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

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  • plot multi graphs with 2 y axis in 1 graph

    - by lytheone
    Hello, Currently I have a a text file with data at the first row is formatted as follow: time;wave height 1;wave height 2;....... I have column until wave height 19 and rows total 4000 rows. Data in the first column is time in second. From 2nd column onwards, it is wave height elevation which is in meter. I would like to plot the follow: ![alt text][1] on the x axis is time. the left hand side is wave height in m and on the right hand side is the distance between each measurment in a model. inside the graph there are 4 plots, each plot is repersent waveight 1, wave height 2etc at a defined distance related to the right hand side y asix. How would you code this in matlab? I am a begineer, please if you could, it will be very useful to give a bit more explain in your answer! Thank you!!!!!!!!!!

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  • How Transport for London Website Works

    - by Subhen
    Hi, Here let me clarify , I have no intentions to peep in to or any evil intention towards tfls database and other related information. But , ofcourse Millions of users are greatly beniftted the way it serves the information. http://journeyplanner.tfl.gov.uk/ So , If we want to create some site like tfl, journeyplanner , what are the basic things we need to keep in mind. Which Architecture We should use? Can We create this website using ASP.NET(Should be able to)? Is TFL integrating it's website with google maps or any other GPS Edit: While you enter the Zip/Pin code or Station name , it creates a map automatically from source to destination and calcculates the distance also. My Question here is , How do they calculate the distance , do they keep help of Maps or GPS or they created there own webservic?

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  • Kerning problems when drawing text character by character

    - by shekel
    I'm trying to draw strings character by character to add lighting effects to shapes composed of text. while (i != line.length()) { c = line.substring(i, i + 1); cWidth = g.getFontMetrics().stringWidth(c); g.drawString(c, xx += cWidth, yy); i++; } The problem is, the width of a character isn't the actual distance it's drawn from another character when those two characters are printed as a string. Is there any way to get the correct distance in graphics2d?

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  • Two parallel line segments intersection

    - by Judarkness
    I know there are many algorithms to verify whether two line segments are intersected. But once they encountered parallel condition, they just tell the user a big "No" and pretend there is no overlap, share end point, or end point collusion. I know I can can calculate the distance between 2 lines segments. If the distance is 0, check the end points located in the other line segments or not. And this means I have to use a lot of if else and && || conditions. This is not difficult, but my question is "Is there a trick( or mathematics) method to calculate this special parallel case?"

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  • How to order by results from 2 seperate tables in PHP and MySQL.

    - by Vafello
    I am trying to output results of 2 sql queries to one JSON file. The problem is that I would like to order them ascending by distance which is the result of equation that takes homelat and homelon from the users table and lat, lng from locations table.(basically it takes lattitude and longitude of one point and another and computes the distance between these points). Is it possible to take some parameters from both select queries, compute it and output the result in ascending order? $wynik = mysql_query("SELECT homelat, homelon FROM users WHERE guid='2'") or die(mysql_error()); ; $query = "SELECT * FROM locations WHERE timestamp"; $result = map_query($query); $points = array(); while ($aaa = mysql_fetch_assoc($wynik)) { while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { array_push($points, array('name'=>$row['name'], 'lat'=>$row['lat'], 'lng'=>$row['lng'], 'description'=>$row['description'], 'eventType'=>$row['eventType'], 'date'=>$row['date'], 'isotime'=>date('c', ($row['timestamp'])), 'homelat'=>$aaa['homelat'], 'homelon'=>$aaa['homelon'])); } echo json_encode(array("Locations"=>$points));

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  • Find recipes that can be cooked from provided ingridients

    - by skaurus
    Sorry for bad English :( Suppose i can preliminary organize recipes and ingredients data in any way. How can i effectively conduct search of recipes by user-provided ingredients, preferably sorted by max match - so, first going recipes that use maximum of provided ingridients and do not contain any other ingrs, after them recipes that uses less of provided set and still not any other ingrs, after them recipes with minimum additional requirements and so on? All i can think about is represent recipe ingridients like bitmasks, and compare required bitmask with all recipes, but it is obviously a bad way to go. And related things like Levenstein distance i don't see how to use here. I believe it should be quite common task...

