displaying distance. I have created a jsfiddle: http://jsfiddle.net/wyZyS/ EDIT: I realize I have nothing calling the "update" function. I am trying to create the array for each marker currently. The calculation you see is converting meters to nautical miles.

    Read the article

  • SAS/R calculate distance between two groups

    - by user976856
    I would like to calculate distance between two groups. I am very confused. I have a two data sets. One is about a company and one is about employees. I would like to find out how their age( a company in which an employee is hired and an employee) are similar or not. I think I need to standarize also.. calcuate euclidean distance between each person and a company. (4-5 people in a company) calculate euclidean distance between each person and a company in industry level. My dataset is like this: person person_age company company_age insustry 1 50 1 5 1 2 40 1 5 1 3 30 2 1 1 4 20 2 1 1 5 25 3 8 2 Please help me. I don't mind using SAS or R. I am very confused.

    Read the article

  • Generate a set of strings with maximum edit distance

    - by Kevin Jacobs
    Problem 1: I'd like to generate a set of n strings of fixed length m from alphabet s such that the minimum Levenshtein distance (edit distance) between any two strings is greater than some constant c. Obviously, I can use randomization methods (e.g., a genetic algorithm), but was hoping that this may be a well-studied problem in computer science or mathematics with some informative literature and an efficient algorithm or three. Problem 2: Same as above except that adjacent characters cannot repeat; the i'th character in each string may not be equal to the i+1'th character. E.g., 'CAT', 'AGA' and 'TAG' are allowed, 'GAA', 'AAT', and 'AAA' are not. Background: The basis for this problem is bioinformatic and involves designing unique DNA tags that can be attached to biologically derived DNA fragments and then sequenced using a fancy second generation sequencer. The goal is to be able to recognize each tag, allowing for random insertion, deletion, and substitution errors. The specific DNA sequencing technology has a relatively low error rate per base (~1%), but is less precise when a single base is repeated 2 or more times (motivating the additional constraints imposed in problem 2).

    Read the article

  • Optimizing Levenshtein Distance Algorithm

    - by Matt
    I have a stored procedure that uses Levenshtein Distance to determine the result closest to what the user typed. The only thing really affecting the speed is the function that calculates the Levenshtein Distance for all the records before selecting the record with the lowest distance (I've verified this by putting a 0 in place of the call to the Levenshtein function). The table has 1.5 million records, so even the slightest adjustment may shave off a few seconds. Right now the entire thing runs over 10 minutes. Here's the method I'm using: ALTER function dbo.Levenshtein ( @Source nvarchar(200), @Target nvarchar(200) ) RETURNS int AS BEGIN DECLARE @Source_len int, @Target_len int, @i int, @j int, @Source_char nchar, @Dist int, @Dist_temp int, @Distv0 varbinary(8000), @Distv1 varbinary(8000) SELECT @Source_len = LEN(@Source), @Target_len = LEN(@Target), @Distv1 = 0x0000, @j = 1, @i = 1, @Dist = 0 WHILE @j <= @Target_len BEGIN SELECT @Distv1 = @Distv1 + CAST(@j AS binary(2)), @j = @j + 1 END WHILE @i <= @Source_len BEGIN SELECT @Source_char = SUBSTRING(@Source, @i, 1), @Dist = @i, @Distv0 = CAST(@i AS binary(2)), @j = 1 WHILE @j <= @Target_len BEGIN SET @Dist = @Dist + 1 SET @Dist_temp = CAST(SUBSTRING(@Distv1, @j+@j-1, 2) AS int) + CASE WHEN @Source_char = SUBSTRING(@Target, @j, 1) THEN 0 ELSE 1 END IF @Dist > @Dist_temp BEGIN SET @Dist = @Dist_temp END SET @Dist_temp = CAST(SUBSTRING(@Distv1, @j+@j+1, 2) AS int)+1 IF @Dist > @Dist_temp SET @Dist = @Dist_temp BEGIN SELECT @Distv0 = @Distv0 + CAST(@Dist AS binary(2)), @j = @j + 1 END END SELECT @Distv1 = @Distv0, @i = @i + 1 END RETURN @Dist END Anyone have any ideas? Any input is appreciated. Thanks, Matt

