matlab求最⼩外接斜矩形minboundrect⽅法
最近在做课程⼤作业时看到opencv函数cv2.minAreaRect()，
但是想⽤matlab实现，于是查到了John D’Errico写的matlab实现求最⼩外接斜矩形函数。（代码贴在最后，仅供学习使⽤）
[rectx,recty,area,perimeter] = minboundrect(c,r,‘a’)
其中a表⽰以⾯积最⼩、如果是p的话则是以边长最⼩
这⾥有个问题，⽤minboundrect函数求得的四个点顺序是什么？
于是做了两张图验证了⼀下
放上结果：
可以得到minboundrect函数得到的结果(rectx,recty)是从最上边的点开始，按照顺时针⽅向索引。minboundrect代码
function[rectx,recty,area,perimeter]=minboundrect(x,y,metric)
% minboundrect: Compute the minimal bounding rectangle of points in the plane
% usage:[rectx,recty,area,perimeter]=minboundrect(x,y,metric)
%
% arguments:(input)
%  x,y - vectors of points, describing points in the plane as
%(x,y) pairs. x and y must be the same lengths.
%
%  metric -(OPTIONAL)- single letter character flag which
%        denotes the use of minimal area or perimeter as the
%        metric to be minimized. metric may be either 'a' or 'p',
%        capitalization is ignored. Any other contraction of'area'minimal
%        or 'perimeter' is also accepted.
%
%DEFAULT:'a'('area')
%
% arguments:(output)
%  rectx,recty -5x1 vectors of points that define the minimal
%        bounding rectangle.
%
%  area -(scalar) area of the minimal rect itself.
%
%  perimeter -(scalar) perimeter of the minimal rect as found
%
%
% Note: For those individuals who would prefer the rect with minimum
% perimeter or area, careful testing convinces me that the minimum area
% rect was generally also the minimum perimeter rect on most problems
%(with one class of exceptions). This same testing appeared to verify my
% assumption that the minimum area rect must always contain at least
% one edge of the convex hull. The exception I refer to above is for
% problems when the convex hull is composed of only a few points,
% most likely exactly 3. Here one may see differences between the
% most likely exactly 3. Here one may see differences between the % two metrics. My thanks to Roger Stafford for pointing out this
%class of counter-examples.
%
% Thanks are also due to Roger for pointing out a proof that the
% bounding rect must always contain an edge of the convex hull,in % both the minimal perimeter and area cases.
%
%
% Example usage:
%  x =rand(50000,1);
%  y =rand(50000,1);
%  tic,[rx,ry,area]=minboundrect(x,y);toc
%
%  Elapsed time is 0.105754 seconds.
%
%[rx,ry]
%  ans =
%0.99994-4.2515e-06
%0.999980.99999
%  2.6441e-051
%-5.1673e-06  2.7356e-05
%0.99994-4.2515e-06
%
%  area
%  area =
%0.99994
%
%
% See also: minboundcircle, minboundtri, minboundsphere
%
%
% Author: John D'Errico
%E-mail:
% Release:3.0
% Release date:3/7/07
%default for metric
if(nargin<3)||isempty(metric)
metric ='a';
elseif ~ischar(metric)
error 'metric must be a character flag if it is supplied.'
else
% check for'a' or 'p'
metric =lower(metric(:)');
ind =strmatch(metric,{'area','perimeter'});
if isempty(ind)
error 'metric does not match either ''area'' or ''perimeter'''
end
% just keep the first letter.
metric =metric(1);
end
% preprocess data
x=x(:);
y=y(:);
% not many error checks to worry about
n =length(x);
if n~=length(y)
error 'x and y must be the same sizes'
end
% start out with the convex hull of the points to
% reduce the problem dramatically. Note that any
% reduce the problem dramatically. Note that any
% points in the interior of the convex hull are
% never needed, so we drop them.
if n>3
edges =convhull(x,y);
%edges =convhull(x,y,{'Qt'});%'Pp' will silence the warnings
% exclude those points inside the hull as not relevant
% also sorts the points into their convex hull as a
% closed polygon
x =x(edges);
y =y(edges);
% probably fewer points now, unless the points are fully convex  nedges =length(x)-1;
elseif n>1
% n must be 2 or 3
nedges = n;
x(end+1)=x(1);
y(end+1)=y(1);
else
% n must be 0 or 1
nedges = n;
end
% now we must find the bounding rectangle of those
% that remain.
% special case small numbers of points. If we trip any
%of these cases, then we are done, so return.
switch nedges
case0
% empty begets empty
rectx =[];
recty =[];
area =[];
perimeter =[];
return
case1
%with one point, the rect is simple.
rectx =repmat(x,1,5);
recty =repmat(y,1,5);
area =0;
perimeter =0;
return
case2
% only two points. also simple.
rectx =x([12211]);
recty =y([12211]);
area =0;
perimeter =2*sqrt(diff(x).^2+diff(y).^2);
return
end
%3 or more points.
% will need a 2x2 rotation matrix through an angle theta
Rmat = @(theta)[cos(theta)sin(theta);-sin(theta)cos(theta)];
%get the angle of each edge of the hull polygon.
ind =1:(length(x)-1);
edgeangles =atan2(y(ind+1)-y(ind),x(ind+1)-x(ind));
% move the angle into the first quadrant.
edgeangles =unique(mod(edgeangles,pi/2));
% now just check each edge of the hull

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