How Do You Create a Sub-Matrix Based on Vector Conditions?

[Saladsamurai]

I am confusing the heck out of myself here.

This is what I have

Square matrix that is M = 'm x m'
A column vector that is C = 'm x 1'
a smaller square matrix that is R ='r x r'Here is what I am trying to accomplish; I usually do this by hand:

The entries of the column vector C are either 1 or 0. If the i^thentry in C is zero then I can "cross out" the i^th row AND column in M. For example, if M is 5 x 5 and the 2nd and 4th entries of C are 0, I draw a line through the 2nd and 4th row AND column of M.

This leaves me with a 3 x 3 sub-matrix that is made up of the un-crossed out entries of M.

This sub-matrix is R. So I need to find a way to assign these values to R from M after crossing out rows and columns according to the entries of C.I have already created a test to initialize a matrix R of the correct dimensions; it is now a matter of getting the entries in there.

Something like:

For i = 1 to M
    For j = 1 to M
    If C(i,1) = 1 Then {R[something,something] = M[Something,Something]

     Next j
Next i

But I think I will need 2 more counter variables in addition to i and j so that if C(i,1)=0 the indices of R do not increment.

I am just a little lost now. Can I get a little help here?

 

[davee123]

I would do something along the lines of:

(Pseudocode)

rx = 0
ry = 0
for(mx = 1 to M) {
    if(C[mx] == 1) {
        ry = 0
        for(my = 1 to M) {
            if(C[my] == 1) {
                R[rx][ry] = M[mx][my]
                ry++
            }
        }
        rx++
    }
}

DaveE

 

[Saladsamurai]

I would do something along the lines of:

(Pseudocode)

rx = 0
ry = 0
for(mx = 1 to M) {
    [B]if(C[mx] == 1) {
        ry = 0[/B]
        for(my = 1 to M) {
            if(C[my] == 1) {
                R[rx][ry] = M[mx][my]
                ry++
            }
        }
        rx++
    }
}

DaveE

Hey there Dave I am kind of bad at this...what is the part on bold doing?

 

[davee123]

It's saying if the value at C[mx] (the mxth value in the single-dimensional matrix C) is equal to 1, then execute the code that's in the following set of braces. The first line in that sub-block of code says to set ry equal to 0.

I should note that actually, there's a little confusion there in terms of indexing. If your arrays are 0-based, then you want:

rx = 0
ry = 0
for(mx = 0 to M-1) {
    if(C[mx] == 1) {
        ry = 0
        for(my = 0 to M-1) {
            if(C[my] == 1) {
                R[rx][ry] = M[mx][my]
                ry++
            }
        }
        rx++
    }
}

And if your arrays are 1-based, then you want:

rx = 1
ry = 1
for(mx = 1 to M) {
    if(C[mx] == 1) {
        ry = 1
        for(my = 1 to M) {
            if(C[my] == 1) {
                R[rx][ry] = M[mx][my]
                ry++
            }
        }
        rx++
    }
}

DaveE

 

[LudusRex]

I can understand how confusing this task may seem. Let me break it down for you.

Firstly, it is important to understand the purpose of the program. It seems like you are trying to create a sub-matrix R from a larger matrix M, based on the entries of a column vector C. The column vector C contains only 1s and 0s, and if the ith entry in C is 0, then the ith row and column in M should be "crossed out". This means that the values in the corresponding row and column should not be included in the sub-matrix R.

To achieve this, you will need to use a nested loop. The outer loop will go through each row of M (i.e. for i = 1 to M) and the inner loop will go through each column of M (i.e. for j = 1 to M). Within this loop, you will need to check the value of C(i,1). If it is 1, then the value in M(i,j) should be included in R. So, you can assign it to R using R[i',j'] = M[i,j], where i' and j' are the indices for R. These indices can simply be the values of i and j, since you are only including values from the un-crossed out rows and columns.

However, if C(i,1) is 0, then you do not want to include the value in M(i,j) in R. In this case, you can simply skip over the assignment step and move on to the next iteration of the inner loop.

So, your code may look something like this:

For i = 1 to M
For j = 1 to M
If C(i,1) = 1 Then
R[i,j] = M[i,j]
End If
Next j
Next i

This way, you will only include the values from the un-crossed out rows and columns in R.

I hope this helps to clarify the logic of your program. Good luck!