Reversing binary stream Conversion According to attached table?
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Hi guys,
function ret=binrestore(m,array)
% returns binary decoded array values from input encoded array
% multiple Mapped Value = 0 Mapped Value= 1
% 1 0 1
% 2 00 11
% 4 1100 0011
% 8 11001100 00110011
assert(mod(numel(array),m)==0,'Input array length not multiple of multipler')
array=reshape(array,m,[]); % separate out digits to compare
switch m
case 1 % nothing else to do at this point
case 2
array=all(array==1);
case 4
array=all(array==[0 0 1 1].');
case 8
array=all(array==[0 0 1 1 0 0 1 1].');
otherwise
error('Multiplier not 1, 2, 4, 8')
end
ret=num2str(array,'%d');
end
this function does the binary converstion according to to the attached table, for example:
binrestore(1,[0 1]) ----> output= [0 1] . because 0 is mapped to zero according to table at m=1, and 1 is mapped to one.
binrestore(2,[0 1]) ----> output= [0 0 1 1 ]. because 0 is mapped to 0 0 and 1 is mapped to 1 1 according to table.
binrestore(4,[0 1]) ----> output= [1 1 0 0 0 0 1 1]. because 0 is mapped to 1 1 0 0 and 1 is mapped to 0 0 1 1 according to table.
what Im trying to implement is the reverse of this function, it means the output of this function is my input to the function that I want to implement which it's called reversebinrestore(m,array) and its output is the input to my function above.
for example of the function that Im trying to implement, If I input to it:
reversebinrestore(1,[0 1]) --> Output=[0 1]
reversebinrestore(2, [0 0 1 1 ]) --> Output=[0 1]
reversebinrestore(4,[1 1 0 0 0 0 1 1]) ---> Output=[0 1] , this because [1 1 0 0] according to table is mapped to zero, and [0 0 1 1 ] is mapped to one.
..etc
exactly the same functionality of binrestore(m ,array) but in opposite/reverse.
Any help please how I can I implement that in matlab? thanks alot!
Antworten (2)
David Hill
am 6 Sep. 2020
function ret=reversebinrestore(m,array)
ret=[];
switch m
case 1
l=[0;1];
for k=1:length(array)
ret=[ret,find(ismember(l,array(k)))-1];
end
case 2
l=[0 0;1 1];
array=reshape(array,2,[])';
for k=1:size(array,1)
ret=[ret,find(ismember(l,array(k,:),'rows'))-1];
end
case 4
l=[1 1 0 0;0 0 1 1];
array=reshape(array,4,[])';
for k=1:size(array,1)
ret=[ret,find(ismember(l,array(k,:),'rows'))-1];
end
case 8
l=[1 1 0 0 1 1 0 0;0 0 1 1 0 0 1 1];
array=reshape(array,8,[])';
for k=1:size(array,1)
ret=[ret,find(ismember(l,array(k,:),'rows'))-1];
end
end
end
5 Kommentare
Bruno Luong
am 6 Sep. 2020
function [output, outputchar] = reversebinrestore(m, array)
if ischar(array)
array = array-'0';
end
switch m
case 1
map0 = [0];
case 2
map0 = [0 0];
case 4
map0 = [1 1 0 0];
case 8
map0 = [1 1 0 0 1 1 0 0];
otherwise
error('Multiplier not 1, 2, 4, 8')
end
map1 = 1-map0;
map = [map0; map1];
array = reshape(array, size(map,2), []).';
[b,output] = ismember(array, map, 'rows');
if ~all(b)
error('invalid array');
end
output = output(:)'-1;
outputchar = char(output+'0');
end
8 Kommentare
Bruno Luong
am 6 Sep. 2020
Bearbeitet: Bruno Luong
am 7 Sep. 2020
>> xxx(2,[0 1])
ans =
0 0 1 1
>> out = xxx(2,[0 1])
out =
0 0 1 1
>> [out, outstr] = xxx(2,[0 1])
out =
0 0 1 1
outstr =
'0011'
>> [~, outstr] = xxx(2,[0 1])
outstr =
'0011'
>> [~, outstr] = xxx(2,'01')
outstr =
'0011'
>>
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