Rosetta Code

Factorial base numbers indexing permutations of a collection

Revision as of 21:03, 9 December 2018 by rosettacode>Craigd (→‎{{header|zkl}}: play with formatting)

You need a random arrangement of a deck of cards, you are sick of lame ways of doing this. This task is a super-cool way of doing this using factorial base numbers. The first 25 factorial base numbers in increasing order are: 0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1,1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0 Observe that the least significant digit is base 2 the next base 3, in general an n-digit factorial base number has digits n..1 in base n+1..2.

Factorial base numbers indexing permutations of a collection is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

I want to produce a 1 to 1 mapping between these numbers and permutations:- 

       0.0.0 -> 0123
       0.0.1 -> 0132
       0.1.0 -> 0213
       0.1.1 -> 0231
       0.2.0 -> 0312
       0.2.1 -> 0321
       1.0.0 -> 1023
       1.0.1 -> 1032
       1.1.0 -> 1203
       1.1.1 -> 1230
       1.2.0 -> 1302
       1.2.1 -> 1320
       2.0.0 -> 2013
       2.0.1 -> 2031
       2.1.0 -> 2103
       2.1.1 -> 2130
       2.2.0 -> 2301
       2.2.1 -> 2310
       3.0.0 -> 3012
       3.0.1 -> 3021
       3.1.0 -> 3102
       3.1.1 -> 3120
       3.2.0 -> 3201
       3.2.1 -> 3210

The following psudo-code will do this: 

      Starting with n=0 and Ω=an array of elements to be permutated, for each digit g starting with the most significant in the factorial base nubmer
         1. if g is greater than zero rotate the elements from n to n+g in Ω (see example)
         2. increment n and goto 1 using the next most significant digit until the factorial base number is exhausted.

Let me work 2.0.1 and 0123 

    step 1 n=0 g=2 Ω=2013
    step 2 n=1 g=0 so no action
    step 3 n=2 g=1 Ω=2031

The task: 

 First use your function to recreate the above table.
 Secondly use your function to generate all permutaions of 11 digits, perhaps count them don't display them, compare this method with
    methods in rc's permutations task.
 Thirdly here following are two ramdom 51 digit factorial base numbers I prepared earlier:
   39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
   51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
   use your function to crate the corresponding permutation of the following shoe of cards:
     A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣
 Finally create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe

F#

The Functıons

<lang fsharp> // Factorial base numbers indexing permutations of a collection

December 7th., 2018

let lN2p (c:int[]) (Ω:'Ω[])= 

 let Ω=Array.copy Ω
 let rec fN i g e l=match l-i with 0->Ω.[i]<-e |_->Ω.[l]<-Ω.[l-1]; fN i g e (l-1)// rotate right
 [0..((Array.length Ω)-2)]|>List.iter(fun n->let i=c.[n] in if i>0 then fN n (i+n) Ω.[i+n] (i+n)); Ω

let lN α = 

 let    n=Array.zeroCreate α
 let fN g=if n.[g]=α-g then n.[g]<-0; false else n.[g]<-n.[g]+1; true
 seq{yield n; while [1..α]|>List.exists(fun g->fN (α-g)) do yield n}

</lang>

Re-create the table

<lang fsharp> lN 3 |> Seq.iter (fun n->printfn "%A -> %A" n (lN2p n [|0;1;2;3|]));; </lang>

