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  • Reverse Engineer Formula

    - by aaronls
    Are there any free programs or web services for reverse engineering a formula given a set of inputs and outputs? Consider if had 3 columns of data. The first two numbers are inputs, and the last one is an output: 3,4,7 1,4,5 4,2,6 The outputs could be produced with simply a+b, but there could be many formulas that would give the same result of course. I am talking about data without any error or deviation, and I think the formula would only need basic operations(divide, multiply, add, sutbract) and possibly use one of floor/ceiling/round.

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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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  • How to use onSensorChanged sensor data in combination with OpenGL

    - by Sponge
    I have written a TestSuite to find out how to calculate the rotation angles from the data you get in SensorEventListener.onSensorChanged(). I really hope you can complete my solution to help people who will have the same problems like me. Here is the code, i think you will understand it after reading it. Feel free to change it, the main idea was to implement several methods to send the orientation angles to the opengl view or any other target which would need it. method 1 to 4 are working, they are directly sending the rotationMatrix to the OpenGl view. all other methods are not working or buggy and i hope someone knows to get them working. i think the best method would be method 5 if it would work, because it would be the easiest to understand but i'm not sure how efficient it is. the complete code isn't optimized so i recommend to not use it as it is in your project. here it is: import java.nio.ByteBuffer; import java.nio.ByteOrder; import java.nio.FloatBuffer; import javax.microedition.khronos.egl.EGL10; import javax.microedition.khronos.egl.EGLConfig; import javax.microedition.khronos.opengles.GL10; import static javax.microedition.khronos.opengles.GL10.*; import android.app.Activity; import android.content.Context; import android.content.pm.ActivityInfo; import android.hardware.Sensor; import android.hardware.SensorEvent; import android.hardware.SensorEventListener; import android.hardware.SensorManager; import android.opengl.GLSurfaceView; import android.opengl.GLSurfaceView.Renderer; import android.os.Bundle; import android.util.Log; import android.view.WindowManager; /** * This class provides a basic demonstration of how to use the * {@link android.hardware.SensorManager SensorManager} API to draw a 3D * compass. */ public class SensorToOpenGlTests extends Activity implements Renderer, SensorEventListener { private static final boolean TRY_TRANSPOSED_VERSION = false; /* * MODUS overview: * * 1 - unbufferd data directly transfaired from the rotation matrix to the * modelview matrix * * 2 - buffered version of 1 where both acceleration and magnetometer are * buffered * * 3 - buffered version of 1 where only magnetometer is buffered * * 4 - buffered version of 1 where only acceleration is buffered * * 5 - uses the orientation sensor and sets the angles how to rotate the * camera with glrotate() * * 6 - uses the rotation matrix to calculate the angles * * 7 to 12 - every possibility how the rotationMatrix could be constructed * in SensorManager.getRotationMatrix (see * http://www.songho.ca/opengl/gl_anglestoaxes.html#anglestoaxes for all * possibilities) */ private static int MODUS = 2; private GLSurfaceView openglView; private FloatBuffer vertexBuffer; private ByteBuffer indexBuffer; private FloatBuffer colorBuffer; private SensorManager mSensorManager; private float[] rotationMatrix = new float[16]; private float[] accelGData = new float[3]; private float[] bufferedAccelGData = new float[3]; private float[] magnetData = new float[3]; private float[] bufferedMagnetData = new float[3]; private float[] orientationData = new float[3]; // private float[] mI = new float[16]; private float[] resultingAngles = new float[3]; private int mCount; final static float rad2deg = (float) (180.0f / Math.PI); private boolean mirrorOnBlueAxis = false; private boolean landscape; public SensorToOpenGlTests() { } /** Called with the activity is first created. */ @Override public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); mSensorManager = (SensorManager) getSystemService(Context.SENSOR_SERVICE); openglView = new GLSurfaceView(this); openglView.setRenderer(this); setContentView(openglView); } @Override protected void onResume() { // Ideally a game should implement onResume() and onPause() // to take appropriate action when the activity looses focus super.onResume(); openglView.onResume(); if (((WindowManager) getSystemService(WINDOW_SERVICE)) .getDefaultDisplay().getOrientation() == 1) { landscape = true; } else { landscape = false; } mSensorManager.registerListener(this, mSensorManager .getDefaultSensor(Sensor.TYPE_ACCELEROMETER), SensorManager.SENSOR_DELAY_GAME); mSensorManager.registerListener(this, mSensorManager .getDefaultSensor(Sensor.TYPE_MAGNETIC_FIELD), SensorManager.SENSOR_DELAY_GAME); mSensorManager.registerListener(this, mSensorManager .getDefaultSensor(Sensor.TYPE_ORIENTATION), SensorManager.SENSOR_DELAY_GAME); } @Override protected void onPause() { // Ideally a game should implement onResume() and onPause() // to take appropriate action when the activity looses focus super.onPause(); openglView.onPause(); mSensorManager.unregisterListener(this); } public int[] getConfigSpec() { // We want a depth buffer, don't care about the // details of the color buffer. int[] configSpec = { EGL10.EGL_DEPTH_SIZE, 16, EGL10.EGL_NONE }; return configSpec; } public void onDrawFrame(GL10 gl) { // clear screen and color buffer: gl.glClear(GL10.GL_COLOR_BUFFER_BIT | GL10.GL_DEPTH_BUFFER_BIT); // set target matrix to modelview matrix: gl.glMatrixMode(GL10.GL_MODELVIEW); // init modelview matrix: gl.glLoadIdentity(); // move camera away a little bit: if ((MODUS == 1) || (MODUS == 2) || (MODUS == 3) || (MODUS == 4)) { if (landscape) { // in landscape mode first remap the rotationMatrix before using // it with glMultMatrixf: float[] result = new float[16]; SensorManager.remapCoordinateSystem(rotationMatrix, SensorManager.AXIS_Y, SensorManager.AXIS_MINUS_X, result); gl.glMultMatrixf(result, 0); } else { gl.glMultMatrixf(rotationMatrix, 0); } } else { //in all other modes do the rotation by hand: gl.glRotatef(resultingAngles[1], 1, 0, 0); gl.glRotatef(resultingAngles[2], 0, 1, 0); gl.glRotatef(resultingAngles[0], 0, 0, 1); if (mirrorOnBlueAxis) { //this is needed for mode 6 to work gl.glScalef(1, 1, -1); } } //move the axis to simulate augmented behaviour: gl.glTranslatef(0, 2, 0); // draw the 3 axis on the screen: gl.glVertexPointer(3, GL_FLOAT, 0, vertexBuffer); gl.glColorPointer(4, GL_FLOAT, 0, colorBuffer); gl.glDrawElements(GL_LINES, 6, GL_UNSIGNED_BYTE, indexBuffer); } public void onSurfaceChanged(GL10 gl, int width, int height) { gl.glViewport(0, 0, width, height); float r = (float) width / height; gl.glMatrixMode(GL10.GL_PROJECTION); gl.glLoadIdentity(); gl.glFrustumf(-r, r, -1, 1, 1, 10); } public void onSurfaceCreated(GL10 gl, EGLConfig config) { gl.glDisable(GL10.GL_DITHER); gl.glClearColor(1, 1, 1, 1); gl.glEnable(GL10.GL_CULL_FACE); gl.glShadeModel(GL10.GL_SMOOTH); gl.glEnable(GL10.GL_DEPTH_TEST); gl.glEnableClientState(GL10.GL_VERTEX_ARRAY); gl.glEnableClientState(GL10.GL_COLOR_ARRAY); // load the 3 axis and there colors: float vertices[] = { 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1 }; float colors[] = { 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1 }; byte indices[] = { 0, 1, 0, 2, 0, 3 }; ByteBuffer vbb; vbb = ByteBuffer.allocateDirect(vertices.length * 4); vbb.order(ByteOrder.nativeOrder()); vertexBuffer = vbb.asFloatBuffer(); vertexBuffer.put(vertices); vertexBuffer.position(0); vbb = ByteBuffer.allocateDirect(colors.length * 4); vbb.order(ByteOrder.nativeOrder()); colorBuffer = vbb.asFloatBuffer(); colorBuffer.put(colors); colorBuffer.position(0); indexBuffer = ByteBuffer.allocateDirect(indices.length); indexBuffer.put(indices); indexBuffer.position(0); } public void onAccuracyChanged(Sensor sensor, int accuracy) { } public void onSensorChanged(SensorEvent event) { // load the new values: loadNewSensorData(event); if (MODUS == 1) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); } if (MODUS == 2) { rootMeanSquareBuffer(bufferedAccelGData, accelGData); rootMeanSquareBuffer(bufferedMagnetData, magnetData); SensorManager.getRotationMatrix(rotationMatrix, null, bufferedAccelGData, bufferedMagnetData); } if (MODUS == 3) { rootMeanSquareBuffer(bufferedMagnetData, magnetData); SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, bufferedMagnetData); } if (MODUS == 4) { rootMeanSquareBuffer(bufferedAccelGData, accelGData); SensorManager.getRotationMatrix(rotationMatrix, null, bufferedAccelGData, magnetData); } if (MODUS == 5) { // this mode uses the sensor data recieved from the orientation // sensor resultingAngles = orientationData.clone(); if ((-90 > resultingAngles[1]) || (resultingAngles[1] > 90)) { resultingAngles[1] = orientationData[0]; resultingAngles[2] = orientationData[1]; resultingAngles[0] = orientationData[2]; } } if (MODUS == 6) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); final float[] anglesInRadians = new float[3]; SensorManager.getOrientation(rotationMatrix, anglesInRadians); if ((-90 < anglesInRadians[2] * rad2deg) && (anglesInRadians[2] * rad2deg < 90)) { // device camera is looking on the floor // this hemisphere is working fine mirrorOnBlueAxis = false; resultingAngles[0] = anglesInRadians[0] * rad2deg; resultingAngles[1] = anglesInRadians[1] * rad2deg; resultingAngles[2] = anglesInRadians[2] * -rad2deg; } else { mirrorOnBlueAxis = true; // device camera is looking in the sky // this hemisphere is mirrored at the blue axis resultingAngles[0] = (anglesInRadians[0] * rad2deg); resultingAngles[1] = (anglesInRadians[1] * rad2deg); resultingAngles[2] = (anglesInRadians[2] * rad2deg); } } if (MODUS == 7) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in x y z * order Rx*Ry*Rz */ resultingAngles[2] = (float) (Math.asin(rotationMatrix[2])); final float cosB = (float) Math.cos(resultingAngles[2]); resultingAngles[2] = resultingAngles[2] * rad2deg; resultingAngles[0] = -(float) (Math.acos(rotationMatrix[0] / cosB)) * rad2deg; resultingAngles[1] = (float) (Math.acos(rotationMatrix[10] / cosB)) * rad2deg; } if (MODUS == 8) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in z y x */ resultingAngles[2] = (float) (Math.asin(-rotationMatrix[8])); final float cosB = (float) Math.cos(resultingAngles[2]); resultingAngles[2] = resultingAngles[2] * rad2deg; resultingAngles[1] = (float) (Math.acos(rotationMatrix[9] / cosB)) * rad2deg; resultingAngles[0] = (float) (Math.asin(rotationMatrix[4] / cosB)) * rad2deg; } if (MODUS == 9) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in z x y * * note z axis looks good at this one */ resultingAngles[1] = (float) (Math.asin(rotationMatrix[9])); final float minusCosA = -(float) Math.cos(resultingAngles[1]); resultingAngles[1] = resultingAngles[1] * rad2deg; resultingAngles[2] = (float) (Math.asin(rotationMatrix[8] / minusCosA)) * rad2deg; resultingAngles[0] = (float) (Math.asin(rotationMatrix[1] / minusCosA)) * rad2deg; } if (MODUS == 10) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in y x z */ resultingAngles[1] = (float) (Math.asin(-rotationMatrix[6])); final float cosA = (float) Math.cos(resultingAngles[1]); resultingAngles[1] = resultingAngles[1] * rad2deg; resultingAngles[2] = (float) (Math.asin(rotationMatrix[2] / cosA)) * rad2deg; resultingAngles[0] = (float) (Math.acos(rotationMatrix[5] / cosA)) * rad2deg; } if (MODUS == 11) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in y z x */ resultingAngles[0] = (float) (Math.asin(rotationMatrix[4])); final float cosC = (float) Math.cos(resultingAngles[0]); resultingAngles[0] = resultingAngles[0] * rad2deg; resultingAngles[2] = (float) (Math.acos(rotationMatrix[0] / cosC)) * rad2deg; resultingAngles[1] = (float) (Math.acos(rotationMatrix[5] / cosC)) * rad2deg; } if (MODUS == 12) { SensorManager.getRotationMatrix(rotationMatrix, null, accelGData, magnetData); rotationMatrix = transpose(rotationMatrix); /* * this assumes that the rotation matrices are multiplied in x z y */ resultingAngles[0] = (float) (Math.asin(-rotationMatrix[1])); final float cosC = (float) Math.cos(resultingAngles[0]); resultingAngles[0] = resultingAngles[0] * rad2deg; resultingAngles[2] = (float) (Math.acos(rotationMatrix[0] / cosC)) * rad2deg; resultingAngles[1] = (float) (Math.acos(rotationMatrix[5] / cosC)) * rad2deg; } logOutput(); } /** * transposes the matrix because it was transposted (inverted, but here its * the same, because its a rotation matrix) to be used for opengl * * @param source * @return */ private float[] transpose(float[] source) { final float[] result = source.clone(); if (TRY_TRANSPOSED_VERSION) { result[1] = source[4]; result[2] = source[8]; result[4] = source[1]; result[6] = source[9]; result[8] = source[2]; result[9] = source[6]; } // the other values in the matrix are not relevant for rotations return result; } private void rootMeanSquareBuffer(float[] target, float[] values) { final float amplification = 200.0f; float buffer = 20.0f; target[0] += amplification; target[1] += amplification; target[2] += amplification; values[0] += amplification; values[1] += amplification; values[2] += amplification; target[0] = (float) (Math .sqrt((target[0] * target[0] * buffer + values[0] * values[0]) / (1 + buffer))); target[1] = (float) (Math .sqrt((target[1] * target[1] * buffer + values[1] * values[1]) / (1 + buffer))); target[2] = (float) (Math .sqrt((target[2] * target[2] * buffer + values[2] * values[2]) / (1 + buffer))); target[0] -= amplification; target[1] -= amplification; target[2] -= amplification; values[0] -= amplification; values[1] -= amplification; values[2] -= amplification; } private void loadNewSensorData(SensorEvent event) { final int type = event.sensor.getType(); if (type == Sensor.TYPE_ACCELEROMETER) { accelGData = event.values.clone(); } if (type == Sensor.TYPE_MAGNETIC_FIELD) { magnetData = event.values.clone(); } if (type == Sensor.TYPE_ORIENTATION) { orientationData = event.values.clone(); } } private void logOutput() { if (mCount++ > 30) { mCount = 0; Log.d("Compass", "yaw0: " + (int) (resultingAngles[0]) + " pitch1: " + (int) (resultingAngles[1]) + " roll2: " + (int) (resultingAngles[2])); } } }

