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  • bash: expanding variables with spaces

    - by adam n
    i have a file called physics 1b.sh in bash, if i try x="physics 1b" grep "string" "$x".sh grep complains: grep: physics 1b: No such file or directory. However, when i do grep "string" physics\ 1b.sh it works fine. So i guess the problem is something to do with the variable not being expanded to include the backslash that grep needs to recognize the space. How do i get this to work? using bash 3.2, mac os 10.6.

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  • How can you transform a set of numbers into mostly whole ones?

    - by Alice
    Small amount of background: I am working on a converter that bridges between a map maker (Tiled) that outputs in XML, and an engine (Angel2D) that inputs lua tables. Most of this is straight forward However, Tiled outputs in pixel offsets (integers of absolute values), while Angel2D inputs OpenGL units (floats of relative values); a conversion factor between these two is needed (for example, 32px = 1gu). Since OpenGL units are abstract, and the camera can zoom in or out if the objects are too small or big, the actual conversion factor isn't important; I could use a random number, and the user would merely have to zoom in or out. But it would be best if the conversion factor was selected such that most numbers outputted were small and whole (or fractions of small whole numbers), because that makes it easier to work with (and the whole point of the OpenGL units is that they are easy to work with). How would I find such a conversion factor reliably? My first attempt was to use the smallest number given; this resulted in no fractions below 1, but often lead to lots of decimal places where the factors didn't line up. Then I tried the mode of the sequence, which lead to the largest number of 1's possible, but often lead to very long floats for background images. My current approach gets the GCD of the whole sequence, which, when it works, works great, but can easily be thrown off course by a single bad apple. Note that while I could easily just pass the numbers I am given along, or pick some fixed factor, or use one of the conversions I specified above, I am looking for a method to reliably scale this list of integers to small, whole numbers or simple fractions, because this would most likely be unsurprising to the end user; this is not a one off conversion. The end users tend to use 1.0 as their "base" for manipulations (because it's simple and obvious), so it would make more sense for the sizes of entities to cluster around this.

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  • Deployment of SQL Server: installing a second instance?

    - by Workshop Alex
    Simple problem. I'm working on a Delphi 2007/WIN32 application which now uses MS Access as simple data store. I have to modify it to support SQL Server Express, which is easy. These modifications are working so the application can be deployed using either SQL Server or MS Access. (Whatever the user prefers.) I did consider deploying the whole application together with the SQL Compact but this is not practicak. Using SQL Server Express 2008 instead of 2005 is an option, but also has a few nasty side-effects which we don't want to resolve for now. The problem is deploying the whole project. The installation with SQL Server would need a quiet installation so the user won't notice it. SQL Server is mentioned in the documentation so they know it's there. We just don't want to bother them with technical issues. In most cases, such an installation will go just fine. But what if the user already has an SQL Server (2005) installation which is used for something else? Personally, I would prefer to just install a second instance of SQL Server on their system so it won't conflict with the other installation. (Thus, if they uninstall the other app, the SQL instance will just stay installed.) While SQL Server 2005 and 2008 can be installed on the same system simply by using two different names for the instance, I wonder if it's also possible to install SQL Server 2005 twice on a single system to get two instances. And if possible, how?

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  • Multiple left joins, how to output in php

    - by Dan
    I have 3 tables I need to join. The contracts table is the main table, the 'jobs' and 'companies' table are extra info that can be associated to the contracts table. so, since I want all entries from my 'contracts' table, and the 'jobs' and 'companies' data only if it exists, I wrote the query like this.... $sql = "SELECT * FROM contracts LEFT JOIN jobs ON contracts.job_id = jobs.id LEFT JOIN companies ON contracts.company_id = companies.id ORDER BY contracts.end_date"; Now how would I output this in PHP? I tried this but kept getting an undefined error "Notice: Undefined index: contracts.id"... $sql_result = mysql_query($sql,$connection) or die ("Fail."); if(mysql_num_rows($sql_result) > 0){ while($row = mysql_fetch_array($sql_result)) { $contract_id = stripslashes($row['contracts.id']); $job_number = stripslashes($row['jobs.job_number']); $company_name = stripslashes($row['companies.name']); ?> <tr id="<?=$contract_id?>"> <td><?=$job_number?></td> <td><?=$company_name?></td> </tr> <? } }else{ echo "No records found"; } Any help is appreciated.

