How do you do a range of numbers in livecode? I tried using "-" like this
if the loc of the image "yellowdisck.png" is 0 50,0-640 then
But this doesn't work.
If you use a loc (location), you need to check the horizontal and vertical values separately. Assuming your horizontal range is 0 to 50 and the vertical range is 0 to 640 (inclusive):
put the loc of img "yellowdisk.png" into theLoc
put item 1 of theLoc into x
put item 2 of theLoc into y
if x>=0 and x<=50 and y>=0 and y<=640 then
-- true: do my stuff here
else
-- false: do something else
end if
It looks like you're trying to determine whether an object's location (its center point) falls within a rectangular area. Try using the is within operator:
if the loc of image "yellowdisk.png" is within the rect of graphic "targetRect" then
# do true stuff
else
# do false stuff
end if
Do you want to find out if the loc of an image is within certain bounds? You would do this:
put item 1 of the loc of ing "yourImage" into x
put item 2 of the loc of ing "yourImage" into y
if x > 100 and x < 200 and y > 100 and y < 200 then...
That sort of thing.
Craig Newman
Related
I have a Matlab script that creates a Model Block for each element i found in a text file.
The problem is that all Models are created on each other in the window. So i'm trying to make a loop like:
for each element in text file
I add a Model block
I place right to the previous one
end
So it can look like this:
As you can see on the left, all models are on each other and I would like to place them like the one on the right.
I tried this:
m = mdlrefCountBlocks(diagrammeName)+500;
add_block('simulink/Ports & Subsystems/Model',[diagrammeName '/' component_NameValue]);
set_param(sprintf('%s/%s',diagrammeName,component_NameValue), 'ModelFile',component_NameValue);
size_blk = get_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position');
X = size_blk(1,1);
Y = size_blk(1,2);
Width = size_blk(1,3);
Height = size_blk(1,4);
set_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position',[X+m Y X+Width Y+Height]);
Inside the loop but it returns an error Invalid definition of rectangle. Width and height should be positive.
Thanks for helping!
The position property of a block does actually not contain its width and height, but the positions of the corners on the canvas (see Common Block Properties):
vector of coordinates, in pixels: [left top right bottom]
The origin is the upper-left corner of the Simulink Editor canvas before any canvas resizing. Supported coordinates are between -1073740824 and 1073740823, inclusive. Positive values are to the right of and down from the origin. Negative values are to the left of and up from the origin.
So change your code to e.g.:
size_blk = get_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position');
set_param(sprintf('%s/%s',diagrammeName,component_NameValue),'Position', size_blk + [m 0 0 0]);
I am facing the same problem as mentioned in this post, however, I am not facing it with OpenGL, but simply with MATLAB. Depth as distance to camera plane in GLSL
I have a depth image rendered from the Z-Buffer from 3ds max. I was not able to get an orthographic representation of the z-buffer. For a better understanding, I will use the same sketch as made by the previous post:
* |--*
/ |
/ |
C-----* C-----*
\ |
\ |
* |--*
The 3 asterisks are pixels and the C is the camera. The lines from the
asterisks are the "depth". In the first case, I get the distance from the pixel to the camera. In the second, I wish to get the distance from each pixel to the plane.
The settins of my camera are the following:
WIDTH = 512;
HEIGHT = 424;
FOV = 89.971;
aspect_ratio = WIDTH/HEIGHT;
%clipping planes
near = 500;
far = 5000;
I calulate the frustum settings like the following:
%calculate frustums settings
top = tan((FOV/2)*5000)
bottom = -top
right = top*aspect_ratio
left = -top*aspect_ratio
And set the projection matrix like this:
%Generate matrix
O_p = [2/(right-left) 0 0 -((right+left)/(right-left)); ...
0 2/(top-bottom) 0 -((top+bottom)/(top-bottom));...
0 0 -2/(far-near) -(far+near)/(far-near);...
0 0 0 1];
After this I read in the depth image, which was saved as a 48 bit RGB- image, where each channel is the same, thus only one channel has to be used.
%Read in image
img = imread('KinectImage.png');
%Throw away, except one channel (all hold the same information)
c1 = img(:,:,1);
The pixel values have to be inverted, since the closer the values are to the camera, the brigher they were. If a pixel is 0 (no object to render available) it is set to 2^16, so , that after the bit complementation, the value is still 0.
