I used the following simple code to check the properties of elseif command in MATLAB:
x = 10;
if x < 0
y = x;
elseif 0 <= x < 2
y = 3*x;
elseif x >= 2
y = 8*x;
end
y
I would have expected the result of this to be 80 since x >= 2. But something amazing happened! The result is 30, not 80!
But why? Any ideas?
Update: when I change the code (as recommended in the answers) to
x = 10;
if x < 0
y = x;
elseif (0 <= x) && ( x < 2 )
y = 3*x;
elseif x >= 2
y = 8*x;
end
y
I get 80 as expected. It's that double condition that threw me off.
When you write
if 0 <= x < 2
you are really writing (without realizing it)
if (0 <= x) < 2
Which is evaluated as
if (true) < 2
which is true, since true evaluates to 1.
So here is what is happening, line by line:
x = 10; % x is set to 10
if x < 0 % false
y = x; % not executed
elseif 0 <= x < 2 % condition is (true) < 2, thus true
y = 3*x; % y is set to 3*x, i.e. 30
elseif x >= 2 % since we already passed a condition, this is not evaluated
y = 8*x; % skipped
end % end of composite if
y % display the value of y - it should be 30
As an aside, when you use scalars, you should really use the && operator, not the & operator. See for example What's the difference between & and && in MATLAB?
0 <= x < 2 doesn't behave as you may expect. Use (0 <= x) & (x < 2)
How does 0 <= x < 2 behave? It is evaluated from left to right. First 0 <= x gives 1 (true), and then 1 < 2 gives 1 (true). So the condition gives true, not false as you would expect.
Since the second condition in your code (0 <= x < 2) is true, you get a value 30 for y. Changing the condition to (0 <= x) & (x < 2) you get a value 80 for y.
Related
Following up on a previous question, I have code that I think should limit the number of randomly generated points in each quadrant of the total tile; however, it is not working.
n = 4;
used = [];
k = 0;
a1_count = 0;
a2_count = 0;
a3_count = 0;
a4_count = 0;
min = 1;
max = 1;
while k<n
x = rand*2;
y = rand*2;
notvalid = 0;
if (0 <= x) && (x <= 1) && (0 <= y) && (y <= 1)
a1_count = a1_count + 1;
end
if (1 < x) && (x <= 2) && (0 <= y) && (y <= 1)
a2_count = a2_count + 1;
end
if (0 <= x) && (x <= 1) && (1 < y) && (y <= 2)
a3_count = a3_count + 1;
end
if (1 < x) && (x <= 2) && (1 < y) && (y <= 2)
a4_count = a4_count + 1;
end
%%%
if (min <= a1_count) && (a1_count <= max) && (min <= a2_count) && (a2_count <= max)...
&& (min <= a3_count) && (a3_count <= max) && (min <= a4_count) && (a4_count <= max)
notvalid=1;
end
if notvalid
continue
end
used(end+1,:) = [x;y];
k = k+1;
end
I wish to generate 4 random points, and have one in each quadrant of the total area. To do this, I have a maximum and minimum number of points in each quadrant (in this case 1), and an if statement to check that the count for each tile falls within the min and max. If it doesn't, then notvalid = 0 and the loop should begin again. This function doesn't seem to work however, as the loop finishes with 4 points total and is completely random (all the counts should = 1).
Can anyone spot where I'm going wrong?
I may be missing something, but the easiest approach would probably be something like
Select N random numbers within the x/y range of the first grid cell
Repeat for all grid cells
Here is some basic code that should create N random x/y points per grid cell
% Define the grid (for demonstration purposes)
dx = 1; dy = 1;
xrange = 0:dx:2;
yrange = 0:dy:2;
% Number of points per cell
N = 1;
[lowerx, lowery] = meshgrid(xrange(1:end-1), yrange(1:end-1));
% Store all those random numbers in a cell
data = cell(size(lowerx));
for k = 1:numel(lowerx);
% Generate 4 random points within the x/y range
xcoord = (rand(N, 1) * dx) + lowerx(k);
ycoord = (rand(N, 1) * dy) + lowery(k);
data{k} = [xcoord, ycoord];
end
disp(data)
data =
[1x2 double] [1x2 double]
[1x2 double] [1x2 double]
EDIT
To address your question directly using the code that you have provided, the logic in the code in your question is a little wonky. I have rewritten your while loop to be a little clearer so we can talk through it.
while k < n
x = rand * 2;
y = rand * 2;
if y >= 0 && y < 1
if x >= 0 && x < 1
a1_count = a1_count + 1;
else
a2_count = a2_count + 1;
end
else
if x >= 0 && x < 1
a3_count = a3_count + 1;
else
a4_count = a4_count + 1;
end
end
counts = [a1_count, a2_count, a3_count, a4_count];
notValid = all(counts >= minimum) && all(counts <= maximum);
if notValid
continue;
end
used(end+1,:) = [x;y];
k = k+1;
end
So the biggest thing is your notValid check. If you actually look at what you're checking (that all your *_count variables are within the pre-specified limits), I believe that if all of those conditions are true, then the current point is valid; however you state just the opposite.