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  • video streaming

    - by mawia
    Hi! all I am designing an application for streaming video.I have developed a model in which a server wait for incoming request.The server it self is serving to a good number of clients and it can't afford to serve any more clients.Now when the new connection comes,the server chooses from among it's clients a candidate client who will serve the request of the incoming client.Now the thing is that this choice should be very intelligent.Now I am using various heuristic like bandwidth of the selected client,it's location,distance from the requesting client to come at a decision.Now my question is,IS THERE AVAILABLE ANY TOOL TO FIND OUT BANDWIDTH,LOCATION of a host,and DISTANCE(my be in hop number)?for hop number I can use traceroute but that will be too expensive as it take long time sending reply from every intermediate router. Any help will be appreciated. Thanks!

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  • OpenGL index buffer object with additional data

    - by muksie
    I have a large set of lines, which I render from a vertex buffer object using glMultiDrawArrays(GL_LINE_STRIP, ...); This works perfectly well. Now I have lots of vertex pairs which I also have to visualize. Every pair consists of two vertices on two different lines, and the distance between the vertices is small. However, I like to have the ability to draw a line between all vertex pairs with a distance less than a certain value. What I like to have is something like a buffer object with the following structure: i1, j1, r1, i2, j2, r2, i3, j3, r3, ... where the i's and j's are indices pointing to vertices and the r's are the distances between those vertices. Thus every vertex pair is stored as a (i, j, r) tuple. Then I like to have a (vertex) shader which only draws the vertex pairs with r < SOME_VALUE as a line. So my question is, what is the best way to achieve this?

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  • Why am I not getting correct result when I calculate exponent with ^ in C++?

    - by xbonez
    I am using Bode's formuala to calculate distance of nth planet from sun dist = (4 + 3*(2^(n-2)))/10 If I calculate the distance this way, I get the right values: dist[2] = ((4 + 3*1)/10.0) ; dist[3] = ((4 + 3*2)/10.0) ; dist[4] = ((4 + 3*4)/10.0) ; But doing it this way, gives me incorrect values: vector <double> dist(5); for (unsigned int i = 2; i < 5; i++) { dist[i] = ((4 + 3*(2^(3-2)))/10.0) ; } Why so?

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  • How to fetch managed objects sorted by calculated value

    - by Marcin Zbijowski
    Hello, I'm working on the app that uses CoreData. There is location entity that holds latitude and longitude values. I'd like to fetch those entities sorted by distance to the user's location. I tried to set sort descriptor to distance formula sqrt ((x1 - x2)^2 + (y1 - y2)^2) but it fails with exception "... keypath ... not found in entity". NSString *distanceFormula = [NSString stringWithFormat:@"sqrt(((latitude - %f) * (latitude - %f)) + ((longitude - %f) * (longitude - %f)))", location.coordinate.latitude, location.coordinate.latitude, location.coordinate.longitude, location.coordinate.longitude]; NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:distanceFormula ascending:YES]; [fetchRequest setSortDescriptors:[NSArray arrayWithObject:sortDescriptor]]; NSError *error; NSArray *result = [[self managedObjectContext] executeFetchRequest:fetchRequest error:&error]; I'd like to fetch already sorted objects rather then fetch them all and then sort in the code. Any tips appreciated.

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  • Android string formatiing from xml

    - by mann
    I am parsing xml from server. One of its node contains data like <distance>16.3432434324354324km</distance> I am putting it into hashmap like for (int i = 0; i < nl.getLength(); i++) { map.put(KEY_DISTANCE, parser.getValue(e, KEY_DISTANCE)); // adding HashList to ArrayList menuItems.add(map); } Everything is nice. But i want it two decimal places for example i want its value should be 16.34km rather then 16.343234324342342km I tried with DecimalFormat twoDForm=new DecimalFormat("##.00"); try{ Double StartVTwo=Double.valueOf(KEY_DISTANCE); Double resultDouble1 = Double.valueOf(twoDForm.format(StartVTwo)); Log.e("check", String.valueOf(resultDouble1)); }catch (NumberFormatException e){ Log.e("error"," This is error "); } But it shows exception and prints this message. Any help would be appreciated!!

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  • conditions without repeats

    - by Luca Romagnoli
    Hi i'm using this for getting data: Topic.find(:all, :include => ..., :conditions => @core ? ["cores_topics.id = ? AND visible = 1 AND (distance < ? OR cores.id IN (?))",@core.id, @user_location[3].to_i, @user_friends] : ["visible = 1 AND (distance < ? OR cores.id IN (?))", @user_location[3].to_i, @user_friends], ... how can i rewrite the conditions shorter? thanks

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  • Javascript function returning number with two decimals...?

