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  • Algorithm for finding symmetries of a tree

    - by Paxinum
    I have n sectors, enumerated 0 to n-1 counterclockwise. The boundaries between these sectors are infinite branches (n of them). The sectors live in the complex plane, and for n even, sector 0 and n/2 are bisected by the real axis, and the sectors are evenly spaced. These branches meet at certain points, called junctions. Each junction is adjacent to a subset of the sectors (at least 3 of them). Specifying the junctions, (in pre-fix order, lets say, starting from junction adjacent to sector 0 and 1), and the distance between the junctions, uniquely describes the tree. Now, given such a representation, how can I see if it is symmetric wrt the real axis? For example, n=6, the tree (0,1,5)(1,2,4,5)(2,3,4) have three junctions on the real line, so it is symmetric wrt the real axis. If the distances between (015) and (1245) is equal to distance from (1245) to (234), this is also symmetric wrt the imaginary axis. The tree (0,1,5)(1,2,5)(2,4,5)(2,3,4) have 4 junctions, and this is never symmetric wrt either imaginary or real axis, but it has 180 degrees rotation symmetry if the distance between the first two and the last two junctions in the representation are equal. Edit: This is actually for my research. I have posted the question at mathoverflow as well, but my days in competition programming tells me that this is more like an IOI task. Code in mathematica would be excellent, but java, python, or any other language readable by a human suffices. Here are some examples (pretend the double edges are single and we have a tree) http://www2.math.su.se/~per/files.php?file=contr_ex_1.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_2.pdf http://www2.math.su.se/~per/files.php?file=contr_ex_5.pdf Example 1 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,3). Example 2 is described as (0,1,4)(1,2,4)(2,3,4)(0,4,5) with distances (2,1,1). Example 5 is described as (0,1,4,5)(1,2,3,4) with distances (2). So, given the description/representation, I want to find some algorithm to decide if it is symmetric wrt real, imaginary, and rotation 180 degrees. The last example have 180 degree symmetry. (These symmetries corresponds to special kinds of potential in the Schroedinger equation, which has nice properties in quantum mechanics.)

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  • rotate a plane around a diagonal

    - by compie
    I would like to rotate a plane, not around a single (X or Y) axis, but around the diagonal (45 degrees between X and Y). How do I calculate the Rx and Ry given the Rdiagonal? (Rdiagonal is the amount of rotation I would like to achieve around the diagonal axis).

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  • Send/cache multidimensional array to php with jQuery

    - by robertdd
    hy, i have a little, big problem here :) after i upload some images i get a list with all the images. I have some jQuery function for rotate, duplicate, delete, shuffle images! when i select a image and hit delete i send a post to php with the alt="" value of the image,i identify the picture and edit. I want to make a save button, Instead of sending a post every time i rotate a image, better send a post after editing the list of images with an array that contains all data? my php array after upload looks like this: [files] => Array ( [lcxkijgr] => lcxkijgr.jpg [xcewxpfv] => xcewxpfv.jpg [rtiurwxf] => rtiurwxf.jpg [gsbxdsdc] => gsbxdsdc.jpg ) say that I uploaded 4 pictures, firs picture i rotate 90 degrees second i want to duplicate third i rotate 270 degrees and the fourth image i delete i can do all this only with jQuery, but on the server the images are the same, after a refresh the images are the same this is the list with the images: <div class="upimage"> <ul id="upimagesQueue"> <li id="upimagesHPVEJM"> <a href="javascript:jQuery('#upimagesHPVEJM').showlargeimage('HPVEJM')"> <img alt="lcxkijgr" src="uploads/s6id9r9icnp8q9102h8md9kfd7/lcxkijgr.jpg?1272087830477" id="HPVEJM" style="display: block;" > </a> </li> <li id="upimagesSTCSAV"> <a href="javascript:jQuery('#upimagesSTCSAV').showlargeimage('STCSAV')"> <img alt="xcewxpfv" src="uploads/s6id9r9icnp8q9102h8md9kfd7/xcewxpfv.jpg?1272087831360" id="STCSAV" style="display: block;" > </a> </li> <li id="upimagesBFPUEQ"> <a href="javascript:jQuery('#upimagesBFPUEQ').showlargeimage('BFPUEQ')"> <img alt="rtiurwxf" src="uploads/s6id9r9icnp8q9102h8md9kfd7/rtiurwxf.jpg?1272087832162" id="BFPUEQ" style="display: block;" > </a> </li> <li id="upimagesRKXNSV"> <a href="javascript:jQuery('#upimagesRKXNSV').showlargeimage('RKXNSV')"> <img alt="gsbxdsdc" src="uploads/s6id9r9icnp8q9102h8md9kfd7/gsbxdsdc.jpg?1272087832957" id="RKXNSV" style="display: block;"> </a> </li> <ul> </div> is ok if i make one array like this: array{ imgFromLi = array(img1,img2,img3,img4,img5,img6) rotate = array{img1=90, img2=270, img3=90} delete = array{img4,img5,img6} duplicate = array{img2, img3} } how i can send/cache this array?? sorry for my very bad english