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  • Move penetrating OBB out of another OBB to resolve collision

    - by Milo
    I'm working on collision resolution for my game. I just need a good way to get an object out of another object if it gets stuck. In this case a car. Here is a typical scenario. The red car is in the green object. How do I correctly get it out so the car can slide along the edge of the object as it should. I tried: if(buildings.size() > 0) { Entity e = buildings.get(0); Vector2D vel = new Vector2D(); vel.x = vehicle.getVelocity().x; vel.y = vehicle.getVelocity().y; vel.normalize(); while(vehicle.getRect().overlaps(e.getRect())) { vehicle.setCenter(vehicle.getCenterX() - vel.x * 0.1f, vehicle.getCenterY() - vel.y * 0.1f); } colided = true; } But that does not work too well. Is there some sort of vector I could calculate to use as the vector to move the car away from the object? Thanks Here is my OBB2D class: public class OBB2D { // Corners of the box, where 0 is the lower left. private Vector2D corner[] = new Vector2D[4]; private Vector2D center = new Vector2D(); private Vector2D extents = new Vector2D(); private RectF boundingRect = new RectF(); private float angle; //Two edges of the box extended away from corner[0]. private Vector2D axis[] = new Vector2D[2]; private double origin[] = new double[2]; public OBB2D(Vector2D center, float w, float h, float angle) { set(center,w,h,angle); } public OBB2D(float left, float top, float width, float height) { set(new Vector2D(left + (width / 2), top + (height / 2)),width,height,0.0f); } public void set(Vector2D center,float w, float h,float angle) { Vector2D X = new Vector2D( (float)Math.cos(angle), (float)Math.sin(angle)); Vector2D Y = new Vector2D((float)-Math.sin(angle), (float)Math.cos(angle)); X = X.multiply( w / 2); Y = Y.multiply( h / 2); corner[0] = center.subtract(X).subtract(Y); corner[1] = center.add(X).subtract(Y); corner[2] = center.add(X).add(Y); corner[3] = center.subtract(X).add(Y); computeAxes(); extents.x = w / 2; extents.y = h / 2; computeDimensions(center,angle); } private void computeDimensions(Vector2D center,float angle) { this.center.x = center.x; this.center.y = center.y; this.angle = angle; boundingRect.left = Math.min(Math.min(corner[0].x, corner[3].x), Math.min(corner[1].x, corner[2].x)); boundingRect.top = Math.min(Math.min(corner[0].y, corner[1].y),Math.min(corner[2].y, corner[3].y)); boundingRect.right = Math.max(Math.max(corner[1].x, corner[2].x), Math.max(corner[0].x, corner[3].x)); boundingRect.bottom = Math.max(Math.max(corner[2].y, corner[3].y),Math.max(corner[0].y, corner[1].y)); } public void set(RectF rect) { set(new Vector2D(rect.centerX(),rect.centerY()),rect.width(),rect.height(),0.0f); } // Returns true if other overlaps one dimension of this. private boolean overlaps1Way(OBB2D other) { for (int a = 0; a < axis.length; ++a) { double t = other.corner[0].dot(axis[a]); // Find the extent of box 2 on axis a double tMin = t; double tMax = t; for (int c = 1; c < corner.length; ++c) { t = other.corner[c].dot(axis[a]); if (t < tMin) { tMin = t; } else if (t > tMax) { tMax = t; } } // We have to subtract off the origin // See if [tMin, tMax] intersects [0, 1] if ((tMin > 1 + origin[a]) || (tMax < origin[a])) { // There was no intersection along this dimension; // the boxes cannot possibly overlap. return false; } } // There was no dimension along which there is no intersection. // Therefore the boxes overlap. return true; } //Updates the axes after the corners move. Assumes the //corners actually form a rectangle. private void computeAxes() { axis[0] = corner[1].subtract(corner[0]); axis[1] = corner[3].subtract(corner[0]); // Make the length of each axis 1/edge length so we know any // dot product must be less than 1 to fall within the edge. for (int a = 0; a < axis.length; ++a) { axis[a] = axis[a].divide((axis[a].length() * axis[a].length())); origin[a] = corner[0].dot(axis[a]); } } public void moveTo(Vector2D center) { Vector2D centroid = (corner[0].add(corner[1]).add(corner[2]).add(corner[3])).divide(4.0f); Vector2D translation = center.subtract(centroid); for (int c = 0; c < 4; ++c) { corner[c] = corner[c].add(translation); } computeAxes(); computeDimensions(center,angle); } // Returns true if the intersection of the boxes is non-empty. public boolean overlaps(OBB2D other) { if(right() < other.left()) { return false; } if(bottom() < other.top()) { return false; } if(left() > other.right()) { return false; } if(top() > other.bottom()) { return false; } if(other.getAngle() == 0.0f && getAngle() == 0.0f) { return true; } return overlaps1Way(other) && other.overlaps1Way(this); } public Vector2D getCenter() { return center; } public float getWidth() { return extents.x * 2; } public float getHeight() { return extents.y * 2; } public void setAngle(float angle) { set(center,getWidth(),getHeight(),angle); } public float getAngle() { return angle; } public void setSize(float w,float h) { set(center,w,h,angle); } public float left() { return boundingRect.left; } public float right() { return boundingRect.right; } public float bottom() { return boundingRect.bottom; } public float top() { return boundingRect.top; } public RectF getBoundingRect() { return boundingRect; } public boolean overlaps(float left, float top, float right, float bottom) { if(right() < left) { return false; } if(bottom() < top) { return false; } if(left() > right) { return false; } if(top() > bottom) { return false; } return true; } };

