array

Question

anyone please explain me how to find loc in lower triangular matrix , i am getting little bit confuse 

suppose matrix is          

                                               1     2       3        4

                                    1         a₁₁   a₁₂   a13   a₁₄

                                     2       a₂₁   a₂₂   a₂₃   a₂₄

                                     3      a₃₁   a₃₂   a₃₃   a₃₄

                                     4     a₄₁    a₄₂   a₄₃   a₄₄

 

 

now i have to find location of a[4] b[3]:- given BA=1000 C=1

=1000+ [(4-1) * ((3)(4))/2 +(3-1)]*1

by this way i we get answer but  something wrong please exolplain  

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Comments

i hope that formula is wrong.

Answer 1

lower triangular matrix means elements which are above the diagonal are 0 ===> we need not to store them.

let your lower bound = 1, your Array is nXn, you want address of A[i][j].

1 row contains ------------>1 element ( A[1][1] )

2 row contains ------------>2 elements ( A[2][1], A[2][2] )

3 row contains ------------>3 elements ( A[3][1], A[3][2], A[3][3] )

.

.

(i-1)^th row contains ------------>i-1 elements ( A[i-1][1], A[i-1][2], ............A[i-1][i-1] )

(i)^th row contains ------------>i elements ( A[i][1], A[i][2], ............A[i][i] )

 

 

for reaching A[i][j], you have to cross i-1 rows ===> you are at required row

total elements up to (i-1)^th row = 1+2+...+(i-1) = $\frac{(i-1)(i)}{(2)}$, you have to cross these elements to get required row

 

to get particular element A[i][j] in that row, you have to skip (j-1) elements.

Therefore total elements you have to skip =  $\frac{(i-1)(i)}{(2)}$ + (j-1)

to get that memory = BA + ( $\frac{(i-1)(i)}{(2)}$ + (j-1) ) * ( size of element )

Comments

thanks bro ..