function V=bubvol(I,ostr,ps)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Copyright 2014 Washington State University: Tom Asaki (tasaki@wsu.edu)
%
% Revision: 12 May 2014
%
% bubvol is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
%(at your option) any later version.
%
% bubvol is distributed in the hope that it will be useful,
% but WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
% GNU General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with sinthgunt. If not, see http://www.gnu.org/licenses/
%
%% Description
% function BUBVOL computes the volume of a surface rotation given an image 
% of the object profile.  This routine is *not* general to all shapes.
% The assumption is that the axis of symmetry is one of the two principle
% directions.
% 
%
% GENERAL USAGE:
% V=bubvol(I,ostr,ps)
%
% INPUTS:
%  I        is the object profile image or image file path/name.  The image 
%           should be nearly binary in value.  The image will be forced 
%           binary by this function by thresholding at the average of the 
%           maximum and minimum image intensities.  The object can be 
%           either light or dark (1 or 0).
%  ostr     is the orientation of the object in the image.  The user 
%           supplies one of four characters that indicate the rough
%           orientation of the axis of symmetry: '|' or '/' or '-' or '\'.
%           There is not default value, orient must be specified.
%  ps       pxiel size in real units.  The pixels are assumed to be square.
%           If ps is not supplied, then the pixel size is assumed to be 1 
%           and the reported volume is in cubic pixels.
%
%
% EXAMPLE USAGE:
% V=bubvol(I,'/',4.35)  computes the volume of the solid of revolution
%                       indicated by the profile image I, whose axis of
%                       symmetry in the image is roughly from bottom left
%                       to top right, and whose pixel size is 4.35 units
%                       (where the meaning of "units" is e.g. meter).
%
% EXAMPLE USAGE FROM FIGURE 2
% V=bubvol('10x.tif','-',0.92e-6)  computes the volume of the solid of 
%                       revolution indicated by the profile image found in 
%                       the file 10x.tif, showing a 10x image of an 
%                       uncompressed pico gauge tip. The edge length of a 
%                       pixel is 0.92 micrometer. The volume is returned in 
%                       units of cubic meter.
%                       The axis of symmetry in the image is roughly 
%                       horizontal.  
%
% DeltaV=bubvol('63x.tif','-',0.146e-6)  computes the volume of the solid 
%                       of revolution indicated by the profile image found 
%                       in the file 63x.tif, showing a 63x image of a 
%                       compressed pico gauge tip. The edge length of a 
%                       pixel is 0.146 micrometer. The volume is returned 
%                       in units of cubic meter.
%                       The axis of symmetry in the image is roughly 
%                       horizontal. 
%
%                       To compute the pressure using the images 10x.tif
%                       and 63x.tif, we use Equation 1:
%
% P = 0.852e9*bubvol('63x.tif','-',0.146e-6)/bubvol('10x.tif','-',0.92e-6)
%
%                      where the bulk modulus of the oil is 0.852 GPa.
%                      In this case, we find P = 1.40 MPa.
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% input checking
% set default volume
V=0;
% check for image and 'fix' it
if ~exist('I','var') || isempty(I)
    disp('You must supply an input image or file name.');
    return
end
if ischar(I)
    I=imread(I);
    if ndims(I)>2
        I=rgb2gray(I);
    end
    I=double(I);
else
    if ndims(I)>2
        disp('Image must be two-dimensional.');
        return
    end
end
I=(I-min(I(:)))/(max(I(:))-min(I(:)));
I=round(I);
if I(1), I=~I; end
% check for orientation and 'fix' it
if ~exist('ostr','var') || isempty(ostr)
    disp('You must supply an orientation string.');
    return 
end
if ischar(ostr)
    if ~strcmp(ostr,{'|' '/' '-' '\'})  
        disp('Unrecognized orientation string.')
        return
    end
else
    disp('Input orient should be a string.');
    return
end
% check for pixel size input
if ~exist('ps','var') || isempty(ps)
    ps=1;
end