Output:
[|0; 0; 0|] -> [|0; 1; 2; 3|]
[|0; 0; 1|] -> [|0; 1; 3; 2|]
[|0; 1; 0|] -> [|0; 2; 1; 3|]
[|0; 1; 1|] -> [|0; 2; 3; 1|]
[|0; 2; 0|] -> [|0; 3; 1; 2|]
[|0; 2; 1|] -> [|0; 3; 2; 1|]
[|1; 0; 0|] -> [|1; 0; 2; 3|]
[|1; 0; 1|] -> [|1; 0; 3; 2|]
[|1; 1; 0|] -> [|1; 2; 0; 3|]
[|1; 1; 1|] -> [|1; 2; 3; 0|]
[|1; 2; 0|] -> [|1; 3; 0; 2|]
[|1; 2; 1|] -> [|1; 3; 2; 0|]
[|2; 0; 0|] -> [|2; 0; 1; 3|]
[|2; 0; 1|] -> [|2; 0; 3; 1|]
[|2; 1; 0|] -> [|2; 1; 0; 3|]
[|2; 1; 1|] -> [|2; 1; 3; 0|]
[|2; 2; 0|] -> [|2; 3; 0; 1|]
[|2; 2; 1|] -> [|2; 3; 1; 0|]
[|3; 0; 0|] -> [|3; 0; 1; 2|]
[|3; 0; 1|] -> [|3; 0; 2; 1|]
[|3; 1; 0|] -> [|3; 1; 0; 2|]
[|3; 1; 1|] -> [|3; 1; 2; 0|]
[|3; 2; 0|] -> [|3; 2; 0; 1|]
[|3; 2; 1|] -> [|3; 2; 1; 0|]
Shuffles

<lang fsharp> let shoe=[|"A♠";"K♠";"Q♠";"J♠";"10♠";"9♠";"8♠";"7♠";"6♠";"5♠";"4♠";"3♠";"2♠";"A♥";"K♥";"Q♥";"J♥";"10♥";"9♥";"8♥";"7♥";"6♥";"5♥";"4♥";"3♥";"2♥";"A♦";"K♦";"Q♦";"J♦";"10♦";"9♦";"8♦";"7♦";"6♦";"5♦";"4♦";"3♦";"2♦";"A♣";"K♣";"Q♣";"J♣";"10♣";"9♣";"8♣";"7♣";"6♣";"5♣";"4♣";"3♣";"2♣";|] //Random Shuffle let N=System.Random() in lc2p [|for n in 52..-1..2 do yield N.Next(n)|] shoe|>Array.iter (printf "%s ");printfn "" //Task Shuffles lN2p [|39;49;7;47;29;30;2;12;10;3;29;37;33;17;12;31;29;34;17;25;2;4;25;4;1;14;20;6;21;18;1;1;1;4;0;5;15;12;4;3;10;10;9;1;6;5;5;3;0;0;0|] shoe|>Array.iter (printf "%s ");printfn "" lN2p [|51;48;16;22;3;0;19;34;29;1;36;30;12;32;12;29;30;26;14;21;8;12;1;3;10;4;7;17;6;21;8;12;15;15;13;15;7;3;12;11;9;5;5;6;6;3;4;0;3;2;1|] shoe|>Array.iter (printf "%s ");printfn "" </lang>

Output:
J♣ Q♦ 10♣ 10♠ 3♥ 7♠ 8♥ 7♥ 8♦ 10♦ 4♥ 9♥ 8♠ K♥ 4♣ 5♥ K♣ Q♥ 9♠ A♦ Q♠ 6♦ K♦ K♠ 2♣ 6♠ 7♦ J♦ 2♥ 5♠ 4♦ 3♦ 6♣ J♥ 9♦ 4♠ 3♣ 2♠ 3♠ 10♥ Q♣ A♥ 2♦ A♠ 7♣ A♣ 9♣ 6♥ 5♦ 5♣ J♠ 8♣

A♣ 3♣ 7♠ 4♣ 10♦ 8♦ Q♠ K♥ 2♠ 10♠ 4♦ 7♣ J♣ 5♥ 10♥ 10♣ K♣ 2♣ 3♥ 5♦ J♠ 6♠ Q♣ 5♠ K♠ A♦ 3♦ Q♥ 8♣ 6♦ 9♠ 8♠ 4♠ 9♥ A♠ 6♥ 5♣ 2♦ 7♥ 8♥ 9♣ 6♣ 7♦ A♥ J♦ Q♦ 9♦ 2♥ 3♠ J♥ 4♥ K♦

