Solve Large Sorted Matrix in MATLAB

[hoffmann]

I have restated my question even further:

Suppose I have a sorted matrix:

[ 0 0 0 1 1 1 1 3 3 3 5 6 7 8 9 9 ]

I need to remove the last value in the matrix until the number of entries that are greater than one equal the number of entries that are either 0 or 1. For the example given above, there are 16 entries. Nine entries are greater than 1 and 7 entries are less than or equal to one. I would need to remove the last two entries such that the number of entries that are greater than one equal the number of entries that are either 0 or 1.

What if I have a much larger set of values? How do code this in MATLAB?

Thanks!

 

[zyh]

I think the algorithm's can be like this:

1. sum the numbers of type A and type B in your string, for example, type A means if x<=1,and type B means if x>1, and after that, we get the number of A(NoA) and number of B(NoA) in your string.
2. move the pointer from the last entry,and remove these last entries, and caculate the remaining number of A(RoA) and number of B(RoB), you know, it's very simple to get the number of A and B you removed.
3. Each time, you remove the last entry and check if RA==RB, then you get the solution.

I hope you can understand my algorithm.

 

[oswaler]

I would approach this problem by first understanding the logic behind the desired outcome. In this case, we want to remove the last values in the matrix until the number of entries greater than 1 equals the number of entries that are either 0 or 1. This can also be rephrased as wanting the number of entries greater than 1 to be equal to half of the total number of entries in the matrix.

To solve this problem in MATLAB, we can use a combination of logical indexing and the built-in function "sum". First, we can create a logical index for the entries in the matrix greater than 1 by using the ">" operator. This will return a matrix of 1s and 0s, with 1s representing entries greater than 1. We can then use the "sum" function to count the number of 1s in this logical index.

Next, we can compare this number to half of the total number of entries in the matrix using the "==" operator. If the number of 1s is greater than half of the total number of entries, we can remove the last value in the matrix using the "end" keyword and re-run the count and comparison process. This can be done in a while loop until the condition is met.

The code might look something like this:

% given sorted matrix
A = [0 0 0 1 1 1 1 3 3 3 5 6 7 8 9 9];

% initialize logical index for values greater than 1
greater_than_1 = A > 1;

% count number of entries greater than 1
num_greater_than_1 = sum(greater_than_1);

% compare to half of total number of entries
while num_greater_than_1 > numel(A)/2
% remove last value in matrix
A(end) = [];
% re-calculate number of entries greater than 1
num_greater_than_1 = sum(greater_than_1);
end

% final matrix after removing values
disp(A)

This code can be easily adapted to work for larger matrices by simply changing the initial matrix and adjusting the comparison criteria. Overall, the key is to understand the problem and break it down into smaller logical steps that can be implemented in MATLAB.