I have been given an exercise for my course and could use some help with it. We have to turn a pixel (row x column) into its memory address and print it. $gp is pixel 0x0 and every pixel is 32 bits. How would I go about calculating let's say pixel 0,1?
(width = 32px, height = 16px)
I've looked everywhere in our course information and can't seem to find anything to help me out.
Firstly you do y * width + x = index. Then you have to multiply the index by the size of the pixel to get the offset and finally $gp + offset.
Related
When convolution uses a kernel size of 4 and stride size of 4, meanwhile, the input size is only 10, it will be fail when trying to do third convolution operation on the boundary of input, so, should the input padded with zeros on boundary implicitly to avoid this problem? Is there any problem when I padded with other real numbers? Is it equals to increase the input size automatically?
Besides, if I expected to get a same size output feature map, usually kernel size of 3 and pad size of 1 can be used, but when kernel size is a odd number, how to decide the pad size on each side of input?
Yes, the input must be padded with zeros to overcome the small input image size problem. To compute the output feature maps at each level use the following formula:
H_out = ( H_in + 2 x Padding_Height - Kernel_Height ) / Stride_Height + 1
W_out = (W_in + 2 x Padding_Width - Kernel_Width) / Stride_Width + 1
You may keep the padding in accordance with the above formula.
I've been struggling to calculate the accelerator. I've spend a whole day in searching, trial & error but all in vain. I've one horizontal line on the stage (AS3) of let say 200 width. Center-point of that line is on 60 (if it was 100, I would have surely done it by just calculating the percentage). Now I need to know the width of given percentage. For example, total width of 60% or where will 30% (or any other percentage) start from?
What I know is the total width, and the center-point (either in percentage or in width).
Your help will be highly appreciated. In case if there is any formula, please give me details, don't just mention a/b/c as I'd never been a student of physics :(
Edit:
I don't have 10 reputations, so I can't post image directly here. Please click the following link to see the image.
Link: http://oi62.tinypic.com/11sk183.jpg
Edit:
Here is what I want exactly: I want to travel n% from any point (A/B/C/D) to its relative point (A->B/A->D ...) (Link)
http://i59.tinypic.com/2wp2lbl.jpg
If I understand correctly, you want a non-linear scale, so that pixel 1 on the line is 0%, pixel 100 on the line is 60% and pixel 200 is 100%?
If x=pixelpos/200 is the relative position on the line, one easy variation of the linear scale y=x*100% is y=(x+a*x*(1-x))*100%.
For x=0.5 the value is y=0.5+a*0.25, so for that to be 0.6=60% one needs a=0.4.
To get in the reverse direction the x for y=0.3=30%, one needs to solve a quadratic equation y=x*(1+a*(1-x)) or a*x^2-(1+a)*x+y=0. With the general solution formula, this gives
x = (1+a)/(2*a)-sqrt((1+a)^2-4*a*y)/(2*a)
= (2*y) / ( (1+a) + sqrt((1+a)^2-4*a*y) )
= (2*y) / ( (1+a) + sqrt((1-a)^2+4*a*(1-y)) )
and with a=0.4 and y=0.3
x = 0.6/( 1.4 + sqrt(1.98-0.48) )
approx 0.6/2.6=3/13=231/1001 approx 0.23
corresponding to pixel 46.
This will only work for a between -1 and 1, since for other values the slope at x=0 or x=1 will not be positive.
Another simple formula uses hyperbola instead of parabola,
y=a*x/(1+(a-1)*x)
with the inversion by
y+(a-1)*x*y = a*x <=> y = (a-(a-1)*y)*x
x = (y/a)/(1+(1/a-1)*y)
and
a = (y*(1-x))/(x*(1-y))
here there is no problem with monotonicity as long as there is no pole for x in [0,1], which is guaranteed for a>0.
How can I set the text size (inside TextField) in standart CSS/printable points? According to the manual:
fontSize - Only the numeric part of the value is used. Units (px, pt)
are not parsed; pixels and points are equivalent.
As far as I understand, 1 pixel may be equal to 1 point only in 72 PPI case. So, actionscript just operating pixels (not the real points). My trouble is to get the actual text size that I can print. Any advices or solutions are welcome.
SWF is measured in pixels, moreover, is scalable, so 1 pixel can be 1 point now, 2 points a bit later (scaleY=scaleX=2), and an undefined number another bit later (removed from stage without dereferencing). In short, for AS there are NO "real points" since it does not know a thing about printers, while it knows about displays.
I can generate a 400x400 image of the Mandelbrot set from minReal to maxReal and from minImaginary to maxImaginary. So,
makeMandel(minReal, maxReal, minImaginary, maxImaginary);
I need to modify it so that I can have,
makeMandel(centerX, centerY, Zoomlevel);
// generates a region of the mandelbrot set centered at centerX,centerY at a zoom level of Zoomlevel
(Considering zoom level represents the distance between the pixels and is given by the formula Zoom level n = 2 ^ (-n) so that zoom level 1 means pixels are 0.5 units apart, zoom level 2, 0.25 and so on...)
My question is how do I calculate the arguments of the first makeMandel function from the arguments of the second one?
I know that the first function is capable of zooming and moving around but I don't know how to calculate the correct numbers for any given center and zoom level.
I've been trying to get this working for more than three days now and I'm really confused. I tried drawing tables, etc... on paper and working it out.
