Question

我想在一条线上安排3个红球，2个蓝球和2个绿球，这样就不会有两个相同颜色的球相邻。如果没有这样的限制，那么这种独特安排的数量是

。

下面是我用来解决这个问题的R中的代码。我们的想法是获得所有210个独特的排列，然后计算连续两个相邻单元格具有相同颜色的情况。使用这个算法，我得到38作为答案。我觉得我的代码是如此丑陋的黑客。你会如何用R（或你的首选语言）做到这一点？

colors <- c("R", "R", "R", "B", "B", "G", "G")
n <- 1:10000
x <- matrix(rep(NA, times=70000), ncol=7)
for (i in n) x[i,] <- sample(colors)
x <- unique(x)
rows <- nrow(x)
y <- rep(NA, times =rows)
for (i in 1:rows) {
  y[i] <- x[i,1]==x[i,2] | 
    x[i,2]==x[i,3] | 
    x[i,3]==x[i,4] | 
    x[i,4]==x[i,5] | 
    x[i,5]==x[i,6] | 
    x[i,6] == x[i,7]
  }
table(y)
## y
## FALSE  TRUE 
##    38   172

Answer 1

这应该以更简单的格式为您提供相同的值：

library(DescTools)
library(Hmisc)

out = Permn(colors)
table(apply(out,1,function(x) any(x==Lag(x))))

Answer 2

@timfaber的答案很棒。我将添加一些更复杂的答案，使您能够从过程中获得更多信息。它将告诉您颜色系列（排列）的外观以及相邻颜色的次数：

library(tidyverse)
library(DescTools)

# vector of colors
colors <- c("R", "R", "R", "B", "B", "G", "G")

Permn(colors) %>%                    # permute colors
  tbl_df() %>%                       # save it as dataframe
  mutate(id = row_number()) %>%      # create permutation id
  gather(v, color, -id) %>%          # reshape data
  select(-v) %>%                     # remove unecessary column
  group_by(id) %>%                   # for each permutation id
  mutate(color_series = paste0(color, collapse = "_"),                       # create the series of colors
         color_lag = lag(color),                                             # get the previous color
         IsSameColor = if_else(color == color_lag, 1, 0, missing = 0)) %>%   # check if you have adjacent colors
  group_by(id, color_series) %>%                                             # for each permutation id and color series
  summarise(CountSameColor = sum(IsSameColor)) %>%                           # count number of adjacent colors
  ungroup()

# # A tibble: 210 x 3
#      id  color_series CountSameColor
#   <int>         <chr>          <dbl>
# 1     1 R_R_R_B_B_G_G              4
# 2     2 R_R_B_R_B_G_G              2
# 3     3 R_B_R_R_B_G_G              2
# 4     4 B_R_R_R_B_G_G              3
# 5     5 R_R_B_B_R_G_G              3
# 6     6 R_B_R_B_R_G_G              1
# 7     7 B_R_R_B_R_G_G              2
# 8     8 R_B_B_R_R_G_G              3
# 9     9 B_R_B_R_R_G_G              2
# 10    10 B_B_R_R_R_G_G             4
# # ... with 200 more rows

如果您只想要使用现在相邻颜色的情况（排列），最后添加一个过滤器filter(CountSameColor == 0)。

Answer 3

你也可以使用游程长度编码：

library(DescTools)
# 3 red balls, 2 blue balls, and 2 green
m <- Permn(rep(c("R","B","G"), times=c(3,2,2)))

# use run length encoding to find adjacent elements
table(apply(m, 1, function(x) 
            (length(rle(x)$lengths)!=7)))

安排3个红球，2个蓝球和2个绿球，使两个相邻的球没有相同的颜色

3 个答案: