Finding the maximum square sub-matrix with all equal elements - r

Question

Does anybody know how it could be possible to subset the maximum K such that K x K is a submatrix with all identical elements, i.e., all the elements in this submatrix must be the same from a given a N x N matrix?
I found many examples in other programming languages except R. I also prefer dplyr if you know.
There is a link to the solution with other languages:
https://www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/
But this link provides a special case when all identical elements are next to each other. It retrieves a maximum block of the same elements, not a submatrix in general. I do not want to limit subsetting with this condition.

Answer 1

Here is a base R implementation to make it.
If you want to search the maximum square sub-matrix within a non-square matrix, you can try the code below:
r <- list()
for (w in rev(seq(min(dim(M))))) {
for (rs in seq(nrow(M)-w+1)) {
for (cs in seq(ncol(M)-w+1)) {
mat <- M[rs-1+(1:w),cs-1+(1:w)]
u <- unique(c(mat))
if (all(u!=0) &length(u)==1) r[[length(r)+1]] <- mat
}
}
if (length(r)>0) break
}
such that
> r
[[1]]
[,1] [,2]
[1,] 3 3
[2,] 3 3
[[2]]
[,1] [,2]
[1,] 2 2
[2,] 2 2
[[3]]
[,1] [,2]
[1,] 3 3
[2,] 3 3
[[4]]
[,1] [,2]
[1,] 2 2
[2,] 2 2
[[5]]
[,1] [,2]
[1,] 1 1
[2,] 1 1
[[6]]
[,1] [,2]
[1,] 1 1
[2,] 1 1
[[7]]
[,1] [,2]
[1,] 3 3
[2,] 3 3
[[8]]
[,1] [,2]
[1,] 3 3
[2,] 3 3
DATA
M <- structure(c(1L, 3L, 1L, 2L, 1L, 3L, 3L, 2L, 2L, 3L, 3L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L,
2L, 2L, 1L, 3L, 1L, 3L, 2L, 2L, 2L, 2L, 3L, 2L, 1L, 3L, 2L, 1L,
1L, 3L, 2L, 2L, 3L, 3L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L,
3L, 3L, 2L, 3L, 3L, 2L, 3L, 3L, 1L, 1L, 1L, 1L, 3L, 2L, 3L, 1L,
1L, 2L, 1L, 1L, 1L, 1L, 3L, 2L, 1L, 1L, 3L, 3L, 3L, 2L, 2L, 2L,
3L, 2L, 2L, 3L, 3L, 3L, 1L, 2L, 2L, 1L, 3L, 3L, 2L, 3L, 2L, 1L,
2L, 1L, 3L, 3L, 1L, 2L, 1L, 3L, 2L, 3L, 3L, 1L, 1L, 2L, 2L, 2L,
1L, 1L, 1L, 2L, 1L, 3L, 2L, 3L, 3L, 2L, 3L, 3L, 1L, 1L, 2L, 2L,
1L, 2L, 3L, 3L, 3L, 3L, 3L, 1L, 3L), .Dim = c(15L, 10L))
> M
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 2 2 1 1 3 2 2 1 3
[2,] 3 2 1 3 3 1 2 3 1 3
[3,] 1 2 3 2 3 1 2 2 2 1
[4,] 2 3 1 2 2 2 3 1 2 1
[5,] 1 1 3 3 3 1 2 2 2 2
[6,] 3 3 2 3 3 1 2 1 1 2
[7,] 3 1 2 2 2 1 3 3 1 1
[8,] 2 1 2 2 3 1 3 3 1 2
[9,] 2 1 2 2 3 3 3 1 2 3
[10,] 3 1 3 2 1 2 1 2 1 3
[11,] 3 2 2 1 1 1 2 1 3 3
[12,] 1 1 1 2 1 1 2 3 2 3
[13,] 1 1 3 2 1 3 1 2 3 3
[14,] 1 2 2 2 3 3 3 3 3 1
[15,] 2 2 1 2 2 3 3 3 2 3
EDIT
The approach above is inefficient when with large matrix since all combinations checked. The method below is a R implementation of algorithm stated in https://www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/, which is is far more efficient.
M <- unname(as.matrix(read.csv(file = "test2.csv")))
S <- matrix(0,nrow = nrow(M),ncol = ncol(M))
S[,1] <- M[,1]
for (i in 1:nrow(S)) {
for (j in 2:ncol(S)) {
if (M[i,j]==1) {
if (i==1) {
S[i,j] <- M[i,j]
} else {
S[i,j] <- min(c(S[i,j-1],S[i-1,j],S[i-1,j-1]))+1
}
}
}
}
inds <- which(S == max(S),arr.ind = TRUE)
w <- seq(max(S))-1
res <- lapply(seq(nrow(inds)), function(k) M[inds[k,"row"]-w,inds[k,"col"]-w])

Answer 2

I found the following answer to this question using dplyr:
M1 <- M %>% data.frame %>% mutate(sumVar = rowSums(.)) %>%
arrange(desc(sumVar)) %>% dplyr::select(-sumVar)
M2 <- M1 %>% as.matrix %>% t %>% data.frame %>%
mutate(sumVar = rowSums(.)) %>% arrange(desc(sumVar)) %>%
dplyr::select(-sumVar) %>% as.matrix %>% t %>% data.frame %>%
arrange_all(funs(desc(.)))
i <- 1
j <- 1
while(sum(M2[1:i,1:j]) == i*j){
i <- i+1
j <- j+1
M3 <- M2[1:i-1,1:j-1]
}
This is a toy data as #ThomasIsCoding proposed:
M <- structure(c(1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Dim = c(5L,
5L))
and this is the result:
> M
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 0 1
[2,] 1 1 1 1 1
[3,] 1 1 1 1 1
[4,] 1 1 1 1 1
[5,] 0 1 1 1 1
> M1
X1 X2 X3 X4 X5
1 1 1 1 1 1
2 1 1 1 1 1
3 1 1 1 1 1
4 1 1 1 0 1
5 0 1 1 1 1
> M2
X1 X2 X3 X4 X5
1 1 1 1 1 1
2 1 1 1 1 1
3 1 1 1 1 1
4 1 1 1 1 0
5 1 1 1 0 1
> M3
X1 X2 X3 X4
1 1 1 1 1
2 1 1 1 1
3 1 1 1 1
4 1 1 1 1
Note that some more functions should be added to keep the variable names and find them after using arrange!