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  • Are products like SQL Server and Oracle are "ORDBMS"?

    - by n10i
    According to wikipedia! http://en.wikipedia.org/wiki/ORDBMS IBM's DB2, Oracle database, and Microsoft SQL Server, make claims to support this technology and do so with varying degrees of success So, are these products true "ORDBMS" like PostgreSQL? or they are they are long way from it? can someone plz! point me to any link where i can read about the features still to be implemented by these RDBMS to become true ORDBMS!

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  • How do you rotate a two dimensional array?

    - by swilliams
    Inspired by Raymond Chen's post, say you have a 4x4 two dimensional array, write a function that rotates it 90 degrees. Raymond links to a solution in pseudo code, but I'd like to see some real world stuff. [1][2][3][4] [5][6][7][8] [9][0][1][2] [3][4][5][6] Becomes: [3][9][5][1] [4][0][6][2] [5][1][7][3] [6][2][8][4] Update: Nick's answer is the most straightforward, but is there a way to do it better than n^2? What if the matrix was 10000x10000?

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  • Great Circle & Rhumb line intersection

    - by Karl T
    I have a Latitude, Longitude, and a direction of travel in degrees true north. I would like to calculate if I will intersect a line defined by two more Lat/Lon points. I figure the two points defining the line would create my great circle and my location and azimuth would define my Rhumb line. I am only interested in intersections that will occur with a few hundred kilometers so I do not need every possible solution. I have no idea where to begin.

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  • Calculating rotation and translation matrices between two odometry positions for monocular linear triangulation

    - by user1298891
    Recently I've been trying to implement a system to identify and triangulate the 3D position of an object in a robotic system. The general outline of the process goes as follows: Identify the object using SURF matching, from a set of "training" images to the actual live feed from the camera Move/rotate the robot a certain amount Identify the object using SURF again in this new view Now I have: a set of corresponding 2D points (same object from the two different views), two odometry locations (position + orientation), and camera intrinsics (focal length, principal point, etc.) since it's been calibrated beforehand, so I should be able to create the 2 projection matrices and triangulate using a basic linear triangulation method as in Hartley & Zissermann's book Multiple View Geometry, pg. 312. Solve the AX = 0 equation for each of the corresponding 2D points, then take the average In practice, the triangulation only works when there's almost no change in rotation; if the robot even rotates a slight bit while moving (due to e.g. wheel slippage) then the estimate is way off. This also applies for simulation. Since I can only post two hyperlinks, here's a link to a page with images from the simulation (on the map, the red square is simulated robot position and orientation, and the yellow square is estimated position of the object using linear triangulation.) So you can see that the estimate is thrown way off even by a little rotation, as in Position 2 on that page (that was 15 degrees; if I rotate it any more then the estimate is completely off the map), even in a simulated environment where a perfect calibration matrix is known. In a real environment when I actually move around with the robot, it's worse. There aren't any problems with obtaining point correspondences, nor with actually solving the AX = 0 equation once I compute the A matrix, so I figure it probably has to do with how I'm setting up the two camera projection matrices, specifically how I'm calculating the translation and rotation matrices from the position/orientation info I have relative to the world frame. How I'm doing that right now is: Rotation matrix is composed by creating a 1x3 matrix [0, (change in orientation angle), 0] and then converting that to a 3x3 one using OpenCV's Rodrigues function Translation matrix is composed by rotating the two points (start angle) degrees and then subtracting the final position from the initial position, in order to get the robot's straight and lateral movement relative to its starting orientation Which results in the first projection matrix being K [I | 0] and the second being K [R | T], with R and T calculated as described above. Is there anything I'm doing really wrong here? Or could it possibly be some other problem? Any help would be greatly appreciated.