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  • How to calculate an angle from three points?

    - by HelloMoon
    Lets say you have this: P1 = (x=2, y=50) P2 = (x=9, y=40) P3 = (x=5, y=20) Assume that P1 is the center point of a circle. It is always the same. I want the angle that is made up by P2 and P3, or in other words the angle that is next to P1. The inner angle to be precise. It will be always a sharp angle, so less than -90 degrees. I thought: Man, that's simplest geometry maths. But I looked for a formula for like 6 hours now and people talk about most complicated NASA stuff like arcos and vector scalar product stuff. My head feels like in a fridge. Some math gurus here that think this is a simple problem? I think the programing language doesn't matter here but for those who think it does: java and objective-c. need that for both. haven't tagged it for these, though.

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  • Is there a calculator with LaTeX-syntax?

    - by Jørgen Fogh
    When I write math in LaTeX I often need to perform simple arithmetic on numbers in my LaTeX source, like 515.1544 + 454 = ???. I usually copy-paste the LaTeX code into Google to get the result, but I still have to manually change the syntax, e.g. \frac{154,7}{25} - (289 - \frac{1337}{42}) must be changed to 154,7/25 - (289 - 1337/42) It seems trivial to write a program to do this for the most commonly used operations. Is there a calculator which understand this syntax? EDIT: I know that doing this perfectly is impossible (because of the halting problem). Doing it for the simple cases I need is trivial. \frac, \cdot, \sqrt and a few other tags would do the trick. The program could just return an error for cases it does not understand.

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  • How do i break a number down into a percentage (0 - 100%)? Details inside...

    - by AJ
    I am using a JS progress bar that is set using a percentage: 0 to 100 (percent). I need the progress bar to reach 100% when 160,000 people have signed a certain form. I have the total number of signers set in a PHP variable but am lost on how to do the math to convert that into a percentage that fits within 1 - 100 (so that the progress bar actually reflects the goal of 160,000). I may be missing something obvious here (i suck at anything number-related) so does anyone here have a clue as to how to do this?

