can anybody help me?

function  shannon_fano(h)
i=input('Enter the message ensembles :','s');
h=unique(sort(i));
j=histc(i,h);
e=fliplr(sort(j));
display (h);
display (e);
n=length(i);
m=e/n;
display (n);
display (m);
probabilities_calculation = zeros(size(j));
for i = 1:length(h)  %find the probabilities of the symbols/occurence of each letter
probabilities_calculation(i) = sum(j==h(i))/length(h);
end
p = sort(probabilities_calculation(:),'descend');
% create the cell array of codes
 codes = cell(size(h));
 % call the recursive encoder function
 codes = shannon_encoder(1,length(p),p,codes);
 display (p);
      function shannon_encoder(begin_point,end_point,p)
high_point = begin_point;
low_point = end_point;
high_sum = p(begin_point);
low_sum = p(end_point);
while(high_point ~= low_point-1)
    if (high_sum > low_sum)
        low_point = low_point - 1;
        low_sum = low_sum + p(low_point);
    else
        high_point = high_point + 1;
        high_sum = high_sum + p(high_point);
    end
end
for i = begin_point:high_point
    p(i) = 0;
end
for j = low_point:end_point
    p(j) = 1;
end
if(high_point-begin_point+1 > 1)
      shannon_encoder(begin_point,high_point,p);
  end
if(end_point-low_point+1 > 1)
      shannon_encoder(low_point,end_point,p);
end

[KSSV]

don't use clear all inside the function. What is the problem? You have to specify the error.

[John D'Errico]

Let me guess. You don't even use the variable keyword in the function, yet you pass it in?

The fact that you then immediately clear, makes the variable even more useless.

There is absolutely no need to use clear at the beginning of a function. In fact, it is the worst thing you can do, since you are clearing any arguments you passed into the fcuntino, before they could have been used.

If there is something else wrong, then only you can know, since you don't tell anypone what the code should do, or what it does that you think is wrong.

[Arthur Richardo Samderubun]

??? Error using ==> Untitled22>shannon_encoder

Too many input arguments.

Error in ==> Untitled22 at 20

codes = shannon_encoder(1,length(p),p,codes);

this is the error

[Walter Roberson]

The line of code from the error message does not occur in the code you posted.

[Arthur Richardo Samderubun]

sorry im fix that now.

done.

can u help me now?

[Ioannis Andreou]

function shannon_encoder(begin_point,end_point,p)

In this definition of your function you allow 3 inputs but

codes = shannon_encoder(1,length(p),p,codes);

here you give 4 inputs. Thus you receive an error

[Walter Roberson]

codes = shannon_encoder(1,length(p),p,codes);

That implies that the function accepts a codes parameter, but it does not.

The line also implies that the function returns a value, but it does not return anything.

When you have a function that does not return anything, it is only useful for its "side effects", or sometimes for debugging.

Is it correct that your function is intended to be recursive?

Answer 1

[Arthur Richardo Samderubun]

Somebody can help me??

i want to ask, about shannon fano code

I already input character and get the results but entropy and coding efficency the result is NAN if i try to input the small number of characters all the results appear

this is the coding and you can try this coding, if somebody know that error i hope to inbox me on email ( [e-mail]). thank you so much

clc
clear
format short g
keyword = input('INPUT KEYWORD : \n','s');
tic; 
fprintf('\n %s \n\n\n',keyword);


symbols = sort(unique(keyword)) %,'descend'
probabilities_calculation = zeros(size(symbols))';
for i = 1:length(symbols)  %find the probabilities of the symbols/occurence of each letter
    probabilities_calculation(i,1) = sum(keyword== symbols(i))/length(keyword);
    probabilities_calculation(i,2) = symbols(i);
    probabilities_calculation(i,3) = sum(keyword== symbols(i));
end



rank  = sortrows(probabilities_calculation);
new_rank = zeros(size(rank));
for i = 1:length(symbols)  %find the probabilities of the symbols/occurence of each letter
    fprintf('\n %f || %s\n',rank(length(symbols) - i+1,1),char(rank(length(symbols) - i+1,2)));
    new_rank(i,:) = rank(length(symbols) - i+1,:);
end

new1 = sort(unique(new_rank(:,1)),'descend');

