图像分块数目

ClydeZhang 2007-12-13 02:42:16

一个nxm二值画图像，如果把相邻(8 临点)的颜色相同的地方看成一块区域，请问这个图像的最大区域数目是什么？数学理论证明？

例如：下面的4x4图像的块数被认为是3而不是4.
1 0 1 0
0 1 0 0
0 0 0 0
1 1 1 1
那么对任意的nxm二值画图像，这个块数的精确上限的表达是是什么？

...全文

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ClydeZhang 2008-03-10

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Thanks for Iorikingdom.I have implemented a function like bwlabel in C in N, and now it works fine.

iorikingdom 2008-02-13

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不需要膨胀收缩的,
估计类似于Matlab的
bwlabel
可以满足你的要求



function [L,numComponents] = bwlabel(BW,mode)

%BWLABEL Label connected components in 2-D binary image.

%   L = BWLABEL(BW,N) returns a matrix L, of the same size as BW, containing

%   labels for the connected components in BW. N can have a value of either

%   4 or 8, where 4 specifies 4-connected objects and 8 specifies

%   8-connected objects; if the argument is omitted, it defaults to 8.

%

%   The elements of L are integer values greater than or equal to 0.  The

%   pixels labeled 0 are the background.  The pixels labeled 1 make up one

%   object, the pixels labeled 2 make up a second object, and so on.

%

%   [L,NUM] = BWLABEL(BW,N) returns in NUM the number of connected objects

%   found in BW.

%

%   Note: Comparing BWLABEL and BWLABELN

%   ------------------------------------

%   BWLABEL supports 2-D inputs only, whereas BWLABELN support any 

%   input dimension.  In some cases you might prefer to use BWLABELN even

%   for 2-D problems because it can be faster.  If you have a 2-D input

%   whose objects are relatively "thick" in the vertical direction,

%   BWLABEL will probably be faster; otherwise BWLABELN will probably be

%   faster.

%

%   Class Support

%   -------------

%   BW can be logical or numeric, and it must be real, 2-D, and

%   nonsparse.  L is double.

%

%   Example

%   -------

%       BW = logical([1 1 1 0 0 0 0 0

%                     1 1 1 0 1 1 0 0

%                     1 1 1 0 1 1 0 0

%                     1 1 1 0 0 0 1 0

%                     1 1 1 0 0 0 1 0

%                     1 1 1 0 0 0 1 0

%                     1 1 1 0 0 1 1 0

%                     1 1 1 0 0 0 0 0]);

%       L = bwlabel(BW,4);

%       [r,c] = find(L == 2);

%

%   See also BWAREAOPEN, BWEULER, BWLABELN, BWSELECT, LABEL2RGB.



%   Copyright 1993-2005 The MathWorks, Inc.

%   $Revision: 1.29.4.5 $  $Date: 2006/06/15 20:08:28 $



iptchecknargin(1,2,nargin,mfilename);

iptcheckinput(BW, {'logical' 'numeric'}, {'real', '2d', 'nonsparse'}, ...

              mfilename, 'BW', 1);



if (nargin < 2)

    mode = 8;

else

    iptcheckinput(mode, {'double'}, {'scalar'}, mfilename, 'N', 2);

end



if ~islogical(BW)

    BW = BW ~= 0;

end





[M,N] = size(BW);



% Compute run-length encoding and assign initial labels.

[sr,er,sc,labels,i,j] = bwlabel1(BW,mode);

if (isempty(labels))

    numLabels = 0;

else

    numLabels = max(labels);

end



% Create a sparse matrix representing the equivalence graph.

tmp = (1:numLabels)';

A = sparse([i;j;tmp], [j;i;tmp], 1, numLabels, numLabels);



% Determine the connected components of the equivalence graph

% and compute a new label vector.



% Find the strongly connected components of the adjacency graph

% of A.  dmperm finds row and column permutations that transform

% A into upper block triangular form.  Each block corresponds to

% a connected component; the original source rows in each block

% correspond to the members of the corresponding connected

% component.  The first two output% arguments (row and column

% permutations, respectively) are the same in this case because A

% is symmetric.  The vector r contains the locations of the

% blocks; the k-th block as indices r(k):r(k+1)-1.

[p,p,r] = dmperm(A);



% Compute vector containing the number of elements in each

% component.

sizes = diff(r);

numComponents = length(sizes);  % Number of components.



blocks = zeros(1,numLabels);

blocks(r(1:numComponents)) = 1;

blocks = cumsum(blocks);

blocks(p) = blocks;

labels = blocks(labels);



% Given label information, create output matrix.

L = bwlabel2(sr, er, sc, labels, M, N);

上面的代码凑合点参考下,详细可以看看matlab里面的

legendwin 2008-02-04

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直觉：你可以用腐蚀，膨胀算法处理，然后去计算大体的块状，这样就能求一个近似的值了。我想依照这条路也是可以去证明的。

ClydeZhang 2008-01-05

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我的问题不是如何编程找到符合要求的图像分块，而是估计这个块数的上限的数学证明。

ClydeZhang 2008-01-03

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感谢大家对这个问题感兴趣，
1 0 1 0
0 1 0 0
0 0 0 0
1 1 1 1
这个图中，三个块分别是,*是无关内容，数字是属于同一个块区域的像素。凡是跟一个点A能通过8临点联通的点B都与A属于同一个块；
1 * 1 *
* 1 * *
* * * *
//////////////////////////////////
* 0 * 0
0 * 0 0
0 0 0 0
* * * *
///////////////////////////////
* * * *
* * * *
1 1 1 1

ClydeZhang 2008-01-03