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  • Mapping A Sphere To A Cube

    - by petrocket
    There is a special way of mapping a cube to a sphere described here: http://mathproofs.blogspot.com/2005/07/mapping-cube-to-sphere.html It is not your basic "normalize the point and your done" approach and gives a much more evenly spaced mapping. I've tried to do the inverse of the mapping going from sphere coords to cube coords and have been unable to come up the working equations. It's a rather complex system of equations with lots of square roots. Any math geniuses want to take a crack at it? Here's the equations in c++ code: sx = x * sqrtf(1.0f - y * y * 0.5f - z * z * 0.5f + y * y * z * z / 3.0f); sy = y * sqrtf(1.0f - z * z * 0.5f - x * x * 0.5f + z * z * x * x / 3.0f); sz = z * sqrtf(1.0f - x * x * 0.5f - y * y * 0.5f + x * x * y * y / 3.0f); sx,sy,sz are the sphere coords and x,y,z are the cube coords.

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  • Inverse Logistic Function / Reverse Sigmoid Function

    - by Chanq
    I am currently coding up a fuzzy logic library in java. I have found the equations for all the standard functions - Grade, inverseGrade, Triangle, Trapezoid, Gaussian. However, I can't find the inverse of the sigmoid/ logistic function. The way I have written the logistic function is java is : //f(x) = 1/(1+e(-x)) public double logistic(double x){ return (1/(1+(Math.exp(-x))); } But I can't work out or find the inverse anywhere. My algebraic/calculus abilities are fairly limited, hence why I haven't been able to work out the inverse of the function. Any hints or pointers would be a big help. Thanks

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  • width of a frustum at a given distance from the near plane

    - by structolite
    I'm using CML to manage the 3D math in an OpenGL-based interface project I'm making for work. I need to know the width of the viewing frustum at a given distance from the eye point, which is kept as a part of a 4x4 matrix that represents the camera. My goal is to position gui objects along the apparent edge of the viewport, but at some distance into the screen from the near clipping plane. CML has a function to extract the planes of the frustum, giving them back in Ax + By + Cz + D = 0 form. This frustum is perpendicular to the camera, which isn't necessarily aligned with the z axis of the perspective projection. I'd like to extract x and z coordinates so as to pin graphical elements to the sides of the screen at different distances from the camera. What is the best way to go about doing it? Thanks!

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  • Union of complex polygons

    - by grenade
    Given two polygons: POLYGON((1 0, 1 8, 6 4, 1 0)) POLYGON((4 1, 3 5, 4 9, 9 5, 4 1),(4 5, 5 7, 6 7, 4 4, 4 5)) How can I calculate the union (combined polygon)? Dave's example uses SQL server to produce the union, but I need to accomplish the same in code. I'm looking for a mathematical formula or code example in any language that exposes the actual math. I am attempting to produce maps that combine countries dynamically into regions. I asked a related question here: http://stackoverflow.com/questions/2653812/grouping-geographical-shapes

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  • Make character escape from shot

    - by M28
    Hello all math masters, I got a problem for you: I have a 2D game (top down), and I would like to make the character escape from a shot, but not just walk away from the shot (I mean, don't be pushed by the shot), I want it to have a good dodging skills. The variables are: shotX - shot x position shotY - shot y position shotSpeedX - shot x speed shotSpeedY - shot x speed charX - character x position charY - character y position keyLeft - Set to true to make the character press the to left key keyRight - Set to true to make the character press the to right key keyUp - Set to true to make the character press the to up key keyDown - Set to true to make the character press the down key I can understand the following languages: C/C++ Java Actionscript 2/3 Javascript

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  • Java: confirm method Binary division and find remainder is correct?

    - by cadwag
    I am parsing binary files and have to implement a CRC algorithm to ensure the file is not corrupted. Problem is that I can't seem to get the binary math working when using larger numbers. The example I'm trying to get working: BigInteger G = new BigInteger("11001", 2); BigInteger M = new BigInteger("1110010000", 2); BigInteger R = M.remainder(G); I am expecting: R = "0101" But I am getting: R = "1100" I am assuming the remainder of 0101 is correct since it is given to me in this book I am using as a reference for the CRC algorithm (it's not based in Java), but I can't seem to get it working. I can get small binary calculations to work that I have solved by hand, but not the larger ones. I'll admit that I haven't worked the larger ones by hand yet, that is my next step, but I wanted to see if someone could point out a glaring flaw I have in my code. Can anyone confirm or deny that my methodology is correct? Thanks

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  • Inaccurate Logarithm in Python

    - by Avihu Turzion
    I work daily with Python 2.4 at my company. I used the versatile logarithm function 'log' from the standard math library, and when I entered log(2**31, 2) it returned 31.000000000000004, which struck me as a bit odd. I did the same thing with other powers of 2, and it worked perfectly. I ran 'log10(2**31) / log10(2)' and I got a round 31.0 I tried running the same original function in Python 3.0.1, assuming that it was fixed in a more advanced version. Why does this happen? Is it possible that there are some inaccuracies in mathematical functions in Python?

