Sign reverse does not work - algorithm

For some reason when I try to reverse the sign of my current velocity in MATLAB, it just won't do it.
For example, I start off with velocity_x = 3 and velocity_y = 3 (I am drawing circle collisions).
Now inside of checking conditions I need to reverse the sign and I do the following:
% This doesn't work:
velocity_x = -velocity_x;
velocity_y = -velocity_y;
These expressions don't seem to work. Even though in the variable list it still shows as -3, the ball is just twitching and not going in the opposite direction. But when I simply put numbers there, it works fine!
% This works perfectly fine:
velocity_x = -3;
velocity_y = -3;
Here's the whole loop:
velocity_x = 3;
velocity_y = 3;
% While is not commanded to exit the loop
while exit_loop == false
[b1_x_c, b1_y_c] = getCenter(b1);
xMove(b1, velocity_x);
yMove(b1, velocity_y);
if ((b1_x_c + radius + 1) >= WINDOW_WIDTH) || ((b1_y_c + radius + 1) >= WINDOW_HEIGHT)
velocity_x = -1 * velocity_x;
velocity_y = -1 * velocity_y;
elseif ((b1_x_c - radius - 1) <= 0) || ((b1_y_c - radius - 1) <= 0)
velocity_x = (-1) * velocity_x;
velocity_y = (-1) * velocity_y;
end
redraw;
end % of the while loop

When you come in region where if or elseif condition fulfills, sign could change every cycle turn - velocity value 3 -3 3 -3 and so on...
You have to use some flag to indicate that the sign has already been changed and don't change it until that region will be leaved (a kind of hysteresis)

Related

subscript indices must be either positiveintegers less than 2^31 or logicals

SOS i keep getting errors in the loop solving by finite difference method.
I either get the following error when i start with i = 2 : N :
diffusion: A(I,J): row index out of bounds; value 2 out of bound 1
error: called from
diffusion at line 37 column 10 % note line change due to edit!
or, I get the following error when i do i = 2 : N :
subscript indices must be either positive integers less than 2^31 or logicals
error: called from
diffusion at line 37 column 10 % note line change due to edit!
Please help
clear all; close all;
% mesh in space
dx = 0.1;
x = 0 : dx : 1;
% mesh in time
dt = 1 / 50;
t0 = 0;
tf = 10;
t = t0 : dt : tf;
% diffusivity
D = 0.5;
% number of nodes
N = 11;
% number of iterations
M = 10;
% initial conditions
if x <= .5 && x >= 0 % note, in octave, you don't need parentheses around the test expression
u0 = x;
elseif
u0 = 1-x;
endif
u = u0;
alpha = D * dt / (dx^2);
for j = 1 : M
for i = 1 : N
u(i, j+1) = u(i, j ) ...
+ alpha ...
* ( u(i-1, j) ...
+ u(i+1, j) ...
- 2 ...
* u(i, j) ...
) ;
end
u(N+1, j+1) = u(N+1, j) ...
+ alpha ...
* ( ...
u(N, j) ...
- 2 ...
* u(N+1, j) ...
+ u(N, j) ...
) ;
% boundary conditions
u(0, :) = u0;
u(1, :) = u1;
u1 = u0;
u0 = 0;
end
% exact solution with 14 terms
%k=14 % COMMENTED OUT
v = (4 / ((k * pi) .^ 2)) ...
* sin( (k * pi) / 2 ) ...
* sin( k * pi * x ) ...
* exp .^ (D * ((k * pi) ^ 2) * t) ;
exact = symsum( v, k, 1, 14 );
error = exact - u;
% plot stuff
plot( t, error );
xlabel( 'time' );
ylabel( 'error' );
legend( 't = 1 / 50' );
Have a look at the edited code I cleaned up for you above and study it.
Don't underestimate the importance of clean, readable code when hunting for bugs.
It will save you more time than it will cost. Especially a week from now when you will need to revisit this code and you will not remember at all what you were trying to do.
Now regarding your errors. (all line references are with respect to the cleaned up code above)
Scenario 1:
In line 29 you initialise u as a single value.
If you start your loop in line 35 starting with i = 2, then as soon as you try to do u(i, j+1), i.e. u(2,2) in the next line, octave will complain that you're trying to index the second row, in an array that so far only contains one row. (in fact, the same will apply for j at this point, since at this point you only have one column as well)
Scenario 2:
I assume the second scenario was a typo and you meant to say i = 1 : N.
If you start with i=1 in the loop, then have a look at line 38: you are trying to get element u(i-1, j), i.e. u(0,1). Therefore octave will complain that you're trying to get the zero element, but in octave arrays start from one and zero is not defined. Attempting to access any array with a zero will result in the error you see (try it in a terminal!).
UPDATE
Also, now that the code is clean, you can spot another bug, which octave helpfully warns you about if you try to run the code.
Look at line 26. There is NO condition in the elseif leg, so octave looks for the next statement as the test condition.
This means that the elseif condition will always succeed as long as the result of u0 = 1-x is non-zero.
This is clearly a bug. Either you forgot to put the condition for the elseif, or more likely, you probably just meant to say else, rather than elseif.

