[R] help with apply, please

[Gabor Grothendieck]

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

Although the above is not wrong I should have removed the rbind
which is no longer needed and simplifying it further, as it seems
that lp will do the rep for you itself for certain arguments, gives:

lp("min", rep(1,3), t(mtrx), ">=", 1)$solution  # 1 0 1


>
>
> >
> >
> >
> > 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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> >
>