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  • Select all points in a matrix within 30m of another point

    - by pinnacler
    So if you look at my other posts, it's no surprise I'm building a robot that can collect data in a forest, and stick it on a map. We have algorithms that can detect tree centers and trunk diameters and can stick them on a cartesian XY plane. We're planning to use certain 'key' trees as natural landmarks for localizing the robot, using triangulation and trilateration among other methods, but programming this and keeping data straight and efficient is getting difficult using just Matlab. Is there a technique for sub-setting an array or matrix of points? Say I have 1000 trees stored over 1km (1000m), is there a way to say, select only points within 30m radius of my current location and work only with those? I would just use a GIS, but I'm doing this in Matlab and I'm unaware of any GIS plugins for Matlab. I forgot to mention, this code is going online, meaning it's going on a robot for real-time execution. I don't know if, as the map grows to several miles, using a different data structure will help or if calculating every distance to a random point is what a spatial database is going to do anyway. I'm thinking of mirroring two arrays, one sorted by X and the other by Y. Then bubble sorting to determine the 30m range in that. I do the same for both arrays, X and Y, and then have a third cross link table that will select the individual values. But I don't know, what that's called, how to program that and I'm sure someone already has so I don't want to reinvent the wheel. Cartesian Plane GIS

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  • How to determine the size of a project (lines of code, function points, other)

    - by sixtyfootersdude
    How would you evaluate project size? Part A: Before you start a project. Part B: For a complete project. I am interested in comparing unrelated projects. Here are some options: 1) Lines of code. I know that this is not a good metric of productivity but is this a reasonable measure of project size? If I wanted to estimate how long it would take to recreate a project would this be a reasonable way to do it? How many lines of code should I estimate a day? 2) Function Points. Functions points are defined as the number of: inputs outputs inquires internal files external interfaces Anyone have a veiw point on whether this is a good measure? Is there a way to **actually do this? Does anyone have another solution? Hours taken seems like it could be a useful metric but not solely. If I ask you what is a "bigger program" and give you two programs how would you approach the question? I have seen several discussions of this on stackover flow but most discuss how to measure programmer productivity. I am more interested in project size.

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  • Incorrect output on changing sequence of declarations

