morton scanning for image size of 256*256 .

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vishnu
vishnu 2012-3-2
i found out this as for 4*4 matrix from matlab central
clc;
clear all;
close all ;
n=2;
ind=morton(n);
a=reshape(1:4^n,2^n,2^n);
disp(a)
here is the matlab function
function ind=morton(n);
% MORTON(N) return the morton permutation order for array of size 2^N
% e.g.
% n=2;
% matrix size is 2^n
% ind=morton(n);
% d=fix(rand(2^n,2^n)*10);
% disp(d)
% disp(d(ind))
linind4=(1:4^n)-1; %start index count for array at zero
ind4str=dec2base(linind4,4); %convert indices to base-4
b1=dec2bin(str2num(ind4str(:,1))); %split each base-4 into two base-2numbers
b2=dec2bin(str2num(ind4str(:,2)));
rb=[b1(:,1) b2(:,1)]; %the rows are given by the first bits of b1 and
b2
cb=[b1(:,2) b2(:,2)]; %the columns are given by the second bits of b1 and b2
r=bin2dec(rb)+1; %convert the row from bit to decimal
c=bin2dec(cb)+1; %convert column
ind=[2^n*(c-1)+r]'; %make a linear row index into array for easyaddressing
%morton.m ends
----------------------------------------------------
i require to do this for whole image with out looping infinitely ..
can you please any one give the way by block processing
  2 个评论
vishnu
vishnu 2012-3-2
i get infinite loop and can able to proceed it .. please any one help ?

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回答（1 个）

Walter Roberson
Walter Roberson 2012-3-2
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vishnu
vishnu 2012-3-7
but i get error like this .. Maximum variable size allowed by the program is exceeded.
Error in ==> morton at 10
linind4=(1:4^n)-1; %start index count for array at zero
.. i am totally confused sir.. i take a image and those image values to be an array representation according to morton order .. its an assignment for me .. please help me .. i want to learn about this
Walter Roberson
Walter Roberson 2012-3-7
For the purpose of your morton code, n should be 8.
But that doesn't have anything to do with the code I referred you to, which is the code you asked about in your comment here. Do not use both routines together: only use one of them.

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