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  • Would it be possible to swap system restore points on different brands of computers?

    - by P'sao
    So I have a ThinkPad computer that I've installed a program called "Deep Freeze" which restores your computer to the "Frozen" state. I also have a Toshiba computer with no Deep Freeze on it. My question is that would it be possible to creat a system restore point on my Toshiba and then replace the system volume information folder with the one on the ThinkPad so I could restore to a point with no Deep Freeze? Would this work?

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  • Reading graph inputs for a programming puzzle and then solving it

    - by Vrashabh
    I just took a programming competition question and I absolutely bombed it. I had trouble right at the beginning itself from reading the input set. The question was basically a variant of this puzzle http://codercharts.com/puzzle/evacuation-plan but also had an hour component in the first line(say 3 hours after start of evacuation). It reads like this This puzzle is a tribute to all the people who suffered from the earthquake in Japan. The goal of this puzzle is, given a network of road and locations, to determine the maximum number of people that can be evacuated. The people must be evacuated from evacuation points to rescue points. The list of road and the number of people they can carry per hour is provided. Input Specifications Your program must accept one and only one command line argument: the input file. The input file is formatted as follows: the first line contains 4 integers n r s t n is the number of locations (each location is given by a number from 0 to n-1) r is the number of roads s is the number of locations to be evacuated from (evacuation points) t is the number of locations where people must be evacuated to (rescue points) the second line contains s integers giving the locations of the evacuation points the third line contains t integers giving the locations of the rescue points the r following lines contain to the road definitions. Each road is defined by 3 integers l1 l2 width where l1 and l2 are the locations connected by the road (roads are one-way) and width is the number of people per hour that can fit on the road Now look at the sample input set 5 5 1 2 3 0 3 4 0 1 10 0 2 5 1 2 4 1 3 5 2 4 10 The 3 in the first line is the additional component and is defined as the number of hours since the resuce has started which is 3 in this case. Now my solution was to use Dijisktras algorithm to find the shortest path between each of the rescue and evac nodes. Now my problem started with how to read the input set. I read the first line in python and stored the values in variables. But then I did not know how to store the values of the distance between the nodes and what DS to use and how to input it to say a standard implementation of dijikstras algorithm. So my question is two fold 1.) How do I take the input of such problems? - I have faced this problem in quite a few competitions recently and I hope I can get a simple code snippet or an explanation in java or python to read the data input set in such a way that I can input it as a graph to graph algorithms like dijikstra and floyd/warshall. Also a solution to the above problem would also help. 2.) How to solve this puzzle? My algorithm was: Find shortest path between evac points (in the above example it is 14 from 0 to 3) Multiply it by number of hours to get maximal number of saves Also the answer given for the variant for the input set was 24 which I dont understand. Can someone explain that also. UPDATE: I get how the answer is 14 in the given problem link - it seems to be just the shortest path between node 0 and 3. But with the 3 hour component how is the answer 24 UPDATE I get how it is 24 - its a complete graph traversal at every hour and this is how I solve it Hour 1 Node 0 to Node 1 - 10 people Node 0 to Node 2- 5 people TotalRescueCount=0 Node 1=10 Node 2= 5 Hour 2 Node 1 to Node 3 = 5(Rescued) Node 2 to Node 4 = 5(Rescued) Node 0 to Node 1 = 10 Node 0 to Node 2 = 5 Node 1 to Node 2 = 4 TotalRescueCount = 10 Node 1 = 10 Node 2= 5+4 = 9 Hour 3 Node 1 to Node 3 = 5(Rescued) Node 2 to Node 4 = 5+4 = 9(Rescued) TotalRescueCount = 9+5+10 = 24 It hard enough for this case , for multiple evac and rescue points how in the world would I write a pgm for this ?

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  • How can you replace the src of an image that points to mjpeg?

    - by user165623
    I have an application that displays images from network cameras. The application will show multiple images on a single page. Each image container has a button on it that causes jQuery to scale that image up to fill a 64x480 div vs much smaller when multiple images are present. I change the src in the image with jquery as you'd expect: $("#imageId").attr("src", "http://cameraUrl.com:6000?width=640&height=480&d="+date.getTime()); the date.getTime() is intended to override the caching in a browser. It works for retrieving still images. Sometimes the video changes to the larger resolution. Other times it just sticks with the original image scaled up to fit the larger div as if it is not loading the new URL. This is evident by the fact it's grainy and the text overlay from the camera is scaled up. Occasionally the image finally loads in the higher resolution version and it works evidenced by the image clarity and the text overlay scaling to what it should be. Is there a correct way to force the setting of an image src where the src points to a motion jpeg stream to reload?