    - by muzzledBYbrass
    I have a function to return points along a line and my return comes back with two decimal points...? For example, a return of my variable px will be something like -88.4029.032940598. vx is the x vector and mult is the distance of the line plus distance to calculate the point. Here is the operation that is returning these values: var mult = parseFloat(mag + theUnit); var px = coord_one.x_point + (vx * mult); console.log(px); Never have seen this before- I appreciate any and all help!

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  • Canvas scalable arc position

    - by Amay
    http://jsfiddle.net/cs5Sg/11/ I want to do the scalable canvas. I created two circles (arcs) and one line, when you click on circle and move it, the line will follow and change position. The problem is when I added code for resize: var canvas = document.getElementById('myCanvas'), context = canvas.getContext('2d'), radius = 12, p = null, point = { p1: { x:100, y:250 }, p2: { x:400, y:100 } }, moving = false; window.addEventListener("resize", OnResizeCalled, false); function OnResizeCalled() { var gameWidth = window.innerWidth; var gameHeight = window.innerHeight; var scaleToFitX = gameWidth / 800; var scaleToFitY = gameHeight / 480; var currentScreenRatio = gameWidth / gameHeight; var optimalRatio = Math.min(scaleToFitX, scaleToFitY); if (currentScreenRatio >= 1.77 && currentScreenRatio <= 1.79) { canvas.style.width = gameWidth + "px"; canvas.style.height = gameHeight + "px"; } else { canvas.style.width = 800 * optimalRatio + "px"; canvas.style.height = 480 * optimalRatio + "px"; } } function init() { return setInterval(draw, 10); } canvas.addEventListener('mousedown', function(e) { for (p in point) { var mouseX = e.clientX - 1, mouseY = e.clientY - 1, distance = Math.sqrt(Math.pow(mouseX - point[p].x, 2) + Math.pow(mouseY - point[p].y, 2)); if (distance <= radius) { moving = p; break; } } }); canvas.addEventListener('mouseup', function(e) { moving = false; }); canvas.addEventListener('mousemove', function(e) { if(moving) { point[moving].x = e.clientX - 1; point[moving].y = e.clientY - 1; } }); function draw() { context.clearRect(0, 0, canvas.width, canvas.height); context.beginPath(); context.moveTo(point.p1.x,point.p1.y); context.lineTo(point.p2.x,point.p2.y); context.closePath(); context.fillStyle = '#8ED6FF'; context.fill(); context.stroke(); for (p in point) { context.beginPath(); context.arc(point[p].x,point[p].y,radius,0,2*Math.PI); context.fillStyle = 'red'; context.fill(); context.stroke(); } context.closePath(); } init(); The canvas is scalable, but the problem is with the points (circles). When you change the window size, they still have the same position on the canvas area, but the distance change (so the click option fails). How to fix that?

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  • Shuffling a list with a constraint

    - by 500
    Preparing a new psychophysic experiment, I have 48 original stimuli displayed 4 times (4 conditions). Resulting in 192 trials. Trying to randomize the order of presentation during the experiment, I need to maximize the distance between the 4 display of the same original stimuli. Please Consider : Table[{j, i}, {j, Range[48]}, {i, Range[4]}] Where j is the original stimuli number and i the condition Output Sample : {{1, 1}, {1, 2}, {1, 3}, {1, 4}, {2, 1}, {2, 2}, {2, 3}, {2, 4}, ... {47, 1}, {47, 2}, {47, 3},{47, 4}, {48, 1}, {48, 2}, {48, 3}, {48, 4}} How could I shuffle the order of presentation of those 192 items, maximizing the distance between identical item with regard to j the original stimuli number ?

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  • [C++] Why am I not getting the same values?

    - by xbonez
    I am using Bode's formuala to calculate distance of nth planet from sun dist = (4 + 3*(2^(n-2)))/10 If I calculate the distance this way, I get the right values: dist[2] = ((4 + 3*1)/10.0) ; dist[3] = ((4 + 3*2)/10.0) ; dist[4] = ((4 + 3*4)/10.0) ; But doing it this way, gives me incorrect values: vector double> dist(5); for (unsigned int i = 2; i < 5; i++) { dist[i] = ((4 + 3*(2^(3-2)))/10.0) ; } ` Why so?

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  • How does LocationManager work

    - by user2511882
    I have been going through the google docs for the locationManager API and specifically over the method locationManager.requestLocationUpdates(LocationManager.NETWORK_PROVIDER, 60000, 10, this); In the above line, the manager will check for location updates after each minute and distance of 10 meters. My question is does it actually take both the parameters into consideration? For example if you are driving along how will the method work? Will it start looking for updated locations since you are over the minimum distance between the two updates or will it wait for a minute irrespective? If someone can tell me its behavior, it would be great. Thanks.!!

