For loop for equation

Hi

I have a vector of 42*1 values and need to implement two other vectors(l and u) of the same size. The two vectors have a starting value of l(1)=0 and u(1)=1 but need to follow the equation as shown in the image.

My attempt at the code was

l=zeros(size(prob));
u=zeros(size(prob));
u(1,:)=1;
l(1,:)=0;

for i=2:length(u)
    for j=2:length(l)
        l(i)=l(i-1)+(u(j-1)-l(i-1))*prob(i-1);
        u(j)=l(i-1)+(u(j-1)-l(i-1))*prob(j);
    end 
end 

but this did not work. Any help is much appreciated

[Chris]

I don't think you need two loops--you would only want to count 2:n once.

This code sets i to 2, and runs through j==2:42. Then it moves to i==3 and does j==2:42 again. The image only has n as the single iterator.

[Maaz Madha]

I tried that as well but it only repeated the first value in the probability vector for every value in both l and u

[Chris]

If you post what you tried, we might be able to find something wrong with it.

[Chris]

Or, eventually someone will come along and answer your homework question for you.

[Maaz Madha]

F = fopen('Compression.txt','r');
Data=fread(F);%equivalent to str

CharData=char(Data);
disp(Data)
len=length(Data);
C=unique(CharData);
D=unique(Data);
E=[histc(Data,D)];
S=sum(E);
prob=E./S;
u=C;
fprintf('The unique characters are : %s\n',u);
len_unique=length(u);

%% General lookup table


z=zeros(1,len_unique);

cpr=cumsum(p);
newcpr=[0 cpr'];

display(newcpr)
interval=zeros(size(len_unique));
for i=1:len_unique
    interval(i,1)=newcpr(i);
    interval(i,2)=cpr(i);
end 

%% Encoder table
low=0;
high=1;
[tbh,idk]=ismember(D,Data);

% 
% pos=idk;
% 
% % displaying tag value
% %% Low
% 
% 
%% Part I'm struggling with
l=zeros(size(prob));
u=zeros(size(prob));
u(1,:)=high;
l(1,:)=low;

for i=2:length(u)
        l(i)=l(i-1)+(u(i-1)-l(i-1))*prob(i-1
        u(i)=l(i-1)+(u(i-1)-l(i-1))*prob(i);
end  

T

The table for my probability is

 a

and my vector l and u are showing the same thing

I've even attached my compression text if you want to run the code for yourself.

Answer 1

[Voss]

l=zeros(size(prob));
u=zeros(size(prob));
u(1,:)=1;
l(1,:)=0;
for i=2:length(u)
    l(i)=l(i-1)+(u(i-1)-l(i-1))*prob(i-1);
    u(i)=l(i-1)+(u(i-1)-l(i-1))*prob(i);
end