Then you basically say, that if the current x y is valid, then add it to the used list. Well this logic is fine except that you define validity backwards as I stated before.
Ok so that aside, let's look at when you think that a point is not valid. Well, then you (correctly) go to the next iteration, but you never decrement the *_count variable. So say you had 1 point in quadrant 1 already and the second iteration through the loop it's in quadrant 1 again. Well you'd add 1 to a1_count and then see that it isn't valid (a1_count exceeds max) and go to the next loop, but a1_count stays at 2 despite really only having 1 because you just rejected it.
Now, all of that aside. Let's consider the first time through your loop and look at your validity check. Since you only add one point. Your validity check can never pass (if implemented correctly) because all *_count variables except the one that was just incremented will be less than the min.
So I think what happened is you probably did the validity check correctly at first, ended up with an infite while loop, and then negated that check, didn't get an infinite loop, but as a result got an incorrect solution.
The solution that you are getting currently, is literally the first 4 times through the while loop due to the incorrect logic.
If you really like your current approach, we can clean up the code to be correct.
n = 4;
used = zeros(0,2);
minimum = 1;
maximum = 1;
counts = [0 0 0 0];
while true
x = rand * 2;
y = rand * 2;
if y >= 0 && y < 1
if x >= 0 && x < 1
quadrant = 1;
else
quadrant = 2;
end
else
if x >= 0 && x < 1
quadrant = 3;
else
quadrant = 4;
end
end
% Check to see if we can even add this point
if counts(quadrant) + 1 > maximum
continue;
end
counts(quadrant) = counts(quadrant) + 1;
used = cat(1, used, [x, y]);
isComplete = all(counts >= minimum & counts <= maximum) && ...
size(used, 1) == n;
if isComplete
break
end
end
I am trying to create a set of random points, but limit the number in each say, 1/4th, of the total area. So imagining my x & y ranged from 0 to 2, and 0 to 2, I would only get a certain number of randomly generated points within the tile (0
So far, I thought I could just create an if statement for each tile, then if the randomly generated point falls within the tile, it gets added to a count, which then if the count exceeds or is not great enough for that tile, another point will be generated and checked. This count method doesn't seem to work, instead of each of the counts going up, I get only the count for tile 4 to increase.
n = 8;
used = [];
k = 0;
a1_count = 0;
a2_count = 0;
a3_count = 0;
a4_count = 0;
while k<n
x = rand*2;
y = rand*2;
if 0 < x < 1 && 0 < y < 1
a1_count = a1_count + 1;
end
if 1 < x < 2 && 0 < y < 1
a2_count = a2_count + 1;
end
if 0 < x < 1 && 1 < y < 2
a3_count = a3_count + 1;
end
if 1 < x < 2 && 1 < y < 2
a4_count = a4_count + 1;
end
used(end+1,:) = [x;y];
k = k+1;
end
If the counts worked correctly, I would then have a min & max, and then use an if statement to check if a count is outwith the min & max, and if so use the continue statement to go on or not.
Can anyone tell me why the count is not increasing for each area? If I run this code, I get a1_count, a2_count, a3_count = 0, while a4_count = 8, even though the points lie within the a1, a2 and a3 boundaries.
An expression of the form 0 < x < 1 needs to be written as 0 < x && x < 1.
0 < x < 1 is syntactically valid so therefore does not flag up any warning or error: it is the same as (0 < x) < 1. Note that (0 < x) is itself either 0 or 1.
This explains why 1 < x < 2 && 1 < y < 2 always evaluates to 1 and therefore a4_count is always increased: 1 < x < 2 is (1 < x) < 2 which is always 1, irrespective of the value of x. Similarly for 1 < y < 2.
I'm trying to solve a piecewise function, but I am getting an error. The following is the code.
script:
syms x
y_intercept = 2;
answerr = solve(pw_f(x) == y_intercept, x);
piecewise function (in a separate file within the same folder):
function y = pw_f(x)
if x < 0
y = x;
elseif (x >=0) && (x <= 20)
y = 2*x;
elseif x > 20
y = 4*x - 40;
else
end
end
The error I'm getting after running the script is:
Conversion to logical from sym is not possible.
Error in pw_f (line 3)
if x < 0
Error in solve_test
answerr = fsolve(pw_f(x) == y_intercept, x);
I know that the error is because Matlab can't perform the comparison x < 0 because x is a symbolic variable, so it does not know what x is yet. I also tried using fsolve, and vpasolve but I'm still getting the same error. Do you know how to solve this in Matlab or get around this error?
Of course, this is an easy problem that I can do in my head (x = 1 is the solution) so Matlab should be able to do this!! However, I want to make this generic for any y-intercept (maybe some random number that is not such a nice whole number) that I choose. PLEASE HELP!!!! Thanks :)
FYI, I am using Matlab R2013a.