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  • How to calculate where bullet hits

    - by lkjoel
    I have been trying to write an FPS in C/X11/OpenGL, but the issue that I have encountered is with calculating where the bullet hits. I have used a horrible technique, and it only sometimes works: pos size, p; size.x = 0.1; size.z = 0.1; // Since the game is technically top-down (but in a 3D perspective) // Positions are in X/Z, no Y float f; // Counter float d = FIRE_MAX + 1 /* Shortest Distance */, d1 /* Distance being calculated */; x = 0; // Index of object to hit for (f = 0.0; f < FIRE_MAX; f += .01) { // Go forwards p.x = player->pos.x + f * sin(toRadians(player->rot.x)); p.z = player->pos.z - f * cos(toRadians(player->rot.x)); // Get all objects that collide with the current position of the bullet short* objs = _colDetectGetObjects(p, size, objects); for (i = 0; i < MAX_OBJECTS; i++) { if (objs[i] == -1) { continue; } // Check the distance between the object and the player d1 = sqrt( pow((objects[i].pos.x - player->pos.x), 2) + pow((objects[i].pos.z - player->pos.z), 2)); // If it's closer, set it as the object to hit if (d1 < d) { x = i; d = d1; } } // If there was an object, hit it if (x > 0) { hit(&objects[x], FIRE_DAMAGE, explosions, currtime); break; } } It just works by making a for-loop and calculating any objects that might collide with where the bullet currently is. This, of course, is very slow, and sometimes doesn't even work. What would be the preferred way to calculate where the bullet hits? I have thought of making a line and seeing if any objects collide with that line, but I have no idea how to do that kind of collision detection. EDIT: I guess my question is this: How do I calculate the nearest object colliding in a line (that might not be a straight 45/90 degree angle)? Or are there any simpler methods of calculating where the bullet hits? The bullet is sort of like a laser, in the sense that gravity does not affect it (writing an old-school game, so I don't want it to be too realistic)

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  • How to use GPS data like the double value returned by getLatitude()?

    - by Dan
    I have been searching quite a bit for an answer, but maybe I'm just not using the correct terminology. I am creating an app that will access a database to return a list of other users that are within a certain distance of the users location. I've never worked with this type of data, and I don't really know what the values mean. I'd like to do all the calculations on the backend with either MySQL or PHP. Currently, I am storing the latitude and longitude as doubles within the database. I can access them and store them, but I have no idea how I might be able to sort them based on distance. Perhaps I should be using a different type or some technique that is common in this area. TIA.

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  • How can I get penetration depth from Minkowski Portal Refinement / Xenocollide?

