How to build a matrix like this

if we have a number n, we want a matrix like this:

So one simple example is:

What is the best way to build this?

Answer 1

[Stephen23]

>> N = 4;
>> V = N:-1:1;
>> bsxfun(@min,V,repmat(V(:),1,N))
ans =
     4     3     2     1
     3     3     2     1
     2     2     2     1
     1     1     1     1

Answer 2

[José-Luis]

n = 4;
result = repmat((n:-1:1),n,1) - tril(cumsum(tril(ones(n)))-1)

Answer 3

[Kuifeng]

The following works, not the 'best' way...

n = 5;
A = [n:-1:1]';
for i = 2:n
    A(:,i) = A(:,i-1);
    A([1:i],i) = n-(i-1);
end

[José-Luis]

It's somewhat of a bad habit to use square brackets when declaring arrays.

[Stephen23]

It is better to use parentheses around an already existing vector:

A = (n:-1:1)';

rather than using the concatenation operator. The editor has warning for this:

[Kuifeng]

Thank you Jose-Luis and Stephen, I noticed this red wave below for some time already, but didnt know why. You cleared my doubt. Good to learn.

Answer 4

[Robert U]

Hi Rui Zhu,

another possibility:

function [ A ] = SpecMatrice( n )
 tic
 A = zeros(n);
 for ik = 1:n
  A(1:ik,1:ik) = A(1:ik,1:ik) + 1;
 end
 t = toc;
 sprintf('MySolution took %.2f µs',t*1e6)
end

On my PC that solution is for low values of n faster than the presented above. Here values for n=4:

 Solution 1 took 806.82 µs
 Solution 2 took 888.93 µs
 Solution 3 took 441.34 µs
 MySolution took 109.33 µs

For high values of n solution 2 & 3 gain speed a lot while solution 1 and MySolution get slowlier due to looping.

Kind regards,

Robert