    Read the article

  • Distance between Long Lat coord using SQLITE

    - by munchine
    I've got an sqlite db with long and lat of shops and I want to find out the closest 5 shops. So the following code works fine. if(sqlite3_prepare_v2(db, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) { while (sqlite3_step(compiledStatement) == SQLITE_ROW) { NSString *branchStr = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 0)]; NSNumber *fLat = [NSNumber numberWithFloat:(float)sqlite3_column_double(compiledStatement, 1)]; NSNumber *fLong = [NSNumber numberWithFloat:(float)sqlite3_column_double(compiledStatement, 2)]; NSLog(@"Address %@, Lat = %@, Long = %@", branchStr, fLat, fLong); CLLocation *location1 = [[CLLocation alloc] initWithLatitude:currentLocation.coordinate.latitude longitude:currentLocation.coordinate.longitude]; CLLocation *location2 = [[CLLocation alloc] initWithLatitude:[fLat floatValue] longitude:[fLong floatValue]]; NSLog(@"Distance i meters: %f", [location1 getDistanceFrom:location2]); [location1 release]; [location2 release]; } } I know the distance from where I am to each shop. My question is. Is it better to put the distance back into the sqlite row, I have the row when I step thru the database. How do I do that? Do I use the UPDATE statement? Does someone have a piece of code to help me. I can read the sqlite into an array and then sort the array. Do you recommend this over the above approach? Is this more efficient? Finally, if someone has a better way to get the closest 5 shops, love to hear it.

    Read the article

  • Constrained/penalized distance function

    - by sigma.z.1980
    Assume a character is located on a n by n grid and has to reach a certain entry on that grid. Its current position is (x1,y1). Also on the same grid is an enemy with coordinates (x2,y2). Each step algorithm randomly generates new candidate locations for the hero (if there are k candidates then there is a kx2 matrix of new potential locations. What I need is some distance objective function to compare the candidates. I'm currently using d1 - c * d2, where d1 is distance to the objective (measure in terms of number of pixels for each axis), d2 is distance to the enemy and c is some coefficient (this is very much like a set-up for Lagrangian). It's not working very well though. I'd be quite keen to learn how what constrained distance function are used for similar cases. Any suggestions are very much appreciated.

    Read the article

  • finding distance between two UK addresses

    - by effkay
    Hello; I need to write an application which is to calculate the estimated driving distance between two UK addresses; I think I can use Google as following: http://maps.google.com/maps/nav?q=from:London%20to:Dover However, anyone knows what is the daily/monthly limit of querying the database from a single IP address? I need to implement in a commercial application (freight services), is there any reliable alternative? Development language/platform: C#, .NET Regards,

    Read the article

  • Calculating the maximum distance between elements of vector in Matlab

    - by lhahne
    Lets assume that we have a vector like x = -1:0.05:1; ids = randperm(length(x)); x = x(ids(1:20)); I would like to calculate the maximum distance between the elements of x in some idiomatic way. It would be easy to just iterate over all possible combinations of x's elements but I feel like there could be a way to do it with Matlab's built-in functions in some crazy but idiomatic way. Any ideas?

    Read the article

  • Distance by sea calculator, intermediate coordinates?

    - by Lucian2k
    How do I calculate distance between 2 coordinates by sea? I also want to be able to draw a route between the two coordinates. Only solution I found so far is to split a map into pixels, identify each pixel as LAND or SEA and then try to find the path using A* algorithm. Then transform pixels to relative coordinates. There are some software packages I could buy but none have online extensions. A service that calculates distances between sea ports and plots the path on a map is searates.com