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  • C++ calculation evaluated to 0

    - by Alexander Stolz
    I'm parsing a file and trying to decode coordinates to the right unit. What happens is that this code is evaluated to 0. If I type it into gdb the result is correct. int pLat = (int)( (argv[6].data() == "plus" ? 1 : -1) * ( atoi(argv[7].data()) + atoi(argv[8].data()) / 60. + atoi(argv[9].data()) / 36000.) * 2.145767 * 0.0001); P.s. If someone knows a better title for this question, please change it

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  • Fetching a result twice with deferend name

    - by user1799842
    I have 1 table with tasks named opentask:columns: id,title,description,expires,creator_id,creator_name, executer_id, executer_name, priority_id, status_id1 table with users named user: with columns: user_id, username What I want is to create a query where there will be all columns from opentask table where the executer_id will be equal to the user_id of user table AND the creator_id will be equal to the user_id again. This creates a confusion because the first equality excludes the second.So I need somehow to create a query where I will include the usernames for the executer with something like where "opentask.executer_id=user_user_id" and at the same time I will include the username again (as a differend name?) for the creator with something like "where opentask.executer_id=user_user_id"So I try this, which of course I know that is missing something, can you help? SELECT DISTINCT id, title, description, expires, creator_id, executer_id, oc_opentask.priority_id, oc_opentask.status_id, priority_name, status_name, user_id, username, (SELECT username FROM oc_opentask, oc_user WHERE oc_opentask.creator_id=oc_user.user_id) AS username2 FROM oc_opentask, oc_opentask_priority, oc_user, oc_opentask_status WHERE oc_opentask.priority_id=oc_opentask_priority.priority_id AND oc_opentask.executer_id=oc_user.user_id AND oc_opentask.status_id=oc_opentask_status.status_id ORDER BY oc_opentask.expires DESC

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  • C++ Compound Interest Exercise

    - by Lameste
    I'm a beginner trying to learn C++ using "C++ Primer Plus Sixth Edition". I'm on Chapter 5, going over loops. Anyways I was doing this programming exercise from the book, the problem is: Daphne invests $100 at 10% simple interest.That is, every year, the investment earns 10% of the original investment, or $10 each and every year: interest = 0.10 × original balance At the same time, Cleo invests $100 at 5% compound interest.That is, interest is 5% of the current balance, including previous additions of interest: interest = 0.05 × current balance Cleo earns 5% of $100 the first year, giving her $105.The next year she earns 5% of $105, or $5.25, and so on.Write a program that finds how many years it takes for the value of Cleo’s investment to exceed the value of Daphne’s investment and then displays the value of both investments at that time. Here is the code I have written for this exercise, I'm not getting good results though. #include <iostream> #include <array> double Daphne(int, double, double); double Chleo(double, double); int main() { using namespace std; int p = 100; //Principle double i1 = 0.1; // 10% interest rate double i2 = 0.05; // 5% interest rate double dInv = 0; //Daphnes investment double cInv = 0; // Chleos investment int t=1; //Starting at year 1 double s1 = 0; //Sum 1 for Daphne double s2 = 0; // Sum 2 for Chleo s1 = p + 10; //Initial interest (base case after year 1) for Daphne s2 = p + (i2*p); //Initial interest (base case after year 1) for Chleo /*cout << s1 << endl; cout << s2 << endl;*/ while (cInv < dInv) { dInv = Daphne(p, i1, s1); cInv = Chleo(i2, s2); t++; } cout << "The time taken for Chleos investment to exceed Daphnes was: " << t << endl; cout << "Daphnes investment at " << t << " years is: " << dInv << endl; cout << "Chleos invesment at " << t << " years is: " << cInv << endl; system("pause"); return 0; } double Daphne(int p, double i, double s1) { s1 = s1 + (p*i); return s1; } double Chleo(double i, double s2){ s2 = s2 + (s2*i); return s2; } Output from console: The time taken for Chleos investment to exceed Daphnes was: 1 Daphnes investment at 1 years is: 0 Chleos invesment at 1 years is: 0 Press any key to continue . . . Can anyone explain why I'm getting this current result? The while loop is supposed to continue executing statements until Chleo's investment exceeds Daphnes.

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  • Are doubles faster than floats in c#?

    - by Trap
    I'm writing an application which reads large arrays of floats and performs some simple operations with them. I'm using floats because I thought it'd be faster than doubles, but after doing some research I've found that there's some confusion about this topic. Can anyone elaborate on this? Thanks.

