how to implement in matlab

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Pooja 2013-8-7

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编辑： Walter Roberson 2017-1-24

n=1011(binary) 11 in decimal and b= 3(decimal) and finally getting answer as 011(binary). the b LSB of n to produce the subexponential code of n. How to implement it in matlab?

n=1011 b=3 z=1|011 ans=011

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Answer 1

dpb 2013-8-7

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编辑：dpb 2013-8-7

在 MATLAB Online 中打开

Ohh....I think I finally figured out what want--the (variable) number of LSB(its) of the input value...ok, if interpretation is correct.

r=bitand(n, bin2dec(repmat('1',1,b)));  % value 
r=dec2bin(bitand(n, bin2dec(repmat('1',1,b))),b);  % bit representation

Use whatever integer is appropriate for the problem space; Matlab will work on default doubles if all are integers.

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Pooja 2013-8-8

在 MATLAB Online 中打开

This is used in decoding part of subexponential codes used for image compression.n is binary which is i/p binary stream,k be a parameter.There are two conditions for decoding n<2^k and n>=2^k.

if n>2^k for eg. n=110011,k=3 need to split the first part upto the stopbit 0.This is the unary part(110) and also want to separate 011 as binary part(another part). Stop bit is 0,the first 0 we see after 1,eg 110,0 is the stop bit.

After splitting we take the binary part and do bitor with 2^b ,where b we calculate by an equation.

Second case is n<2^k,here start bit will be 0. for eg.n=010011,k=3 i need to decode with k bits ie.3 bits needed to be taken as the decoded value.

My problem is to implement the first case and second case in matlab.I am copy pasting the C code of the decoding portion of subexponential coding.

if true
  % uintmax_t result =0;
// Subexponential has two different cases
if (( membuff -> GetBit ())==0){
// Case n <(2^ k)
return ( int ) membuff -> Get (k );
} else {
// Case n >(2^ k)
int b ,u;
uintmax_t base ;
// Count the number of 1' s until the stop bit .
u =1; // We already have read the first 1
while ( membuff - > GetBit ()){
u ++;
}
// Now the least b significant bits have to be read .
b=u +k -1;
base =1 < < b;
result = ( base | membuff - > Get (b ));
return ( int ) result ;
}
end

Thank you so much for helping me.

dpb 2013-8-10

在 MATLAB Online 中打开

n>2^k.
b = floor(log2(n)) = floor(log2(11)) = 3 = 011
u = b-k+1 = 3-2+1 = 2 --> 110  (????)

How do you get '110' from u=2???

As for the C code, it might be clear(er) if one had the header file. I presume there's a GetBit macro and there has to be a definition for the data type.

If it is clear to you, I suggest translating it directly; it isn't clear to me, at least w/o the missing pieces (but I'm no c-lizard).

for decoding

ip= 110011

if the first bit is 1,...

But in an 8-bit integer, there are 3 (unshown) leading zeros so how's one to know the "first" bit is the one?

... Above mentioned is the second case of n>2^k

And again we're back to how do you determine which group of bits to use?

i/p like 011101, the first bit is 0, ...

But how is there any indication in the input uint8 that the first bit isnt the '1' of the '11101' portion?

I can't see any unequivocal way to interpret an integer from the description. I'm sure there must be but there's something missing in the definition to remove that.

IOW, if one takes any input integer (other than identically 0) then there's always a leading 1 bit at some point. If you can't mask off k LSBs and use the rest, how does one decide?

Counting bits in a true bitstream would work (I think), but that's not what you're saying your input is (unless you've translated one into something else). I'm guessing that's what the C code macro(?) GetBit() must be doing is scanning the memory buffer but it's not possible to say for absolute certain (but it seems it must be else't afaict there is no unique solution).

So now we're back to the original problem/question -- what's the real datastream?

dpb 2013-8-10

编辑：dpb 2013-8-11

在 MATLAB Online 中打开

do you have the rest of the C code to disambiguate it?

OK, I went to your link and I see that's where you got the code from--after reading it again, I'm now positive you have to create your input as a bitstream to be able to decode it--saving it as an integer array is worthless.

It falls back now onto your generation mechanism--once you figure that out and generate a useful data source, then we can talk about how to process it in Matlab.

And let's use your example above to illustrate why you have a problem--you say the coded value is to be 110011

>> ip=uint8(bin2dec('110011'));     % store as an uint8
>> dec2bin(ip,8)                    % display internal storage
ans =
00110011
>>

Now, we're supposed to read the first bit and decide based on whether it's set or not whether there's a u or this is it...but, unless the code width is always precisely 8 bits, the first bit is always going to be zero. There's simply no way to know which is actually the first bit in the code inside the integer not knowing a priori that the code is M bits--but your algorithm generates a variable-length code dependent on the magnitude of the value--Catch 22.

