Generating a particular sequnce of numbers

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Adi gahlawat
Adi gahlawat am 27 Jul. 2013
Hi,
given a variable natural number d, I'm trying to generate a sequence of the form:
[1 2 1 3 2 1 4 3 2 1.......d d-1 d-2......3 2 1].
I don't want to use for loop for this process, does anyone know a better (faster) method. I tried the colon operator without any success.
Thank you.
Adi

Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 27 Jul. 2013
Bearbeitet: Azzi Abdelmalek am 27 Jul. 2013
d=4
cell2mat(arrayfun(@(x) x:-1:1,1:d,'un',0))
  7 Kommentare
Youssef  Khmou
Youssef Khmou am 27 Jul. 2013
ok thank you for the explanation .
Jan
Jan am 28 Jul. 2013
Azzi's for loop approach is faster.

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Weitere Antworten (6)

Roger Stafford
Roger Stafford am 27 Jul. 2013
Here's another method to try:
N = d*(d+1)/2;
A = zeros(1,N);
n = 1:d;
A((n.^2-n+2)/2) = n;
A = cumsum(A)-(1:N)+1;
  1 Kommentar
Adi gahlawat
Adi gahlawat am 27 Jul. 2013
Bearbeitet: Adi gahlawat am 27 Jul. 2013
Hi Roger,
your method is excellent. It's about 2 times faster than my for loop based code. Much obliged.
Adi

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Azzi Abdelmalek
Azzi Abdelmalek am 28 Jul. 2013
Bearbeitet: Azzi Abdelmalek am 28 Jul. 2013
Edit
This is twice faster then Stafford's answer
A4=zeros(1,d*(d+1)/2); % Pre-allocate
c=0;
for k=1:d
A4(c+1:c+k)=k:-1:1;
c=c+k;
end
  1 Kommentar
Jan
Jan am 28 Jul. 2013
Bearbeitet: Jan am 28 Jul. 2013
Yes, this is exactly the kind of simplicity, which runs fast. While the one-liners with anonymous functions processed by cellfun or arrayfun look sophisticated, such basic loops hit the point. +1
I'd replace sum(1:d) by: d*(d+1)/2 . Anbd you can omit idx.

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Richard Brown
Richard Brown am 29 Jul. 2013
Even faster:
k = 1;
n = d*(d+1)/2;
out = zeros(n, 1);
for i = 1:d
for j = i:-1:1
out(k) = j;
k = k + 1;
end
end
  7 Kommentare
Jan
Jan am 29 Jul. 2013
Bearbeitet: Jan am 29 Jul. 2013
Under R2011b I get for d=1000 and 500 repetitions:
Elapsed time is 3.466296 seconds. Azzi's loop
Elapsed time is 3.765340 seconds. Richard's double loop
Elapsed time is 1.897343 seconds. C-Mex (see my answer)
Richard Brown
Richard Brown am 29 Jul. 2013
Bearbeitet: Richard Brown am 29 Jul. 2013
I checked again, and I agree with Azzi. My method was running faster because of another case I had in between his and mine. The JIT was doing some kind of unanticipated optimisation between cases.
I get similar orders of magnitude results to Azzi for R2012a if I remove that case, and if I run in R2013a (Linux), his method is twice as fast.
Shame, I like it when JIT brings performance of completely naive loops up to vectorised speed :)

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Jan
Jan am 29 Jul. 2013
An finally the C-Mex:
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray*prhs[]) {
mwSize d, i, j;
double *r;
d = (mwSize) mxGetScalar(prhs[0]);
plhs[0] = mxCreateDoubleMatrix(1, d * (d + 1) / 2, mxREAL);
r = mxGetPr(plhs[0]);
for (i = 1; i <= d; i++) {
for (j = i; j != 0; *r++ = j--) ;
}
}
And if your number d can be limited to 65535, the times shrink from 1.9 to 0.34 seconds:
#include "mex.h"
void mexFunction(int nlhs, mxArray *plhs[], int nrhs, const mxArray*prhs[]) {
uint16_T d, i, j, *r;
d = (uint16_T) mxGetScalar(prhs[0]);
plhs[0] = mxCreateNumericMatrix(1, d * (d + 1) / 2, mxUINT16_CLASS, mxREAL);
r = (uint16_T *) mxGetData(plhs[0]);
for (i = 1; i <= d; i++) {
for (j = i; j != 0; *r++ = j--) ;
}
}
For UINT32 0.89 seconds are required.
  1 Kommentar
Richard Brown
Richard Brown am 29 Jul. 2013
Nice. I imagine d would be limited to less than 65535, that's a pretty huge vector otherwise

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Richard Brown
Richard Brown am 29 Jul. 2013
Bearbeitet: Richard Brown am 29 Jul. 2013
Also comparable, but not (quite) faster
n = 1:(d*(d+1)/2);
a = ceil(0.5*(-1 + sqrt(1 + 8*n)));
out = a.*(a + 1)/2 - n + 1;
  3 Kommentare
Jan
Jan am 29 Jul. 2013
@Richard: How did you find this formula?
Richard Brown
Richard Brown am 29 Jul. 2013
If you look at the sequence, and add 0, 1, 2, 3, 4 ... you get
n: 1 2 3 4 5 6 7 8 9 10
1 3 3 6 6 6 10 10 10 10
Note that these are the triangular numbers, and that the triangular numbers 1, 3, 6, 10 appear in their corresponding positions, The a-th triangular number is given by
n = a (a + 1) / 2
So if you solve this quadratic for a where n is a triangular number, you get the index of the triangular number. If you do this for a value of n in between two triangular numbers, you can round this up, and invert the formula to get the nearest triangular number above (which is what the sequence is). Finally, you just subtract the sequence 0, 1, 2, ... to recover the original one.

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Andrei Bobrov
Andrei Bobrov am 27 Jul. 2013
Bearbeitet: Andrei Bobrov am 30 Jul. 2013
out = nonzeros(triu(toeplitz(1:d)));
or
out = bsxfun(@minus,1:d,(0:d-1)');
out = out(out>0);
or
z = 1:d;
z2 = cumsum(z);
z1 = z2 - z + 1;
for jj = d:-1:1
out(z1(jj):z2(jj)) = jj:-1:1;
end
or
out = ones(d*(d+1)/2,1);
ii = cumsum(d:-1:1) - (d:-1:1) + 1;
out(ii(2:end)) = 1-d : -1;
out = flipud(cumsum(out));

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