MATLAB: ??? Input argument “des2” is undefined. Error in ==> do_match at 13 des2t = des2; % Precompute matrix transpose

error

I run the following codes ,the matlab is getting error I1=imreadbw('B2.jpg') ; I2=imreadbw('A1.jpg') ;
I1=imresize(I1, [240 320]); I2=imresize(I2, [240 320]);
I1=I1-min(I1(:)) ; I1=I1/max(I1(:)) ; I2=I2-min(I2(:)) ; I2=I2/max(I2(:)) ;
fprintf('Computing frames and descriptors.\n') ; [frames1,descr1,gss1,dogss1] = do_sift( I1, 'Verbosity', 1, 'NumOctaves', 4, 'Threshold', 0.2/3/2 ) ; %0.04/3/2 [frames2,descr2,gss2,dogss2] = do_sift( I2, 'Verbosity', 1, 'NumOctaves', 4, 'Threshold', 0.2/3/2 ) ;
fprintf('Computing matches.\n') ;
descr1 = descr1'; descr2 = descr2'; tic ; matches=do_match( descr1, descr2 ) ; fprintf('Matched in %.3f s\n', toc) ;
figure(3) ; clf ; plotmatches(I1,I2,frames1(1:2,:),frames2(1:2,:),matches) ; drawnow ;
function do_match.m function num = do_match(im1, des1, loc1, im2, des2, loc2)
% This function matchings the SIFT keys from two iamges % distRatio: Only keep matches in which the ratio of vector angles from the % nearest to second nearest neighbor is less than distRatio. % Postprocessing: check each matching point, eliminate false matches by voting from % neighbouring area % Author: Yantao Zheng. Nov 2006. For Project of CS5240
distRatio = 0.75; matched_points_img1 =[]; % For each descriptor in the first image, select its match to second image. des2t = des2; % Precompute matrix transpose for i = 1 : size(des1,1) %对全部特征点处理 dotprods = des1(i,:) * des2t; % 第i点的特征向量与第二幅图的特征向量做内积 [vals,indx] = sort(acos(dotprods)); % 做反余弦并排序
   % Check if nearest neighbor has angle less than distRatio times 2nd.
   %检查是否为最近邻（角度低于0.75倍的第二个点）
   if (vals(1) < distRatio * vals(2))
      match(i) = indx(1);
      matched_points_img1 = [matched_points_img1; i];
   else
      match(i) = 0;
   end
end
%——————————————————— % check each matching point, eliminate false matches by voting from % neighbouring area %检查每个匹配点，通过从近邻区域的投票去除错误匹配点 %—— dis_thres = 0.3; orien_thres = 0.3; num = sum(match > 0); final_match = zeros(1,length(match)); % if == 1, the corresponding match(i) is accepted. if ==0, the corresponding match(i) is rejected dis_img_1= zeros(num ,num); % each row contains the spatial distance vector that descript the correctness of the point dis_img_2= zeros(num ,num);
orien_diff_img_1= zeros(num ,num); % each row contains the orientation difference vector that descript the correctness of the point orien_diff_img_2= zeros(num ,num); for k = 1: num dis_img_1(k, k) = 0; dis_img_2(k, k) = 0; % the distance between the point and distance is 0
    orien_diff_img_1(k,k) = 0;
    orien_diff_img_2(k,k) = 0;
    for j = k+ 1: num
           dis_img_1(k, j) = sqrt( (loc1(matched_points_img1(k), 1) - loc1(matched_points_img1(j), 1))^2 ... 
                    + (loc1(matched_points_img1(k), 2) - loc1(matched_points_img1(j), 2))^2 );
           dis_img_1(j, k) =  dis_img_1(k, j); % dis_img_1 is a symmetric matrix



           % compute the corresponding distances of the matching points in
           % image 2
           dis_img_2(k, j) = sqrt( ((loc2(match(matched_points_img1(k)), 1) - loc2(match(matched_points_img1(j)), 1))^2 + (loc2(match(matched_points_img1(k)), 2) - loc2(match(matched_points_img1(j)), 2))^2 ));
           dis_img_2(j, k) =  dis_img_2(k, j); % dis_img_1 is a symmetric matrix
           orien_diff_img_1(k, j) =  loc1(matched_points_img1(k), 4) - loc1(matched_points_img1(j), 4);
           orien_diff_img_1(j, k) =  orien_diff_img_1(k, j); % dis_img_1 is a symmetric matrix
           orien_diff_img_2(k, j) =  loc2(match(matched_points_img1(k)), 4) - loc2(match(matched_points_img1(j)), 4);
           orien_diff_img_2(j, k) =  orien_diff_img_2(k, j); % dis_img_1 is a symmetric matrix
    end
end
% normalize the distance and orein_diff vector for ii =1: num dis_img_1(ii, 🙂 = dis_img_1(ii, 🙂 ./ norm(dis_img_1(ii, 🙂 ); dis_img_2(ii, 🙂 = dis_img_2(ii, 🙂 ./ norm(dis_img_2(ii, 🙂 );
    orien_diff_img_1(ii,:) = orien_diff_img_1(ii,:) ./( eps + norm(orien_diff_img_1(ii,:)));
    orien_diff_img_2(ii,:) = orien_diff_img_2(ii,:) ./( eps +norm(orien_diff_img_2(ii,:)));
end
for m = 1: num dis_coherence = dot(dis_img_1(m,:), dis_img_2(m,:)); orein_coh = dot(orien_diff_img_1(m,:), orien_diff_img_2(m,:)); if dis_coherence > dis_thres && (orein_coh > orien_thres ( sum(orien_diff_img_1(m,:)>0) == 0 && sum(orien_diff_img_2(m,:)>0) == 0 ) ) % if the orientations are the same, then orein_coh will be 0, so % cope with this with another condition final_match(matched_points_img1(m)) = 1; end end
num = sum(match > 0); fprintf('找到 %d 个匹配点.\n', num);
% Create a new image showing the two images side by side. im3 = appendimages(im1,im2);
% Show a figure with lines joining the accepted matches. figure('Position', [100 100 size(im3,2) size(im3,1)]); colormap('gray'); imagesc(im3); hold on; cols1 = size(im1,2); for i = 1: size(des1,1) if (match(i) > 0) line([loc1(i,1) loc2(match(i),1)+cols1], … [loc1(i,2) loc2(match(i),2)], 'Color', 'c'); end end hold off;

Best Answer

Set a breakpoint on that line and see what des2 is. You'll find out that you never set it equal to anything. http://blogs.mathworks.com/videos/2010/09/02/using-debugger-to-walk-through-code/
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