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  • How to authenticate multiple entry points in a facebook app?

    - by Simon_Weaver
    I am using an IFrame application with XFBML and the new Javascript API. I'd like to have a facebook application with multiple entry points. These will most likely represent different links coming from a fan page tab. I can do this quite easily if the pages don't require authentication - for instance I can create several pages under the app and if a new user comes I can send them to any page: http://apps.facebook.com/myapp/offers http://apps.facebook.com/myapp/game http://apps.facebook.com/myapp/products The problem is that if I need to have authentication then once the user is authenticated they get redirected to my default post-authorization url. Is there a way for a user that comes to /game to stay on /game after they are authenticated without redirecting. I thought I could do it with the AJAX login form - but I cannot find out how to do that in a Facebook IFrame application. I think the example using requirelogin only works for FBML. <a href="http://apps.facebook.com/mysmiley" requirelogin=1> Welcome to my app</a>. Is there a way to accomplish this with Facebook APIs - or will I have to do some kind of clever cookie handling?

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  • What is the most efficient way to handle points / small vectors in JavaScript?

    - by Chris
    Currently I'm creating an web based (= JavaScript) application thata is using a lot of "points" (= small, fixed size vectors). There are basically two obvious ways of representing them: var pointA = [ xValue, yValue ]; and var pointB = { x: xValue, y: yValue }; So translating my point a bit would look like: var pointAtrans = [ pointA[0] + 3, pointA[1] + 4 ]; var pointBtrans = { x: pointB.x + 3, pointB.y + 4 }; Both are easy to handle from a programmer point of view (the object variant is a bit more readable, especially as I'm mostly dealing with 2D data, seldom with 3D and hardly with 4D - but never more. It'll allways fit into x,y,z and w) But my question is now: What is the most efficient way from the language perspective - theoretically and in real implementations? What are the memory requirements? What are the setup costs of an array vs. an object? ... My target browsers are FireFox and the Webkit based ones (Chromium, Safari), but it wouldn't hurt to have a great (= fast) experience under IE and Opera as well...

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  • Adding visible "Markers" to represent Geopoints to a MapView using ItemizedOverlay in Android

    - by LordSnoutimus
    Hello, I am building an application which stores GPS locations in a SQLite database and then outputs the data onto a MapView using an Overlay by drawing a red line between the points. I want to be able to show graphical markers (images) for each of these points as well as the red line. My code is as follows: public class MyOverlay extends ItemizedOverlay { // private Projection projection; private Paint linePaint; private Vector points; public MyOverlay(Drawable defaultMarker) { super(defaultMarker); points = new Vector<GeoPoint>(); //set colour, stroke width etc. linePaint = new Paint(); linePaint.setARGB(255, 255, 0, 0); linePaint.setStrokeWidth(3); linePaint.setDither(true); linePaint.setStyle(Style.FILL); linePaint.setAntiAlias(true); linePaint.setStrokeJoin(Paint.Join.ROUND); linePaint.setStrokeCap(Paint.Cap.ROUND); } public void addPoint(GeoPoint point) { populate(); points.addElement(point); } //public void setProjection(Projection projection) { // this.projection = projection; // } public void draw(Canvas canvas, MapView view, boolean shadow) { populate(); int size = points.size(); Point lastPoint = new Point(); if(size == 0) return; view.getProjection().toPixels(points.get(0), lastPoint); Point point = new Point(); for(int i = 1; i<size; i++){ view.getProjection().toPixels(points.get(i), point); canvas.drawLine(lastPoint.x, lastPoint.y, point.x, point.y, linePaint); lastPoint = point; } } @Override protected OverlayItem createItem(int arg0) { // TODO Auto-generated method stub return null; } @Override public int size() { // TODO Auto-generated method stub return 0; } } What would be the easiest way to implement adding markers for each GeoPoint? Thanks