In file called pw_f.m
function y = pw_f(x)
if x < 0
y = x;
elseif (x >=0) && (x <= 20)
y = 2*x;
elseif x > 20
y = 4*x - 40;
else
end
end
In command window
>> y_intercept = 2; % set object value
>> x0 = 0; % initial guess
>> answerr = fzero(#(x)pw_f(x) - y_intercept, x0) % solve
answerr =
1
>> pw_f(answerr) % test solution
ans =
2
I have a function like this:
f(x) = { x if 0 < x < n
{ n-x if n < x < 2*n
How to enter this function in MATLAB?
Best way is to put this in a sub-function or nested function, or in a separate m-file:
function y = f(x)
n = 4; %// Or whatever your N is
if x <= 0 || x >= 2*n
y = 0;
elseif x < n
y = x;
else
y = n-x;
end
end
or, more generally, when x is a vector/matrix,
function y = f(x)
y = x;
y(x >= n) = n-x(x >= n);
y(x <= 0 | x >= 2*n) = 0;
end
Alternatively, you can of course pass the n as an argument:
function y = f(x, n)
...
end
Alternatively, you can use this anonymous function:
f = #(x) (x>0 & x<n).*x + (x>=n & x<=2*n).*(n-x);
again, optionally, pass the n:
f = #(x,n) ...
This question already has answers here:
Draw 3d Inequality on Matlab
(1 answer)
matlab - plot inequality in 3d with surf
(1 answer)
Closed 1 year ago.
I want to plot a 3d region in MATLAB bounded from a set of inequalities.
For example:
0 <= x <= 1
sqrt(x) <= y <= 1
0 <= z <= 1 - y
I found a 2d example that someone has done on this site but I'm not sure how to convert that to 3d. How to plot inequalities.
Edit:
From #Tobold's help I modified the code to restrict the points that are plotted to those that are defined by all three regions, but it plots only 2 or 3 points. It looks like the points in the vectors X1, Y1 and Z1 are right but for some reason its plotting only a few. Any ideas why it is only plotting a few points from the X1, Y1 and Z1 vectors instead of all of them?
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1); % Make a grid of points between 0 and 1
p1=0.1; p2=0.2; % Choose some parameters
X1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
Y1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
Z1 = (X >= 0 & X <= 1) & (Y >= sqrt(X) & Y <= 1) & (Z >= 0 & Z <= 1 - Y);
ineq1 = (X >= 0 & X <= 1) * 2;
ineq2 = (Y >= sqrt(X) & Y <= 1) * 4;
ineq3 = (Z >= 0 & Z <= 1 - Y) * 8;
all = ineq1 & ineq2 & ineq3;
colors = zeros(size(X))+ineq1+ineq2+ineq3;
scatter3(X1(:),Y1(:),Z1(:),3,colors(:)','filled')
You can do almost the same thing as in the 2d case that you linked to. Just write down your three inequalities, use a 3d meshgrid, multiply each inequality with a number from a set of three numbers that has unique subset sums (e.g. 2, 4, 8) and use scatter3:
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1); % Make a grid of points between 0 and 1
p1=0.1; p2=0.2; % Choose some parameters
ineq1 = (X >= 0 & X <= 1) * 2;
ineq2 = (X >= sqrt(X) & Y <= 1) * 4;
ineq3 = (Z >= 0 & Z <= 1 - Y) * 8;
colors = zeros(size(X))+ineq1+ineq2+ineq3;
scatter3(X(:),Y(:),Z(:),3,colors(:),'filled')
I've been trying to figure out the same thing, and the trick is to make the size of everything not in the intersection 0. Tobold's scatter3 line uses '3' as the option for size, meaning all points will show up as point 3. This can be substituted for a matrix of equal size to X1 with the set of sizes. The easiest way to do this is just make s = 3*all:
all = ineq1 & ineq2 & ineq3;
colors = zeros(size(X))+all;
sizes = 3 * all;
scatter3(X1(:),Y1(:),Z1(:),sizes,colors(:)','filled')
That should get you just the area in the intersection.
--
edit: The color variable needs to change too. You just want the intersection, not the other inequalities.
I don't understand several things in the code that you wrote as modification of #Tobold's help. For example what are the parameters p1 and p2 doing in your code?
Anyway, The code plotting only the points of your grid satisfying all of your inequalities;
[X,Y,Z]=meshgrid(0:0.1:1,0:0.1:1,0:0.1:1);
ineq1 = (X >= 0 & X <= 1);
ineq2 = (Y >= sqrt(X) & Y <= 1);
ineq3 = (Z >= 0 & Z <= 1 - Y);
all = ineq1 & ineq2 & ineq3;
scatter3(X(all),Y(all),Z(all),'b','filled')
The result is brought in the following image.