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feixiaolin

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Сľ³æ: ½ð±Ò+0.5, ¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
ÒýÓûØÌû:
10Â¥: Originally posted by ppt1210 at 2014-05-08 17:11:45
ÒÑÔÞÖú£¬Ð»Ð»´Í½Ì¡£
¿É·ñ¸æÖª£¬ËùµÃÐÅÏ¢ÄÜ˵Ã÷ʲôÎÊÌ⣿ÒÔ¼°Èç¹û½âÎöµÃµ½£¬Ð»Ð»£¡...

C code
//  ¶¨Òå´ýÓñäÁ¿
#define int i, j, k, i1, j1;
                int kk[6], ll[6], gmax, gmin;
                int s[6]={2,3,4,6,8,12};
                double  M=24, D[6], Nr[6];

// Ëã·¨²¿·Ö
for(k=0; k<6; k++)
{
        kk[k]=0;
        ll[k]=0;
        Nr[k]=0;
   
        // read image
        for(i=0; i<nWidth-1-s[k], i+=s[k])
                for(j=0; j<nHight-1-s[k], j+=s[k])
                {
                        max=0;
                        min=255;
                        for(i1=i; i1<i+s[k], i1+=s[k])
                                for(j1=j; j1<j+s[k], j1+=s[k])
                                {
                                        if(gray[i1, j1]<min)
                                                min=gray[i1, j1];
                                        if(gray[i1, j1]>max)
                                                max=gray[i1, j1];
                                }
                        kk[k]+=floor((float)min/s[k]);
                        ll[k]+=floor((float)max/s[k] + 0.5);
                        Nr[k]+=ll[k] - kk[k] + 1;                               
                }
        D[k]=log(N[k])/log(M/s[k]);
}

for(k=1; k<6; k++)
        D[0]+=D[k];
D[0]/=6.0;

Matlab code
% ¸Ã³ÌÐòÖ»ÄܼÆËãsizeΪNxN£¨N=2^n£©µÄͼÐΡ£
function fd=box_frac_dem(X);
% ²î·ÖºÐάÊý
% Example:
%    X=double(imread('rice.tif'));
%    fd=box_frac_dem(X);
% Author email of the program:% zjliu2001@163.com
%
% Reference:
% Sarkar N,Chaudhuri B B. An efficient approach to estimate
% fractal dimensionof textural images [J].Pattern Recognition,
% 1992,25(9):1035-1041

if size(X,1)~=size(X,2);
   error('The size of X must NxN.');
end
B=size(X,1);
if mod(log2(B),1)>0;
   error('The size of X must 2^n');
end
t=log2(B);
s=2.^(1:t);
Nr=zeros(1,t);
for k=1:t;
   d=s(k);
   h=256/d;
   for m=1:h;
       for n=1:h;
           A=X(d*(m-1)+[1:d],d*(n-1)+[1:d]);
           mn=min(A(1:end));
           mx=max(A(1:end));
           nr=fix(mx/d)-fix(mn/d)+1;
           Nr(k)=Nr(k)+nr;
       end
   end
end
r=B./s;
p=polyfit(log10(r),log10(Nr),1);
fd=p(1);

» ±¾Ìû¸½¼þ×ÊÔ´Áбí

11Â¥2014-05-08 17:17:46
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû
²é¿´È«²¿ 41 ¸ö»Ø´ð

mathstudy

½ð³æ (ÕýʽдÊÖ)

¡ï ¡ï
Сľ³æ: ½ð±Ò+0.5, ¸ø¸öºì°ü£¬Ð»Ð»»ØÌû
ppt1210: ½ð±Ò+1, ºÇºÇ£¬ÎÒÒ²ÊDz»¶®£¬Ö»ÖªµÀÕâÑùÒ»¸ö¸ÅÄ¾ÍÊÇ¿´×źÜìÅ 2014-05-08 12:39:31
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2Â¥2014-05-07 14:48:17
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

feixiaolin

ÈÙÓþ°æÖ÷ (ÎÄ̳¾«Ó¢)

ÓÅÐã°æÖ÷

¡ï ¡ï
ppt1210: ½ð±Ò+2 2014-05-08 12:38:23
ÍêÈ«ÐС£

[ ·¢×ÔÊÖ»ú°æ http://muchong.com/3g ]
3Â¥2014-05-07 16:09:58
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

ppt1210

ÖÁ×ðľ³æ (ÖøÃûдÊÖ)

ÒýÓûØÌû:
3Â¥: Originally posted by feixiaolin at 2014-05-07 16:09:58
ÍêÈ«ÐС£

ÇëÎÊ£¬ÎÒÏÖÔÚÕ⼸·ùͼ£¬¿É·ñ°ïæ·ÖÎöÒ»ÏÂÄØ£¿½«À´³öpaperµÄʱºò¿ÉÒÔ¹ÒÃû
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4Â¥2014-05-08 12:38:57
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