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

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Ming
Ming 2015-8-20
编辑： David Alejandro Ramirez Cajigas 2021-8-18
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?

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Guillaume
Guillaume 2015-8-20
编辑：Guillaume 2015-8-20
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')

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Sebastian Castro
Sebastian Castro 2015-8-20
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
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Walter Roberson
Walter Roberson 2021-8-18
Ran in:
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
David Alejandro Ramirez Cajigas 2021-8-18
编辑：David Alejandro Ramirez Cajigas 2021-8-18
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

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