    Read the article

  • Shortest Distance From An Array

    - by notyou61
    I have an ajax function which returns the latitudes and longitudes of locations stored in a database. These are returned and placed in an array. A calculation is performed to return their distance from the users current location based on the latitude/longitude. I would like to return only the record with the shortest calculated distance. My code is as follows: Ajax Success // Success success: function (data) { // Obtain Log/Lat navigator.geolocation.getCurrentPosition(function(position) { // Obtain Current Position Lat/Lon glbVar.latitude = position.coords.latitude; glbVar.longitude = position.coords.longitude; // Console Log //console.log('Lat: ' + glbVar.latitude + ' Lon: ' + glbVar.longitude); // Obtain Location Distances for ( var i = 0; i < data.length; i++ ) { // Location Instances var varLocation = data[i]; // Location Distance varLocation.distance = calculateDistance(glbVar.longitude, glbVar.latitude, varLocation.locationLongitude, varLocation.locationLatitude); } // Sort Locations By Distance var sortedData = data.sort(function(a, b) { // Return Locations return a.distance - b.distance; }); // Output Results $.map(sortedData, function(item) { // Obtain Location Distance varLocationsDistance = calculateDistance(glbVar.longitude, glbVar.latitude, item.locationLongitude, item.locationLatitude); // Obtain Location Radius Assignment if (varLocationsDistance <= varLocationRadius) { // Function Return functionReturn = $({locationID : item.locationID + ', Distance : ' + varLocationsDistance + ' m'}); // Return // Function to get the Min value in Array Array.min = function( sortedData ){ functionReturn = Math.min.apply( Math, sortedData ); // console.log(functionReturn); }; } }); }); } The calculateDistance function returns the distance from the users current location and those from the database. The varLocationsDistance <= varLocationRadius "If" statement returns records within a certain distance radius (100 meters), within that statement I would like to return the shortest distance. I am a self taught amateur web developer and as a result may not have provide enough information for an answer, please let me know. Thanks,

    Read the article

  • Find points whose pairwise distances approximate a given distance matrix

    - by Stephan Kolassa
    Problem. I have a symmetric distance matrix with entries between zero and one, like this one: D = ( 0.0 0.4 0.0 0.5 ) ( 0.4 0.0 0.2 1.0 ) ( 0.0 0.2 0.0 0.7 ) ( 0.5 1.0 0.7 0.0 ) I would like to find points in the plane that have (approximately) the pairwise distances given in D. I understand that this will usually not be possible with strictly correct distances, so I would be happy with a "good" approximation. My matrices are smallish, no more than 10x10, so performance is not an issue. Question. Does anyone know of an algorithm to do this? Background. I have sets of probability densities between which I calculate Hellinger distances, which I would like to visualize as above. Each set contains no more than 10 densities (see above), but I have a couple of hundred sets. What I did so far. I did consider posting at math.SE, but looking at what gets tagged as "geometry" there, it seems like this kind of computational geometry question would be more on-topic here. If the community thinks this should be migrated, please go ahead. This looks like a straightforward problem in computational geometry, and I would assume that anyone involved in clustering might be interested in such a visualization, but I haven't been able to google anything. One simple approach would be to randomly plonk down points and perturb them until the distance matrix is close to D, e.g., using Simulated Annealing, or run a Genetic Algorithm. I have to admit that I haven't tried that yet, hoping for a smarter way. One specific operationalization of a "good" approximation in the sense above is Problem 4 in the Open Problems section here, with k=2. Now, while finding an algorithm that is guaranteed to find the minimum l1-distance between D and the resulting distance matrix may be an open question, it still seems possible that there at least is some approximation to this optimal solution. If I don't get an answer here, I'll mail the gentleman who posed that problem and ask whether he knows of any approximation algorithm (and post any answer I get to that here).

    Read the article

  • scaling point sprites with distance

    - by Will
    How can you scale a point sprite by its distance from the camera? GLSL fragment shader: gl_PointSize = size / gl_Position.w; seems along the right tracks; for any given scene all sprites seem nicely scaled by distance. Is this correct? How do you compute the proper scaling for my vertex attribute size? I want each sprite to be scaled by the modelview matrix. I had played with arbitrary values and it seems that size is the radius in pixels at the camera, and is not in modelview scale. I've also tried: gl_Position = pMatrix * mvMatrix * vec4(vertex,1.0); vec4 v2 = pMatrix * mvMatrix * vec4(vertex.x,vertex.y+0.5*size,vertex.z,1.0); gl_PointSize = length(gl_Position.xyz-v2.xyz) * gl_Position.w; But this makes the sprites be bigger in the distance, rather than smaller:

    Read the article

1 2 3 4 5 6 7 8 9 10 11 12  | Next Page >