Diamond-Square algorithm: How do I determine what tiles I need to run a diamond function and what tiles I need to run a square function on?

I'm working on a diamond-square heightmap generator and I've been stuck on a certain part for a while now.
I'm having trouble determining which tiles I need to run a square() function on and which tiles I need to run a diamond() function on.
I took a look at this guide: http://www.playfuljs.com/realistic-terrain-in-130-lines/ and I took their for loop they're using as an example, but it didn't seem to work at all.
The preferred language for the answer is Lua (or just kindly point me in the right direction). I just need someone to tell me what I need to do to get a for loop that works for both diamond and square functions.
-- height constraints
local min_height = 10
local max_height = 100
-- the grid
local K = 4
local M = 2^K -- the field is cyclic integer grid 0 <= x,y < M (x=M is the same point as x=0)
local heights = {} -- min_height <= heights[x][y] <= max_height
for x = 0, M-1 do
heights[x] = {}
end
-- set corners height (all 4 corners are the same point because of cyclic field)
heights[0][0] = (min_height + max_height) / 2
local delta_height = (max_height - min_height) * 0.264
local side = M
local sqrt2 = 2^0.5
repeat
local dbl_side = side
side = side/2
-- squares
for x = side, M, dbl_side do
for y = side, M, dbl_side do
local sum =
heights[(x-side)%M][(y-side)%M]
+ heights[(x-side)%M][(y+side)%M]
+ heights[(x+side)%M][(y-side)%M]
+ heights[(x+side)%M][(y+side)%M]
heights[x][y] = sum/4 + (2*math.random()-1) * delta_height
end
end
delta_height = delta_height / sqrt2
-- diamonds
for x = 0, M-1, side do
for y = (x+side) % dbl_side, M-1, dbl_side do
local sum =
heights[(x-side)%M][y]
+ heights[x][(y-side)%M]
+ heights[x][(y+side)%M]
+ heights[(x+side)%M][y]
heights[x][y] = sum/4 + (2*math.random()-1) * delta_height
end
end
delta_height = delta_height / sqrt2
until side == 1
-- draw field
for x = 0, M-1 do
local s = ''
for y = 0, M-1 do
s = s..' '..tostring(math.floor(heights[x][y]))
end
print(s)
end

How to calculate the mean of 3D matrices in an image without NaN?

