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  1. # 列舉抽出且放回1-5的所有情況，一列一輪，共5^3種情況
  2. m <- gtools::permutations(5, 3, repeats.allowed = T)
  3.  
  4. # 保留左右相鄰不等的列，剩80種
  5. ri1 <-
  6.   apply(m, 1, function(x) {
  7.     foo <- vector("logical", length(x) - 1)
  8.     for (i in 1:(length(x) - 1)) {
  9.       if (x[i] != x[i + 1]) {
  10.         foo[i] <- T
  11.       }
  12.     }
  13.     all(foo)
  14.   })
  15. m <- m[ri1, ]
  16.  
  17. # 取出二個列的所有可能排列序號
  18. ri2 <- gtools::permutations(nrow(m), 2)
  19. # 用來儲存符合題意的容器
  20. res <- vector("list", nrow(ri2))
  21. # 上下相鄰全不相等的情況存入容器中
  22. for (i in 1:nrow(ri2)) {
  23.   m.this <- m[ri2[i, ], ]
  24.   if (all(m.this[1, ] != m.this[2, ])) {
  25.     res[[i]] <- m.this
  26.   }
  27. }
  28.  
  29. # 去除NULL後為所求，共3380種情況
  30. ans <- res[!sapply(res, is.null)]
  31. length(ans)
  32. print(ans[1:10])
  33.  
Success #stdin #stdout 0.22s 188992KB
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[1] 3380
[[1]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    1    2

[[2]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    1    3

[[3]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    1    4

[[4]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    1    5

[[5]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    3    2

[[6]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    3    4

[[7]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    3    5

[[8]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    4    2

[[9]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    4    3

[[10]]
     [,1] [,2] [,3]
[1,]    1    2    1
[2,]    2    4    5
https://ideone.com/ZVizRL
language:
R (R 3.5.2)
created:
6 years ago
visibility:
public

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