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  • Get most left|right|top|bottom point contained in box

    - by skyman
    I'm storing Points Of Interest (POI) in PostgreSQL database, and retrieve them via PHP script to Android application. To reduce internet usage I want my mobile app to know if there are any points in the neighborhood of currently displayed area. My idea is to store bounds of the rectangle containing all points already retrieved (in other words: nearest point on the left (West) of most west already retrieved, nearest point above (North) of most north already retrieved etc.) and I will make next query when any edge of screen goes outside of this bounds. Currently I can retrieve points which are in "single screen" (in the area covered by currently displayed map) using: SELECT * FROM ch WHERE loc <@ (box '((".-$latSpan.", ".$lonSpan."),(".$latSpan.", ".-$lonSpan."))' + point '".$loc."') Now I need to know four most remote points in each direction, than I will be able to retrieve next four "more remote" points. Is there any possibility to get those points (or box) directly from PostgreSQL (maybe using some "aggregate points to box" function)?

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  • Hibernate limitations on using variables in queries

    - by sammichy
    I had asked the following question I have the following table structure for a table Player Table Player { Long playerID; Long points; Long rank; } Assuming that the playerID and the points have valid values, can I update the rank for all the players based on the number of points in a single query? If two people have the same number of points, they should tie for the rank. And received the answer from Daniel Vassalo (thank you). UPDATE player JOIN (SELECT p.playerID, IF(@lastPoint <> p.points, @curRank := @curRank + 1, @curRank) AS rank, IF(@lastPoint = p.points, @curRank := @curRank + 1, @curRank), @lastPoint := p.points FROM player p JOIN (SELECT @curRank := 0, @lastPoint := 0) r ORDER BY p.points DESC ) ranks ON (ranks.playerID = player.playerID) SET player.rank = ranks.rank; When I try to execute this as a native query in Hibernate, the following exception is thrown. java.lang.IllegalArgumentException: org.hibernate.QueryException: Space is not allowed after parameter prefix ':' Apparently this has been an open issue for the last couple of years, I want to know if the ranking query can be made to work either Without using any variables in the SQL query OR Using any workaround for Hibernate.

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  • [Ruby] Modifying object inside a loop doesn't change object outside of the loop?

    - by Jergason
    I am having problems with modifying objects inside blocks and not getting the expected values outside the blocks. This chunk of code is supposed to transform a bunch of points in 3d space, calculate a score (the rmsd or root mean squared deviation), and store both the score and the set of points that produced that score if it is lower than the current lowest score. At the end, I want to print out the best bunch of points. first = get_transformed_points(ARGV[0]) second = get_transformed_points(ARGV[1]) best_rmsd = first.rmsd(second) best_points = second #transform the points around x, y, and z and get the rmsd. If the new points # have a smaller rmsd, store them. ROTATION = 30 #rotate by ROTATION degrees num_rotations = 360/ROTATION radians = ROTATION * (Math::PI/180) num_rotations.times do |i| second = second * x_rotate num_rotations.times do |j| second = second * y_rotate num_rotations.times do |k| second = second * z_rotate rmsd = first.rmsd(second) if rmsd < best_rmsd then best_points = second best_rmsd = rmsd end end end end File.open("#{ARGV[1]}.out", "w") {|f| f.write(best_points.to_s)} I can print out the points that are getting stored inside the block, and they are getting transformed and stored correctly. However, when I write out the points to a file at the end, they are the same as the initial set of points. Somehow the best_points = second chunk doesn't seem to be doing anything outside of the block. It seems like there are some scoping rules that I don't understand here. I had thought that since I declared and defined best_points above, outside of the blocks, that it would be updated inside the blocks. However, it seems that when the blocks end, it somehow reverts back to the original value. Any ideas how to fix this? Is this a problem with blocks specifically?

