CareerCup

Lets say we have m x n matrix.

In row-major order we have
to access (i,j) we have ((i * no_of_col) + j).
we have ((i * n) + j) in our case.

In col-major order we have
to access (j,i) we have ((j * no_of_row) + i).
we have ((j * m) + i).
since (i,j)th element in row-major form is the same as (j,i) in col_major_order and m x n will now be n x m.

so equating we have ((i * n) + j) = ((j * m) + i)
we get ni + j = mj + i i.e i / j = (m - 1)/(n -1).


Can somebody please verify this.

Can you give an example of what you're talking about? I don't understand. Thanks!

Khoa,

You would be knowing that any n-dimensional matrix (array) finally boils down to a sequential representation in memory.

Eg.:

a[0,0] a[o,1]
a[1,0] a[1,1]

would be stored as:

1-D Index 0 1 2 3
Address: 1000 1002 1004 1006
Item: a[0,0] a[0,1] a[1,0] a[1,1]

which is row major order.

Here, a[1,0] = a[2] (not literally, but in memory)

So, the Q was to write a general expression to represent this relation for both row and column major order representations of an array.

I hope this clarifies. Sorry, I typed the Q in a hurry so, it is ambiguous.

I misinterpreted the question. Thanks!

for row major

((no. of columns *i) +j)

where i=row no
j=colomn no

http://en.wikipedia.org/wiki/Row-major_order

If you want to find the actual position in the memory -

Location in memory - rownum * nocolumns * bytes + columnnum *bytes

Where bytes = No of bytes required to store that variable. E.g. int is 4 bytes, char is one byte etc.