On Saturday 19 November 2005 22:09, Gabor Grothendieck wrote:
> Getting back to your original question of using apply, solving the LP
> gives us the number of components in any minimal solution and
> exhaustive search of all solutions with that many components can
> be done using combinations from gtools and apply like this:
>
> library(gtools) # needed for combinations
> soln <- lp("min", rep(1,3), rbind(t(mtrx)), rep(">=", 4),
> rep(1,4))$solution k <- sum(soln)
> m <- nrow(mtrx)
> combos <- combinations(m,k)
> combos[apply(combos, 1, function(idx) all(colSums(mtrx[idx,]))),]
>
> In the example we get:
>
> [,1] [,2]
> [1,] 1 3
> [2,] 2 3
>
> which says that rows 1 and 3 of mtrx form one solution
> and rows 2 and 3 of mtrx form another solution.
I'm speechless.
It is exactly what I needed.
A billion of thanks!
Adrian
--
Adrian DUSA
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