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  • Const references when dereferencing iterator on set, starting from Visual Studio 2010

    - by Patrick
    Starting from Visual Studio 2010, iterating over a set seems to return an iterator that dereferences the data as 'const data' instead of non-const. The following code is an example of something that does compile on Visual Studio 2005, but not on 2010 (this is an artificial example, but clearly illustrates the problem we found on our own code). In this example, I have a class that stores a position together with a temperature. I define comparison operators (not all them, just enough to illustrate the problem) that only use the position, not the temperature. The point is that for me two instances are identical if the position is identical; I don't care about the temperature. #include <set> class DataPoint { public: DataPoint (int x, int y) : m_x(x), m_y(y), m_temperature(0) {} void setTemperature(double t) {m_temperature = t;} bool operator<(const DataPoint& rhs) const { if (m_x==rhs.m_x) return m_y<rhs.m_y; else return m_x<rhs.m_x; } bool operator==(const DataPoint& rhs) const { if (m_x!=rhs.m_x) return false; if (m_y!=rhs.m_y) return false; return true; } private: int m_x; int m_y; double m_temperature; }; typedef std::set<DataPoint> DataPointCollection; void main(void) { DataPointCollection points; points.insert (DataPoint(1,1)); points.insert (DataPoint(1,1)); points.insert (DataPoint(1,2)); points.insert (DataPoint(1,3)); points.insert (DataPoint(1,1)); for (DataPointCollection::iterator it=points.begin();it!=points.end();++it) { DataPoint &point = *it; point.setTemperature(10); } } In the main routine I have a set to which I add some points. To check the correctness of the comparison operator, I add data points with the same position multiple times. When writing the contents of the set, I can clearly see there are only 3 points in the set. The for-loop loops over the set, and sets the temperature. Logically this is allowed, since the temperature is not used in the comparison operators. This code compiles correctly in Visual Studio 2005, but gives compilation errors in Visual Studio 2010 on the following line (in the for-loop): DataPoint &point = *it; The error given is that it can't assign a "const DataPoint" to a [non-const] "DataPoint &". It seems that you have no decent (= non-dirty) way of writing this code in VS2010 if you have a comparison operator that only compares parts of the data members. Possible solutions are: Adding a const-cast to the line where it gives an error Making temperature mutable and making setTemperature a const method But to me both solutions seem rather 'dirty'. It looks like the C++ standards committee overlooked this situation. Or not? What are clean solutions to solve this problem? Did some of you encounter this same problem and how did you solve it? Patrick

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  • How to convert this query to a "django model query" ?

    - by fabriciols
    Hello ! What i want is simple : models : class userLastTrophy(models.Model): user = models.ForeignKey(userInfo) platinum = models.IntegerField() gold = models.IntegerField() silver = models.IntegerField() bronze = models.IntegerField() level = models.IntegerField() rank = models.IntegerField() perc_level = models.IntegerField() date_update = models.DateTimeField(default=datetime.now, blank=True) total = models.IntegerField() points = models.IntegerField() class userTrophy(models.Model): user = models.ForeignKey(userInfo) platinum = models.IntegerField() gold = models.IntegerField() silver = models.IntegerField() bronze = models.IntegerField() total = models.IntegerField() level = models.IntegerField() perc_level = models.IntegerField() date_update = models.DateTimeField(default=datetime.now, blank=True) rank = models.IntegerField(default=0) total = models.IntegerField(default=0) points = models.IntegerField(default=0) last_trophy = models.ForeignKey(userLastTrophy, default=0) I have this query : select t2.user_id as id, t2.platinum - t1.platinum as plat, t2.gold - t1.gold as gold, t2.silver - t1.silver as silver, t2.bronze - t1.bronze as bronze, t2.points - t1.points as points from myps3t_usertrophy t2, myps3t_userlasttrophy t1 where t1.id = t2.last_trophy_id order by points; how to do this with django models ?

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  • Django unable to update model

    - by user292652
    i have the following function to override the default save function in a model match def save(self, *args, **kwargs): if self.Match_Status == "F": Team.objects.filter(pk=self.Team_one.id).update(Played=F('Played')+1) Team.objects.filter(pk=self.Team_two.id).update(Played=F('Played')+1) if self.Winner !="": Team.objects.filter(pk=self.Winner.id).update(Win=F('Win')+1, Points=F('Points')+3) else: return if self.Match_Status == "D": Team.objects.filter(pk=self.Team_one.id).update(Played=F('Played')+1, Draw = F('Draw')+1, Points=F('Points')+1) Team.objects.filter(pk=self.Team_two.id).update(Played=F('Played')+1, Draw = F('Draw')+1, Points=F('Points')+1) super(Match, self).save(*args, **kwargs) I am able to save the match model just fine but Team model does not seem to be updating at all and no error is being thrown. am i missing some thing here ?