    - by max
    Writing C++ code to implement Sutherland-Hodgeman polygon clipping. This order of declaration of these 2 statements gives correct output, reverse does not. int numberOfVertices = 5; Point pointList[] = { {50,50}, {200,300}, {310,110}, {130,90}, {70,40} }; I am passing the polygon vertex set to clippers in order - LEFT, RIGHT, TOP, BOTTOM. The exact error which comes when the declarations are reversed is that the bottom clipper, produces an empty set of vertices so no polygon is displayed after clipping. Correct: Incorrent: Confirmed by outputting the number of vertices produced after each pass: Correct: Incorrect: What is the reason for this error? Code: #include <iostream> #include <GL/glut.h> #define MAXVERTICES 10 #define LEFT 0 #define RIGHT 1 #define TOP 2 #define BOTTOM 3 using namespace std; /* Clipping window */ struct Window { double xmin; double xmax; double ymin; double ymax; }; struct Point { double x; double y; }; /* If I interchange these two lines, the code doesn't work. */ /**************/ int numberOfVertices = 5; Point pointList[] = { {50,50}, {200,300}, {310,110}, {130,90}, {70,40} }; /**************/ const Window w = { 100, 400, 60, 200 }; /* Checks whether a point is inside or outside a window side */ int isInside(Point p, int side) { switch(side) { case LEFT: return p.x >= w.xmin; case RIGHT: return p.x <= w.xmax; case TOP: return p.y <= w.ymax; case BOTTOM: return p.y >= w.ymin; } } /* Calculates intersection of a segment and a window side */ Point intersection(Point p1, Point p2, int side) { Point temp; double slope, intercept; bool infinite; /* Find slope and intercept of segment, taking care of inf slope */ if(p2.x - p1.x != 0) { slope = (p2.y - p1.y) / (p2.x - p1.x); infinite = false; } else { infinite = true; } intercept = p1.y - p1.x * slope; /* Calculate intersections */ switch(side) { case LEFT: temp.x = w.xmin; temp.y = temp.x * slope + intercept; break; case RIGHT: temp.x = w.xmax; temp.y = temp.x * slope + intercept; break; case TOP: temp.y = w.ymax; temp.x = infinite ? p1.x : (temp.y - intercept) / slope; break; case BOTTOM: temp.y = w.ymin; temp.x = infinite ? p1.x : (temp.y - intercept) / slope; break; } return temp; } /* Clips polygon against a side, updating the point list (called once for each side) */ void clipAgainstSide(int sideToClip) { int i, j=0; Point s,p; Point outputList[MAXVERTICES]; /* Main algorithm */ s = pointList[numberOfVertices-1]; for(i=0 ; i<numberOfVertices ; i++) { p = pointList[i]; if(isInside(p, sideToClip)) { /* p inside */ if(!isInside(s, sideToClip)) { /* p inside, s outside */ outputList[j] = intersection(p, s, sideToClip); j++; } outputList[j] = p; j++; } else if(isInside(s, sideToClip)) { /* s inside, p outside */ outputList[j] = intersection(s, p, sideToClip); j++; } s = p; } /* Updating number of points and point list */ numberOfVertices = j; /* ERROR: In last call with BOTTOM argument, numberOfVertices becomes 0 */ /* all earlier 3 calls have correct output */ cout<<numberOfVertices<<endl; for(i=0 ; i<numberOfVertices ; i++) { pointList[i] = outputList[i]; } } void SutherlandHodgemanPolygonClip() { clipAgainstSide(LEFT); clipAgainstSide(RIGHT); clipAgainstSide(TOP); clipAgainstSide(BOTTOM); } void init() { glClearColor(1,1,1,0); glMatrixMode(GL_PROJECTION); gluOrtho2D(0,1000,0,500); } void display() { glClear(GL_COLOR_BUFFER_BIT); /* Displaying ORIGINAL box and polygon */ glColor3f(0,0,1); glBegin(GL_LINE_LOOP); glVertex2i(w.xmin, w.ymin); glVertex2i(w.xmin, w.ymax); glVertex2i(w.xmax, w.ymax); glVertex2i(w.xmax, w.ymin); glEnd(); glColor3f(1,0,0); glBegin(GL_LINE_LOOP); for(int i=0 ; i<numberOfVertices ; i++) { glVertex2i(pointList[i].x, pointList[i].y); } glEnd(); /* Clipping */ SutherlandHodgemanPolygonClip(); /* Displaying CLIPPED box and polygon, 500px right */ glColor3f(0,0,1); glBegin(GL_LINE_LOOP); glVertex2i(w.xmin+500, w.ymin); glVertex2i(w.xmin+500, w.ymax); glVertex2i(w.xmax+500, w.ymax); glVertex2i(w.xmax+500, w.ymin); glEnd(); glColor3f(1,0,0); glBegin(GL_LINE_LOOP); for(int i=0 ; i<numberOfVertices ; i++) { glVertex2i(pointList[i].x+500, pointList[i].y); } glEnd(); glFlush(); } int main(int argc, char** argv) { glutInit(&argc, argv); glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB); glutInitWindowSize(1000,500); glutCreateWindow("Sutherland-Hodgeman polygon clipping"); init(); glutDisplayFunc(display); glutMainLoop(); return 0; }

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  • Code Golf: Collatz Conjecture

    - by Earlz
    Inspired by http://xkcd.com/710/ here is a code golf for it. The Challenge Given a positive integer greater than 0, print out the hailstone sequence for that number. The Hailstone Sequence See Wikipedia for more detail.. If the number is even, divide it by two. If the number is odd, triple it and add one. Repeat this with the number produced until it reaches 1. (if it continues after 1, it will go in an infinite loop of 1 -> 4 -> 2 -> 1...) Sometimes code is the best way to explain, so here is some from Wikipedia function collatz(n) show n if n > 1 if n is odd call collatz(3n + 1) else call collatz(n / 2) This code works, but I am adding on an extra challenge. The program must not be vulnerable to stack overflows. So it must either use iteration or tail recursion. Also, bonus points for if it can calculate big numbers and the language does not already have it implemented. (or if you reimplement big number support using fixed-length integers) Test case Number: 21 Results: 21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1 Number: 3 Results: 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 Also, the code golf must include full user input and output.

