[R] help with apply, please

[Gabor Grothendieck]

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



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
>
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