Matrix formation by applying some operation using loop logic.

 This first matrix table1 contains 5 names and 5 scenes.I need to perform some operations on this  matrix and I have to obtain second matrix as shown in table2.

 Diagonal elements of table2 should obtained by performing addition of 1st row,2nd row,3rd row..and so on.Remaining elements obtained by comparing 2 rows and summing them.Suppose take table1 as A matrix and table2 as B matrix.

 OPERATION1:For diagonal elements

 B(1,1)=7+7+7+1+0=22
 B(2,2)=6+0+0+0+0=6
 B(3,3)=0+6+0+4+0=10…….and so on

 OPERATION2:For remaining elements

 B(1,2)=MIN(A(1,1),A(2,1))+ MIN(A(1,2),A(2,2))+ MIN(A(1,4),A(2,4))+ MIN(A(1,5),A(2,5));
 B(1,3)=
 B(1,4)=
 B(1,5)=

 Table1:

              Scene1   Scene2 Scene3  Scene4   Scene5  
              

 BASAVARAJ  7  7  7  1  0

 MANOJ    6  0  0  0  0

 NATESH    0  6  0  4  0

VIJAY    0  0  0  4  2

GOWDA    0  0  6  0  2

 Table2:

            BASAVARAJ   MANOJ   NATESH    VIJAY    GOWDA
        

BASAVARAJ  22    6   7    1     6

MANOJ    6    6   0    0     0

 NATESH    7    0  10    4     0

 VIJAY    1    0   4    6     2

 GOWDA    6    0   0    2     8

Answer 1

[Chandrasekhar]

A = [7 7 7 1 0;6 0 0 0 0; 0 6 0 4 0;0 0 0 4 2;0 0 6 0 2];
for i = 1:5
    for j = 1:5
        if i == j
           B(i,j) = sum(A(i,1:end));
        else
            minval = 0;
            for k = 1:5
            minval = minval + min(A(i,k),A(j,k));
            end
            B(i,j)= minval;
        end
    end
end

[prashanth]

thanx thanx thank you so much.

Answer 2

[Andrei Bobrov]

s = size(A);
b = nchoosek(1:s(1),2);
B = zeros(s);
B(tril(true(s),-1)) = sum(min(A(b(:,1),:),A(b(:,2),:)),2);
B = B + B.';
B(eye(s)>0) = sum(A,2);

[Chandrasekhar]

excellent..Really simple.. thanks