Question
Trying to follow Gradients, Gradient Plots and Tangent Planes.
The gradient vectors of (X^2 + Y^2) do not show up correctly in Octave 4.2.0 on Windows. With the code, expected the gradients of a circle diverge from center outwards. However the actual is diagonal.
Please assist to understand what is wrong.
syms x y
f1 = x^2 + y^2;
gradf1 = jacobian(f1,[x,y]);
f1fun = function_handle(f1);
f1xfun = function_handle(gradf1(1));
f1yfun = function_handle(gradf1(2));
[xx, yy] = meshgrid(-1:.1:1,-1:.1:1);
hold on
contour(xx, yy, f1fun(xx, yy), 10)
quiver(xx, yy, f1xfun(xx, yy), f1yfun(xx, yy), 0.5)
axis equal tight
hold off
Expected
Actual
When you perform:
f1xfun = function_handle(gradf1(1));
f1yfun = function_handle(gradf1(2));
The output is:
f1xfun =
#(x) 2 * x % note: single-argument function
f1yfun =
#(y) 2 * y % note: single-argument function
that is AS OPPOSED TO
f1xfun =
#(x,y) 2 * x % two-argument function
f1yfun =
#(x,y) 2 * y % two-argument function
which is what you seem to think was happening. (i.e. the resulting functions actually only take a single input, not both x and y).
Therefore later on when you call f1yfun with two inputs, the second input (i.e. y) is simply silently discarded, and you are essentially calculating 2*x in both axes, hence the diagonal arrows.
tl;dr Your call to quiver should be:
quiver(xx, yy, f1xfun(xx), f1yfun(yy), 0.5);
I think you have a bug in your code and the call to quiver should be
quiver(xx, yy, f1xfun(xx), f1yfun(yy), 0.5)
which then gives (with colormap("jet"))
I have a 2-D 512x512 image and I want to display the middle section i.e. A(256-50:256+50,256-50:256+50) but cannot get to work using imshow.
You can use imcrop function
pkg load image
x = 256-50; // starting position (x cordinate)
y = 256-50; // starting position (y cordinate)
a = 100;
b = 100;
B = imcrop(A, [x y a b]);
imshow(B);
This will crop the image starting from x,y by 100 pixels
I'm new to flash and I'm trying to create a board game with actionscript 3.0
I have already created the background (checker squares) for the board and now I have to partition the background by each box. What are the ways I can achieve that? I want to logically put numbers for each square as seen in the picture.
I realized its possible to do it using lasso tool and convert each to symbols. But is there any "lazy" way of doing that? There are lots of cuts I'd have to make in order to do that.
We can use some simple calculations to map some (x, y) value to a number. Lets say:
widht = width of the image
height = height of the image
gridCount = 8
gridWidth = width / gridCount
gridHeight = height / gridCount
Now first we would like to map user click point (x, y) to some integer index i, j to the logical 8 x 8 matrix where top left is index 0, 0.
i = x / gridWidth
j = y / gridHeight
For example, if gridWidth = 60, gridHeight = 50 and user clicks on (10, 15) then i = 0, j = 0.
Now we have to map this i, j to the specified numbers. As bottom line contains 11, 21, 31, ... and every column is increasing, the final number will be:
num = (11 + i * 10) + (gridCount - j - 1)
Converting these equations to AS3 code is straight forward, so I'm not adding them.
ok so i have a character called character_mc and i want it to move towards the mouse when you press the forward arrow and strafe relative to right angles of that.
i am quite new to actionscript so could you please include and example of your code in my original code
Here is my current code:
import flash.events.MouseEvent;
//Event Listners
stage.addChild(crosshair_mc);
crosshair_mc.mouseEnabled = false;
crosshair_mc.addEventListener(Event.ENTER_FRAME, fl_CustomMouseCursor);
function fl_CustomMouseCursor(event:Event)
{
crosshair_mc.x = stage.mouseX;
crosshair_mc.y = stage.mouseY;
}
Mouse.hide();
stage.addEventListener(MouseEvent.MOUSE_MOVE,facecursor);
stage.addEventListener(KeyboardEvent.KEY_DOWN, fl_KeyboardDownHandler);
//Functions
function facecursor(event):void
{
character_mc.rotation = (180 * Math.atan2(mouseY - character_mc.y,mouseX - character_mc.x))/Math.PI + 90;
}
function fl_KeyboardDownHandler(event:KeyboardEvent):void
{
trace("Key Code Pressed: " + event.keyCode);
if (event.keyCode == 38)
{
character_mc.y = character_mc.y - 5;
}
if (event.keyCode == 40)
{
character_mc.y = character_mc.y + 5;
}
if (event.keyCode == 39)
{
character_mc.x = character_mc.x + 5;
}
if (event.keyCode == 37)
{
character_mc.x = character_mc.x - 5;
}
}
I can tell you the basic concept of how you could do this, but you'll have to apply it to your own code. To involves converting your movement code to use a vector, then modifying the vector to get a direction facing the mouse (or at right angles to that direction) and a little bit of math.