I need to calculate the mean of a 3D matrices (last step in the code). However, there are many NaNs in the (diff_dataframe./dataframe_vor) calculation. So when I use this code, some results will be NaN. How could I calculate the mean of this matrix by ignoring the NaNs? I attached the code as below.
S.amplitude = 1:20;%:20;
S.blocksize = [1 2 3 4 5 6 8 10 12 15 20];
S.frameWidth = 1920;
S.frameHeight = 1080;
S.quality=0:10:100;
image = 127*ones(S.frameHeight,S.frameWidth,3);
S.yuv2rgb = [1 0 1.28033; 1 -0.21482 -0.38059; 1 2.12798 0];
i_bs = 0;
for BS = S.blocksize
i_bs = i_bs + 1;
hblocks = S.frameWidth / BS;
vblocks = S.frameHeight / BS;
i_a = 0;
dataU = randi([0 1],vblocks,hblocks);
dataV = randi([0 1],vblocks,hblocks);
dataframe_yuv = zeros(S.frameHeight, S.frameWidth, 3);
for x = 1 : hblocks
for y = 1 : vblocks
dataframe_yuv((y-1)*BS+1:y*BS, ...
(x-1)*BS+1:x*BS, 2) = dataU(y,x) * 2 - 1;
dataframe_yuv((y-1)*BS+1:y*BS, ...
(x-1)*BS+1:x*BS, 3) = dataV(y,x) * 2 - 1;
end
end
dataframe_rgb(:,:,1) = S.yuv2rgb(1,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(1,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(1,3) * dataframe_yuv(:,:,3);
dataframe_rgb(:,:,2) = S.yuv2rgb(2,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(2,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(2,3) * dataframe_yuv(:,:,3);
dataframe_rgb(:,:,3) = S.yuv2rgb(3,1) * dataframe_yuv(:,:,1) + ...
S.yuv2rgb(3,2) * dataframe_yuv(:,:,2) + ...
S.yuv2rgb(3,3) * dataframe_yuv(:,:,3);
for A = S.amplitude
i_a = i_a + 1;
i_q = 0;
image1p = round(image + dataframe_rgb * A);
image1n = round(image - dataframe_rgb * A);
dataframe_vor = ((image1p-image1n)/2)/255;
for Q = S.quality
i_q = i_q + 1;
namestrp = ['greyjpegs/Img_BS' num2str(BS) '_A' num2str(A) '_Q' num2str(Q) '_1p.jpg'];
namestrn = ['greyjpegs/Img_BS' num2str(BS) '_A' num2str(A) '_Q' num2str(Q) '_1n.jpg'];
imwrite(image1p/255,namestrp,'jpg', 'Quality', Q);
imwrite(image1n/255,namestrn,'jpg', 'Quality', Q);
error_mean(i_bs, i_a, i_q) = mean2((abs(diff_dataframe./dataframe_vor)));
end
end
end
mean2 is a shortcut function that's part of the image processing toolbox that finds the entire average of a 2D region which doesn't include handling NaN. In that case, simply remove all values that are NaN and find the resulting average. Note that the removal of NaN unrolls the 2D region into a 1D vector, so we can simply use mean in this case. As an additional check, let's make sure there are no divide by 0 errors, so also check for Inf as well.
Therefore, replace this line:
error_mean(i_bs, i_a, i_q) = mean2((abs(diff_dataframe./dataframe_vor)));
... with:
tmp = abs(diff_dataframe ./ dataframe_vor);
mask = ~isnan(tmp) | ~isinf(tmp);
tmp = tmp(mask);
if isempty(tmp)
error_mean(i_bs, i_a, i_q) = 0;
else
error_mean(i_bs, i_a, i_q) = mean(tmp);
We first assign the desired operation to a temporary variable, use isnan and isinf to remove out the offending values, then find the average of the rest. One intricacy is that if your entire region is NaN or Inf, then the removal of all these entries in the region results in the empty vector, and finding the mean of this undefined. A separate check is there to be sure that if it's empty, simply assign the value of 0 instead.

