Recurrence Relation for a Multidimensional Matrix
$begingroup$
Let us consider a $2$ dimensional matrix. Suppose we store all the elements of this matrix in a columnar fashion in a single one dimensional matrix or an array, then we would basically store the matrix in order $A[0][0] A[1][0] A[2][0]...A[n_1-1][0] ... A[0][1] A[1][1] ... A[n_1-1][1] .... A[0][n_2-1]...A[n_1-1][n_2-1]$
For a particular element $A[i_1][i_2]...[i_m]$ of an $m$ dimensional matrix $A[n_1][n_2]...[n_m]$ if we denote $e_m$ as the number of elements in column wise fashion before the given element, we can see the following
$$ e_1 = i_1$$
$$ e_2 = i_1 + i_2 times n_1$$
$$ e_3 = i_1 + i_2 times n_1 + i_3 times n_1 times n_2$$
How to show that generally
$$ e_m = e_{m-1} + i_m times prod_{j=1}^{m-1}{n_j}$$
If we look at the two dimensional matrix, the number of elements before a particular matrix location can be calculated as the column number of the element we are looking for summing with the $row times column$ number of elements. How does the above recurrence relation work and does it relate to this multiplication logic?
linear-algebra matrices recurrence-relations
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
$endgroup$
add a comment |
$begingroup$
Let us consider a $2$ dimensional matrix. Suppose we store all the elements of this matrix in a columnar fashion in a single one dimensional matrix or an array, then we would basically store the matrix in order $A[0][0] A[1][0] A[2][0]...A[n_1-1][0] ... A[0][1] A[1][1] ... A[n_1-1][1] .... A[0][n_2-1]...A[n_1-1][n_2-1]$
For a particular element $A[i_1][i_2]...[i_m]$ of an $m$ dimensional matrix $A[n_1][n_2]...[n_m]$ if we denote $e_m$ as the number of elements in column wise fashion before the given element, we can see the following
$$ e_1 = i_1$$
$$ e_2 = i_1 + i_2 times n_1$$
$$ e_3 = i_1 + i_2 times n_1 + i_3 times n_1 times n_2$$
How to show that generally
$$ e_m = e_{m-1} + i_m times prod_{j=1}^{m-1}{n_j}$$
If we look at the two dimensional matrix, the number of elements before a particular matrix location can be calculated as the column number of the element we are looking for summing with the $row times column$ number of elements. How does the above recurrence relation work and does it relate to this multiplication logic?
linear-algebra matrices recurrence-relations
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
$endgroup$
add a comment |
$begingroup$
Let us consider a $2$ dimensional matrix. Suppose we store all the elements of this matrix in a columnar fashion in a single one dimensional matrix or an array, then we would basically store the matrix in order $A[0][0] A[1][0] A[2][0]...A[n_1-1][0] ... A[0][1] A[1][1] ... A[n_1-1][1] .... A[0][n_2-1]...A[n_1-1][n_2-1]$
For a particular element $A[i_1][i_2]...[i_m]$ of an $m$ dimensional matrix $A[n_1][n_2]...[n_m]$ if we denote $e_m$ as the number of elements in column wise fashion before the given element, we can see the following
$$ e_1 = i_1$$
$$ e_2 = i_1 + i_2 times n_1$$
$$ e_3 = i_1 + i_2 times n_1 + i_3 times n_1 times n_2$$
How to show that generally
$$ e_m = e_{m-1} + i_m times prod_{j=1}^{m-1}{n_j}$$
If we look at the two dimensional matrix, the number of elements before a particular matrix location can be calculated as the column number of the element we are looking for summing with the $row times column$ number of elements. How does the above recurrence relation work and does it relate to this multiplication logic?
linear-algebra matrices recurrence-relations
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
$endgroup$
Let us consider a $2$ dimensional matrix. Suppose we store all the elements of this matrix in a columnar fashion in a single one dimensional matrix or an array, then we would basically store the matrix in order $A[0][0] A[1][0] A[2][0]...A[n_1-1][0] ... A[0][1] A[1][1] ... A[n_1-1][1] .... A[0][n_2-1]...A[n_1-1][n_2-1]$
For a particular element $A[i_1][i_2]...[i_m]$ of an $m$ dimensional matrix $A[n_1][n_2]...[n_m]$ if we denote $e_m$ as the number of elements in column wise fashion before the given element, we can see the following
$$ e_1 = i_1$$
$$ e_2 = i_1 + i_2 times n_1$$
$$ e_3 = i_1 + i_2 times n_1 + i_3 times n_1 times n_2$$
How to show that generally
$$ e_m = e_{m-1} + i_m times prod_{j=1}^{m-1}{n_j}$$
If we look at the two dimensional matrix, the number of elements before a particular matrix location can be calculated as the column number of the element we are looking for summing with the $row times column$ number of elements. How does the above recurrence relation work and does it relate to this multiplication logic?
linear-algebra matrices recurrence-relations
linear-algebra matrices recurrence-relations
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
asked Jan 6 at 17:35
Kaustabha RayKaustabha Ray
18610
18610
add a comment |
add a comment |
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