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  • Java Slick2d - Mouse picking how to take into account camera

    - by Corey
    When I move it it obviously changes the viewport so my mouse picking is off. My camera is just a float x and y and I use g.translate(-cam.cameraX+400, -cam.cameraY+300); to translate the graphics. I have the numbers hard coded just for testing purposes. How would I take into account the camera so my mouse picking works correctly. double mousetileX = Math.floor((double)mouseX/tiles.tileWidth); double mousetileY = Math.floor((double)mouseY/tiles.tileHeight); double playertileX = Math.floor(playerX/tiles.tileWidth); double playertileY = Math.floor(playerY/tiles.tileHeight); double lengthX = Math.abs((float)playertileX - mousetileX); double lengthY = Math.abs((float)playertileY - mousetileY); double distance = Math.sqrt((lengthX*lengthX)+(lengthY*lengthY)); if(input.isMousePressed(Input.MOUSE_LEFT_BUTTON) && distance < 4) { if(tiles.map[(int)mousetileX][(int)mousetileY] == 1) { tiles.map[(int)mousetileX][(int)mousetileY] = 0; } } That is my mouse picking code

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  • Routes on a sphere surface - Find geodesic?

    - by CaNNaDaRk
    I'm working with some friends on a browser based game where people can move on a 2D map. It's been almost 7 years and still people play this game so we are thinking of a way to give them something new. Since then the game map was a limited plane and people could move from (0, 0) to (MAX_X, MAX_Y) in quantized X and Y increments (just imagine it as a big chessboard). We believe it's time to give it another dimension so, just a couple of weeks ago, we began to wonder how the game could look with other mappings: Unlimited plane with continous movement: this could be a step forward but still i'm not convinced. Toroidal World (continous or quantized movement): sincerely I worked with torus before but this time I want something more... Spherical world with continous movement: this would be great! What we want Users browsers are given a list of coordinates like (latitude, longitude) for each object on the spherical surface map; browsers must then show this in user's screen rendering them inside a web element (canvas maybe? this is not a problem). When people click on the plane we convert the (mouseX, mouseY) to (lat, lng) and send it to the server which has to compute a route between current user's position to the clicked point. What we have We began writing a Java library with many useful maths to work with Rotation Matrices, Quaternions, Euler Angles, Translations, etc. We put it all together and created a program that generates sphere points, renders them and show them to the user inside a JPanel. We managed to catch clicks and translate them to spherical coords and to provide some other useful features like view rotation, scale, translation etc. What we have now is like a little (very little indeed) engine that simulates client and server interaction. Client side shows points on the screen and catches other interactions, server side renders the view and does other calculus like interpolating the route between current position and clicked point. Where is the problem? Obviously we want to have the shortest path to interpolate between the two route points. We use quaternions to interpolate between two points on the surface of the sphere and this seemed to work fine until i noticed that we weren't getting the shortest path on the sphere surface: We though the problem was that the route is calculated as the sum of two rotations about X and Y axis. So we changed the way we calculate the destination quaternion: We get the third angle (the first is latitude, the second is longitude, the third is the rotation about the vector which points toward our current position) which we called orientation. Now that we have the "orientation" angle we rotate Z axis and then use the result vector as the rotation axis for the destination quaternion (you can see the rotation axis in grey): What we got is the correct route (you can see it lays on a great circle), but we get to this ONLY if the starting route point is at latitude, longitude (0, 0) which means the starting vector is (sphereRadius, 0, 0). With the previous version (image 1) we don't get a good result even when startin point is 0, 0, so i think we're moving towards a solution, but the procedure we follow to get this route is a little "strange" maybe? In the following image you get a view of the problem we get when starting point is not (0, 0), as you can see starting point is not the (sphereRadius, 0, 0) vector, and as you can see the destination point (which is correctly drawn!) is not on the route. The magenta point (the one which lays on the route) is the route's ending point rotated about the center of the sphere of (-startLatitude, 0, -startLongitude). This means that if i calculate a rotation matrix and apply it to every point on the route maybe i'll get the real route, but I start to think that there's a better way to do this. Maybe I should try to get the plane through the center of the sphere and the route points, intersect it with the sphere and get the geodesic? But how? Sorry for being way too verbose and maybe for incorrect English but this thing is blowing my mind! EDIT: This code version is related to the first image: public void setRouteStart(double lat, double lng) { EulerAngles tmp = new EulerAngles ( Math.toRadians(lat), 0, -Math.toRadians(lng)); //set route start Quaternion qtStart.setInertialToObject(tmp); //do other stuff like drawing start point... } public void impostaDestinazione(double lat, double lng) { EulerAngles tmp = new AngoliEulero( Math.toRadians(lat), 0, -Math.toRadians(lng)); qtEnd.setInertialToObject(tmp); //do other stuff like drawing dest point... } public V3D interpolate(double totalTime, double t) { double _t = t/totalTime; Quaternion q = Quaternion.Slerp(qtStart, qtEnd, _t); RotationMatrix.inertialQuatToIObject(q); V3D p = matInt.inertialToObject(V3D.Xaxis.scale(sphereRadius)); //other stuff, like drawing point ... return p; } //mostly taken from a book! public static Quaternion Slerp(Quaternion q0, Quaternion q1, double t) { double cosO = q0.dot(q1); double q1w = q1.w; double q1x = q1.x; double q1y = q1.y; double q1z = q1.z; if (cosO < 0.0f) { q1w = -q1w; q1x = -q1x; q1y = -q1y; q1z = -q1z; cosO = -cosO; } double sinO = Math.sqrt(1.0f - cosO*cosO); double O = Math.atan2(sinO, cosO); double oneOverSinO = 1.0f / senoOmega; k0 = Math.sin((1.0f - t) * O) * oneOverSinO; k1 = Math.sin(t * O) * oneOverSinO; // Interpolate return new Quaternion( k0*q0.w + k1*q1w, k0*q0.x + k1*q1x, k0*q0.y + k1*q1y, k0*q0.z + k1*q1z ); } A little dump of what i get (again check image 1): Route info: Sphere radius and center: 200,000, (0.0, 0.0, 0.0) Route start: lat 0,000 °, lng 0,000 ° @v: (200,000, 0,000, 0,000), |v| = 200,000 Route end: lat 30,000 °, lng 30,000 ° @v: (150,000, 86,603, 100,000), |v| = 200,000 Qt dump: (w, x, y, z), rot. angle°, (x, y, z) rot. axis Qt start: (1,000, 0,000, -0,000, 0,000); 0,000 °; (1,000, 0,000, 0,000) Qt end: (0,933, 0,067, -0,250, 0,250); 42,181 °; (0,186, -0,695, 0,695) Route start: lat 30,000 °, lng 10,000 ° @v: (170,574, 30,077, 100,000), |v| = 200,000 Route end: lat 80,000 °, lng -50,000 ° @v: (22,324, -26,604, 196,962), |v| = 200,000 Qt dump: (w, x, y, z), rot. angle°, (x, y, z) rot. axis Qt start: (0,962, 0,023, -0,258, 0,084); 31,586 °; (0,083, -0,947, 0,309) Qt end: (0,694, -0,272, -0,583, -0,324); 92,062 °; (-0,377, -0,809, -0,450)