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  • Convert rank-per-candidate format to OpenSTV BLT format

    - by kibibu
    I recently gathered, using a questionnaire, a set of opinions on the importance of various software components. Figuring that some form of Condorcet voting method would be the best way to obtain an overall rank, I opted to use OpenSTV to analyze it. My data is in tabular format, space delimited, and looks more or less like: A B C D E F G # Candidates 5 2 4 3 7 6 1 # First ballot. G is ranked first, and E is ranked 7th 4 2 6 5 1 7 3 # Second ballot etc In this format, the number indicates the rank and the sequence order indicates the candidate. Each "candidate" has a rank (required) from 1 to 7, where a 1 means most important and a 7 means least important. No duplicates are allowed. This format struck me as the most natural way to represent the output, being a direct representation of the ballot format. The OpenSTV/BLT format uses a different method of representing the same info, conceptually as follows: G B D C A F E # Again, G is ranked first and E is ranked 7th E B G A D C F # etc The actual numeric file format uses the (1-based) index of the candidate, rather than the label, and so is more like: 7 2 4 3 1 6 5 # Same ballots as before. 5 2 7 1 4 3 6 # A -> 1, G -> 7 In this format, the number indicates the candidate, and the sequence order indicates the rank. The actual, real, BLT format also includes a leading weight and a following zero to indicate the end of each ballot, which I don't care too much about for this. My question is, what is the most elegant way to convert from the first format to the (numeric) second?

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  • Designing a database for a user/points system? (in Django)

    - by AP257
    First of all, sorry if this isn't an appropriate question for StackOverflow. I've tried to make it as generalisable as possible. I want to create a database (MySQL, site running Django) that has users, who can be allocated a certain number of points for various types of action - it's a collaborative game. My requirements are to obtain: the number of points a user has the user's ranking compared to all other users and the overall leaderboard (i.e. all users ranked in order of points) This is what I have so far, in my Django models.py file: class SiteUser(models.Model): name = models.CharField(max_length=250 ) email = models.EmailField(max_length=250 ) date_added = models.DateTimeField(auto_now_add=True) def points_total(self): points_added = PointsAdded.objects.filter(user=self) points_total = 0 for point in points_added: points_total += point.points return points_total class PointsAdded(models.Model): user = models.ForeignKey('SiteUser') action = models.ForeignKey('Action') date_added = models.DateTimeField(auto_now_add=True) def points(self): points = Action.objects.filter(action=self.action) return points class Action(models.Model): points = models.IntegerField() action = models.CharField(max_length=36) However it's rapidly becoming clear to me that it's actually quite complex (in Django query terms at least) to figure out the user's ranking and return the leaderboard of users. At least, I'm finding it tough. Is there a more elegant way to do something like this? This question seems to suggest that I shouldn't even have a separate points table - what do people think? It feels more robust to have separate tables, but I don't have much experience of database design.

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  • Booting sequence. Ubuntu 12.04 installation and cohabitation with former OSes

    - by Stephane Rolland
    I am on the brink of installing Ubuntu 12.04 Precise Pengolin on the first primary partition of my hard-drive. (A day in History for me since I had always kept a MS windows at this first place). But I have some fears: This is my last computer available (In the past I used to have 2 or even 3 machines so I could always un/plug HDs for recovery operations and rescue) The current booting sequence is not straight forard. So as to explain the boot sequence let me briefly sum-up the history of this laptop computer. It was a dedicated Windows Vista computer. 1st and only Primary partition. Then I added Windows 7 (on the 2nd primary partition) letting the Windows Vista Boot Loader manage the boot sequence. Then I added Ubuntu 10.04 Lucid Lynx on the 1st sub-partition of the Extended Partitionm asking Grub to be the boot loader. But when I ask Grub to launch windows it launches the Vista BootLoader that manages the choice betzeen Vista and 7. So in theory Grub is on the MasterBootRecord - though I understand where the Vista BootLoader remains. Now, I will no longer use the Ubuntu 10.04 ( on extended partition) and also the Windows vista (on the first primary partition). I will install Ubuntu 12.04 on the First Primary, asking it to install a new bootloader. I want to keep the Windows 7 that is already on the Second Primary partition. And I want it to be loaded by the Ubuntu Boot loader(I don4t knoz zhich is included in this version)... And I am afraid the last point will not work.