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  • Python: Filter a dictionary

    - by Adam Matan
    Hi, I have a dictionary of points, say: >>> points={'a':(3,4), 'b':(1,2), 'c':(5,5), 'd':(3,3)} I want to create a new dictionary with all the points whose x and y value is smaller than 5, i.e. points 'a', 'b' and 'd'. According to the the book, each dictionary has the items() function, which returns a list of (key, pair) tuple: >>> points.items() [('a', (3, 4)), ('c', (5, 5)), ('b', (1, 2)), ('d', (3, 3))] So I have written this: >>> for item in [i for i in points.items() if i[1][0]<5 and i[1][1]<5]: ... points_small[item[0]]=item[1] ... >>> points_small {'a': (3, 4), 'b': (1, 2), 'd': (3, 3)} Is there a more elegant way? I was expecting Python to have some super-awesome dictionary.filter(f) function... Adam

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  • How would you sample a real-time stream of coordinates to create a Speed Graph?

    - by Andrew Johnson
    I have a GPS device, and I am receiving continuous points, which I store in an array. These points are time stamped. I would like to graph distance/time (speed) vs. distance in real-time; however, I can only plot 50 of the points because of hardware constraints. How would you select points from the array to graph? For example, one algorithm might be to select every Nth point from the array, where N results in 50 points total. Code: float indexModifier = 1; if (MIN(50,track.lastPointIndex) == 50) { indexModifier = track.lastPointIndex/50.0f; } index = ceil(index*indexModifier); Another algorithm might be to keep an array of 50 points, and throw out the point with the least speed change each time you get a new point.

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  • java quaternion 3D rotation implementation

    - by MRM
    I made a method to rotate a list of points using quaternions, but all i get back as output is the same list i gave to rotate on. Maybe i did not understood corectly the math for 3d rotations or my code is not implemented the right way, could you give me a hand? This is the method i use: public static ArrayList<Float> rotation3D(ArrayList<Float> points, double angle, int x, int y, int z) { ArrayList<Float> newpoints = points; for (int i=0;i<points.size();i+=3) { float x_old = points.get(i).floatValue(); float y_old = points.get(i+1).floatValue(); float z_old = points.get(i+2).floatValue(); double[] initial = {1,0,0,0}; double[] total = new double[4]; double[] local = new double[4]; //components for local quaternion //w local[0] = Math.cos(0.5 * angle); //x local[1] = x * Math.sin(0.5 * angle); //y local[2] = y * Math.sin(0.5 * angle); //z local[3] = z * Math.sin(0.5 * angle); //components for final quaternion Q1*Q2 //w = w1w2 - x1x2 - y1y2 - z1z2 total[0] = local[0] * initial[0] - local[1] * initial[1] - local[2] * initial[2] - local[3] * initial[3]; //x = w1x2 + x1w2 + y1z2 - z1y2 total[1] = local[0] * initial[1] + local[1] * initial[0] + local[2] * initial[3] - local[3] * initial[2]; //y = w1y2 - x1z2 + y1w2 + z1x2 total[2] = local[0] * initial[2] - local[1] * initial[3] + local[2] * initial[0] + local[3] * initial[1]; //z = w1z2 + x1y2 - y1x2 + z1w2 total[3] = local[0] * initial[3] + local[1] * initial[2] - local[2] * initial[1] + local[3] * initial[0]; //new x,y,z of the 3d point using rotation matrix made from the final quaternion float x_new = (float)((1 - 2 * total[2] * total[2] - 2 * total[3] * total[3]) * x_old + (2 * total[1] * total[2] - 2 * total[0] * total[3]) * y_old + (2 * total[1] * total[3] + 2 * total[0] * total[2]) * z_old); float y_new = (float) ((2 * total[1] * total[2] + 2 * total[0] * total[3]) * x_old + (1 - 2 * total[1] * total[1] - 2 * total[3] * total[3]) * y_old + (2 * total[2] * total[3] + 2 * total[0] * total[1]) * z_old); float z_new = (float) ((2 * total[1] * total[3] - 2 * total[0] * total[2]) * x_old + (2 * total[2] * total[3] - 2 * total[0] * total[1]) * y_old + (1 - 2 * total[1] * total[1] - 2 * total[2] * total[2]) * z_old); newpoints.set(i, x_new); newpoints.set(i+1, y_new); newpoints.set(i+2, z_new); } return newpoints; } For rotation3D(points, 50, 0, 1, 0) where points is: 0.0, 0.0, -9.0; 0.0, 0.0, -11.0; 20.0, 0.0, -11.0; 20.0, 0.0, -9.0; i get back the same list.

