Change index of element in matrix with constraint

Hi all!

i have i binary matrix A(180,60), in each column i have 3 "ones" and only "one" by row,the sum of each row equal to 1.

i used this , but i had 3 "ones" in the 3 successive rows. How can i change the index of "1"?

M = 180; N =60;

k = 3;

C = repelem(eye(N), k, 1);

NEED URGENT HELP !

[Riccardo Spica]

Hi,

does this work?

N =60;
k = 3;
C = eye(N);
C = C(repelem(1:N,1,k),:);

Answer 1

[Chunru]

A = repmat(eye(6), 3, 1)   % change 6 to 60

A = 18×6

1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0

% if you need random permutation
B = A(randperm(6*3), :)

B = 18×6

0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0

[Ideth Kara]

i have a constraint :

after 3 "ones" in the 3 successive rows , i can't put "1" in the first column.

after 3"ones" in the next successive rows, i can't put "1" in the second column,...

[Chunru]

Then "A" in above code always satisfies your requirement (since no 3 "ones" in the 3 successive rows). Or do you also have other requirements?

[Ideth Kara]

In A , there is "1" in the 7th row.

i need 3 'ones' in the 3 successive rows but with different position. as condition , after the 3 rows , the rest os first column will be "zero", after the next 3 rows , the rest of second column "zero",...

like this :

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 1 0 0 0

0 1 0 0 0 0

0 0 0 1 0 0

0 0 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

[Chunru]

Then how can you ensure sum of each column is 3?

[Ideth Kara]

the maximum is 3 , it means that i can have 2 "ones" per column or one per column.

[Chunru]

I could not understand your requirements:

"i need 3 'ones' in the 3 successive rows but with different position. as condition , after the 3 rows , the rest os first column will be "zero", after the next 3 rows , the rest of second column "zero",..."

[Ideth Kara]

i will try to explain by using independent matrix :

A [3,6]

1 0 0 0 0 0

1 0 0 0 0 0

0 1 0 0 0 0

B[3,7]

0 1 0 0 0 0 0

0 0 1 0 0 0 0

0 0 0 1 0 0 0

C[3,8]

0 0 1 0 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

D[3,9]

0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 1 0 0

until i get [3,60]. but all this will be in the same matrix [180,60]