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  • How do I set up a double domain like domain2.mydomain.com and mydomain.com?

    - by kdavis8
    I would like to set up a server similar to Google's. Their domain acts like a double domain, like you can use these URLS, "play.Google.com" or "apps.Google.com", to go to different sites.. For example, my domain would now be "my_domain.com" but i would like another one to be "domain2.my_domain.com". My question is,what is this officially called and how do i set it up? I'm not sure if you need two servers or just 1;

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  • Why does my .desktop file execute via double click but not from the menu?

    - by Insperatus
    I've installed FTL: Faster Than Light on my girlfriend's Lubuntu machine and created a .desktop file for it. Strangely, the program won't launch via its menu entry under 'Games'. If I navigate to /home/andi/.local/share/applications/ via pcmanfm and double click on FTL Faster Than Light.desktop the game launches without a problem. I know the menu entry is generated through the .desktop file so why won't it launch from the menu? Here's the .desktop file I created: FTL Faster Than Light.desktop

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  • TechDays 2011 : Microsoft a doublé le nombre de ses clients SaaS en France depuis l'été dernier avec 1,2 million d'utilisateurs

    Microsoft a doublé le nombre de ses clients SaaS en France Depuis l'été dernier avec 1,2 million d'utilisateurs A l'occasion de la deuxième journée des TechDays 2011 (placés sous le signe du Cloud), Microsoft met l'accent sur le rôle d'accélérateur de « l'entreprise numérique » joué par les technologies Microsoft qu'elles soient ou non dans le cloud. « Le cloud computing porte en lui la promesse d'accélérer l'accès aux bénéfices de l'entreprise numérique, aussi bien en termes d'infrastructures que de nouveaux usages. Il favorise la transformation des directions informatiques en centre de services et améliore son alignement sur les enjeux métiers, » explique Marc Jalabert, D...

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  • Why can't `main` return a double or String rather than int or void?

    - by sunny
    In many languages such as C, C++, and Java, the main method/function has a return type of void or int, but not double or String. What might be the reasons behind that? I know a little bit that we can't do that because main is called by runtime library and it expects some syntax like int main() or int main(int,char**) so we have to stick to that. So my question is: why does main have the type signature that it has, and not a different one?

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  • Coup d'envoi d'Imagine Cup 2013 : Microsoft double le montant des primes du championnat de l'innovation numérique pour étudiant

    Coup d'envoi d'Imagine Cup 2013 : Microsoft double les primes et modifie la thématique de son championnat de l'innovation numérique pour étudiant C'est parti pour une nouvelle saison d'Imagine Cup, le championnat du monde étudiant de l'innovation numérique organisé par Microsoft. [IMG]http://rdonfack.developpez.com/images/imaginecup20132.jpg[/IMG] Créé en 2002, Imagine Cup permet à des étudiants du monde entier de « s'affronter » pour révéler leur potentiel créatif et entrepreneurial avec une mission : répondre à un ou plusieurs des « Objectifs du Millénaire pour le Développement ». L'édition 2013 de la compétition est ouverte dan...

