For loop for equation

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Maaz Madha
Maaz Madha 2021-12-14
评论： Maaz Madha 2021-12-15
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
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Chris
Chris 2021-12-15
Or, eventually someone will come along and answer your homework question for you.
Maaz Madha
Maaz Madha 2021-12-15
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.

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回答（1 个）

Voss
Voss 2021-12-15
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

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