Kaczmarz animation

i am asking for help.. I want to animate the Kaczmarz method on Matlab. It's method allows to find solution of system of equations by the serial projecting solution vector on hyperplanes, which which is given by the eqations of system.
And i want make animation of this vector moving (like the point is going on the projected vectors).
%% System of equations
% 2x + 3y = 4;
% x - y = 2;
% 6x + y = 15;
%%
A = [2 3;1 -1; 6 1];
f = [4; 2; 15];
resh = pinv(A)*f
x = -10:0.1:10;
e1 = (1 - 2*x)/3;
e2 = (x - 2);
e3 = 15 - 6*x;
plot(x,e1)
grid on
%
axis([0 4 -2 2])
hold on
plot(x,e2)
hold on
plot(x,e3)
hold on
precision = 0.001; % точность
iteration = 100; % количество итераций
lambda = 0.75; % лямбда
[m,n] = size(A);
x = zeros(n,1);
%count of norms
for i = 1:m
nrm(i) = norm(A(i,:));
end
for i = 1:1:iteration
j = mod(i-1,m) + 1;
if (nrm(j) <= 0), continue, end;
predx = x;
x = x + ((f(j) - A(j,:)*x)*A(j,:)')/(nrm(j))^2;
p = plot(x);
set(p)
%pause 0.04;
hold on;
if(norm(predx - x) <= precision), break, end
end
I wrote the code for this method, by don't imagine how make the animation, how I can use the set function.
In your code there are a lot of redundant and random pieces. Do not call hold on more than once, it does nothing. Also set(p) does nothing, you want to set some ps properties to something, then you use set.
Also, you are plotting the result, but not the "change". The change is a line between the previous and current, and that is the only reason you'd want to have a variable such as predx, to plot. SO USE IT!
Anyway, this following code plots your algorithm. I added a repeated line to plot in green and then delete, so you can see what the last step does. I also changed the plots in the begging to just plot in red so its more clear what is each of the things.
Change your loop for:
for i = 1:1:iteration
j = mod(i-1,m) + 1;
if (nrm(j) <= 0), continue, end;
predx = x;
x = x + ((f(j) - A(j,:)*x)*A(j,:)')/(nrm(j))^2;
plot([predx(1) x(1)],[predx(2) x(2)],'b'); %plot line
c=plot([predx(1) x(1)],[predx(2) x(2)],'g'); %plot it in green
pause(0.1)
children = get(gca, 'children'); %delete the green line
delete(children(1));
drawnow
% hold on;
if(norm(predx - x) <= precision), break, end
end
This will show:

Can someone help me vectorize / speed up this Matlab Loop?

correlation = zeros(length(s1), 1);
sizeNum = 0;
for i = 1 : length(s1) - windowSize - delta
s1Dat = s1(i : i + windowSize);
s2Dat = s2(i + delta : i + delta + windowSize);
if length(find(isnan(s1Dat))) == 0 && length(find(isnan(s2Dat))) == 0
if(var(s1Dat) ~= 0 || var(s2Dat) ~= 0)
sizeNum = sizeNum + 1;
correlation(i) = abs(corr(s1Dat, s2Dat)) ^ 2;
end
end
end
What's happening here:
Run through every values in s1. For every value, get a slice for s1
till s1 + windowSize.
Do the same for s2, only get the slice after an intermediate delta.
If there are no NaN's in any of the two slices and they aren't flat,
then get the correlaton between them and add that to the
correlation matrix.
This is not an answer, I am trying to understand what is being asked.
Take some data:
N = 1e4;
s1 = cumsum(randn(N, 1)); s2 = cumsum(randn(N, 1));
s1(randi(N, 50, 1)) = NaN; s2(randi(N, 50, 1)) = NaN;
windowSize = 200; delta = 100;
Compute correlations:
tic
corr_s = zeros(N - windowSize - delta, 1);
for i = 1:(N - windowSize - delta)
s1Dat = s1(i:(i + windowSize));
s2Dat = s2((i + delta):(i + delta + windowSize));
corr_s(i) = corr(s1Dat, s2Dat);
end
inds = isnan(corr_s);
corr_s(inds) = 0;
corr_s = corr_s .^ 2; % square of correlation coefficient??? Why?
sizeNum = sum(~inds);
toc
This is what you want to do, right? A moving window correlation function? This is a very interesting question indeed …

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