R: Matrix with colSums and rowSum constrained by 2 vectors

Question

Is there more elegant (less code) way of find a matrix OUT,

with colSums(OUT)<=a and rowSums(OUT)<=b,

given ORD = order of filling

sum(OUT) -> maximised

Sudoku-like problem, (numbers are not unique and filling order is given, so not really sudoku). I feel that there is some simpler solution for this problem.

a <- c(4,2,1)
b <- c(3,2,2)
ORD <- matrix(c(1,5,6,8,2,9,7,4,3),ncol=3)

MIN <- outer(a,b,pmin)
OUT <- matrix(0,ncol=ncol(ORD),nrow=nrow(ORD))
L <- cbind(as.vector(row(ORD)),as.vector(col(ORD)))[order(ORD),]
for( i in 1:nrow(L)){
  r <- L[i,1]
  c <- L[i,2]
  OUT[r,c] <- min(a[c],b[r])
  a[c] <- max(a[c] - OUT[r,c],0)
  b[r] <- max(b[r] - OUT[r,c],0)
}

OUT

Edit: Thanks! And finally I've ended with this (quite a long code for a super simple problem ;) :

cs <- c(4,2,1)
rs <- c(3,3,2)
ORD <- matrix(c(1,5,6,8,2,9,7,4,3),ncol=length(cs),byrow=T)

OUT <- matrix(0, nrow = length(rs), ncol = length(cs))
ROW <- row(OUT)
COL <- col(OUT)
for (i in order(ORD)){
  r <- ROW[i]
  c <- COL[i]
  k <- min(cs[c],rs[r])
  if(k>0){
    OUT[i] <- k
    cs[c] <- cs[c] - k
    rs[r] <- rs[r] - k
  }
  if(all(cs==0) | (all(rs==0)))
    break
}

wkmor1 · Answer 1 · 2012-03-26T03:21:50.167

1

Here is an elegant (by your definition), but potentially slow way, that uses control flow to get such a matrix by brute force.

while({OUT <- matrix(sample(0:max(a, b), 9, T), 3)
 !all(colSums(OUT) <= a & rowSums(OUT) <= b)}) {}

OUT

edited Mar 26 '12 at 03:21

answered Mar 26 '12 at 01:09

wkmor1

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score 1 · Accepted Answer · answered Mar 26 '12 at 01:42

Without modifying your algorithm, I can get it a bit shorter:

OUT <- matrix(0, nrow = length(b), ncol = length(a))
ROW <- row(OUT)
COL <- col(OUT)
for (i in order(ORD)) {
  r <- ROW[i]
  c <- COL[i]
  OUT[i] <- min(a[c], b[r])
  a[c] <- max(a[c] - OUT[i], 0)
  b[r] <- max(b[r] - OUT[i], 0)
}

If you only care about the number of lines, then you can do:

OUT <- matrix(0, nrow = length(b), ncol = length(a))
for (i in order(ORD)) {
  OUT[i] <- min(a[col(OUT)[i]], b[row(OUT)[i]])
  a[col(OUT)[i]] <- max(a[col(OUT)[i]] - OUT[i], 0)
  b[row(OUT)[i]] <- max(b[row(OUT)[i]] - OUT[i], 0)
}

but I would highly recommend the first version for its better readability.

R: Matrix with colSums and rowSum constrained by 2 vectors

2 Answers2