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  • 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!

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  • Ivar definitions show 'long' type encoding as 'long long' type encoding

    - by Frank C.
    I've found what I think may be a bug with Ivar and Objective-C runtime. I'm using XCode 3.2.1 and associated libraries, developing a 64 bit app on X86_64 (MacBook Pro). Where I would expect the type encoding for the following "longVal" to be 'l', the Ivar encoding is showing a 'q' (which is a 'long long'). Anyone else seeing this? Simplified code and output follows: Code: #import <Foundation/Foundation.h> #import <objc/runtime.h> @interface Bug : NSObject { long longVal; long long longerVal; } @property (nonatomic,assign) long longVal; @property (nonatomic,assign) long long longerVal; @end @implementation Bug @synthesize longVal,longerVal; @end int main (int argc, const char * argv[]) { NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init]; unsigned int ivarCount=0; Ivar *ivars= class_copyIvarList([Bug class], &ivarCount); for(unsigned int x=0;x<ivarCount;x++) { NSLog(@"Name [%@] encoding [%@]", [NSString stringWithCString:ivar_getName(ivars[x]) encoding:NSUTF8StringEncoding], [NSString stringWithCString:ivar_getTypeEncoding(ivars[x]) encoding:NSUTF8StringEncoding]); } [pool drain]; return 0; } And here is output from debug console: This GDB was configured as "x86_64-apple-darwin".tty /dev/ttys000 Loading program into debugger… sharedlibrary apply-load-rules all Program loaded. run [Switching to process 6048] Running… 2010-03-17 22:16:29.138 ivarbug[6048:a0f] Name [longVal] encoding [q] 2010-03-17 22:16:29.146 ivarbug[6048:a0f] Name [longerVal] encoding [q] (gdb) continue Not a pretty picture! -- Frank

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  • 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:

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  • 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:

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  • Is it possible to add -pedantic to GCC command line, yet have it not warn about 'long long'

    - by doublep
    I'm using mostly GCC to develop my library, but I'd like to ensure cross-compiler compatibility and especially standard conformance as much as possible. For this, I have add several -W... flags to command line. I'd also add -pedantic, but I have a problem with its warning about long long type. The latter is important for my library and is properly guarded with #if code, i.e. is not compiled on compilers that don't know it anyway. In short: can I have GCC in -pedantic mode warn about any extension except long long?

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  • Need help implementing an Android service that does http long polling

    - by Erdal
    I've seen persistent TCP connections implemented (http://devtcg.blogspot.com/2009/01/push-services-implementing-persistent.html), but my needs are a little different. I need an Android service that always runs in the background and keeps a long polling connection to an HTTP server and communicates with it using JSON over POST method. Does anyone have anything similar to this?

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  • 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

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  • Twice as long and half as long

    - by PointsToShare
    We are in a project and we hit some snags. What’s a snag? An activity that takes longer than expected. Actually it takes longer than the time assigned to it by an over pressed PM who accepts an impossible time table and tries his best to make it possible, but I digress (again!).  So we have snags and we also have the opposite. Let’s call these “cinches”. The question is: how does a combination of snags and cinches affect the project timeline? Well, there is no simple answer. It depends on the projects dependencies as we see in the PERT chart. If all the snags are in the critical path and all the cinches are elsewhere then the cinches don’t help at all. In fact any snag in the critical path will delay the project.  Conversely, a cinch in the critical path will expedite it. A snag outside the critical path might be serious enough to even change the critical path. Thus without the PERT chart, we cannot really tell. Still there is a principle involved – Time and speed are non-linear! Twice as long adds a full unit, half as long only takes ½ unit away. Let’s just investigate a simple project. It consists of two activities – S and C - each estimated to take a week. Alas, S is a snag and really needs twice the time allotted and – a sigh of relief – C is a cinch and will take half the time allotted, so everything is Hun-key-dory, or is it?  Even here the PERT chart is important. We have 2 cases: 1: S depends on C (or vice versa) as in when the two activities are assigned to one employee. Here the estimated time was 1 + 1 and the actual time was 2 + ½ and we are ½ week late or 25% late. 2: S and C are done in parallel. Here the estimated time was 1, but the actual time is 2 – we are a whole week or 100% late. Let’s change the equation a little. S need 1.5 and C needs .5 so in case 1, we have the loss fully compensated by the gain, but in case 2 we are still behind. There are cases where this really makes no difference. This is when the critical path is not affected and we have enough slack in the other paths to counteract the difference between its snags and cinches – Let’s call this difference DSC. So if the slack is greater than DSC the project will not suffer. Conclusion: There is no general rule about snags and cinches. We need to examine each case within its project, still as we saw in the 4 examples above; the snag is generally more powerful than the cinch. Long live Murphy! That’s All Folks

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  • 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

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  • It's a Long, Long Way to Tipperary but not that Far to Yak about Apps

    - by linda.fishman.hoyle
    I wanted to let everyone know that my blog URL will be moving to http://blogs.oracle.com/lindafishman/. I will focus my future writings to be about the upgrade and adoption strategies of Oracle E-Business Suite customers. To give you a little preview, here is a link to a book of 60 customers who are live on E-Business Suite Release 12 and 12.1. We have thousands of customers live on Release 12.x and are feverishly trying to write as many stories as we can so those of you who are thinking about upgrading, putting a business case together to move from another ERP application to E-Business Suite or for small and midsize companies who want a better understanding of the benefits E-Business Suite provides organizations of your size, this will be the place to go. See you at the new site! Linda

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  • Interop c# using a "long" from c++

    - by Daniel
    On my System: sizeof(long) in c++ is 4 aka 32bits sizeof(long) in c# is 8 aka 64 bits So in my Interop method declarations I've been substituting c++ longs with c# int's however I get the feeling this isn't safe? Why is a long the same size as an int in c++? And long long is 64bits? What's next a long long long long??

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  • 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.

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  • 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?

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  • 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!

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  • 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....

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  • 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.

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  • 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.

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  • 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

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  • 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.

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  • C++ long long manipulation

    - by Krakkos
    Given 2 32bit ints iMSB and iLSB int iMSB = 12345678; // Most Significant Bits of file size in Bytes int iLSB = 87654321; // Least Significant Bits of file size in Bytes the long long form would be... // Always positive so use 31 bts long long full_size = ((long long)iMSB << 31); full_size += (long long)(iLSB); Now.. I don't need that much precision (that exact number of bytes), so, how can I convert the file size to MiBytes to 3 decimal places and convert to a string... tried this... long double file_size_megs = file_size_bytes / (1024 * 1024); char strNumber[20]; sprintf(strNumber, "%ld", file_size_megs); ... but dosen't seem to work. i.e. 1234567899878Bytes = 1177375.698MiB ??

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  • How long does it take to iterate through a long loop?

    - by Carl Rosenberger
    On a modern 64-Bit machine, how long do you think it takes to iterate through all the positive long numbers? Below is a code snippet in Java to demonstrate the idea. Without running the code yourself, how long do you think this code is going to run? How long will similar code run in other programming languages? public class LongLoop { public static void main(String[] args) { long startTime = System.currentTimeMillis(); for (long i = 0; i < Long.MAX_VALUE; i++) { // do nothing, just loop } long stopTime = System.currentTimeMillis(); long duration = stopTime - startTime; System.out.println("Time taken: " + duration + " milliseconds"); } }

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  • 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.

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