Re: [Boost-users] [boost-users] Cartesian product

Subject: Re: [Boost-users] [boost-users] Cartesian product
From: Larry Evans (cppljevans_at_[hidden])
Date: 2011-10-26 11:34:39

On 10/26/11 09:46, Christoph Heindl wrote:
> Hi,
> 
> I'm on search for an algorithm that produces the cartesian product for
> n-dimension.
> The coordinates of each dimension are not limited to numbers.
> 
> Ideally I'd like to write something along the lines of
> 
> std::vector<int> d0;               // (assume d0 = [0, 3, 2]);
> std::vector<bool> d1;             // (assume d1 = [true]);
> std::vector<std::string> d2;    // (assume d2 = ["hello", "world"])
> 
> typedef boost::fusion::vector<
>   std::vector<int>,
>   std::vector<bool>,
>   std::vector<std::string> > tt;
> 
> tt space(d0, d1, d2);
> 
> boost::cartesian_product<tt> cp(space);
> 
> BOOST_FOREACH(const boost::cartesian_product<tt>::value_type &t, cp)
> {
>   // Invoked for every possible combination of the three dimension.
>   // t is boost::tuple<int, bool, std::string> or similar
> }
> 
> I think that something similar can be accomplished using boost.fusion,
> but I cannot derive
> the algorithm. What I have found so far is a meta cartesian product [1].
> 
> Has anyone such an algorithm lying around? Any help would be greatly
> appreciated.
> 
> Best regards,
> Christoph
> 
> [1] http://organicvectory.com/index.php?option=com_content&view=article&id=75:boostmplcartesianproduct&catid=42:boost&Itemid=78

The index_stack_simple.cpp upload here:

  http://www.datasimfinancial.com/forum/viewtopic.php?t=416#984

shows how to do this where values in the containers are all same type.
It works by simply keeping a vector of loop indices or iterators.
During the operator++, it checks which which iterator to increment,
and if that is at the end, then it moves to the next iterator.
Dereferencing all the iterators should give 1 element in the
cross product.  I would think that simply changing the vector
of loop indices to a fusion vector of those indices  would enable
the same to be done containers of different types.

HTH.