    - by Raven Dreamer
    I recently got an implementation of Minkowski Portal Refinement (MPR) successfully detecting collision. Even better, my implementation returns a good estimate (local minimum) direction for the minimum penetration depth. So I took a stab at adjusting the algorithm to return the penetration depth in an arbitrary direction, and was modestly successful - my altered method works splendidly for face-edge collision resolution! What it doesn't currently do, is correctly provide the minimum penetration depth for edge-edge scenarios, such as the case on the right: What I perceive to be happening, is that my current method returns the minimum penetration depth to the nearest vertex - which works fine when the collision is actually occurring on the plane of that vertex, but not when the collision happens along an edge. Is there a way I can alter my method to return the penetration depth to the point of collision, rather than the nearest vertex? Here's the method that's supposed to return the minimum penetration distance along a specific direction: public static Vector3 CalcMinDistance(List<Vector3> shape1, List<Vector3> shape2, Vector3 dir) { //holding variables Vector3 n = Vector3.zero; Vector3 swap = Vector3.zero; // v0 = center of Minkowski sum v0 = Vector3.zero; // Avoid case where centers overlap -- any direction is fine in this case //if (v0 == Vector3.zero) return Vector3.zero; //always pass in a valid direction. // v1 = support in direction of origin n = -dir; //get the differnce of the minkowski sum Vector3 v11 = GetSupport(shape1, -n); Vector3 v12 = GetSupport(shape2, n); v1 = v12 - v11; //if the support point is not in the direction of the origin if (v1.Dot(n) <= 0) { //Debug.Log("Could find no points this direction"); return Vector3.zero; } // v2 - support perpendicular to v1,v0 n = v1.Cross(v0); if (n == Vector3.zero) { //v1 and v0 are parallel, which means //the direction leads directly to an endpoint n = v1 - v0; //shortest distance is just n //Debug.Log("2 point return"); return n; } //get the new support point Vector3 v21 = GetSupport(shape1, -n); Vector3 v22 = GetSupport(shape2, n); v2 = v22 - v21; if (v2.Dot(n) <= 0) { //can't reach the origin in this direction, ergo, no collision //Debug.Log("Could not reach edge?"); return Vector2.zero; } // Determine whether origin is on + or - side of plane (v1,v0,v2) //tests linesegments v0v1 and v0v2 n = (v1 - v0).Cross(v2 - v0); float dist = n.Dot(v0); // If the origin is on the - side of the plane, reverse the direction of the plane if (dist > 0) { //swap the winding order of v1 and v2 swap = v1; v1 = v2; v2 = swap; //swap the winding order of v11 and v12 swap = v12; v12 = v11; v11 = swap; //swap the winding order of v11 and v12 swap = v22; v22 = v21; v21 = swap; //and swap the plane normal n = -n; } /// // Phase One: Identify a portal while (true) { // Obtain the support point in a direction perpendicular to the existing plane // Note: This point is guaranteed to lie off the plane Vector3 v31 = GetSupport(shape1, -n); Vector3 v32 = GetSupport(shape2, n); v3 = v32 - v31; if (v3.Dot(n) <= 0) { //can't enclose the origin within our tetrahedron //Debug.Log("Could not reach edge after portal?"); return Vector3.zero; } // If origin is outside (v1,v0,v3), then eliminate v2 and loop if (v1.Cross(v3).Dot(v0) < 0) { //failed to enclose the origin, adjust points; v2 = v3; v21 = v31; v22 = v32; n = (v1 - v0).Cross(v3 - v0); continue; } // If origin is outside (v3,v0,v2), then eliminate v1 and loop if (v3.Cross(v2).Dot(v0) < 0) { //failed to enclose the origin, adjust points; v1 = v3; v11 = v31; v12 = v32; n = (v3 - v0).Cross(v2 - v0); continue; } bool hit = false; /// // Phase Two: Refine the portal int phase2 = 0; // We are now inside of a wedge... while (phase2 < 20) { phase2++; // Compute normal of the wedge face n = (v2 - v1).Cross(v3 - v1); n.Normalize(); // Compute distance from origin to wedge face float d = n.Dot(v1); // If the origin is inside the wedge, we have a hit if (d > 0 ) { //Debug.Log("Do plane test here"); float T = n.Dot(v2) / n.Dot(dir); Vector3 pointInPlane = (dir * T); return pointInPlane; } // Find the support point in the direction of the wedge face Vector3 v41 = GetSupport(shape1, -n); Vector3 v42 = GetSupport(shape2, n); v4 = v42 - v41; float delta = (v4 - v3).Dot(n); float separation = -(v4.Dot(n)); if (delta <= kCollideEpsilon || separation >= 0) { //Debug.Log("Non-convergance detected"); //Debug.Log("Do plane test here"); return Vector3.zero; } // Compute the tetrahedron dividing face (v4,v0,v1) float d1 = v4.Cross(v1).Dot(v0); // Compute the tetrahedron dividing face (v4,v0,v2) float d2 = v4.Cross(v2).Dot(v0); // Compute the tetrahedron dividing face (v4,v0,v3) float d3 = v4.Cross(v3).Dot(v0); if (d1 < 0) { if (d2 < 0) { // Inside d1 & inside d2 ==> eliminate v1 v1 = v4; v11 = v41; v12 = v42; } else { // Inside d1 & outside d2 ==> eliminate v3 v3 = v4; v31 = v41; v32 = v42; } } else { if (d3 < 0) { // Outside d1 & inside d3 ==> eliminate v2 v2 = v4; v21 = v41; v22 = v42; } else { // Outside d1 & outside d3 ==> eliminate v1 v1 = v4; v11 = v41; v12 = v42; } } } return Vector3.zero; } }

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  • Evolution Of High Definition TV Viewing