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  • C++ Templates: implicit conversion, no matching function for call to ctor

    - by noname
    template<class T> class test { public: test() { } test(T& e) { } }; int main() { test<double> d(4.3); return 0; } Compiled using g++ 4.4.1 with the following errors: g++ test.cpp -Wall -o test.exe test.cpp: In function 'int main()': test.cpp:18: error: no matching function for call to 'test<double>::test(double) ' test.cpp:9: note: candidates are: test<T>::test(T&) [with T = double] test.cpp:5: note: test<T>::test() [with T = double] test.cpp:3: note: test<double>::test(const test<double>&) make: *** [test.exe] Error 1 However, this works: double a=1.1; test<double> d(a); Why is this happing? Is it possible that g++ cannot implicitly convert literal expression 1.1 to double? Thanks.

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  • Steganography Experiment - Trouble hiding message bits in DCT coefficients

    - by JohnHankinson
    I have an application requiring me to be able to embed loss-less data into an image. As such I've been experimenting with steganography, specifically via modification of DCT coefficients as the method I select, apart from being loss-less must also be relatively resilient against format conversion, scaling/DSP etc. From the research I've done thus far this method seems to be the best candidate. I've seen a number of papers on the subject which all seem to neglect specific details (some neglect to mention modification of 0 coefficients, or modification of AC coefficient etc). After combining the findings and making a few modifications of my own which include: 1) Using a more quantized version of the DCT matrix to ensure we only modify coefficients that would still be present should the image be JPEG'ed further or processed (I'm using this in place of simply following a zig-zag pattern). 2) I'm modifying bit 4 instead of the LSB and then based on what the original bit value was adjusting the lower bits to minimize the difference. 3) I'm only modifying the blue channel as it should be the least visible. This process must modify the actual image and not the DCT values stored in file (like jsteg) as there is no guarantee the file will be a JPEG, it may also be opened and re-saved at a later stage in a different format. For added robustness I've included the message multiple times and use the bits that occur most often, I had considered using a QR code as the message data or simply applying the reed-solomon error correction, but for this simple application and given that the "message" in question is usually going to be between 10-32 bytes I have plenty of room to repeat it which should provide sufficient redundancy to recover the true bits. No matter what I do I don't seem to be able to recover the bits at the decode stage. I've tried including / excluding various checks (even if it degrades image quality for the time being). I've tried using fixed point vs. double arithmetic, moving the bit to encode, I suspect that the message bits are being lost during the IDCT back to image. Any thoughts or suggestions on how to get this working would be hugely appreciated. (PS I am aware that the actual DCT/IDCT could be optimized from it's naive On4 operation using row column algorithm, or an FDCT like AAN, but for now it just needs to work :) ) Reference Papers: http://www.lokminglui.com/dct.pdf http://arxiv.org/ftp/arxiv/papers/1006/1006.1186.pdf Code for the Encode/Decode process in C# below: using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Drawing.Imaging; using System.Drawing; namespace ImageKey { public class Encoder { public const int HIDE_BIT_POS = 3; // use bit position 4 (1 << 3). public const int HIDE_COUNT = 16; // Number of times to repeat the message to avoid error. // JPEG Standard Quantization Matrix. // (to get higher quality multiply by (100-quality)/50 .. // for lower than 50 multiply by 50/quality. Then round to integers and clip to ensure only positive integers. public static double[] Q = {16,11,10,16,24,40,51,61, 12,12,14,19,26,58,60,55, 14,13,16,24,40,57,69,56, 14,17,22,29,51,87,80,62, 18,22,37,56,68,109,103,77, 24,35,55,64,81,104,113,92, 49,64,78,87,103,121,120,101, 72,92,95,98,112,100,103,99}; // Maximum qauality quantization matrix (if all 1's doesn't modify coefficients at all). public static double[] Q2 = {1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1, 