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  • Factorising program not working. Help required.

    - by Ender
    I am working on a factorisation problem using Fermat's Factorization and for small numbers it is working well. I've been able to calculate the factors (getting the answers from Wolfram Alpha) for small numbers, like the one on the Wikipedia page (5959). Just when I thought I had the problem licked I soon realised that my program was not working when it came to larger numbers. The program follows through the examples from the Wikipedia page, printing out the values a, b, a2 and b2; the results printed for large numbers are not correct. I've followed the pseudocode provided on the Wikipedia page, but am struggling to understand where to go next. Along with the Wikipedia page I have been following this guide. Once again, as my Math knowledge is pretty poor I cannot follow what I need to do next. The code I am using so far is as follows: import java.math.BigInteger; /** * * @author AlexT */ public class Fermat { private BigInteger a, b; private BigInteger b2; private static final BigInteger TWO = BigInteger.valueOf(2); public void fermat(BigInteger N) { // floor(sqrt(N)) BigInteger tmp = getIntSqrt(N); // a <- ceil(sqrt(N)) a = tmp.add(BigInteger.ONE); // b2 <- a*a-N b2 = (a.multiply(a)).subtract(N); final int bitLength = N.bitLength(); BigInteger root = BigInteger.ONE.shiftLeft(bitLength / 2); root = root.add(b2.divide(root)).divide(TWO); // while b2 not square root while(!(isSqrt(b2, root))) { // a <- a + 1 a = a.add(BigInteger.ONE); // b2 <- (a * a) - N b2 = (a.multiply(a)).subtract(N); root = root.add(b2.divide(root)).divide(TWO); } b = getIntSqrt(b2); BigInteger a2 = a.pow(2); // Wrong BigInteger sum = (a.subtract(b)).multiply((a.add(b))); //if(sum.compareTo(N) == 0) { System.out.println("A: " + a + "\nB: " + b); System.out.println("A^2: " + a2 + "\nB^2: " + b2); //} } /** * Is the number provided a perfect Square Root? * @param n * @param root * @return */ private static boolean isSqrt(BigInteger n, BigInteger root) { final BigInteger lowerBound = root.pow(2); final BigInteger upperBound = root.add(BigInteger.ONE).pow(2); return lowerBound.compareTo(n) <= 0 && n.compareTo(upperBound) < 0; } public BigInteger getIntSqrt(BigInteger x) { // It returns s where s^2 < x < (s+1)^2 BigInteger s; // final result BigInteger currentRes = BigInteger.valueOf(0); // init value is 0 BigInteger currentSum = BigInteger.valueOf(0); // init value is 0 BigInteger sum = BigInteger.valueOf(0); String xS = x.toString(); // change input x to a string xS int lengthOfxS = xS.length(); int currentTwoBits; int i=0; // index if(lengthOfxS % 2 != 0) {// if odd length, add a dummy bit xS = "0".concat(xS); // add 0 to the front of string xS lengthOfxS++; } while(i < lengthOfxS){ // go through xS two by two, left to right currentTwoBits = Integer.valueOf(xS.substring(i,i+2)); i += 2; // sum = currentSum*100 + currentTwoBits sum = currentSum.multiply(BigInteger.valueOf(100)); sum = sum.add(BigInteger.valueOf(currentTwoBits)); // subtraction loop do { currentSum = sum; // remember the value before subtract // in next 3 lines, we work out // currentRes = sum - 2*currentRes - 1 sum = sum.subtract(currentRes); // currentRes++ currentRes = currentRes.add(BigInteger.valueOf(1)); sum = sum.subtract(currentRes); } while(sum.compareTo(BigInteger.valueOf(0)) >= 0); // the loop stops when sum < 0 // go one step back currentRes = currentRes.subtract(BigInteger.valueOf(1)); currentRes = currentRes.multiply(BigInteger.valueOf(10)); } s = currentRes.divide(BigInteger.valueOf(10)); // go one step back return s; } /** * @param args the command line arguments */ public static void main(String[] args) { Fermat fermat = new Fermat(); //Works //fermat.fermat(new BigInteger("5959")); // Doesn't Work fermat.fermat(new BigInteger("90283")); } } If anyone can help me out with this problem I'll be eternally grateful.

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  • CODE1 at SPOJ - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code on SPOJ, and it's obviously a math problem. The full problem For those who are lazy to go and read, it's like this: a0, a1, a2, ..., an - sequence of N numbers B - a Complex Number (has both real and imaginary components) X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0, a1, ..an. I don't know how or where to start, because not even N is known, just X and B. The problem is not as easy as expressing a number in a base B, because B is a complex number. How can it be solved?