    - by Gopinath
    The following guest post is written by Rob, who is also blogging on entertainment technology topics on iwantsky.com Gone are the days when you need to squint to be able to see the emotions on the faces of Humphrey Bogart and Ingrid Bergman as the lovers bid each other adieu in the classic film Casablanca. These days, watching an ordinary ant painstakingly carry a leaf in Animal Planet can be an exhilarating experience as you get to see not only the slightest movement but also the demarcation line between the insect’s head, thorax and abdomen. The crystal clear imagery was made possible by the sharp minds and the tinkering hands of the scientists that have designed the modern world’s HDTV. What is HDTV and what makes people so agog to have this new innovation in TV watching? HDTV stands for High Definition TV. Television viewing has indeed made a big leap. From the grainy black and whites, TV viewing had moved to colored TVs, progressed to SD TVs and now to HDTV. HDTV is the emerging trend in TV viewing as it delivers bigger and clearer pictures and better audio. Viewers can have a cinema-like TV viewing experience right in the comforts of their own home. With HDTV the viewer is allowed to have a better viewing range. With Standard (SD) TV, the viewer has to be at a distance that is from 3 to 6 times the size of the screen. HDTV allows the viewer to enjoy sharper and clearer images as it is possible to sit at a distance that is 1.5 or 3 times the size of the screen without noticing any image pixilation. Although HDTV appears to be a fairly new innovation, this system has actually existed in various forms years ago. Development of the HDTV was started in Europe as early as 1940s. However, the NTSC and the PAL/SECAM, the two analog TV standards became dominant and became popular worldwide. The analog TV was replaced by the digital TV platform in the 1990s. Even during the analog era, attempts have been made to develop HDTV. Japan has come out with MUSE system. However, due to channel bandwidth requirement concerns, the program was shelved. The entry of four organizations into the HDTV market spurred the development of a beneficial coalition. The AT&T, ATRC, MIT and Zenith HDTV combined forces. In 1993, a Grand Alliance was formed. This group is composed of researchers and HDTV manufacturers. A common standard for the broadcast system of HDTV was developed. In 1995, the system was tested and found successful. With the higher screen resolution of HDTV, viewing has never been more enjoyable. [Image courtesy: samsung] This article titled,Evolution Of High Definition TV Viewing, was originally published at Tech Dreams. Grab our rss feed or fan us on Facebook to get updates from us.

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  • Evolution Of High Definition TV Viewing

    - by Gopinath
    The following guest post is written by Rob, who is also blogging on entertainment technology topics on iwantsky.com Gone are the days when you need to squint to be able to see the emotions on the faces of Humphrey Bogart and Ingrid Bergman as the lovers bid each other adieu in the classic film Casablanca. These days, watching an ordinary ant painstakingly carry a leaf in Animal Planet can be an exhilarating experience as you get to see not only the slightest movement but also the demarcation line between the insect’s head, thorax and abdomen. The crystal clear imagery was made possible by the sharp minds and the tinkering hands of the scientists that have designed the modern world’s HDTV. What is HDTV and what makes people so agog to have this new innovation in TV watching? HDTV stands for High Definition TV. Television viewing has indeed made a big leap. From the grainy black and whites, TV viewing had moved to colored TVs, progressed to SD TVs and now to HDTV. HDTV is the emerging trend in TV viewing as it delivers bigger and clearer pictures and better audio. Viewers can have a cinema-like TV viewing experience right in the comforts of their own home. With HDTV the viewer is allowed to have a better viewing range. With Standard (SD) TV, the viewer has to be at a distance that is from 3 to 6 times the size of the screen. HDTV allows the viewer to enjoy sharper and clearer images as it is possible to sit at a distance that is 1.5 or 3 times the size of the screen without noticing any image pixilation. Although HDTV appears to be a fairly new innovation, this system has actually existed in various forms years ago. Development of the HDTV was started in Europe as early as 1940s. However, the NTSC and the PAL/SECAM, the two analog TV standards became dominant and became popular worldwide. The analog TV was replaced by the digital TV platform in the 1990s. Even during the analog era, attempts have been made to develop HDTV. Japan has come out with MUSE system. However, due to channel bandwidth requirement concerns, the program was shelved. The entry of four organizations into the HDTV market spurred the development of a beneficial coalition. The AT&T, ATRC, MIT and Zenith HDTV combined forces. In 1993, a Grand Alliance was formed. This group is composed of researchers and HDTV manufacturers. A common standard for the broadcast system of HDTV was developed. In 1995, the system was tested and found successful. With the higher screen resolution of HDTV, viewing has never been more enjoyable. [Image courtesy: samsung] This article titled,Evolution Of High Definition TV Viewing, was originally published at Tech Dreams. Grab our rss feed or fan us on Facebook to get updates from us.

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