Right now you have the character moving straight along the x and y axis only in each key press case. Left/Right only move along the X and Up/Down only move along the Y.
To move towards the mouse will require the character to move both along the X and Y when the Up/Down/Left/Right keys are pressed. Clearly you can see if you move both the character's x/y positions by the same amount, say 5, then it'll move exactly at 45 degrees (though it'll actually move a step of 7.07 pixels, hopefully you can see why). You can represent this as a vector: (5,5). You can use a Point object to represent this vector:
var movementVector:Point = new Point(5, 5);
trace(movementVector.x); // gives 5
trace(movementVector.y); // also gives 5
With that in mind, you can also use a vector to represent movement straight up and down on the y axis:
// set the x to 0 and y to 5
movementVector.x = 0; // 0 would mean not to move the character along the x
movementVector.y = 5; // using -5 would move the character up
And to move along the x axis only:
movementVector.x = 5; // using -5 would move the character right
movementVector.y = 0; // 0 would mean not to move the character along the y
To do the actual movement of the character would be the same as you are doing now, except you use the vector's values:
character_mc.x = character_mc.x + movementVector.x;
character_mc.y = character_mc.y + movementVector.y;
Now to figure out the proper vector to move on a diagonal from the character's position to the mouse position is pretty simple. The x value of the vector is the x distance from the character to the mouse, and the y value of the vector is the y distance from the character to the mouse.
Let's say the character is ay 125, 100 and the mouse at 225, 150. This means the distance between the character and mouse is 100, 50 x and y. Thus you'd end up with a vector:
movementVector.x = 100;
movementVector.y = 50;
If you were to apply this vector as it is to the character's position as it is, it would arrive at the mouse instantly (and then go beyond it) as the character is moving 100 pixels along the x and 50 pixels along the y right away. The step size would be 111.8 pixels long -too big. You would need to scale it down to the character's speed. You can do this by calling the normalise() method on the Point class to scale down the vector:
trace(movementVector.x); // gives 100
trace(movementVector.y); // gives 50
// assuming '5' is the max speed of the character
movementVector.normalise(5);
trace(movementVector.x); // gives 4.47213595499958
trace(movementVector.y); // gives 2.23606797749979
This would result in a 'step' size of 5 now. Applying this would make your character move 5 pixels towards a point 100 pixels to the right and 50 pixels down from where it started.
To transform a vector exactly 90 degrees, a quick and simple way is to swap the x and y values around.
If you are curious on what normalise() method mathematically does, is that it takes the x and y values of the vector (or point) and divides it by the length to get a unit vector (or a vector with a step size of 1), then times the input you give it to scale it to the desired length.
To move your character_mc towards the mouse point you only need the direction vector between the two:
var dir:Point = new Point(mouseX - character_mc.x, mouseY - character_mc.y);
dir.Normalize();
// The following should be called when the 'up' or 'forward' arrow is pressed
// to move the character closer to mouse point
character_mc.x += dir.x; // dir can be multiplied by a 'speed' variable
character_mc.y += dir.y;
Strafing left and right around the point is a little more tricky:
// Where radius is the distance between the character and the mouse
character_mc.x = mouseX + radius * Math.cos(rad);
character_mc.y = mouseY + radius * Math.sin(rad);
You should find this tutorial useful as it does everything you describe and more:
http://active.tutsplus.com/tutorials/actionscript/circular-motion-in-as3-make-one-moving-object-orbit-another/
I want to draw a oval in html5 canvas,and i found a good method for it in stackoverflow.but I have another quesition.
function drawEllipse(ctx, x, y, w, h) {
var kappa = 0.5522848;
ox = (w / 2) * kappa, // control point offset horizontal
oy = (h / 2) * kappa, // control point offset vertical
xe = x + w, // x-end
ye = y + h, // y-end
xm = x + w / 2, // x-middle
ym = y + h / 2; // y-middle
ctx.beginPath();
ctx.moveTo(x, ym);
ctx.bezierCurveTo(x, ym - oy, xm - ox, y, xm, y);
ctx.bezierCurveTo(xm + ox, y, xe, ym - oy, xe, ym);
ctx.bezierCurveTo(xe, ym + oy, xm + ox, ye, xm, ye);
ctx.bezierCurveTo(xm - ox, ye, x, ym + oy, x, ym);
ctx.closePath();
ctx.stroke();
}
the method in the above link has use bezierCurveTo to draw a ellipse,but it has draw bezierCurveTo 4 times. but I think just 2 bezierCurveTo can draw a ellipse.like this:
but i'm weak in Mathematics,could someone tell me the relationship of "the control point" and "the oval point" please? or we must draw four bezier Curve to draw a oval?
thanks everybody
My background isn't in mathematics so if I'm wrong I'm sure someone will correct me, but from my understanding we can draw a pretty good approximation of an ellipse with just two cubic bezier curves but the coordinates will be a little tricky.
