Maxk gives an incorrect result

Hi all,

I am trying to get 4 different value from a matrix. I am using maxk to find these values indices. Maxk can find first two maximum value with indices but when I try to find third it gives an incorrect result. for example when ı select row "1" it found 2 maximum value with their column in matrix and gives me 7-15 then I select 7 for my row it should needs to give me 8th column but it gives 1. I cannot understand why

Here is my code:

for i = nodes
    if i == 1 [~,seed] = maxk(m(1:n),2);
    end
end
s1 = seed(1,1);
s2 = seed(1,2);
for i = nodes
    if i == 7 [~,seed2] = maxk(m(7,n),1);
    end
end

Answer 1

[Voss]

For the case that i == 1 you have:

maxk(m(1:n),2)

m(1:n)is the first n elements of m (going in column-major order, which is what MATLAB uses). For example:

m = magic(4)

m = 4×4

16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

n = 6;
m(1:n)

ans = 1×6

16 5 9 4 2 11

For the case that i == 7, you have:

maxk(m(7,n),2)

m(7,n) is the element(s) of m at row 7, column(s) n. If n is a scalar, that's one element. For example:

m = magic(7)

m = 7×7

30 39 48 1 10 19 28 38 47 7 9 18 27 29 46 6 8 17 26 35 37 5 14 16 25 34 36 45 13 15 24 33 42 44 4 21 23 32 41 43 3 12 22 31 40 49 2 11 20

n = 6;
m(7,n)

ans = 11

If you want to use maxk on rows of m, then the syntax would be:

maxk(m(1,:),2) % using first row of m
maxk(m(7,:),2) % using 7th row of m

For example (using m as the 7x7 magic square still):

m(1,:)

ans = 1×7

30 39 48 1 10 19 28

m(7,:)

ans = 1×7

22 31 40 49 2 11 20

[~,seed] = maxk(m(1,:),2)

seed = 1×2

3 2

[~,seed2] = maxk(m(7,:),2)

seed2 = 1×2

4 3

[Kk]

Thank you very much. Is there a iterative way to do that. I mean I would like to create a set named seed which will have 4 value. First I would like to pick max value while row = 1 then I Would like to select a second value while row = first max value. But here is the thing the code should not consider the first selected max value when choosing the second max value. Is there a way to select four value with iterative way ?

[Voss]

If I understand correctly, you are finding the indexes of the 2 maximum values in row 1, so that if the maximum occurs in column 1, you use the 2nd maximum's index instead, since you don't want to use row 1 again. Is that right?

m = magic(4)

m = 4×4

16 2 3 13 5 11 10 8 9 7 6 12 4 14 15 1

current_row = 1;
seed = zeros(1,4);
for ii = 1:4
    [~,idx] = maxk(m(current_row,:),2);
    if idx(1) == current_row
        current_row = idx(2);
    else
        current_row = idx(1);
    end
    seed(ii) = current_row;
end
disp(seed)

     4     3     4     3

I'm not sure that's what you mean. If not, can you provide a concete example where you step through the process?

[Kk]

I could not explain myself cleary very sorry about that. What I am trying to say is actually first step of Grasp Algorithm which is creating a seed set. For example:

I have a distance matrix which name is dist

dist = [ 0 5 10 15
    5 0 21 22
    10 21 0 13
    15 14 13 0];

I try to create a seed set. First value of seed set is:

[~,seed] = maxk(dist(1,:),1);

which is 4 (Value of 15). Second value of seed set needs to be 2 (value of 14).

[~,seed] = maxk(dist(4,:),1);

at the end my seed set should be 4 and 2.

I am trying to this actually. But I try to this in the same part not seperatly.

[Voss]

I think I see. 15 at index 1 in row 4 is the maximum, but you can't use 15 again, so you pick 14 at index 2, the second-largest element of row 4, so seed #2 is 2 not 1. Is that it?

dist = [ 0 5 10 15
    5 0 21 22
    10 21 0 13
    15 14 13 0];

m = dist;
n_seeds = 2;
current_row = 1;
seed = zeros(1,n_seeds);
for ii = 1:n_seeds
    % just use regular max (only need one max value)
    % since m will have NaNs where previously used 
    % values occurred (see below)
    [~,idx] = max(m(current_row,:));
    % set any element of m equal to m(current_row,idx)
    % to NaN, so it cannot be selected again by max
    m(m == m(current_row,idx)) = NaN;
    % next row is index of this value
    current_row = idx;
    % set the seed value
    seed(ii) = current_row;
end
disp(seed)

     4     2

[Kk]

That is exactly the same thing what I was trying to say and do. Thank you very much mate.

[Voss]

You're welcome!