%% Preliminaries

% extract pixels of interest and locations.  L=[col,row] in pixel units.
% Then subrtract the mean -- the axis must go throught this mean point
% anyway.
[r,c]=find(I);
L=[c r];
M=mean(L);
L=L-repmat(M,size(L,1),1);
nr=size(I,2);

% clean up some stuff
clear I c r

%% Get symmetry direction

% w contains the two principle directions (read as either columns or rows)
[U,S,V]=svd(L','econ');     %#ok

% clean up some stuff 
clear S V

% Pick the one most closely aligned with the given orientation.
% Keep in mind that the visual orientation (e.g. '/') is not the same
% as the computed orientation because of the flipped Matlab/octave 
% indexing on the first variable.
th=atan(U(:,2)./U(:,1));
th=[th ; th+pi];
switch ostr
    case '|'
        a=pi/2;
    case '/'
        a=-pi/4;
    case '-'
        a=0;
    case '\'
        a=pi/4;
end
[da idx]=min(abs(th-a));     %#ok
lineangle=th(idx);


%% Compute the volume

% the volume of a rotated pixel, independent of the rotation angle
% of the pixel in the profile plane, is V=2*pi*d*w^2, where d is
% the distance from the pixel center to the line and w is the pixel
% width.  If the pixel overlaps the axis of symmetry then this 
% simple formula does not hold.  Thus, we compute the partial volumes
% for pixel which do not cross the axis then iteratively subdivide
% the pixels into smaller pixels.

% start with zero volume
V=0;
% sa and ca are used to compute distances to the axis
sa=sin(lineangle);  
ca=cos(lineangle);
% depth is the iteration depth
depth=2;      
% subsampling factor: each pixel becomes subsamp^2 pixels at each iteration
subsamp=2;    
% the normalized subsampling position vector
subvec=linspace(-(subsamp-1)/2,(subsamp-1)/2,subsamp)/subsamp*2;
% main iteration
% depth=0 corresponds to given pixel size ps
% depth=k considers pixels of width ps/subsamp^k
for k=0:depth
    % set current pixel size
    w=subsamp^(-k);
    % compute the distance of all pixels to the axis
    t=L(:,1)*ca+L(:,2)*sa;
    d2=(L(:,1)-t*ca).^2+(L(:,2)-t*sa).^2;
    % keep a mask of sufficiently distant pixels
    gomask=sqrt(d2)>=w/sqrt(2);
    % add the partial volume to the total
    Vold=V;
    V=V+sum(sqrt(d2(gomask)))*w^2;
    % downselect to the pixels yet to consider
    L=L(~gomask,:);
    % compute the new finer-grid pixel locations in K, then save to L.
    K=[];
    dx=repmat(w/2*subvec',1,subsamp);
    dy=repmat(w/2*subvec,subsamp,1);
    for j=1:size(L,1)
        x=repmat(L(j,1)',subsamp,subsamp);
        y=repmat(L(j,2)',subsamp,subsamp);
        x=x+dx;
        y=y+dy;
        K=[K ; [x(:) y(:)] ];                       %#ok
    end
    L=K;
end

% final correction based on all pixels now being very close
% to the axis of symmetry.  Basically, each iteration is now
% only adding 1/subsamp^2 of the previous added volume.  So,
% sum the series and add the total.
V=Vold+(V-Vold)*(subsamp^2/(subsamp^2-1));

%% final tweaks

% The volumes computed above double count the volume
% and did not include the factor pi, thus:
V=pi*V;

% now fold in the actual pixel size in real units.
V=V*ps^3;

%% display results to command prompt
disp(' ');
disp('------------------------------------------');
disp(' ');
if ps==1
    disp(['  Volume = ',num2str(V),' cubic pixels.']);
else
    disp(['  Volume = ',num2str(V),' cubic units.']);
end
disp(' ');
disp(['  Symmetry Axis Angle = ',num2str(-lineangle/pi*180),' degrees.']);
disp(['  Symmetry Center [h,v] = [',num2str(M(1)),',',num2str(nr-M(2)+1),'] pixels.']);
disp(' ');
disp('------------------------------------------');
disp(' ');

%% done
return