To answer your question about the relation between the oval point and the control points I think it best be answered by watching this video from the point I've selected if you're familiar with interpolation or from the beginning if you are not. Don't worry it is short.
One problem we're probably going to run into is that when we start from the top and do a bezierCurveTo the bottom of the ellipse with the corners of the rectangle (of the same width and height) as the control points, the ellipses width is going to be smaller than the rectangle. .75 times the size we want. So we can just scale the control points accordingly.
Our control point's x would be adjusted like so (assuming width is the width of the ellipse and we're dividing by two to get its offset from the origin)
var cpx = (width / .75) / 2;
Put together a visualization where you can play with the width and height and see the drawn ellipse.
The red ellipse is how we wanted it to be drawn, with the inner one how it would be drawn if we didnt reposition the control points. The lines illustrate De Casteljau's algorithm that was shown in the video.
Here's a screenshot of the visualization
You only need two cubic bezier curves to draw an ellipse. Here's a simplified version of DerekR's code that uses the original function arguments that you provided--assuming you want x and y to be the center of the ellipse:
jsFiddle
function drawEllipse(ctx, x, y, w, h) {
var width_over_2 = w / 2;
var width_two_thirds = w * 2 / 3;
var height_over_2 = h / 2;
ctx.beginPath();
ctx.moveTo(x, y - height_over_2);
ctx.bezierCurveTo(x + width_two_thirds, y - height_over_2, x + width_two_thirds, y + height_over_2, x, y + height_over_2);
ctx.bezierCurveTo(x - width_two_thirds, y + height_over_2, x - width_two_thirds, y - height_over_2, x, y -height_over_2);
ctx.closePath();
ctx.stroke();
}
Big thanks to BKH.
I used his code with two bezier curves to complete my ellipse drawing with any rotation angle. Also, I created an comparison demo between ellipses drawn by bezier curves and native ellipse() function (for now implemented only in Chrome).
function drawEllipseByBezierCurves(ctx, x, y, radiusX, radiusY, rotationAngle) {
var width_two_thirds = radiusX * 4 / 3;
var dx1 = Math.sin(rotationAngle) * radiusY;
var dy1 = Math.cos(rotationAngle) * radiusY;
var dx2 = Math.cos(rotationAngle) * width_two_thirds;
var dy2 = Math.sin(rotationAngle) * width_two_thirds;
var topCenterX = x - dx1;
var topCenterY = y + dy1;
var topRightX = topCenterX + dx2;
var topRightY = topCenterY + dy2;
var topLeftX = topCenterX - dx2;
var topLeftY = topCenterY - dy2;
var bottomCenterX = x + dx1;
var bottomCenterY = y - dy1;
var bottomRightX = bottomCenterX + dx2;
var bottomRightY = bottomCenterY + dy2;
var bottomLeftX = bottomCenterX - dx2;
var bottomLeftY = bottomCenterY - dy2;
ctx.beginPath();
ctx.moveTo(bottomCenterX, bottomCenterY);
ctx.bezierCurveTo(bottomRightX, bottomRightY, topRightX, topRightY, topCenterX, topCenterY);
ctx.bezierCurveTo(topLeftX, topLeftY, bottomLeftX, bottomLeftY, bottomCenterX, bottomCenterY);
ctx.closePath();
ctx.stroke();
}
You will find this explained slightly more math-based in http://pomax.github.io/bezierinfo/#circles_cubic, but the gist is that using a cubic bezier curve for more than a quarter turn is usually not a good idea. Thankfully, using four curves makes finding the required control points rather easy. Start off with a circle, in which case each quarter circle is (1,0)--(1,0.55228)--(0.55228,1)--(0,1) with scaled coordinates for your ellipse. Draw that four times with +/- signs swapped to effect a full circle, scale the dimensions to get your ellipse, and done.
If you use two curves, the coordinates become (1,0)--(1,4/3)--(-1,4/3)--(-1,0), scaled for your ellipse. It may still look decent enough in your application, it depends a bit on how big your drawing ends up being.
It can be mathematically proven, that circle can not be made with Bézier curve of any degree. You can make "almost circle" by approximating it.
Say you want to draw a quarter of circle around [0,0]. Cubic bézier coordinates are:
[0 , 1 ]
[0.55, 1 ]
[1 , 0.55]
[1 , 0 ]
It is a very good approximation. Transform it linearly to get an ellpise.