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  • How to get all n sets of three consecutives elements in an array or arraylist with a for statement ?

    - by newba
    Hi, I'm trying to do a convex hull approach and the little problem is that I need to get all sets of three consecutive vertices, like this: private void isConvexHull(Ponto[] points) { Arrays.sort(points); for (int i = 0; i <points.length; i++) { isClockWise(points[i],points[i+1],points[i+2]); } //... } I always do something that I don't consider clean code. Could please help me find one or more ways to this? I want it to be circular, i.e., if my fisrt point of the a set is the last element in the array, the 2nd element will be the 3rd in the list and the 3rd in that set will be the the 2nd element in the list, and so on. They must be consecutive, that's all.

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  • T-SQL: How to make a positive value turn into the equivalent negative value (e.g "10.00" to "-10.00"

    - by RPM1984
    Ok so i have a DECIMAL field called "Score". (e.g 10.00) Now, in my SP, i want to increment/decrement the value of this field in update transactions. So i might want to do this: SET @NewScore = @CurrentScore + @Points Where @Points is the value im going to increment/decrement. Now lets say @Points = 10.00. In a certain scenario, i want 10.00 to become -10.00 So the statement would be translated to: SET @NewScore = @CurrentScore + -10.00 How can i do that? I know its a strange question, but basically i want that statement to be dynamic, in that i dont want to have a different statement for incrementing/decrementing the value. I just want something like this: SET @Points = 10.00 IF @ActivityBeingPerformedIsFoo BEGIN -- SET @Points to be equivalent negative value, (e.g -10.00) END SET @NewScore = @CurrentScore + @Points

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  • How to partition a plane

    - by puls200
    Let's say I have a fixed number (X) of points, e.g. coordinates within a given plane (I think you can call it a 2-D point cloud). These points should be partitioned into Y polygons where Y < X. The polygons should not overlap. It would be wonderful if the polygons were konvex (like a Voronoi diagram). Imagine it like locations forming countries. For example, I have 12 points and want to create 3 polygons with 4 points each. I thought about creating a grid which covers the points. Then iterate across the points, assigning them to the closest grid cells. Maybe I miss the obvious? I am sure there are better solutions. Thanks, Daniel I just found an optimization (kmeans++) .Maybe this will yield better results..

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  • Intersection between sets containing different types of variables

    - by Gacek
    Let's assume we have two collections: List<double> values List<SomePoint> points where SomePoint is a type containing three coordinates of the point: SomePoint { double X; double Y; double Z; } Now, I would like to perform the intersection between these two collections to find out for which points in points the z coordinate is eqal to one of the elements of values I created something like that: HashSet<double> hash = new HashSet<double>(points.Select(p=>p.Z)); hash.IntersectWith(values); var result = new List<SomePoints>(); foreach(var h in hash) result.Add(points.Find(p => p.Z == h)); But it won't return these points for which there is the same Z value, but different X and Y. Is there any better way to do it?

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  • Sync Your Pidgin Profile Across Multiple PCs with Dropbox

