I got the problem with polyshape, please help me

2022 12 4
1 个回答
回答已采纳
更新时间：2022 12 5
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Demand: sketch the domain D be the region bounded by
My code:
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
x = [x1(x1<=xcut), x1(end)];
y = [sqrt(x1).*(12-x1)./2 (x1<=xcut), 2-x1(end)];
M = polyshape(x,y);
plot(x1,y1,x1,y2,x,y)
hold on
plot(M)
hold off
xlim([0,30])
ylim([0,30])
Error:
Error using polyshape/getXY
x- and y-coordinates must be vectors of the same size with at least 3 elements.
Error in polyshape/checkInput (line 842)
[X, Y, xy2input, next_arg] = polyshape.getXY(varargin{:});
Error in polyshape (line 169)
[X, Y, tc, simpl, collinear] = polyshape.checkInput(param, varargin{:});
Error in bai4 (line 7)
M = polyshape(x,y);

4 个评论
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Matt J
Matt J 2022-12-4
编辑：Matt J 2022-12-4
The domain of the first function is x>=0 and the domain of y=2-x is all .Neither domain is bounded, and therefore cannot be sketched.
John D'Errico
John D'Errico 2022-12-4
As well, you cannot use polyshape on unbounded regions.
Thai Anh
Thai Anh 2022-12-4
oh I'm sorry.
The demand is sketch the domain D be the region bounded by

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 采纳的回答

Stephan
Stephan 2022-12-4
Ran in:

0 个投票

Ran in:
You have to take both points into account to find a closed region:
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
I had the code ready here - but homework is homework... I leave the rest to you. It should be easy now...

12 个评论
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Thai Anh
Thai Anh 2022-12-4
I appreciate that a lot :3
Thanks
Thai Anh
Thai Anh 2022-12-4
编辑：Thai Anh 2022-12-4
sorry but I still cannot solve it
Is this right ?
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
x = [x1(x1>=xcut1), x1(x1<=xcut2)];
y = [sqrt(x1).*(12-x1)./2 -2+x1(x1>=xcut1),sqrt(x1).*(12-x1)./2 -2+x1(x1<=xcut2)];
M = polyshape(x,y);
plot(x1,y1,x1,y2,x,y)
hold on
plot(M)
hold off
xlim([0,30])
ylim([0,30])
Thai Anh
Thai Anh 2022-12-4
can you explained it clearly or give me some hints please .
Thanks in advance @Stephan
Stephan
Stephan 2022-12-4
编辑：Stephan 2022-12-4
your x used for the polyshape should look like:
x = xcut1:0.1:xcut2;
then think about the corresponding y values...
Hint: Use the false flag for the simplify option of the polyshape function to suppress warnings:
M = polyshape([YOUR X],[YOUR Y], 'simplify', false);
Thai Anh
Thai Anh 2022-12-5
编辑：Thai Anh 2022-12-5
Ran in:
Ran in:
I tried but it still cannot work.
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
x = xcut1:0.1:xcut2;
ycut1 = fzero(@(x)sqrt(xcut1).*(12-xcut1)./2 - 2+xcut1, 0.5);
Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.
ycut2 = fzero(@(x)sqrt(xcut2).*(12-xcut2)./2 - 2+xcut2, 20);
Exiting fzero: aborting search for an interval containing a sign change because no sign change is detected during search. Function may not have a root.
M = polyshape([xcut1 ycut1],[xcut2 ycut2], 'simplify', false);
Warning: Boundaries with less than 3 points were removed.
plot(x1,y1,x1,y2,xcut1,xcut2,ycut1,ycut2)
hold on;
plot(M)
hold on;
grid on;
xlim([0,30])
ylim([0,30])
Stephan
Stephan 2022-12-5
ycut1 and ycut2 dont make sense.
Think about how polyshape works:
  • you need a closed region --> so far we know closing points in xdirection are xcut1 and xcut2
  • inside this interval of x-values of your ROI you need correspondig y-values
  • you need several points for the nonlinear part of your region
  • the linear part of your region should be able to be described using 2 points --> you can guess which points this will be...
  • using this thoughts you can construct a vector pair x,y that describes the ROI properly for the usage of polyshape
Thai Anh
Thai Anh 2022-12-5
I change it into this
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
x = xcut1:0.1:xcut2;
y = 2-x1(x1>=xcut1) :0.1: 2-x1(x1<=xcut2);
M = polyshape([xcut1 y],[xcut2 y], 'simplify', false);
plot(x1,y1,x1,y2,x,y)
hold on;
plot(M)
hold on;
grid on;
xlim([0,30])
ylim([0,30])
But it's still wrong in plot(x1,y1,x1,y2,x,y)
Thai Anh
Thai Anh 2022-12-5
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
x = xcut1:0.1:xcut2;
y = 2-x1(x1>=xcut1) :0.1: 2-x1(x1<=xcut2);
M = polyshape([xcut1 y],[xcut2 y], 'simplify', false);
plot(x1,y1,x1,y2,x,y)
hold on;
plot(M)
hold on;
grid on;
xlim([0,30])
ylim([0,30])
it still doesn't work
Stephan
Stephan 2022-12-5
编辑：Stephan 2022-12-5
Ran in:
Ran in:
inside this interval of x-values of your ROI you need corresponding y-values - that means to calculate them
rectangle example:
x = [1 5 5 1];
y = [1 1 2 2];
M = polyshape(x,y);
plot(M)
xlim([0 6])
ylim([0 6])
Try to find out how it works - then you get it
Thai Anh
Thai Anh 2022-12-5
x1 = 0:1:30;
y1 = 2-x1;
y2 = sqrt(x1).*(12-x1)./2;
xcut1 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 0.5);
xcut2 = fzero(@(x)sqrt(x).*(12-x)./2 - 2+x, 20);
a=[x1(xcut1<=x1),x1(x1<=xcut2)];
b=[2-x1 (x1>=xcut1), 2-x1 (x1<=xcut2)] ;
M = polyshape(a,b);
plot(M)
hold on;
grid on;
xlim([0,30])
ylim([0,30])
..... I tried but I cannot mark the curve region
Stephan
Stephan 2022-12-5
编辑：Stephan 2022-12-5
Ran in:
Ran in:
Sorry, last try - i can not make it more clear without doing your homework:
x = -10:1:10;
y1 = -x.^2 + 100;
y2 = 0.*x + 50;
xcut1 = fzero(@(x)-x.^2 + 100 - 50,-7);
xcut2 = fzero(@(x)-x.^2 + 100 - 50,7);
x_linear = [xcut1 xcut2];
y_linear = [50 50];
x_quad = linspace(xcut1, xcut2, 25);
y_quad = -x_quad.^2 + 100;
M = polyshape([x_linear, x_quad], [y_linear, y_quad],'simplify', false);
plot(x,y1,x,y2)
hold on
plot(M)

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