indices of a spiral

It is simple to geerate a growing spiral matrix vial the spiral function:

S=spiral(3)

which yields a nice spiral matrix:

S =
   7     8     9
   6     1     2
   5     4     3

However it is not straight forward to generate the list of indices for that sequence:

[2,2], [3,2], [3,3], [2,3], [1,3], ...

Does anyone have an idea for a one-liner that can do this?

[Stephen23]

You have swapped the dimensions: from smallest to largest the indices are actually

(2,2) (2,3) (3,3) (3,2) (3,1) (2,1) (1,1) (1,2) (1,3)

Answer 1

[Stephen23]

Not quite one line, but you can use sort and ind2sub for this:

>> M = spiral(3)
M =
   7   8   9
   6   1   2
   5   4   3
>> [~,idx] = sort(M(:));
>> [R,C] = ind2sub([3,3],idx);
>> [R,C]
ans =
   2   2
   2   3
   3   3
   3   2
   3   1
   2   1
   1   1
   1   2
   1   3

Or copy the spiral Mfile to your user directory, give it a new name, and edit it to return the indices. It wouldn't be hard to do this, just remember to preallocate the output matrices! Note that the file is copyright.

[Boris Povazay]

Thanks for the quick reply and the solution. That ind2sub in combination with the sorting indices does the trick - I got stuck after assigning the indices form the sort function. As you say the spiral(n) function is copyrighted, which makes me wonder if this is maintained code and where one might ask for an implementation that also generates the indices directly. Best regards!

[Stephen23]

@Boris Povazay: that specific code is copyright... but I doubt that the algorithm is: online you can find plenty of implementations of the Ulam spiral, e.g.:

https://rosettacode.org/wiki/Ulam_spiral_(for_primes)

which relies on exactly that arrangement of numbers. Note that the Ulam spiral is flipped vertically relative to the MATLAB spiral (in other words: MATLAB is clockwise, Ulam spiral is counter-clockwise). I am sure you could easily implement a version which generates the indices directly.