% MAXLINEDEV - Find max deviation from a line in an edge contour. % % Function finds the point of maximum deviation from a line joining the % endpoints of an edge contour. % % Usage: [maxdev, index, D, totaldev] = maxlinedev(x,y) % % Arguments: % x, y - arrays of x,y (col,row) indicies of connected pixels % on the contour. % Returns: % maxdev - Maximum deviation of contour point from the line % joining the end points of the contour (pixels). % index - Index of the point having maxdev. % D - Distance between end points of the contour so that % one can calculate maxdev/D - the normalised error. % totaldev - Sum of the distances of all the pixels from the % line joining the endpoints. % % See also: EDGELINK, LINESEG % % Peter Kovesi % School of Computer Science & Software Engineering % The University of Western Australia % pk at csse uwa edu au % http://www.csse.uwa.edu.au/~pk % % December 2000 - Original version % February 2003 - Added calculation of total deviation % August 2006 - Avoid degeneracy when endpoints are coincident % February 2007 - Corrected distance calculation when endpoints are % coincident. function [maxdev, index, D, totaldev] = maxlinedev(x,y) Npts = length(x); if Npts == 1 warning('Contour of length 1'); maxdev = 0; index = 1; D = 1; totaldev = 0; return; elseif Npts == 0 error('Contour of length 0'); end % D = norm([x(1) y(1)] - [x(Npts) y(Npts)]); % Distance between end points D = sqrt((x(1)-x(Npts))^2 + (y(1)-y(Npts))^2); % This runs much faster if D > eps % Eqn of line joining end pts (x1 y1) and (x2 y2) can be parameterised by % % x*(y1-y2) + y*(x2-x1) + y2*x1 - y1*x2 = 0 % % (See Jain, Rangachar and Schunck, "Machine Vision", McGraw-Hill % 1996. pp 194-196) y1my2 = y(1)-y(Npts); % Pre-compute parameters x2mx1 = x(Npts)-x(1); C = y(Npts)*x(1) - y(1)*x(Npts); % Calculate distance from line segment for each contour point d = abs(x*y1my2 + y*x2mx1 + C)/D; else % End points are coincident, calculate distances from 1st point d = sqrt((x - x(1)).^2 + (y - y(1)).^2); D = 1; % Now set D to 1 so that normalised error can be used end [maxdev, index] = max(d); if nargout == 4 totaldev = sum(d.^2); end