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資源簡介

在傳統的NCC算法上采用卷積加速

資源截圖

代碼片段和文件信息

clear?all;
close?all;
clc;
Il?=?double(imread(‘left.jpg‘));
left?=?rgb2gray(Il);
Ir?=?double(imread(‘right.jpg‘));
right?=?rgb2gray(Ir);
%?left=double(imread(‘left.bmp‘));
%?right=double(imread(‘right.bmp‘));

[HW]=size(left);

%%
%?根據匹配圖片需要修改??
%窗口半徑
N=5;
%最大視差
dMax=59;???????%cones
%dMax=59;??????%teddy
%dMax=15;??????%tsukuba
%dMax=19;??????%venus
%%
im=zeros(HW);
e=zeros(1dMax-1);

B=zeros(2*N+12*N+1);
B(:)=1;%全1矩陣
Il=left;
Ir=right;

Ir2=Ir.^2;
Il2=Il.^2;

D2=conv2(Il2B‘valid‘);
D3=conv2(Ir2B‘valid‘);?%卷積

[H1W1]=size(D2);
e=zeros(H1W1);
NCCD=zeros(H1W1dMax-1);

for?k=1:1:dMax
????IlIr=Il(1:H1+dMax:W).*Ir(1:H1+dMax-k:W-k);?%?Il(ij)*Ir(ij-k)大小為H*W-dMax(1:H1+dMax:W)
????D1=conv2(IlIrB‘valid‘);
????
????for?i=1:H1
????????for?j=1+dMax:W1
?????
???????????nCCD=D1(ij-dMax)/sqrt(D2(ij)*D3(ij-k));
???????????e(ij)=nCCD;?
????????end
???end
???NCCD(::k)=e;
end

for?i=1:H1
????for?j=1:W1
????????[junkim(ij)]=max(NCCD(ij:));
????end
end
imgn=zeros(HW);
imgn(1+N:H-N1+dMax+N:W1+N)=im(1:H11+dMax:W1);
imgn=medfilt1(imgn5);
%%
%?根據匹配圖片需要修改
imgn=4*imgn;???%cones????
%imgn=4*imgn;??%teddy
%imgn=16*imgn;?%tsukuba
%imgn=8*imgn;??%venus
imshow(imgn[])
imwrite(imgn/255‘result.png‘)

?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----

?????文件?????362946??2015-07-07?18:33??ncc?simple\left.png

?????文件???????1328??2016-05-13?21:14??ncc?simple\NCC.m

?????文件?????364420??2015-07-07?18:33??ncc?simple\right.png

?????目錄??????????0??2017-04-18?20:10??ncc?simple

-----------?---------??----------?-----??----

???????????????728694????????????????????4


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