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  • Can some explain why this wont draw a circle? It is drawing roughly 3/4?

    - by Brandon Shockley
    If we want to use n small lines to outline our circle then we can just divide both the circumference and 360 degrees by n (i.e , (2*pi*r)/n and 360/n). Did I not do that? import turtle, math window = turtle.Screen() window.bgcolor('blue') body = turtle.Turtle() body.pencolor('black') body.fillcolor('white') body.speed(10) body.width(3) body.hideturtle() body.up() body.goto(0, 200) lines = 40 toprad = 40 top_circum = 2 * math.pi * toprad sol = top_circum / lines circle = 360 / lines for stops in range(lines): body.pendown() body.left(sol) body.forward(circle) window.exitonclick()

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  • XNA Track rotated pixel positions

    - by jonny adams
    Hi, Im making a game in xna where a tank has to move over a landscape. I need to be able find the bottom of the tank when it is rotated so I can make it move up and down as the player goes over the landscape. for example if i have a sprite at with its top left corner at 400,300 and i rotate it around its center by 45 degrees around its center, how do i find the new locations of the bottom track. Thanks

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  • Spinning a circle in J2ME using a Canvas.

    - by JohnQPublic
    Hello all! I have a problem where I need to make a multi-colored wheel spin using a Canvas in J2ME. What I need to do is have the user increase the speed of the spin or slow the spin of the wheel. I have it mostly worked out (I think) but can't think of a way for the wheel to spin without causing my cellphone to crash. Here is what I have so far, it's close but not exactly what I need. class MyCanvas extends Canvas{ //wedgeOne/Two/Three define where this particular section of circle begins to be drawn from int wedgeOne; int wedgeTwo; int wedgeThree; int spinSpeed; MyCanvas(){ wedgeOne = 0; wedgeTwo = 120; wedgeThree = 240; spinSpeed = 0; } //Using the paint method to public void paint(Graphics g){ //Redraw the circle with the current wedge series. g.setColor(255,0,0); g.fillArc(getWidth()/2, getHeight()/2, 100, 100, wedgeOne, 120); g.setColor(0,255,0); g.fillArc(getWidth()/2, getHeight()/2, 100, 100, wedgeTwo, 120); g.setColor(0,0,255); g.fillArc(getWidth()/2, getHeight()/2, 100, 100, wedgeThree, 120); } protected void keyPressed(int keyCode){ switch (keyCode){ //When the 6 button is pressed, the wheel spins forward 5 degrees. case KEY_NUM6: wedgeOne += 5; wedgeTwo += 5; wedgeThree += 5; repaint(); break; //When the 4 button is pressed, the wheel spins backwards 5 degrees. case KEY_NUM4: wedgeOne -= 5; wedgeTwo -= 5; wedgeThree -= 5; repaint(); } } I have tried using a redraw() method that adds the spinSpeed to each of the wedge values while(spinSpeed0) and calls the repaint() method after the addition, but it causes a crash and lockup (I assume due to an infinite loop). Does anyone have any tips or ideas how I could automate the spin so you do not have the press the button every time you want it to spin? (P.S - I have been lurking for a while, but this is my first post. If it's too general or asking for too much info (sorry if it is) and I either remove it or fix it. Thank you!)

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  • rotate image with php - any degree

    - by Luis
    How to rotate an image with PHP, any degree, without having the filled space? For instance, if I rotate an image 10 degrees, the image rotates fine but the container around (square) gets filled with black. Is it possible to get rid of this, when for example merging 2 images?

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  • round slider select

    - by mmcgrail
    I am in need of a jquery based 360 rotational slider I want the user to be able to grab the end and rotate it 360 degrees anyone know of anything like that I can't seem to find one

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  • Tool to generated rotated versions of an image

    - by John
    In sprite-based systems, it's common to fake rotation of a sprite by having many different images, each showing it rotated an extra few degrees. Is there any free tool which will take a single image, and output a single image containing several rotations? It should also ideally let us control how many images are in each row. e.g if I have a 32x32 sprite and I want it rotated at 10 degree intervals, the tool might generate a 320x32 file or a 160x64 file

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  • Send multidimensional array to php with jQuery

