minboundrect

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'%        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% 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.%%% See also: minboundcircle, minboundtri, minboundsphere%%% default for metricif (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 datax=x(:);y=y(:);% not many error checks to worry aboutn = 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% points in the interior of the convex hull are% never needed, so we drop them.if n>3  edges = convhull(x,y);  % '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  case 0    % empty begets empty    rectx = [];    recty = [];    area = [];    perimeter = [];    return  case 1    % with one point, the rect is simple.    rectx = repmat(x,1,5);    recty = repmat(y,1,5);    area = 0;    perimeter = 0;    return  case 2    % only two points. also simple.    rectx = x([1 2 2 1 1]);    recty = y([1 2 2 1 1]);    area = 0;    perimeter = 2*sqrt(diff(x).^2 + diff(y).^2);    returnend% 3 or more points.% will need a 2x2 rotation matrix through an angle thetaRmat = @(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 hullnang = length(edgeangles);              area = inf;                           perimeter = inf;met = inf;xy = [x,y];for i = 1:nang                           % rotate the data through -theta   rot = Rmat(-edgeangles(i));  xyr = xy*rot;  xymin = min(xyr,[],1);  xymax = max(xyr,[],1);  % The area is simple, as is the perimeter  A_i = prod(xymax - xymin);  P_i = 2*sum(xymax-xymin);  if metric=='a'    M_i = A_i;  else    M_i = P_i;  end  % new metric value for the current interval. Is it better?  if M_i<met    % keep this one    met = M_i;    area = A_i;    perimeter = P_i;    rect = [xymin;[xymax(1),xymin(2)];xymax;[xymin(1),xymax(2)];xymin];    rect = rect*rot';    rectx = rect(:,1);    recty = rect(:,2);  endend% get the final rect% all doneend % mainline end调用实例% bwim是二值化的图片[r, c]=find(bwim==1)[rectx,recty,area,perimeter] = minboundrect(c,r,'a');imshow(bwim);line(rectx(:),recty(:),'color','r');