fill the entry of a vector with a given distance

Hi there

If I have a vector A of size . I want to put some value, say 1, into some certain entries. Here are the rules:

1. given a distance or guard space denote by d.
2. the first entry must be filled.
3. then the next entry will be filled at from the last filled entry.
4. For the final filled entry, we need to check if the filled entry has enough space larger than d. For example, if and , the first place to be filled is entry 1, the second place to be filled is entry 5, and there is no the third place to be filled because for the entry 9, there is no enough space left, i.e., entry 9 and entry 10 is only has one distance, which is less than .
5. I want to generate multiple vectors, such that each vector only has one place to be filled. For example, if $d=3 and , then for the first vector, its first place will be filled; for the second vector, its 5th place will be filled; and there is no the third vector due to the reason of rule 4.

Is there any simple code to compute this? I am a bit struggling...

Answer 1

[Matt J]

M = 10;
d = 2;

I=1:d+1:M-d;
J=1:nnz(I);

A = accumarray([I(:),J(:)],1,[M,numel(J)])

A = 10×3

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

[Dyuman Joshi]

I have accepted Matt's answer as it provides a robust solution to the question posted.

[Matt J]

@Dyuman Joshi Very kind of you!

[Dyuman Joshi]

You're welcome and thanks for sharing this interesting approach!

Answer 2

[Dyuman Joshi]

M = 10;
A = [1;zeros(M-1,1)];
d = 3;

A(d+1:d+1:end-(d+1))=1

A = 10×1

1 0 0 1 0 0 0 0 0 0

[mingcheng nie]

Thank you so much for your answer. I am sorry that I forget one more rule:

5. I want to generate multiple vectors, such that each vector only has one place to be filled. For example, if and , then for the first vector, its first place will be filled; for the second vector, its 5th place will be filled; and there is no the third vector due to the reason of rule 4.

Could you please update your answer for this new rule? Sorry for the inconvenience.

[Dyuman Joshi]

@mingcheng nie, Sure, no problem.

Here, the columns of A are the vectors, which you can access them by indexing -

M = 10;
d = 3;

vec = [1 d+1:d+1:M-(d+1)]

vec = 1×2

1 4

n = nnz(vec);
A = zeros(M, n);
for k=1:n
    A(vec(k), k) = 1;
end
A

A = 10×2

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