%Inverse bits that are not zero, so that the z-image has the correct values
c1(c1 == 0) = 2^16
c1_cmp = bitcmp(c1);
To apply the matrix, to each z-Buffer value, I lay out the vector one dimensional and build up a vector like this [0 0 z 1] , over every element.
c1_cmp1d = squeeze(reshape(c1_cmp,[512*424,1]));
converted = double([zeros(WIDTH*HEIGHT,1) zeros(WIDTH*HEIGHT,1) c1_cmp1d zeros(WIDTH*HEIGHT,1)]) * double(O_p);
After that, I pick out the 4th element of the result vector and reshape it to a image
img_con = converted(:,4);
img_con = reshape(img_con,[424,512]);
However, the effect, that the Z-Buffer is not orthographic is still there, so did I get sth wrong? Is my calculation flawed ? Or did I make mistake here?
Depth Image coming from 3ds max
After the computation (the white is still "0" , but the color axis has changed)
It would be great to achieve this with 3ds max, which would resolve this issue, however I was not able to find this setting for the z-buffer. Thus, I want to solve this using Matlab.
If I set up my html document as dir=rtl, all absolute positioned elements get moved to the right as expected but the translation is awkward. I need to move the elements to a negative position in order to center them correctly.
Removing the rtl solved the problem and everything retains the same Cartesian origin.
If you normally correct x to the center of the element like this:
x = x * windowWidth - elm.clientWidth / 2; // x is a value in [0..1]
Then you will probably need to do something like:
x = (x -1) * windowWidth + elm.clientWidth / 2;
If you want to position normally then don't divide by 2.
Below is one of the facebook puzzle:
I am not able to understand how to proceed for this.
You are given C containers, B black balls and an unlimited number of white balls. You want to distribute balls between the containers in a way that every container contains at least one ball and the probability of selecting a white ball is greater or equal to P percent. The selection is done by randomly picking a container followed by randomly picking a ball from it.
Find the minimal required number of white balls to achieve that.
INPUT
The first line contains 1 <= T <= 10 - the number of testcases.
Each of the following T lines contain three integers C B P separated by a single space 1<= C <= 1000; 0 <= B <= 1000; 0 <= P <= 100;
OUTPUT
For each testcase output a line containing an integer - the minimal number of white balls required. (The tests will assure that it's possible with a finite number of balls)
SAMPLE INPUT
3
1 1 60
2 1 60
10 2 50
SAMPLE OUTPUT
2
2
8
EXPLANATION
In the 1st testcase if we put 2 white balls and 1 black ball in the box the probability of selecting a white one is 66.(6)% which is greater than 60%
In the 2nd testcase putting a single white ball in one box and white+black in the other gives us 0.5 * 100% + 0.5 * 50% = 75%
For the 3rd testcase remember that we want at least one ball in each of the boxes.
You will probably have to do the following:
Initial no. of white balls Nw = 1.
Given the number of white balls, Nw, find the configuration that gives the maximum probability of picking a white ball.
Check if this probability is greater than P.
If yes, then Nw is your answer, else, increment Nw and goto 1.
Ofcourse the challenge is to find the best configuration in Step 1.
EDIT: The problem now boils down to given W white balls, B black balls, and C containers, find the configuration that gives the maximum probability of picking a white ball.
P = ( w1/(w1+b1) + w2/(w2+b2) + ... + wc/(wc+bc) ) /c.
Max(P) = Max ( w1/(w1+b1) + w2/(w2+b2) + ... + wc/(wc+bc) )
Given: summation(wi) = W, summation(bi) = B, wi + bi >= 1
I am guessing that the configuration shall be such that if there are N containers having white balls, then atleast N-1 shall have only 1 white ball and no black balls and at most 1 container shall have both white balls and black balls. Just a guess though...
I have performed rgb2gray on an image and did a sobel edge detection on the image.
then did
faceEdges = faceNoNoise(:,:) > 50; %binary threshold
so it sets the outline of the image (a picture of a face), to black and white. Values 1 is white pixel, and 0 is black pixel. Someone said I could use this,
mouthsquare = rectangle('position',[recX-mouthBoxBuffer, recY-mouthBoxBuffer, recXDiff*2+mouthBoxBuffer/2, recYDiff*2+mouthBoxBuffer/2],... % see the change in coordinates
'edgecolor','r');
numWhite = sum(sum(mouthsquare));
He said to use two sum()'s because it gets the columns and rows of the contained pixels within the rectangle. numWhite always returns 178 and some decimal numbers.
If you have a 2D matrix M (this being -- for exmple -- an image), the way to count how many elements have the value 1 is:
count_1 = sum(M(:)==1)
or
count_1 = sum(reshape(M,1,[])==1)
If the target values are not exactly 1, but have a Δ-threshold of, let's say, +/- 0.02, then one should ask for:
count_1_pm02 = sum((M(:)>=0.98) & (M(:)<=1.02))
or the equivalent using reshape.