Re: how to make a matrix with all combinations of digits efficiently

[Jean Dubois]

2017-12-29 14:51 GMT+01:00 Francesco Potortì <address@hidden>:
>>2017-12-29 0:07 GMT+01:00 Juan Pablo Carbajal <address@hidden>:
>>> On Thu, Dec 28, 2017 at 9:25 PM, Jean Dubois <address@hidden> wrote:
>>>> I'd like to generate a matrix like this (small version to give you the 
>>>> idea):
>>>>
>>>> 0 0 0 (three columns in this example)
>>>> 0 0 1
>>>> 0 0 2
>>>> 0 0 3
>>>> .
>>>> .
>>>> .
>>>> 9 9 8
>>>> 9 9 9
>>>>
>>>> Could anyone here show me how to do this efficiently?
>
> [...]
>
>>Dear Juan,
>>I bumped into the following problem. I changed your code to a function
>>like this:
>>
>>function [counter] = decomp (base, nc)
>>i            = (0:(base^nc-1)).'; # row index - 1
>>counter = mod( floor (i ./ base.^[(nc-1):-1:0]), base);
>>endfunction
>>
>>It works fine for a lot of cases but the case I want to use is
>>problematic as you can see here:
>>result=decomp(10,9);
>>error: out of memory or dimension too large for Octave's index type
>
> decomp(B,C) creates a matrix with B^C rows and C columns.  In the case
> you tested this means 9e9 elements.  Each element is a double needing 8
> bytes, which means that you need 72GB of memory only for hosting the
> final result.  In practice, you need additional space for hosting
> intermediate variables.
>
I wonder whether it is possible to make octave  store the array more
efficiently knowing that the elements are all
integers which can be represented by 1 byte
> This is the only problem if Octave has been specially compiled for using
> 64-bit indexes, which is not the general case, and most probably not
> your case.
>
> For the general case, have a look at
> https://wiki.octave.org/Enable_large_arrays:_Build_octave_such_that_it_can_use_arrays_larger_than_2Gb.
>
Unfortunately this wiki-page is empty

kind regards




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