1,1,1,1,1,1,1,1}; public static Bitmap Encode(Bitmap b, string key) { Bitmap response = new Bitmap(b.Width, b.Height, PixelFormat.Format32bppArgb); uint imgWidth = ((uint)b.Width) & ~((uint)7); // Maximum usable X resolution (divisible by 8). uint imgHeight = ((uint)b.Height) & ~((uint)7); // Maximum usable Y resolution (divisible by 8). // Start be transferring the unmodified image portions. // As we'll be using slightly less width/height for the encoding process we'll need the edges to be populated. for (int y = 0; y < b.Height; y++) for (int x = 0; x < b.Width; x++) { if( (x >= imgWidth && x < b.Width) || (y>=imgHeight && y < b.Height)) response.SetPixel(x, y, b.GetPixel(x, y)); } // Setup the counters and byte data for the message to encode. StringBuilder sb = new StringBuilder(); for(int i=0;i<HIDE_COUNT;i++) sb.Append(key); byte[] codeBytes = System.Text.Encoding.ASCII.GetBytes(sb.ToString()); int bitofs = 0; // Current bit position we've encoded too. int totalBits = (codeBytes.Length * 8); // Total number of bits to encode. for (int y = 0; y < imgHeight; y += 8) { for (int x = 0; x < imgWidth; x += 8) { int[] redData = GetRedChannelData(b, x, y); int[] greenData = GetGreenChannelData(b, x, y); int[] blueData = GetBlueChannelData(b, x, y); int[] newRedData; int[] newGreenData; int[] newBlueData; if (bitofs < totalBits) { double[] redDCT = DCT(ref redData); double[] greenDCT = DCT(ref greenData); double[] blueDCT = DCT(ref blueData); int[] redDCTI = Quantize(ref redDCT, ref Q2); int[] greenDCTI = Quantize(ref greenDCT, ref Q2); int[] blueDCTI = Quantize(ref blueDCT, ref Q2); int[] blueDCTC = Quantize(ref blueDCT, ref Q); HideBits(ref blueDCTI, ref blueDCTC, ref bitofs, ref totalBits, ref codeBytes); double[] redDCT2 = DeQuantize(ref redDCTI, ref Q2); double[] greenDCT2 = DeQuantize(ref greenDCTI, ref Q2); double[] blueDCT2 = DeQuantize(ref blueDCTI, ref Q2); newRedData = IDCT(ref redDCT2); newGreenData = IDCT(ref greenDCT2); newBlueData = IDCT(ref blueDCT2); } else { newRedData = redData; newGreenData = greenData; newBlueData = blueData; } MapToRGBRange(ref newRedData); MapToRGBRange(ref newGreenData); MapToRGBRange(ref newBlueData); for(int dy=0;dy<8;dy++) { for(int dx=0;dx<8;dx++) { int col = (0xff<<24) + (newRedData[dx+(dy*8)]<<16) + (newGreenData[dx+(dy*8)]<<8) + (newBlueData[dx+(dy*8)]); response.SetPixel(x+dx,y+dy,Color.FromArgb(col)); } } } } if (bitofs < totalBits) throw new Exception("Failed to encode data - insufficient cover image coefficients"); return (response); } public static void HideBits(ref int[] DCTMatrix, ref int[] CMatrix, ref int bitofs, ref int totalBits, ref byte[] codeBytes) { int tempValue = 0; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { if ( (u != 0 || v != 0) && CMatrix[v+(u*8)] != 0 && DCTMatrix[v+(u*8)] != 0) { if (bitofs < totalBits) { tempValue = DCTMatrix[v + (u * 8)]; int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); byte value = (byte)((codeBytes[bytePos] & mask) >> bitPos); // 0 or 1. if (value == 0) { int a = DCTMatrix[v + (u * 8)] & (1 << HIDE_BIT_POS); if (a != 0) DCTMatrix[v + (u * 8)] |= (1 << HIDE_BIT_POS) - 1; DCTMatrix[v + (u * 8)] &= ~(1 << HIDE_BIT_POS); } else if (value == 1) { int a = DCTMatrix[v + (u * 8)] & (1 << HIDE_BIT_POS); if (a == 0) DCTMatrix[v + (u * 8)] &= ~((1 << HIDE_BIT_POS) - 1); DCTMatrix[v + (u * 8)] |= (1 << HIDE_BIT_POS); } if (DCTMatrix[v + (u * 8)] != 0) bitofs++; else DCTMatrix[v + (u * 8)] = tempValue; } } } } } public static void MapToRGBRange(ref int[] data) { for(int i=0;i<data.Length;i++) { data[i] += 128; if(data[i] < 0) data[i] = 0; else if(data[i] > 255) data[i] = 255; } } public static int[] GetRedChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x,y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 16) & 0xff) - 128; } } return (data); } public static int[] GetGreenChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x, y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 8) & 0xff) - 128; } } return (data); } public static int[] GetBlueChannelData(Bitmap b, int sx, int sy) { int[] data = new int[8 * 8]; for (int y = sy; y < (sy + 8); y++) { for (int x = sx; x < (sx + 8); x++) { uint col = (uint)b.GetPixel(x, y).ToArgb(); data[(x - sx) + ((y - sy) * 8)] = (int)((col >> 0) & 0xff) - 128; } } return (data); } public static int[] Quantize(ref double[] DCTMatrix, ref double[] Q) { int[] DCTMatrixOut = new