help with apply, please

On 11/19/05, Gabor Grothendieck <ggrothendieck at gmail.com> wrote:

Try minizing 1'x subject to 1 >= x >= 0 and m'x >= 1 where m is your mtrx
and ' means transpose.  It seems to give an integer solution, 1 0 1,
with linear programming even in the absence of explicit integer
constraints:

library(lpSolve)
lp("min", rep(1,3), rbind(t(mtrx), diag(3)), rep(c(">=", "<="), 4:3),
rep(1,7))$solution

Just one after thought.  The constraint x <= 1 will not be active so,
eliminating it, the above can be reduced to:

lp("min", rep(1,3), rbind(t(mtrx)), rep(">=", 4), rep(1,4))$solution

On 11/19/05, Adrian DUSA <adi at roda.ro> wrote:

Dear list,

I have a problem with a toy example:
mtrx <- matrix(c(1,1,0,1,1,1,0,1,1,0,0,1), nrow=3)
rownames(ma) <- letters[1:3]

I would like to determine which is the minimum combination of rows that
"covers" all columns with at least a 1.
None of the rows covers all columns; all three rows clearly covers all
columns, but there are simpler combinations (1st and the 3rd, or 2nd and 3rd)
which also covers all columns.

I solved this problem by creating a second logical matrix which contains all
possible combinations of rows:
tt <- matrix(as.logical(c(1,0,0,0,1,0,0,0,1,1,1,0,1,0,1,0,1,1,1,1,1)), nrow=3)

and then subset the first matrix and check if all columns are covered.
This solution, though, is highly inneficient and I am certain that a
combination of apply or something will do.

###########################

possibles <- NULL
length.possibles <- NULL
## I guess the minimum solution is has half the number of rows
guesstimate <- floor(nrow(tt)/2) + nrow(tt) %% 2
checked <- logical(nrow(tt))
repeat {
   ifelse(checked[guesstimate], break, checked[guesstimate] <- TRUE)
   partials <- as.matrix(tt[, colSums(tt) == guesstimate])
   layer.solution <- logical(ncol(partials))

   for (j in 1:ncol(partials)) {
       if (length(which(colSums(mtrx[partials[, j], ]) > 0)) == ncol(mtrx)) {
           layer.solution[j] <- TRUE
       }
   }
   if (sum(layer.solution) == 0) {
       if (!is.null(possibles)) break
       guesstimate <- guesstimate + 1
   } else {
       for (j in which(layer.solution)) {
           possible.solution <- rownames(mtrx)[partials[, j]]
           possibles[[length(possibles) + 1]] <- possible.solution
           length.possibles <- c(length.possibles, length(possible.solution))
       }
       guesstimate <- guesstimate - 1
   }
}
final.solution <- possibles[which(length.possibles == min(length.possibles))]

###########################

More explicitely (if useful) it is about reducing a prime implicants chart in
a Quine-McCluskey boolean minimisation algorithm. I tried following the
original algorithm applying row dominance and column dominance, but (as I am
not a computer scientist), I am unable to apply it.

If you have a better solution for this, I would be gratefull if you'd share
it.
Thank you in advance,
Adrian

--
Adrian DUSA
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
         +40 21 3120210 / int.101