And, it's clear in reading the verbal description your reference gives in the steps to decode--

"[If] k is known, subexponential codes created following the directives specified ... can be decoded following these steps:

1. Read the first bit from the code.

2. If the first bit is 0, the following k bits are the decoded value. Otherwise, the unary code of u must be read.

It still would be a_good_thing (tm) if the missing header file(s)/functions/macros of the sample implementation were around somewhere--don't suppose you've any access to any of the author(s)? The paper at the link misses the header file with the specification of the assumed memory buffer structure and the macros(?) GetBit and Get that do the fetching therefrom.

dpb 2013-8-11

编辑：dpb 2013-8-12

As I've said all along, it all depends on the how the input "is coming". What is the storage mechanism of the values? If they're in an integer, the answer as above is "you can't" w/o some additional information as the leading bit will always be zero unless the code length exactly fills the width of the storage word (which it can't in general whatever the word length because the code length is variable). Again, I repeat, "Catch-22".

I spent a little time this morning reading the paper more thoroughly -- the giveaway is on p. 26, the middle paragraph below the numbered steps of how to generate the code. Note carefully the penultimate sentence of that paragraph which I quote --

"Finally, the length of the produced code must be computed in order to properly transfer the coded value to the next stage."

Unfortunately, the author didn't include the code or even a description of that "transfer to the next stage" mechanism although one can presume from the code of the subsequent decoder he built a memory buffer structure and overlaid a continuous bit stream into it w/ the aid of some unpublished helper functions/macros.

You've got to also do the same thing in your mechanism somehow; it's not possible to tell you the coding to do the decoding until you define what that storage transfer mechanism is going to be.

The obvious choices are to

a) build a bitstream as he apparently did, or

b) use a character or byte stream where each element represents a bit

c) use integer storage but pass a second vector of the same length that is the length of the corresponding code element.

The first is quite involved to implement as Matlab code as one doesn't have ready access to the low-level addressing while the latter two are pretty easy but have obvious performance and storage issues for anything other than demonstration purposes.

But, the problem is that you can't get there from here without one of the above or another equivalent way to be able to infer the code length in the decoder.

ADDENDUM:

Perhaps there's another way

d) Prepend a 'start' bit to the code in the integer coding--that would seem w/o a lot of thought given to contain enough information at minimal additional overhead/cost and not destroy the code.

ADDENDUM 2

e) IFF one can assure that the code length is always < word width, one could modify the coding slightly and eliminate the problem of the length. Since there's only a leading '0' for input n < (2^k), instead of coding it as b=0|u=n, store -n for those cases and use the subexponential coding only for n >= (2*k). Then you simply test for <0 and negate those cases in the decoder and use the algorithm for >=0.

dpb 2013-8-11

编辑：dpb 2013-8-11

All in all, I don't think you're either reading or comprehending what I've just spent a great deal of time and effort explaining and your repeating the same thing over and over doesn't help get us anywhere. :(

I understand that conceptually the coded values represent a bitstream. The problem is one of how are you storing these code values in Matlab and getting them to the desired decoder function?

Matlab doesn't have any such data structure as a "bitstream" -- it has various length integers, floats and higher-level abstractions built of the same.

And there's the problem--when you place the code value in one of those you've lost the implied length of the code irretrievably unless you've done something additional to allow it to be recovered.

Show a piece of Matlab coding that computes a coded value and then passes that value to the desired decoder.

You talk in words and give example values to decode but you don't provide the requested/needed information on how they are being manipulated in Matlab to have any way to know how to (or even for certain whether can) decode them in Matlab.

And, unless you use one of the previously-outlined alternatives or some other equivalent it can't be done in general from an integer representation. If you have somehow simulated a bitstream, then must have the details of that representation to know how to access it.

Pooja 2013-8-12

I think i am not able to make out the question.I have stored the output of the encoded binary stream as save temp.txt .And for decoding i am calling as load temp.txt.

dpb 2013-8-12

编辑：dpb 2013-8-15

I think i am not able to make out the question...

Clearly. We've had this problem from the very first question that I can't seem to get across to you the difference between internal representation and the conceptualization of what that representation means.

And the underlying issue is that whatever bits you calculate if they're stored in anything except a sequence of only that number of bits are then irretrievable as to which are/aren't actually data in the larger entity when the code length they represent isn't a fixed length.

So, instead of trying to explain verbally show some source code. It will have a precise meaning in a common language...from that starting point maybe we can finally reach a mutual understanding.

Post your Matlab code that does the calculation of the coding and the write...it can be the few lines needed to read in or generate an input array, the call of the coder that returns the encoded result and the subsequent write/save to the file. That should be able to be brought down to a half-dozen lines or so at most I'd think.

Then, for completeness show the encoding routine and the line or two necessary to read the file. Wouldn't hurt to add the first few lines (3 or 4, say) of the file as well.

I'm pretty sure this will show that you're saving each encoded result in an array element, either an integer or default double, in which case you have the problem I've outlined above.

Do you understand the comment I made time stamped 11 Aug 2013 at 15:50? In particular, do you understand the implications of the sentence from the paper quoted there? If not, we've got a hurdle to get you over to see why there's a problem here.

I'll repeat it here for a reminder--

... p. 26, the middle paragraph below the numbered steps of how to generate the code. Note carefully the penultimate sentence of that paragraph--

"Finally, the length of the produced code must be computed in order to properly transfer the coded value to the next stage." (emphasis added)

It's that length that isn't available if the data are passed as individual array elements--then they're all the length of the (internal) storage unit and the length of the code stored in the array element is unknown. Hence we (as the decoder) don't know and can't tell which is the first encoded bit within all the bits of the array element.

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