if and else loop problem

1 次查看（过去 30 天）

CHANDRABHAN Singh 2021-12-26

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1617685-if-and-else-loop-problem

评论： CHANDRABHAN Singh 2022-1-5

if (loop==false && centre(ii,1)<95 )
    centre(ii,:) = centre(ii,:)+ [0.1 0];  
    
elseif (loop==false && centre(ii,2)<95)
    centre(ii,:) = centre(ii,:)+ [0 0.1];
    
elseif (loop==false && centre(ii,1)>5)
    centre(ii,:) = centre(ii,:)- [0.1 0];
elseif (loop==false && centre(ii,2)>5)
    centre(ii,:) = centre(ii,:)- [0 0.1];
else
    break
end

Lets's say centre(ii,1)=94.95, then the first if increases it to 95.05.

As soon as centre(ii,1)=95.05, the third condition makes it again 94.95 and the process continues. I want, once the first condition is achieved i.e. x>95, it should go to third condition. The similar case is for 2nd and 4th condition. How to do this?

19 个评论
显示 17更早的评论隐藏 17更早的评论

Voss 2021-12-29

在 MATLAB Online 中打开

Maybe try this:

% %% This is for generating the circular aggregate %%

xmin = 5; xmax = 45;

%ymin = 5; ymax = 95;

%the function "randnum" for generating random number is saved in the directory

n = 20;

centre = zeros(n,2);

r = zeros(n,1);

centre(1,:) = xmin + rand(1,2).*(xmax-xmin);

r(1) = (4.75 + rand(1)*(10-4.75))/2;

t = 2*pi.*linspace(0,1,50);

figure

hold on

plot(r(1).*sin(t)+centre(1,1),r(1).*cos(t)+centre(1,2));

no_of_agg=1;

% %% Condition that no circle should overlap %

%This loop has to start with ii= 2 because the r(1) has been obtained in the

%previous section

% this is defined to be used in the conditional statement if else

for ii =2:n

jj = 1;

% eta1 = randnum;

% eta2 = randnum;

eta1 = rand;

eta2 = rand;

centre(ii,:) = xmin + ([eta1 eta2]).*(xmax-xmin);

% cx and cy are defined to control the if else loop at the end for

% shifting of the centre of the aggregate

cx = centre(ii,1);

cy = centre(ii,2);

loop = 0;

% r(ii) = (4.75 + randnum*(10-4.75))/2;

r(ii) = (4.75 + rand*(10-4.75))/2;

% the p matrix is a condition employed to chech the two circles do not

% intersect

while (loop==0)

p=zeros(ii,ii);

for m = 1:ii

p(m,m)=1;

for n = (m+1):ii

p(n,n)=1;

p(m,n)=(1.1*(r(m)+r(n))< sqrt((centre(m,1)-centre(n,1))^2+ (centre(m,2)-centre(n,2))^2 ) );

% p(n,m)=(1.1*(r(m)+r(n))< sqrt((centre(m,1)-centre(n,1))^2+ (centre(m,2)-centre(n,2))^2 ) );

p(n,m) = p(m,n);

end

% while loop parameter is loop and is checked where loop is

% equal to 1 or not

loop = all(p(:));

if (loop==false && centre(ii,1)<45 && jj==1)

% disp('loop1')

centre(ii,:) = centre(ii,:)+ [0.1 0];

if (centre(ii,1)>45)

jj=2;

end

if (loop==false && centre(ii,2)<45 && jj==2)

centre(ii,1)=5;

% disp('loop2')

centre(ii,:) = centre(ii,:)+ [0 0.1];

jj=1;

if centre(ii,2)>45

centre(ii,2)=cy;

centre(ii,1) =cx;

jj=3;

end

if (loop==false && centre(ii,1)>5 && jj==3 )

% disp('loop3')

centre(ii,:) = centre(ii,:)-[0.1, 0];

if (centre(ii,1)<5)

jj=4;

end

if (loop==false && centre(ii,2)>5 && jj==4 )

centre(ii,1)=45;

% disp('loop4')

centre(ii,:) = centre(ii,:)-[0, 0.1];

jj=3;

if (centre(ii,2)<5)

jj=5;

end

if (loop==false && jj==5)

r(ii)=r(ii)-0.1;

centre(ii,:)=[cx, cy];

jj=1;

% disp('radius')

if (r(ii)<(4.5/2))

jj=6;

end

if (jj==6)

break

end

if (loop==true)

no_of_agg=no_of_agg+1;

plot(r(ii).*sin(t)+centre(ii,1),r(ii).*cos(t)+centre(ii,2));

end

Walter Roberson 2021-12-30

"Why did you use 'figure' command before 'plot' command?"

When you figure() in that place, you can be sure that you have an empty figure, and you can be sure that the "hold on" that follows right after the figure() call will not be accidentally holding anything that might have previously be in the display. Another way to achieve the same thing would have been to call cla() instead of figure()

"Is loop = all(p(:)) better then all(all(p))? If yes, how?"

Although technically speaking p(:) involves a call, in the special case that p is numeric and not complex, the (:) operation is the fastest operation in MATLAB (other than constants and assignment of a complete variable to another complete variable.) The (:) operation only requires bashing the shape information but keeping the data pointer; no data needs to be copied at all for it. Then the all() call is working on a vector of all of the elements and can process everything at the same time to come up with a single answer.

The alternative, all(all(p)) requires two calls to all(), with the inner call having to produce one result per column. Although part of that can potentially be run on other cores, producing and saving the partial results is more work than just producing one result. Especially since, in theory, all() could abort processing as soon as it hit a single value that is 0.

Walter Roberson 2021-12-30

switch() would probably be better.

But I suspect that you can be more efficient.

Consider the way that you build p. Suppose that on round 4 you create p(2,4) . And suppose that loop is false that time, so you loop back and re-create p . Now, under what circumstances can the just-recalculated p(2,4) be different than the p(2,4) that you calculated the previous time? Okay, now suppose that you get through and loop becomes true, and you make it through to the next ii value, and you go to rebuild p again. You get to the point where you are going to rebuild p(2,4) . Under what circumstances can the outcome be different than the time you last built p(2,4) for the previous ii value?

I suspect that if you map the flow out, that you will find that you can calculate how far you are going to have to move centre(ii,:) . And as each state is associated with a direction vector, you might even be able to calculate how many iterations within that state are going to be needed to get to the point where you switch states.

Your while loop is determininstic, and that means that instead of looping all the time, you should be able to directly calculate what you need to do in order to change state.

Walter Roberson 2022-1-4

Maybe use polyshape() objects and intersect() ?

https://www.mathworks.com/help/matlab/ref/polyshape.intersect.html

CHANDRABHAN Singh 2022-1-5

在 MATLAB Online 中打开

@Walter Roberson, thank you for your kind support.

Let's say

these are two non interscting ploygons (shown below). How can i get a logical relationship out of this. Something like, if the polygons intersect or touch

paramter = true;

otherwise

parameter = false;

x1 =[-2.6967   -2.0891   -0.0846    1.5544    2.6872   -2.6967];
x2 = [5.6494    6.6386    6.6898    4.0313    1.3002    1.9802    5.6494];
y1 = [ 0.1340   -1.7104   -2.6987   -2.2076   -0.2624    0.1340];
y2 = [1.8624    3.4274    4.2346    6.6998    3.9686    2.2082    1.8624];
plot(x1,y1,x2,y2);
grid on;
poly1 = polyshape(x1',y1');
poly2 = polyshape(x2',y2');
polyout = intersect(poly1,poly2);