How to create a square matrix with consecutive numbers on each row?

Hi everyone,

Given a vector i.e. n=[1 12 25 78], is there any way to create a matrix A, such that

A=[ 1 2 3 4; 11 12 13 14; 23 24 25 26; 75 76 77 78]?

without FOR LOOP?

Answer 1

[Guillaume]

With toeplitz construct a symmetric matrix with 0 on diagonal and increments on the sides and with bsxfun add that to your n:

n = [1 12 25 78];
A = bsxfun(@plus, toeplitz(0:-1:1-numel(n), 0:numel(n)-1), n')

[Ming]

Thanks a lot!

Answer 2

[Sebastian Castro]

Yeah, for sure.

I'm sure there are more efficient ways to do this, but this one will show you a few examples of the "repmat" function to string together vectors and matrices (either row-wise or column-wise).

I first avoided hard-coding parameters by using a variable "nCols" for number of columns, which should be the same as number of rows (or numel(n)). Note that I had to transpose n to n' to meet your desired solution.

>> nCols = numel(n);

>> baseMatrix = repmat(n',[1 nCols])

baseMatrix =

     1     1     1     1
    12    12    12    12
    25    25    25    25
    78    78    78    78

Next, you have to make the pretty complicated matrix to add to that matrix above. I would copy-paste both of the terms below into MATLAB to see what each of those does. Basically, I create a column pattern and then a row pattern, and subtract them.

>> addMatrix = repmat(0:nCols-1,[nCols 1]) - repmat((0:nCols-1)',[1 nCols])

addMatrix =

     0     1     2     3
    -1     0     1     2
    -2    -1     0     1
    -3    -2    -1     0

Finally, add 'em up!

>> A = baseMatrix + addMatrix

A =

     1     2     3     4
    11    12    13    14
    23    24    25    26
    75    76    77    78

- Sebastian

[Ming]

Thanks a lot!

[David Alejandro Ramirez Cajigas]

a matrix of size NxN, whose diagonal is zero and from each "0" there is a consecutive list of numbers up to N

for example the following, after every 0 there is a consecutive numbering up to N

N = 22

but suppose that N is any number

How do I make it in matlab automatically giving only the number N?

[Walter Roberson]

N = 22;
v = [0:N];
M = toeplitz([v(1) fliplr(v(2:end))], v)

M = 23×23

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 18 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 17 18 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 16 17 18 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 16 17 18 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 14 15 16 17 18 19 20 21 22 0 1 2 3 4 5 6 7 8 9 10 11 12 13

result = mod(tril(-tril(M)) + triu(M), N+1)

result = 23×23

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13

[David Alejandro Ramirez Cajigas]

Bingo!

The answer is:

N=22
Top1=N
Top12=repmat(0:Top1-1,[Top1 1]) - repmat((0:Top1-1)',[1 Top1]); %genera matriz 0 hasta n

Top17=(tril(Top12,-1)*-1);
Top18=Top17+Top12;
Top19=Top17+Top18