iterasi = 1;
for i=1:length(new1)
    for j = 1:length(symbols) 
        if new1(i) == new_rank(length(symbols) - j+1,1)
            data_baru(iterasi,1) = pembulatan(new_rank(length(symbols)-j+1,1));
            data_baru(iterasi,2) = new_rank(length(symbols)-j+1,2);
            data_baru(iterasi,3) = new_rank(length(symbols)-j+1,3);
            iterasi = iterasi +1;
            
        end;
    end;
end;


p = data_baru(:,1)';
sta=[];
ar=[];
coded=[];
other=[];

first=0;
final=length(p);
sta=[first final]

for f=1:length(p)              %loop for every step
    other=[];
    ar=[];
    for h=1:(length(sta)-1)   %loop for each sub-group
        first=sta(h)+1
        final=sta(h+1)
        if(first>=final)       %adding invalid numbers(here 2)
            other=[other; 2];  %-when only one element in a sub group
            continue;
        end
        asum=0;
        difmat=[]
        for i=first:final       %loop for finding difference vector
            asum=asum+p(i);
            resum=0;
            for j=i+1:final;    
                resum=resum+p(j);
            end
            dif=abs(asum-resum);
            difmat=[difmat dif]
        end
        small=min(difmat);
        k=1;
        for i=first:final        %loop for finding index of min difference
            if(small==difmat(k))
                break;
            end
            k=k+1;
        end
        index=i
        ar=[ar index]           %storing index in temporary stack
        ind=(index+1)-first     %calulating number of zeros
        remind=final-index      %-and ones for each sub group
        other=[other; zeros(ind,1); ones(remind,1)]
    end
    sta=[ar sta]                %creating final stack for each step
    sta=sort(sta)
    coded=[coded other]        %creating final code word matrix
    if(length(sta)>length(p))  %break when all sub groups have one element
        break;
    end
end


%display codewords
len=[];
for i=1:length(p)
    word=[];
    for(j=1:f)               % 'f' contains max number of bits among codes
        if(coded(i,j)==2)    % break when invalid number(i.e. 2) reached
            break;
        end
        word=[word coded(i,j)];
    end
    len=[len length(word)];
    fprintf('\nSymbol %d code ==> ',i);
    disp(word)

    code_final{i} = concat_array(word);
    fi_final(i) = data_baru(i,1);
    char_final(i) = data_baru(i,2);
    freq_final{i} = strcat(': ',num2str(data_baru(i,3)),'/',num2str(sum(data_baru(:,3)))); 
    li_final(i) = size(word,2);
    si_final{i} = char(data_baru(i,2));
    
    
end

for i=1:length(p)
    expansi_binner_final{i} = expansi_binner(code_final{i},code_final);
end;

cum_pi_final(1)=0.000;
for j=2:length(si_final);
    cum_pi_final(j)=cum_pi_final(j-1)+fi_final(j-1);
end

clc;
fprintf('\nKEYWORD = %s',keyword);
fprintf('\nKEYWORD LENGTH  = %i ',length(char(keyword)));
fprintf('\nUNIQUE  = %s ',symbols);
fprintf('\nUNIQUE LENGTH  = %i \n\n',length(char(symbols)));


DataHead = {'Si', 'Pi','Li','kode'};
%DataBody(:,1) = si_final'
%DataBody(:,2) = pi_final' 
%DataBody(:,3) = code_final' 
%DataBody(:,4) = li_final'

fprintf(1, '  \t %s\t\t\t\t %s\t\t\t %s\t\t \t%s\n', DataHead{:})

for i=1:length(si_final)
    fprintf(1, '\t%s %s \t\t %.3f\t\t  %d\t\t\t %s\n', si_final{i},freq_final{i},fi_final(i),li_final(i),code_final{i})
end;    






ent=0;
avginfo=0;
for i=1:length(p)
    ent=ent+(p(i)*log2(1/p(i)));
    avginfo=avginfo+len(i)*p(i);
end
eff=(ent/avginfo)*100;

fprintf('\nENTROPY = %.3f',ent);
fprintf('\nAVERAGE CODEWORD LEGHT = %.3f',avginfo);
fprintf('\nCODING EFFICIENCY = %.3f%s \n',eff,'%');
disp(['Duration = ',num2str(toc),' second']) 

[Walter Roberson]

pembulation is not a MATLAB function.

It looks like it might be an Indonesian word for rounding:

http://its-matematika.blogspot.com/2011/06/pembulatan-pada-matlab.html