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  • 3D effect to distort paper

    - by donpal
    This may be a little hard to describe since I don't have a sample. I'm trying to find a math function or full 3d function in php or a similar language that can help me with the following effect: imagine if you were to take a flat sheet or paper and glue it on a glass of water. It wouldn't be flat any more. It would have a curve, and one of its sides might end up being slightly hidden. Anyone can refer me to a good library or resource on the web where such functions can be found?

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  • Are all the system's floating points operations the same?

    - by Jj
    We're making this web app in PHP and when working in the reports we have Excel files to compare our results to make sure our coding is doing the right operations. Now we're running into some differences due floating point arithmetics. We're doing the same divisions and multiplications and running into slightly different numbers, that add up to a notable difference. My question is if Excel is delegating it's floating point arithmetic to the CPU and PHP is also relying in the CPU for it's operations. Or does each application implements its own set of math algorithms?

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  • Explicitly multiplying values as longs

    - by Bill Szerdy
    I understand that all math is done as the largest data type required to handle the current values but when you transverse a loop how do you explicitly multiply longs? The following code returns 0, I suspect, because of an overflow. long result = 0L; List<Long> temp = (List<Long>) getListOfIntegers(); for (int i = 0; i < temp.size(); i++) { result *= temp.get(i).longValue(); } System.out.println(result);

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  • What's a better choice for SQL-backed number crunching - Ruby 1.9, Python 2, Python 3, or PHP 5.3?

    - by Ivan
    Crterias of 'better': fast im math and simple (little of fields, many records) db transactions, convenient to develop/read/extend, flexible, connectible. The task is to use a common web development scripting language to process and calculate long time series and multidimensional surfaces (mostly selectint/inserting sets of floats and dong maths with rhem). The choice is Ruby 1.9, Python 2, Python 3, PHP 5.3, Perl 5.12, JavaScript (node.js). All the data is to be stored in a relational database (due to its heavily multidimensional nature), all the communication with outer world is to be done by means of web services.

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  • Obtain Latitude and Longitude from a GeoTIFF File

    - by Mikee
    Using GDAL in Python, how do you get the latitude and longitude of a GeoTIFF file? GeoTIFF's do not appear to store any coordinate information. Instead, they store the XY Origin coordinates. However, the XY coordinates do not provide the latitude and longitude of the top left corner and bottom left corner. It appears I will need to do some math to solve this problem, but I don't have a clue on where to start. What procedure is required to have this performed? I know that the GetGeoTransform() method is important for this, however, I don't know what to do with it from there.

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  • Calculate proportional width of object (proportion: 1600x1080)

    - by Hans Stauden
    Hello, this jquery question: when I set a specific height to a "#div", i want to set the width of a inner object automatically too (cause i need a width to read it out) [ "#div" ["object"] ] example: (object).css(width: [ CALCULATION ], height: ($("#div").height())+'px' ) the original proportion of the image is: 1600x1080 here's the link to the attachment, take a look at it (tinypic): link text the heights "500px", "600px" and "700px" you can see in the attachment are just examples, the heigth could also be "711px", "623px", "998px" etc. cause the "#div" scales with the browser (i can read out the height of window, that works) my math skills aren't really good, would be great if someone could help me out :-)

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  • How do you create a formula that has diminishing returns?

    - by egervari
    I guess this is a math question and not a programming question, but what is a good way to create a formula that has diminishing returns? Here are some example points on how I want the curve to look like. f(1) = 1 f(1.5)= .98 f(2) = .95 f(2.5) = .9 f(3) = .8 f(4) = .7 f(5) = .6 f(10) = .5 f(20) = .25 Notice that as the input gets higher, the percentage decreases rapidly. Is there any way to model a function that has a very smooth and accurate curve that says this? Another way to say it is by using a real example. You know in Diablo II they have Magic Find? There are diminishing returns for magic find. If you get 100%, the real magic find is still 100%. But the more get, your actual magic find goes down. So much that say if you had 1200, your real magic find is probably 450%. So they have a function like: actualMagicFind(magicFind) = // some way to reduced magic find

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  • Fastest implementation of the frac function in C#

    - by user349937
    I would like to implement a frac function in C# (just like the one in hsl here http://msdn.microsoft.com/en-us/library/bb509603%28VS.85%29.aspx) but since it is for a very processor intensive application i would like the best version possible. I was using something like public float Frac(float value) { return value - (float)Math.Truncate(value); } but I'm having precision problems, for example for 2.6f it's returning in the unit test Expected: 0.600000024f But was: 0.599999905f I know that I can convert to decimal the value and then at the end convert to float to obtain the correct result something like this: public float Frac(float value) { return (float)((decimal)value - Decimal.Truncate((decimal)value)); } But I wonder if there is a better way without resorting to decimals...

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