int[8*8]; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { DCTMatrixOut[v + (u * 8)] = (int)Math.Round(DCTMatrix[v + (u * 8)] / Q[v + (u * 8)]); } } return(DCTMatrixOut); } public static double[] DeQuantize(ref int[] DCTMatrix, ref double[] Q) { double[] DCTMatrixOut = new double[8*8]; for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { DCTMatrixOut[v + (u * 8)] = (double)DCTMatrix[v + (u * 8)] * Q[v + (u * 8)]; } } return(DCTMatrixOut); } public static double[] DCT(ref int[] data) { double[] DCTMatrix = new double[8 * 8]; for (int v = 0; v < 8; v++) { for (int u = 0; u < 8; u++) { double cu = 1; if (u == 0) cu = (1.0 / Math.Sqrt(2.0)); double cv = 1; if (v == 0) cv = (1.0 / Math.Sqrt(2.0)); double sum = 0.0; for (int y = 0; y < 8; y++) { for (int x = 0; x < 8; x++) { double s = data[x + (y * 8)]; double dctVal = Math.Cos((2 * y + 1) * v * Math.PI / 16) * Math.Cos((2 * x + 1) * u * Math.PI / 16); sum += s * dctVal; } } DCTMatrix[u + (v * 8)] = (0.25 * cu * cv * sum); } } return (DCTMatrix); } public static int[] IDCT(ref double[] DCTMatrix) { int[] Matrix = new int[8 * 8]; for (int y = 0; y < 8; y++) { for (int x = 0; x < 8; x++) { double sum = 0; for (int v = 0; v < 8; v++) { for (int u = 0; u < 8; u++) { double cu = 1; if (u == 0) cu = (1.0 / Math.Sqrt(2.0)); double cv = 1; if (v == 0) cv = (1.0 / Math.Sqrt(2.0)); double idctVal = (cu * cv) / 4.0 * Math.Cos((2 * y + 1) * v * Math.PI / 16) * Math.Cos((2 * x + 1) * u * Math.PI / 16); sum += (DCTMatrix[u + (v * 8)] * idctVal); } } Matrix[x + (y * 8)] = (int)Math.Round(sum); } } return (Matrix); } } public class Decoder { public static string Decode(Bitmap b, int expectedLength) { expectedLength *= Encoder.HIDE_COUNT; uint imgWidth = ((uint)b.Width) & ~((uint)7); // Maximum usable X resolution (divisible by 8). uint imgHeight = ((uint)b.Height) & ~((uint)7); // Maximum usable Y resolution (divisible by 8). // Setup the counters and byte data for the message to decode. byte[] codeBytes = new byte[expectedLength]; byte[] outBytes = new byte[expectedLength / Encoder.HIDE_COUNT]; int bitofs = 0; // Current bit position we've decoded too. int totalBits = (codeBytes.Length * 8); // Total number of bits to decode. for (int y = 0; y < imgHeight; y += 8) { for (int x = 0; x < imgWidth; x += 8) { int[] blueData = ImageKey.Encoder.GetBlueChannelData(b, x, y); double[] blueDCT = ImageKey.Encoder.DCT(ref blueData); int[] blueDCTI = ImageKey.Encoder.Quantize(ref blueDCT, ref Encoder.Q2); int[] blueDCTC = ImageKey.Encoder.Quantize(ref blueDCT, ref Encoder.Q); if (bitofs < totalBits) GetBits(ref blueDCTI, ref blueDCTC, ref bitofs, ref totalBits, ref codeBytes); } } bitofs = 0; for (int i = 0; i < (expectedLength / Encoder.HIDE_COUNT) * 8; i++) { int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); List<int> values = new List<int>(); int zeroCount = 0; int oneCount = 0; for (int j = 0; j < Encoder.HIDE_COUNT; j++) { int val = (codeBytes[bytePos + ((expectedLength / Encoder.HIDE_COUNT) * j)] & mask) >> bitPos; values.Add(val); if (val == 0) zeroCount++; else oneCount++; } if (oneCount >= zeroCount) outBytes[bytePos] |= mask; bitofs++; values.Clear(); } return (System.Text.Encoding.ASCII.GetString(outBytes)); } public static void GetBits(ref int[] DCTMatrix, ref int[] CMatrix, ref int bitofs, ref int totalBits, ref byte[] codeBytes) { for (int u = 0; u < 8; u++) { for (int v = 0; v < 8; v++) { if ((u != 0 || v != 0) && CMatrix[v + (u * 8)] != 0 && DCTMatrix[v + (u * 8)] != 0) { if (bitofs < totalBits) { int bytePos = (bitofs) >> 3; int bitPos = (bitofs) % 8; byte mask = (byte)(1 << bitPos); int value = DCTMatrix[v + (u * 8)] & (1 << Encoder.HIDE_BIT_POS); if (value != 0) codeBytes[bytePos] |= mask; bitofs++; } } } } } } } UPDATE: By switching to using a QR Code as the source message and swapping a pair of coefficients in each block instead of bit manipulation I've been able to get the message to survive the transform. However to get the message to come through without corruption I have to adjust both coefficients as well as swap them. For example swapping (3,4) and (4,3) in the DCT matrix and then respectively adding 8 and subtracting 8 as an arbitrary constant seems to work. This survives a re-JPEG'ing of 96 but any form of scaling/cropping destroys the message again. I was hoping that by operating on mid to low frequency values that the message would be preserved even under some light image manipulation.

