How can I build a Matrix dependencies of a variabel and a matrix? Easy question..

*Here is an example,

N3 uses N2 and one "k" variable(k is 3 for N3). But I entered k variable by hand. Since I want to find N4 using N3 and "k" (k is 4 for N4), I want "k" goes automaticly...

k starts from 2 and ends whenever I want. I don't want to create another for loop because I have already 2 for loops.And I know there is an easy way.But I couldn't remember it. Something like= N3(:,:,3).*What we_ call *_for these matrices? If you don't know how to this way, can you tell me how to do with for loops?

Thanks and Best Regards..

------------------

N2=zeros(rr-3,rr-2);

for j=1:rr-3

for i=1:rr-2

     A1=(X(i+1)-X(i));

     A2=(X(i+2)-X(i+1));
      if A1==0 && A2==0
          N2(j,i)=0;
      elseif A1==0 && A2~=0
         N2(j,i)=((X(i+2)-t(j))/(A2))*N1(j,i+1);
      elseif A2==0 && A1~=0
          N2(j,i)=((t(j)-X(i))/(A1))*N1(j,i);
      elseif A2~=0 && A1~=0
          N2(j,i)=((X(i+2)-t(j))/(A2))*N1(j,i+1)+((t(j)-X(i))/(A1))*N1(j,i);
      end
  end

------------------------

N3=zeros(rr-3,rr-3);

for j=1:rr-3

for i=1:rr-3
    A1=(X(i+2)-X(i));
    A2=(X(i+3)-X(i+1));
    if A1==0 && A2==0
        N3(j,i)=0;
    elseif A1==0 && A2~=0
       N3(j,i)=((X(i+3)-t(j))/(A2))*N2(j,i+1);
    elseif A2==0 && A1~=0
        N3(j,i)=((t(j)-X(i))/(A1))*N2(j,i);
    elseif A2~=0 && A1~=0
        N3(j,i)=((t(j)-X(i))/(A1))*N2(j,i)+((X(i+3)-t(j))/(A2))*N2(j,i+1);
    end
end
end

------------------------

Answer 1

[Jan]

If you use N{3} instead of N3 an additional loop would be easy. At least I assume that it would. Unfortunately I cannot understand the first sentence already:

N3 uses N2 and one "k" variable(k is 3 for N3).

What is "one k variable" and where does this appear in the code? What is rr?

You could avoid the inner loop:

N = zeros(rr-3, rr-3);
for j = 1:rr-3
  A1 = X(2:rr-1) - X(1:rr-3);
  A2 = X(4:rr) - X(2:rr-2);
  % A1==0 && A2==0 :  N(j,i)=0;  Is the default already

    index = find(A1==0 & A2~=0);
    N(j, index) = ((X(index + 3) - t(j)) ./ A2(index)) * N2(j,index + 1);

    index = find(~A1~=0 & A2==0);
    N(j, index) = ((t(j) - X(index)) ./ A1(index)) * N2(j,index);

    index = find(A1~=0 & ~A2~=0);
    N(j, index) = ((t(j) - X(index)) ./ A1(index)) * N2(j,index) + ...
                  ((X(index + 3) - t(j)) / A2(index)) * N2(j, index+1);
  end
  N{3} = N;

Another approach would be to calculate both matrices and replace NaNs by zeros:

% !For demonstration only! Indices are most likely wrong!
A1 = X(2:rr-1) - X(1:rr-3);
A2 = X(4:rr) - X(2:rr-2);
c1 = bsxfun(@minus, (t ./ A1)' - X ./ A1) .* N2;
c2 = bsxfun(@minus, (X ./ A2)' - t ./ A2) .* N2;
c1(insan(c1)) = 0;
c2(insan(c2)) = 0;
N{3} = c1 + c2;

I did not care about the pile of indices here and I cannot run the versions due to the absence of test data. So treat this as an idea only and implement it by your own.

Answer 2

[STamer]

Thank you so much sir!

Finally,

N{1} is known, k is integer, rr is integer.However I couldn't apply "find" built-in function instead of elseif statements.Can you help me on it? And what's this bsxfun. Does it do all stuff in one row?

for kk=2:k
N{kk}=zeros(rr-3,rr-kk);
    for j=1:rr-3
    for i=1:rr-kk
        A1=(X(i+kk-1)-X(i));
        A2=(X(i+kk)-X(i+1));
        if A1==0 && A2==0
            N{kk}(j,i)=0;
        elseif A1==0 && A2~=0
            N{kk}(j,i)=((X(i+kk)-t(j))/(A2))*N{kk-1}(j,i+1);
        elseif A2==0 && A1~=0
             N{kk}(j,i)=((t(j)-X(i))/(A1))*N{kk-1}(j,i);
        elseif A2~=0 && A1~=0
            N{kk}(j,i)=((t(j)-X(i))/(A1))*N{kk-1}(j,i)...
            +((X(i+kk)-t(j))/(A2))*N{kk-1}(j,i+1);
        end
    end
    end 
end

[Jan]

Give me a chance to run this function by providing test data for N{1}, rr, t and X. Perhaps simple RAND data are sufficient, but most likely something more tricky is required to ensure that some A1 and A2 are zero.

[STamer]

   N{1}=  
       0     0     1     0     0     0     0
       0     0     1     0     0     0     0
       0     0     0     1     0     0     0
       0     0     0     0     1     0     0
       0     0     0     0     1     1     1

rr =8;
t=[  0    0.3063    0.5000    0.6937    1.0000];
X=[0         0         0    0.3333    0.6667   1.0000    1.0000    1.0000];

When k=3,

Answer N{3} should be;

N{3}=      1.0000         0         0         0         0
      0.0066    0.5712    0.4222         0         0
           0    0.1250    0.7500    0.1250         0
           0         0    0.4222    0.5712    0.0066
           0         0         0         0    1.0000

I checked it works.

[Jan]

[Ignore this] This is my magic FLAGFLAGFLAG because the forum does not allow to set a bookmark.

[STamer]

Can't you see my message? Why did it flag? Maybe you can send me a private message.

[Cedric]

[Ignore this] This is my magic FLAGFLAGFLAG because the forum does not allow to set a bookmark.

=> http://www.mathworks.com/matlabcentral/answers/42413#answer_86241

:-)

C

[Jan]

@STamer: I added a personal flag such that I can let Google find it again, when I'm back on my Matlab computer. I see so many messages per day, that I cannot remember where I wanted to add something after a test with Matlab. Sorry for the confusion.

Actually I do not like to sync my bookmarks through the internet. But of course I could sync a list of my visited links by Prism.

Unfortunately I'm too tired now for further experiments. I take a look into the problem tomorrow, if no other contributor has been faster.

[STamer]

Thank you sir.I'm waiting for your response.