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  • Add kilometers to a map point

    - by proveyourselfthom
    Good morning. I would like to know how do I add kilometers to a map point (latitude / longitude). For example: The city Jaraguá do Sul is in latitude -26.462049, longitude -49.059448. I want to add 100 kilometers up, down, and on the sides. I want to do a square and get the new points. How do I do that? I tried it: <?php $distance = 100; $earthRadius = 6371; $lat1 = -26.4853239150483; $lon1 = -49.075927734375; $bearing = 0; $lat2 = asin(sin($lat1) * cos($distance / $earthRadius) + cos($lat1) * sin($distance / $earthRadius) * cos($bearing)); $lon2 = $lon1 + atan2(sin($bearing) * sin($distance / $earthRadius) * cos($lat1), cos($distance / $earthRadius) - sin($lat1) * sin($lat2)); echo 'LAT: ' . $lat2 . '<br >'; echo 'LNG: ' . $lon2; ?> But it's returning wrong cordinates. Thank you! Thank you very much.

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  • Find distance between two points using MKMapKit

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

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  • How do I increase the number of evaluation points in geom_smooth for ggplot2 in R

    - by Halpo
    I'm creating a plot and adding a basic loess smooth line to it. qplot(Age.GTS2004., X.d18O,data=deepsea, geom=c('point')) + geom_smooth(method="loess",se=T,span=0.01, alpha=.5, fill='light blue',color='navy') The problem is that the line is coming out really choppy. I need more evaluation point for the curve in certain areas. Is there a way to increase the number of evaluation points without having to reconstruct geom_smooth?

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  • Value object or not for 3d points ?

    - by Stefano Borini
    I need to develop a geometry library in python, describing points, lines and planes in 3d space, and various geometry operations. Related to my previous question. The main issue in the design is if these entities should have identity or not. I was wondering if there's a similar library out there (developed in another language) to take inspiration from, what is the chosen design, and in particular the reason for one choice vs. the other.

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  • CDN with South America peering points / edge nodes

    - by Bill
    Hello, Does anyone one know of a CDN (Content Delivery Network) with true South American peering points or edge nodes? This seems to be quite rare. It seems that most CDNs serve Central and South America from Texas. However, our application requires low latency in Brazil, so this is not a good solution for us. We would like to avoid having to set up servers in South America just for this piece of the application, but may end up doing that. Thanks, Bill

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  • How to draw a line between points in google map in Android

    - by wafa.cs1
    Hello I already wrote a program that read locations from android GPS ; each locatin(long , lat) will be sent to remote server to save it and display it in a website map. what I'm trying to do now is to display my path in android by drawing line between the points I didn't find any sufficient answer until this moment! so how this can be done?

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  • C# Drawing Arc with 3 Points

    - by Keeper
    Hi, I need to draw an arc using GraphicsPath and having initial, median and final points. The arc has to pass on them. I tried .DrawCurve and .DrawBezier but the result isn't exactly an arc. What can I do?

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  • Bug in the reputation points [closed]

    - by harigm
    I have answered the question and I have been voted "2", but the reputation points has not been awarded. http://stackoverflow.com/questions/2711281/naming-convention-for-number-of/2711409#2711409 This is to bring to your notice, if any bug on this. Please check and clarify me about it

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  • How to get points that intersect the treadline?

    - by chutsu
    Basically I did the Cavendish experiment, and I have a damped sinusoidal wave plotted on Excel. With Position (mm) against Time (s). My problem is that I have added a tread line through the wave function, and wish to calculate the points of which the wave function intersects the tread line. From this I will then be able to calculate the time period. At the moment I'm just having difficulty getting the intersects.. Thanks

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  • ReSharper: find derived types constructor usages points

    - by Roman
    I have some base class ControlBase and many derived classes which also have derived classes... ControlBase and derived classes have parameterless constructor. How can I easily find all derived classes constructor invocation points? ReSharper find usages on ControlBase constructor shows only usages of this base class constructor but not derived classes constructors. Thanks.

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