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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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  • Annoying flickering of vertices and edges (possible z-fighting)

    - by Belgin
    I'm trying to make a software z-buffer implementation, however, after I generate the z-buffer and proceed with the vertex culling, I get pretty severe discrepancies between the vertex depth and the depth of the buffer at their projected coordinates on the screen (i.e. zbuffer[v.xp][v.yp] != v.z, where xp and yp are the projected x and y coordinates of the vertex v), sometimes by a small fraction of a unit and sometimes by 2 or 3 units. Here's what I think is happening: Each triangle's data structure holds the plane's (that is defined by the triangle) coefficients (a, b, c, d) computed from its three vertices from their normal: void computeNormal(Vertex *v1, Vertex *v2, Vertex *v3, double *a, double *b, double *c) { double a1 = v1 -> x - v2 -> x; double a2 = v1 -> y - v2 -> y; double a3 = v1 -> z - v2 -> z; double b1 = v3 -> x - v2 -> x; double b2 = v3 -> y - v2 -> y; double b3 = v3 -> z - v2 -> z; *a = a2*b3 - a3*b2; *b = -(a1*b3 - a3*b1); *c = a1*b2 - a2*b1; } void computePlane(Poly *p) { double x = p -> verts[0] -> x; double y = p -> verts[0] -> y; double z = p -> verts[0] -> z; computeNormal(p -> verts[0], p -> verts[1], p -> verts[2], &p -> a, &p -> b, &p -> c); p -> d = p -> a * x + p -> b * y + p -> c * z; } The z-buffer just holds the smallest depth at the respective xy coordinate by somewhat casting rays to the polygon (I haven't quite got interpolation right yet so I'm using this slower method until I do) and determining the z coordinate from the reversed perspective projection formulas (which I got from here: double z = -(b*Ez*y + a*Ez*x - d*Ez)/(b*y + a*x + c*Ez - b*Ey - a*Ex); Where x and y are the pixel's coordinates on the screen; a, b, c, and d are the planes coefficients; Ex, Ey, and Ez are the eye's (camera's) coordinates. This last formula does not accurately give the exact vertices' z coordinate at their projected x and y coordinates on the screen, probably because of some floating point inaccuracy (i.e. I've seen it return something like 3.001 when the vertex's z-coordinate was actually 2.998). Here is the portion of code that hides the vertices that shouldn't be visible: for(i = 0; i < shape.nverts; ++i) { double dist = shape.verts[i].z; if(z_buffer[shape.verts[i].yp][shape.verts[i].xp].z < dist) shape.verts[i].visible = 0; else shape.verts[i].visible = 1; } How do I solve this issue? EDIT I've implemented the near and far planes of the frustum, with 24 bit accuracy, and now I have some questions: Is this what I have to do this in order to resolve the flickering? When I compare the z value of the vertex with the z value in the buffer, do I have to convert the z value of the vertex to z' using the formula, or do I convert the value in the buffer back to the original z, and how do I do that? What are some decent values for near and far? Thanks in advance.

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  • Make Ms Word activate window of file if it is already open when I double click the file's icon?

    - by barlop
    When I try to open a word document that is already open, I want it to just activate the window where the file is open. How can I do that? sometimes it takes time to check if an ms word document is already open, I don't to have to have to check through a bunch of open word documents to see if it's in there or not, I just want to double click the icon of a word document, and if it's open then go to it, if not then open it. With the doc/docx or shortcut to the doc/docx , I have some files where when I double click , it activates it when already open. I have other files where double clicking will bring up a "file in use" dialog box. I can't find what is the cause. I want it to always activate the window rather than reopen it. update- maybe that is default behaviour to activate when already open, and after a crash I had that stopped. i'll try deleting the working files and starting ms word again, idea from here

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