minboundrect.m

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.
%
%
% 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
%      0.99998      0.99999
%   2.6441e-05            1
%  -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: woodchips@rochester.rr.com
% 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
% points in the interior of the convex hull are
% never needed, so we drop them.
if n>3
  edges = convhull(x,y);

  % 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);
    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
nang = 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);
  end
end
% get the final rect

% all done

end % mainline end