    - by robertdd
    hy, i have a little, big problem here :) after i upload some images i get a list with all the images. I have some jQuery function for rotate, duplicate, delete, shuffle images! when i select a image and hit delete i send a post to php with the alt="" value of the image,i identify the picture and edit. I want to make a save button, Instead of sending a post every time i rotate a image, better send a post after editing the list of images with an array that contains all data? my php array after upload looks like this: [files] => Array ( [lcxkijgr] => lcxkijgr.jpg [xcewxpfv] => xcewxpfv.jpg [rtiurwxf] => rtiurwxf.jpg [gsbxdsdc] => gsbxdsdc.jpg ) say that I uploaded 4 pictures, firs picture i rotate 90 degrees second i want to duplicate third i rotate 270 degrees and the fourth image i delete i can do all this only with jQuery, but on the server the images are the same, after a refresh the images are the same this is the list with the images: <div class="upimage"> <ul id="upimagesQueue"> <li id="upimagesHPVEJM"> <a href="javascript:jQuery('#upimagesHPVEJM').showlargeimage('HPVEJM')"> <img alt="lcxkijgr" src="uploads/s6id9r9icnp8q9102h8md9kfd7/lcxkijgr.jpg?1272087830477" id="HPVEJM" style="display: block;" > </a> </li> <li id="upimagesSTCSAV"> <a href="javascript:jQuery('#upimagesSTCSAV').showlargeimage('STCSAV')"> <img alt="xcewxpfv" src="uploads/s6id9r9icnp8q9102h8md9kfd7/xcewxpfv.jpg?1272087831360" id="STCSAV" style="display: block;" > </a> </li> <li id="upimagesBFPUEQ"> <a href="javascript:jQuery('#upimagesBFPUEQ').showlargeimage('BFPUEQ')"> <img alt="rtiurwxf" src="uploads/s6id9r9icnp8q9102h8md9kfd7/rtiurwxf.jpg?1272087832162" id="BFPUEQ" style="display: block;" > </a> </li> <li id="upimagesRKXNSV"> <a href="javascript:jQuery('#upimagesRKXNSV').showlargeimage('RKXNSV')"> <img alt="gsbxdsdc" src="uploads/s6id9r9icnp8q9102h8md9kfd7/gsbxdsdc.jpg?1272087832957" id="RKXNSV" style="display: block;"> </a> </li> <ul> </div> is ok if i make one array like this: array{ imgFromLi = array(img1,img2,img3,img4,img5,img6) rotate = array{img1=90, img2=270, img3=90} delete = array{img4,img5,img6} duplicate = array{img2, img3} } how i can make/send/cache this array?? sorry for my very bad english

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  • How to get iPhones current orientation?

    - by Mike Rychev
    Is there a special method to get iPhones orientation? I don't need it in degrees or radians, I want it to return an UIInterfaceOrientation object. I just need it for an if-else construction like if(currentOrientation==UIInterfaceOrientationPortrait ||currentOrientation==UIInterfaceOrientationPortraitUpsideDown) { //Code } if (currentOrientation==UIInterfaceOrientationLandscapeRight ||currentOrientation==UIInterfaceOrientationLandscapeLeft ) { //Code } Thanks in advance!

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  • Did "general education" classes make you a better programmer?

    - by Big Johnson
    I'm surprised by the number of general education classes computer science students must complete to get their bachelors. For example, I must take: three English classes two history classes public speaking economics biology I hardly think these general education requirements are unique to the university I attend. My question is (for those of you who have degrees), in what ways have these general education requirements improved your career as a programmer?

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  • Rotate text in XSL-FO

    - by Shekar_XSL
    Hi, I am generating the xsl-fo document for my XML content and then passing this content to one of the third party DLL that will generate the PDF. I have a requirement to display a test in 45 degrees angle. How to achive this? Thanks

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  • Rotating blocks in JavaScript.

    - by Alvin SMith
    I'd like to create a JavaScript web app that makes blocks appear on the page that can be dragged around by the user. If I used DIVs with background colors, it would be easy to rotate them by 90 degrees at a time. However, if I wanted to rotate them arbitrarily, how could I accomplish this? I'd rather not have to resort to Flash, images, or HTML5.

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  • Find Upper Right Point of Rotated Rectangle in AS3 (Flex)

    - by coltech
    I have a rectangle of any arbitrary width and height. I know X,Y, width, and height. How do I solve the upper right hand coordinates when the rectangle is rotated N degrees? I realized if it were axis aligned I would simply solve for (x,y+width). Unforunatly this doesn't hold true when I apply a transform matrix on the rectangle to rotate it around its center.

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