寻找所有元素相等的最大平方子矩阵

Question

有谁知道如何对最大 K 进行子集化，使得 K x K 是具有所有相同元素的子矩阵，即该子矩阵中的所有元素必须与给定的 N x N 矩阵相同？ 我在除 R 之外的其他编程语言中找到了许多示例。如果您知道，我也更喜欢dplyr 。

有其他语言的解决方案链接： https : //www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/

但是当所有相同的元素彼此相邻时，此链接提供了一种特殊情况。 它检索相同元素的最大块，而不是一般的子矩阵。 我不想用这种条件限制子集。

Answer 1

这是一个基本的 R 实现。

如果要搜索非方阵内的最大方阵子矩阵，可以试试下面的代码：

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
}

以至于

> 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

数据

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

编辑

由于检查了所有组合，因此当矩阵较大时，上述方法效率低下。 下面的方法是https://www.geeksforgeeks.org/maximum-size-sub-matrix-with-all-1s-in-a-binary-matrix/中所述算法的R实现，效率更高.

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

我使用dplyr找到了这个问题的以下答案：

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]
}

这是@ThomasIsCoding 提出的玩具数据：

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))

这是结果：

> 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

注意，应该增加一些函数来保存变量名并在使用arrange后找到它们！