A triangular creation in a matrix with the rest zeros

I want to write a code which does the following:

Say its input is 'n = 12' to create a matrix arranged triangularly with the rest zeros (Only use loop), for example:

    % code 
Input:  n = 12
Output: mat = [1  0  0  0 
               2  3  0  0 
               4  5  6  0 
               7  8  9  10
               11 12 0  0]

Thank you

[Walter Roberson]

https://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

[Jan]

The shape of the non-zero value is not triangular here.

The question sounds like a homework. The forum will not solve it, but if you ask a specific question and post, what you have done so far, you will get assistance here for sure.

[Emilia]

Ok I'm sorry, thanks for sending a link.

Yes it was taken from the homework (Problem 44457), I wrote a code for a triple matrix (without zeros) and here i need help how to place zeros (rest) within the triangular matrix to get a new matrix.

    % code
 n=input('Enter a number');
s = 1;
  for k = 1 : n
    g=[s : s + k - 1];
    s = s + k;
    b=(g<=n);
    m=g(b);
  end

After receiving code

     1

     2     3

     4     5     6

     7     8     9    10

    11    12

[dpb]

What if you were to start with an array of the proper final size containing (let me guess...) zeros, maybe?

Answer 1

[Jan]

you do not have to run the loop from 1 to n, but only until all rows have been created.

n   = input('Enter a number');
s   = 1;
mat = [];
k   = 1;
while s <= n
  g = s : s + k - 1;  % No square brackets needed. a:b:c is a vector already
  g = g(g <= n);
  s = s + k;
  k = k + 1;
  mat(k, 1:length(g)) = g;
end

Or create g with the maximum length directly - then omit g=g(g<=n):

    g = s : min(n, s + k - 1);

You could pre-allocate mat by zeros() also. This would be more efficient that letting the array grow iteratively, but run time is not the problem in this question. But you could replace the while loop by a simpler for loop, if you know the number of rows in advance.

rows = ceil((sqrt(8 * n + 1) - 1) / 2)

Try this to create a for k = 1:rows loop.

[Emilia]

Thank you!