circular permutation of similar objects

Question

I need to an R code for circular permutation of similar objects which defines this code exactly. The number of circular permutations that can be formed using 'n' objects out of which 'p' are identical and of one kind and 'q' are identical and of another kind.

(n-1)!/p!q!

This is the best code which I found but it is not exactly what I want

library(arrangements)
permutations(x =  c("A","B","C"), freq = c(2,1,1))

output:


    [,1] [,2] [,3] [,4]
 [1,] "A"  "A"  "B"  "C" 
 [2,] "A"  "A"  "C"  "B" 
 [3,] "A"  "B"  "A"  "C" 
 [4,] "A"  "B"  "C"  "A" 
 [5,] "A"  "C"  "A"  "B" 
 [6,] "A"  "C"  "B"  "A" 
 [7,] "B"  "A"  "A"  "C" 
 [8,] "B"  "A"  "C"  "A" 
 [9,] "B"  "C"  "A"  "A" 
[10,] "C"  "A"  "A"  "B"     
[11,] "C"  "A"  "B"  "A" 
[12,] "C"  "B"  "A"  "A"

I do not want "A" "A" are beside each other.

Do you mean how many different ways you can visit 'n' objects and then return to the start? Like a travelling salesman? — Simon Woodward, Jul 20 '19 at 21:41
Dear @SimonWoodward this problem is like Travelling Salesman but with one kind and another kind cities. e.g. you have 5 cities among which 2 London, 2 Hamilton and 1 Auckland. I can not go from London to London. — amir, Jul 21 '19 at 07:58
@SimonWoodward Thank you for your code. Yes, I need so. consider sitting around a table for 10 people from 3 countries. People from the same country should not sit beside each other, but all of them should sit at the table. — amir, Jul 22 '19 at 10:06
Do you need the first and last people to be different too? I didn't consider that. — Simon Woodward, Jul 22 '19 at 11:13
woooow, @SimonWoodward this is almost what I need to, Thank you. however unfortunately yes, I need the first and the last people to be different. — amir, Jul 22 '19 at 11:58
just change it to `if (length(remaining)==0 && head(visited,1) != tail(visited,1)){` — Simon Woodward, Jul 22 '19 at 17:00
Dear @SimonWoodward Thank you for your code. This is exactly what I need. Best Wishes — amir, Jul 23 '19 at 08:30

Simon Woodward · Accepted Answer · 2019-07-22T11:08:50.943

It turns out that a recursive function works well for this problem. The function takes the journey so far, figures out which remaining towns are possible to visit next, and then calls itself for each of these. If there are no remaining towns it reports the route.

# recursive function to visit remaining towns
journey <- function(remaining, visited){
  # possible towns to visit next
  possible <- setdiff(remaining, tail(visited, 1))
  if (length(possible)==0){
    if (length(remaining)==0){
      # report and store journey
      print(visited)
      routei <<- routei + 1
      routes[[routei]] <<- visited
    } else {
      # route failed to visit all towns
    }
  } else {
    # loop through options
    for (i in possible){
      # continue journey
      journey(remaining[-match(i, remaining)], c(visited, i))
    }
  }
}

remaining <- c("A", "A", "B", "B", "C")
visited <- character(0)

routes <- vector("list", length(remaining)^2)
routei <- 0

journey(remaining, visited)
#> [1] "A" "B" "A" "B" "C"
#> [1] "A" "B" "A" "C" "B"
#> [1] "A" "B" "C" "A" "B"
#> [1] "A" "B" "C" "B" "A"
#> [1] "A" "C" "B" "A" "B"
#> [1] "B" "A" "B" "A" "C"
#> [1] "B" "A" "B" "C" "A"
#> [1] "B" "A" "C" "A" "B"
#> [1] "B" "A" "C" "B" "A"
#> [1] "B" "C" "A" "B" "A"
#> [1] "C" "A" "B" "A" "B"
#> [1] "C" "B" "A" "B" "A"

^{Created on 2019-07-22 by the reprex package (v0.3.0)}

circular permutation of similar objects

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