how to return to a back step by checking back steps . problem in 4 and 5 step

Question

Sultan Mehmood 2019-6-13

编辑： madhan ravi 2019-6-14

step 2:

x=0.3;
x=0.3;
p=0.343;
for n=2:65536;
if x(n-1)>=0 & x(n-1)<=p
 x(n)=x(n-1)/p;
else
 x(n)=(1-x(n-1))/(1-p);
end 
end

step 3:

K(n)= mod(floor((x(n)*10^2-floor(x(n)*10^2))*10^3);,256);
if K(n)<3
    K(n)=K(n)+3;
end 

step 4. Checking ?(n), if ?(n) < 3, then ?(n) = ?(n) + 3.

Step 5. Let n ← n+ 1, return to Step 3 until n reaches ?.

Answer 1

Raj 2019-6-14

After running the code upto here:

x=0.3;
x=0.3;
p=0.343;
for n=2:65536;
if x(n-1)>=0 & x(n-1)<=p
 x(n)=x(n-1)/p;
else
 x(n)=(1-x(n-1))/(1-p);
end 
end

You get x as an 1x65536 matrix. Then you have to calculate K like this:

for ii=1:numel(x)
K(ii)= mod(floor((x(ii)*10^2-floor(x(ii)*10^2))*10^3),256);
end

Then you have to check for each element of K w.r.t to the condition like this:

for ii=1:numel(K) % I assume L is total number of elements in K
if K(ii)<3
    K(ii)=K(ii)+3;
end 
end

I am not sure but is this what you are looking for? Where is the problem?