    - by Matthew Guay
    Pidgin is definitely our favorite universal chat client, but adding all of your chat accounts to multiple computers can be frustrating.  Here’s how you can easily transfer your Pidgin settings to other computers and keep them in sync using Dropbox. Getting Started Make sure you have both Pidgin and Dropbox installed on any computers you want to sync.  To sync Pidgin, you need to: Move your Pidgin profile folder on your first computer to Dropbox Create a symbolic link from the new folder in Dropbox to your old profile location Delete the default pidgin profile on your other computer, and create a symbolic link from your Dropbox Pidgin profile to the default Pidgin profile location This sounds difficult, but it’s actually easy if you follow these steps.  Here we already had all of our accounts setup in Pidgin in Windows 7, and then synced this profile with an Ubuntu and a XP computer with fresh Pidgin installs.  Our instructions for each OS are based on this, but just swap the sync order if your main Pidgin install is in XP or Ubuntu. Please Note:  Please make sure Pidgin isn’t running on your computer while you are making the changes! Sync Your Pidgin Profile from Windows 7 Here is Pidgin with our accounts already setup.  Our Pidgin profile has a Gtalk, MSN Messenger, and Facebook Chat account, and lots of log files. Let’s move this profile to Dropbox to keep it synced.  Exit Pidgin, and then enter %appdata% in the address bar in Explorer or press Win+R and enter %appdata%.  Select the .purple folder, which is your Pidgin profiles and settings folder, and press Ctrl+X to cut it. Browse to your Dropbox folder, and press Ctrl+V to paste the .purple folder there. Now we need to create the symbolic link.  Enter  “command” in your Start menu search, right-click on the Command Prompt shortcut, and select “Run as administrator”. We can now use the mklink command to create a symbolic link to the .purple folder.  In Command Prompt, enter the following and substitute username for your own username. mklink /D “C:\Users\username\Documents\My Dropbox\.purple” “C:\Users\username\AppData\Roaming\.purple” And that’s it!  You can open Pidgin now to make sure it still works as before, with your files being synced with Dropbox. Please Note:  These instructions work the same for Windows Vista.  Also, if you are syncing settings from another computer to Windows 7, then delete the .purple folder instead of cutting and pasting it, and reverse the order of the file paths when creating the symbolic link. Add your Pidgin Profile to Ubuntu Our Ubuntu computer had a clean install of Pidgin, so we didn’t need any of the information in its settings.  If you’ve run Pidgin, even without creating an account, you will need to first remove its settings folder.  Open your home folder, and click View, and then “Show Hidden Files” to see your settings folders. Select the .purple folder, and delete it. Now, to create the symbolic link, open Terminal and enter the following, substituting username for your username: ln –s /home/username/Dropbox/.purple /home/username/ Open Pidgin, and you will see all of your accounts that were on your other computer.  No usernames or passwords needed; everything is setup and ready to go.  Even your status is synced; we had our status set to Away in Windows 7, and it automatically came up the same in Ubuntu. Please Note: If your primary Pidgin account is in Ubuntu, then cut your .purple folder and paste it into your Dropbox folder instead.  Then, when creating the symbolic link, reverse the order of the folder paths. Add your Pidgin Profile to Windows XP In XP we also had a clean install of Pidgin.  If you’ve run Pidgin, even without creating an account, you will need to first remove its settings folder.  Click Start, the Run, and enter %appdata%. Delete your .purple folder. XP does not include a way to create a symbolic link, so we will use the free Junction tool from Sysinternals.  Download Junction (link below) and unzip the folder. Open Command Prompt (click Start, select All Programs, then Accessories, and select Command Prompt), and enter cd followed by the path of the folder where you saved Junction.   Now, to create the symbolic link, enter the following in Command Prompt, substituting username with your username. junction –d “C:\Documents and Settings\username\Application Data\.purple” “C:\Documents and Settings\username\My Documents\My Dropbox\.purple” Open Pidgin, and you will see all of your settings just as they were on your other computer.  Everything’s ready to go.   Please Note: If your primary Pidgin account is in Windows XP, then cut your .purple folder and paste it into your Dropbox folder instead.  Then, when creating the symbolic link, reverse the order of the folder paths. Conclusion This is a great way to keep all of your chat and IM accounts available from all of your computers.  You can easily access logs from chats you had on your desktop from your laptop, or if you add a chat account on your work computer you can use it seamlessly from your home computer that evening.  Now Pidgin is the universal chat client that is always ready whenever and wherever you need it! Links Downlaod Pidgin Download and signup for Dropbox Download Junction for XP Similar Articles Productive Geek Tips Add "My Dropbox" to Your Windows 7 Start MenuUse Multiple Firefox Profiles at the Same TimeEasily Add Facebook Chat to PidginPut Your Pidgin Buddy List into the Windows Vista SidebarBackup and Restore Firefox Profiles Easily TouchFreeze Alternative in AutoHotkey The Icy Undertow Desktop Windows Home Server – Backup to LAN The Clear & Clean Desktop Use This Bookmarklet to Easily Get Albums Use AutoHotkey to Assign a Hotkey to a Specific Window Latest Software Reviews Tinyhacker Random Tips DVDFab 6 Revo Uninstaller Pro Registry Mechanic 9 for Windows PC Tools Internet Security Suite 2010 Download Free iPad Wallpapers at iPad Decor Get Your Delicious Bookmarks In Firefox’s Awesome Bar Manage Photos Across Different Social Sites With Dropico Test Drive Windows 7 Online Download Wallpapers From National Geographic Site Spyware Blaster v4.3