I read most documents that you find on Google when searching for the mandelbrot set and a couple of past stackoverflow questions but I still don't understand. Please help me out.
You may solve it the following way. If you have the two definitions
centerX = (minReal + maxReal)/2
sizeX = maxReal - minReal
you can calculate extends on the axis via
minReal = centerX - sizeX/2
maxReal = centerX + sizeX/2
The size then is calculated using the zoomLevel:
sizeX = 2^(-zoomLevel) * baseSize
The same formulas hold for y and imaginary axis.
sizeY = 2^(-zoomLevel) * baseSize
minImaginary = centerY - sizeY/2
maxImaginary = centerY + sizeY/2
The only thing to define as a constant is your baseSize, i.e. the extend in real and imaginary axis when zoomLevel is zero. You may consider different baseSize in real and imaginary direction to cover an non-square aspect ratio of your image.
I am creating a 3D sphere gallery with ActionScript 3 and the Flash 10 3D (2.5D) APIs. I have found a method that works but is not ideal. I would like to see if there is a better method.
My algorithm goes like this:
Let n = the number of images
h = the height of each image
w = the width of each image
Approximate the radius of the circle by assuming (incorrectly) that the surface area of the images is equal to the surface area of the sphere we want to create.To calculate the radius solve for r in nwh = 4πr2. This is the part that needs to be improved.
Calculate the angle between rows. rowAngle = 2atan(h / 2 / r).
Calculate the number of rows.rows = floor(π / rowAngle).
Because step one is an approximation, the number of rows will not fit perfectly, so for presentation add padding rowAngle.rowAngle += (π - rowAngle * rows) / rows.
For each i in rows:
Calculate the radius of the circle of latitude for the row.latitudeRadius = radius * cos(π / 2 - rowAngle * i.
Calculate the angle between columns.columnAngle = atan(w / 2 / latitudeRadius) * 2.
Calculate the number of colums.columns = floor(2 * π / columnAngle)
Because step one is an approximation, the number of columns will not fit perfectly, so for presentation add padding to columnAngle.columnAngle += (2 * π - columnAngle * column) / column.
For each j in columns, translate -radius along the Z axis, rotate π / 2 + rowAngle * i around the X axis, and rotate columnAngle * j around the Y axis.
To see this in action, click here. alternate link. Notice that with the default settings, the number of items actually in the sphere are less by 13. I believe is the error introduced by my approximation in the first step.
I am not able to figure out a method for determining what the exact radius of such a sphere should be. I'm hoping to learn either a better method, the correct method, or that what I am trying to do is hard or very hard (in which case I will be happy with what I have).
I would divide this problem into two connected problems.
Given a radius, how do you pack things on to the sphere?
Given a number of things, how do you find the right radius?
If you have a solution to the first problem, the second is easy to solve. Here it is in pseudo-code.
lowerRadius = somethingTooSmall
fittedItems = itemsForRadius(lowerRadius)
while fittedItems < wantedItems:
lowerRadius *= 2
fittedItems = itemsForRadius(lowerRadius)
upperRadius = 2 * lowerRadius
while threshold < upperRadius - lowerRadius:
middleRadius = (upperRadius + lowerRadius)/2
if itemsForRadius(middleRadius) < wantedItems:
lowerRadius = middleRadius
else:
upperRadius = middleRadius
This will find the smallest radius that will pack the desired number of things with your packing algorithm. If you wish you could start with a better starting point - your current estimate is pretty close. But I don't think that an analytic formula will do it.
Now let's turn to the first problem. You have a very reasonable approach. It does have one serious bug though. The bug is that your columnAngle should not be calculated for the middle of your row. What you need to do is figure out the latitude which your items are in that is closest to the pole, and use that for the calculation. This is why when you try to fit 10 items you find a packing that causes the corners to overlap.
If you want a denser packing, you can try squishing rows towards the equator. This will result in sometimes having room for more items in a row so you'll get more things in a smaller sphere. But visually it may not look as nice. Play with it, and decide whether you like the result.
BTW I like the idea. It looks nice.
In the case of squares, it seems to be an approximate formula for knowing the relationship between the radius, the square's side and the number of squares embedded.
Following this, the number of squares is:
Floor[4 Pi/Integrate[(x^2 + y^2 + r^2)^(-3/2), {x, -a/2, a/2}, {y, -a/2, a/2}]]
or
Floor[(Pi r)/ArcCot[(2 Sqrt[2] r Sqrt[a^2+2 r^2])/a^2]]
where
r = Radius
a = Square side
If you plot for r=1, as a function of a:
Where you can see the case a=2 is the boundary for n=6, meaning a cube:
Still working to see if it can be extended to the case of a generic rectangle.
Edit
For rectangles, the corresponding formula is:
Floor[4 Pi/Integrate[(x^2 + y^2 + r^2)^(-3/2), {x, -a/2, a/2}, {y, -b/2, b/2}]]
which gives:
Floor[(2 Pi r)/(Pi-2 ArcTan[(2 r Sqrt[a^2+b^2+4 r^2])/(a b)])]
where
r = Radius
a,b = Rectangle sides
Let's suppose we want rectangles with one side half of the other (b = a/2) and a sphere of radius 1.
So, the number of rectangles as a function of a gives:
Where you may see that a rectangle with a "large" side of size 2 allows 10 rectangles in the sphere, while a rectangle of "large" side 4 allows only 4 rectangles.