Hello! I have a problem with a while condition. Can you please help me?

Hello! I have a problem with my while condition. I have a project where i must make a square with it's points on a loop. The code i have so far includes the making of the loop and that of the square, but it's not a square with all it's points on the loop. It has at least 2 points and the rest are either on the outside of the loop or on the inside. The thing is that the code creating my square is choosing a starting point randomly, so every time i run the code it gives a different square. My problem comes here: i want to make a while that has as condition this: the other 2 points must be on the loop as well. If they are then there must be a message box appearing and if they are not then the code creating the square must run until it gets them all on the loop. This is my code: Is it a possible thing? If so then how can i do it? Can you, plase, help me?

fontSize = 15;
x = [5.5, 6, 4.5, 3.5, 5, 5, 3, 2, 2.5, 3, 2, 2.5, 4, 5, 4.5]  
y = [3, 2, 2.5, 1.5, 1.5, 1, 0.5, 1, 2, 3, 4, 4.5, 5, 4, 3.5]
% Append first point to last to close the curve
x = [x, x(1)];
y = [y, y(1)];
plot(x, y, 'r*');
grid on;
knots = [x; y];
areaOfPolygon = polyarea(x,y);
numberOfPoints = length(x);
% Interpolate with a spline curve and finer spacing.
originalSpacing = 1 : numberOfPoints;
% Make 9 points in between our original points that the user clicked on.
finerSpacing = 1 : 0.1 : numberOfPoints;
% Do the spline interpolation.
splineXY = spline(originalSpacing, knots, finerSpacing);
% Plot the interpolated curve.
hold off;
plot(knots(1, :), knots(2, :), 'ro', 'LineWidth', 2, 'MarkerSize', 16);
hold on;
plot(splineXY(1, :), splineXY(2, :), 'b+-', 'LineWidth', 2, 'MarkerSize', 8);
title('Blue Spline Between Red Knots', 'FontSize', fontSize);
legend('Knots', 'Spline');
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
grid on;
hold on;
% Get a known index.  "known" because it's one of the training points.
knownIndex = randperm(length(x), 1)
% Get a unknown index.  "unknown" because it's one of the interpolated points.
unknownIndex = randperm(length(finerSpacing), 1)
% Get the x,y coordinates for these indexes.
xKnown = knots(1, knownIndex)
yKnown = knots(2, knownIndex)
xUnknown = splineXY(1, unknownIndex)
yUnknown = splineXY(2, unknownIndex)
A=[xKnown, yKnown]
B=[xUnknown, yUnknown]
AB=sqrt((xKnown-splineXY(1, unknownIndex))^2+(yKnown-splineXY(2, unknownIndex))^2)
% Now draw a line between them in dark green.
darkGreen = [0, 0.5, 0];
plot([xKnown, xUnknown], [yKnown, yUnknown], 'o-', ...
  'Color', darkGreen, 'MarkerSize', 24, 'LineWidth', 3);
legend('Knots', 'Spline', 'Line between random knot and random point')unkownIndex1= randperm(length(finerSpacing), 1)
hold on
% slope of oldL
slope = (yKnown - yUnknown)./(xKnown - xUnknown);
% slope and tilt angle of newL 
perSlope = -1/slope;
perTheta = atan(perSlope);
% set the lenght of line segment you want to add to the figure
lineLength = 2*AB; 
halfLineLength = lineLength/2;
% calculate the end points of a newL based on the Known points of oldL
xPerKnown = [xKnown + halfLineLength*cos(perTheta), ...
                   xKnown - halfLineLength*cos(perTheta)];
yPerKnown = [yKnown + halfLineLength*sin(perTheta), ...
                   yKnown - halfLineLength*sin(perTheta)];
% calculate the ends points of another newL based on the Unknown points of oldL
xPerUnknown = [xUnknown + halfLineLength*cos(perTheta), ...
                      xUnknown - halfLineLength*cos(perTheta)];
yPerUnknown = [yUnknown + halfLineLength*sin(perTheta), ...
                      yUnknown - halfLineLength*sin(perTheta)];
xPerKnown(1) = [xKnown + halfLineLength*cos(perTheta)]
xPerKnown(2)=[xKnown - halfLineLength*cos(perTheta)];
yPerKnown(1)= [yKnown + halfLineLength*sin(perTheta)]
yPerKnown(2)=[yKnown - halfLineLength*sin(perTheta)];
% calculate the ends points of another newL based on the Unknown points of oldL
xPerUnknown(1) = [xUnknown + halfLineLength*cos(perTheta)]
xPerUnkown(2)=[xUnknown - halfLineLength*cos(perTheta)];
yPerUnknown(1) = [yUnknown + halfLineLength*sin(perTheta)]
yPerUnknown(2)=[yUnknown - halfLineLength*sin(perTheta)];
                      % draw these two newL
plot([xPerKnown(1), xKnown], [yPerKnown(1), yKnown], 'o-', ...
  'Color', darkGreen, 'MarkerSize', 24, 'LineWidth', 3);
plot([xPerUnknown(1), xUnknown], [yPerUnknown(1), yUnknown], 'o-', ...
  'Color', darkGreen, 'MarkerSize', 24, 'LineWidth', 3);
% set the unit length of two axis to be equal, to get a clear visualization of 90 degree
axis equal
plot([xPerKnown(1),xUnknown + halfLineLength*cos(perTheta)],[yPerKnown(1),yUnknown + halfLineLength*sin(perTheta)],'o-','Color', darkGreen, 'MarkerSize', 24, 'LineWidth', 3)

[Image Analyst]

[Stany Stone]

Yes, this is how it looks. But i need it to run untul it can find the square with all it's points on the loop. Can you, please, help?

Answer 1

[Image Analyst]

There is no guarantee that there will be a square possible on your finite set of spline points. I've asked John D'Errico for an "interior" verion of his Suite of minimal bounding shapes, but he says it's really complicated. So much so that he didn't want to tackle it. But that would be a solution. You could try a smart brute force search but that would take a long time to find the shape that had all 4 corners as close to 90 as possible and all 4 sides as close to the same length as possible.

[Stany Stone]

[in,on] = inpolygon(x,y,xPerKnown(1),yPerKnown(1)) What if i put the condition like this: While [in,on] = inpolygon(x,y,xPerKnown(1),yPerKnown(1)) If [in,on]=0 f=msgbox('three points on loop') else &here should be the code of the square, so it can run again End End

Is it ok like this? Is it right? Because i tried to run it with this condition and it doesn't create another square and i don't know why.

[Image Analyst]

Why do you think there should be a square? Show a diagram and point out where you think a square could be. And say what coordinates the corners should have.

[Stany Stone]

Because that's the request of this project, that there should be an inscribed square that must be found.

[Image Analyst]

I suggest you do a search for "minimum perimeter polygon" and specify that there should be 4 sides. I'm attaching a paper on it. I do not have MATLAB code for it.

[Stany Stone]

Ok! Thank you for your help!

[Image Analyst]

There is now a function called reducepoly in the Image Processing Toolbox.

reducepoly

Reduce density of points in ROI using Ramer–Douglas–Peucker algorithm

Syntax

P_reduced = reducepoly(P)

P_reduced = reducepoly(P,tolerance)

Description

example

P_reduced = reducepoly(P) reduces the density of points in array P. The reducepoly function uses the Ramer-Douglas-Peucker line simplification algorithm, removing points along straight lines and leaving only knickpoints (points where the line curves).

P_reduced = reducepoly(P,tolerance) reduces the density of points in array P, where tolerance specifies how much a point can deviate from a straight line.