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  • Finding the shortest path between two points on a grid, using Haskell.

    - by esperantist
    This is a problem that I can easily enough solve in a non-functional manner. But solving it in Haskell is giving me big problems. Me being inexperienced when it comes to functional programming is surely a reason. The problem: I have a 2D field divided into rectangles of equal size. A simple grid. Some rectangles are empty space (and can be passed through) while others are impassable. Given a starting rectangle A and a destination rectangle B, how would I calculate the shortest path between the two? Movement is possible only vertically and horizontally, in steps a single rectangle large. How would I go about accomplishing this in Haskell? Code snippets certainly welcome, but also certainly not neccessary. And links to further resources also very welcome! Thanks!

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  • Why does division yield a vastly different result than multiplication by a fraction in floating points.

    - by Avram
    I understand why floating point numbers can't be compared, and know about the mantissa and exponent binary representation, but I'm no expert and today I came across something I don't get: Namely lets say you have something like: float denominator, numerator, resultone, resulttwo; resultone = numerator / denominator; float buff = 1 / denominator; resulttwo = numerator * buff; To my knowledge different flops can yield different results and this is not unusual. But in some edge cases these two results seem to be vastly different. To be more specific in my GLSL code calculating the Beckmann facet slope distribution for the Cook-Torrance lighitng model: float a = 1 / (facetSlopeRMS * facetSlopeRMS * pow(clampedCosHalfNormal, 4)); float b = clampedCosHalfNormal * clampedCosHalfNormal - 1.0; float c = facetSlopeRMS * facetSlopeRMS * clampedCosHalfNormal * clampedCosHalfNormal; facetSlopeDistribution = a * exp(b/c); yields very very different results to float a = (facetSlopeRMS * facetSlopeRMS * pow(clampedCosHalfNormal, 4)); facetDlopeDistribution = exp(b/c) / a; Why does it? The second form of the expression is problematic. If I say try to add the second form of the expression to a color I get blacks, even though the expression should always evaluate to a positive number. Am I getting an infinity? A NaN? if so why?

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  • OpenGL texture on sphere

    - by Cilenco
    I want to create a rolling, textured ball in OpenGL ES 1.0 for Android. With this function I can create a sphere: public Ball(GL10 gl, float radius) { ByteBuffer bb = ByteBuffer.allocateDirect(40000); bb.order(ByteOrder.nativeOrder()); sphereVertex = bb.asFloatBuffer(); points = build(); } private int build() { double dTheta = STEP * Math.PI / 180; double dPhi = dTheta; int points = 0; for(double phi = -(Math.PI/2); phi <= Math.PI/2; phi+=dPhi) { for(double theta = 0.0; theta <= (Math.PI * 2); theta+=dTheta) { sphereVertex.put((float) (raduis * Math.sin(phi) * Math.cos(theta))); sphereVertex.put((float) (raduis * Math.sin(phi) * Math.sin(theta))); sphereVertex.put((float) (raduis * Math.cos(phi))); points++; } } sphereVertex.position(0); return points; } public void draw() { texture.bind(); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glVertexPointer(3, GL10.GL_FLOAT, 0, sphereVertex); gl.glDrawArrays(GL10.GL_TRIANGLE_FAN, 0, points); gl.glDisableClientState(GL10.GL_VERTEX_ARRAY); } My problem now is that I want to use this texture for the sphere but then only a black ball is created (of course because the top right corner s black). I use this texture coordinates because I want to use the whole texture: 0|0 0|1 1|1 1|0 That's what I learned from texturing a triangle. Is that incorrect if I want to use it with a sphere? What do I have to do to use the texture correctly?

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