2♣ 5♣ J♥ 4♥ Jâ™  Aâ™  5♥ A♣ 6♦ Qâ™  9♣ 3♦ Q♥ J♣ 10♥ K♣ 10♣ 5♦ 7♥ 10♦ 3â™  8♥ 10â™  7â™  6♥ 5â™  K♥ 4♦ A♥ 4♣ 2♥ 9♦ Q♣ 8♣ 7♦ 6♣ 3♥ 6â™  7♣ 2♦ J♦ 9♥ A♦ Q♦ 8♦ 4â™  K♦ Kâ™  3♣ 2â™  8â™  9â™
Comparıson wıth [Permutations(F#)]

<lang fsharp> let g=[|0..10|] lC 10 |> Seq.map(fun n->lc2p n g) |> Seq.length </lang>

Output:
Real: 00:01:08.430, CPU: 00:01:08.970, GC gen0: 9086, gen1: 0
val it : int = 39916800

8GB of memory is insufficient for rc's perm task

Go

<lang go>package main

import ( 

   "fmt"
   "math/rand"
   "strconv"
   "strings"
   "time"

)

func factorial(n int) int { 

   fact := 1
   for i := 2; i <= n; i++ {
       fact *= i
   }
   return fact

}

func genFactBaseNums(size int, countOnly bool) ([][]int, int) { 

   var results [][]int
   count := 0
   for n := 0; ; n++ {
       radix := 2
       res := make([]int, size)
       k := n
       for k > 0 {
           div := k / radix
           rem := k % radix
           if !countOnly {
               if radix <= size+1 {
                   res[size-radix+1] = rem
               }
           }
           k = div
           radix++
       }
       if radix > size+2 {
           break
       }
       count++
       if !countOnly {
           results = append(results, res)
       }
   }
   return results, count

}

func mapToPerms(factNums [][]int) [][]int { 

   var perms [][]int
   psize := len(factNums[0]) + 1
   start := make([]int, psize)
   for i := 0; i < psize; i++ {
       start[i] = i
   }
   for _, fn := range factNums {
       perm := make([]int, psize)
       copy(perm, start)
       for m := 0; m < len(fn); m++ {
           g := fn[m]
           if g == 0 {
               continue
           }
           first := m
           last := m + g
           for i := 1; i <= g; i++ {
               temp := perm[first]
               for j := first + 1; j <= last; j++ {
                   perm[j-1] = perm[j]
               }
               perm[last] = temp
           }
       }
       perms = append(perms, perm)
   }
   return perms

}

func join(is []int, sep string) string { 

   ss := make([]string, len(is))
   for i := 0; i < len(is); i++ {
       ss[i] = strconv.Itoa(is[i])
   }
   return strings.Join(ss, sep)

}

func undot(s string) []int { 

   ss := strings.Split(s, ".")
   is := make([]int, len(ss))
   for i := 0; i < len(ss); i++ {
       is[i], _ = strconv.Atoi(ss[i])
   }
   return is

}

func main() { 

   rand.Seed(time.Now().UnixNano())

   // Recreate the table.
   factNums, _ := genFactBaseNums(3, false)
   perms := mapToPerms(factNums)
   for i, fn := range factNums {
       fmt.Printf("%v -> %v\n", join(fn, "."), join(perms[i], ""))
   }

   // Check that the number of perms generated is equal to 11! (this takes a while).
   _, count := genFactBaseNums(10, true)
   fmt.Println("\nPermutations generated =", count)
   fmt.Println("compared to 11! which  =", factorial(11))
   fmt.Println()

   // Generate shuffles for the 2 given 51 digit factorial base numbers.
   fbn51s := []string{
       "39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0",
       "51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1",
   }
   factNums = [][]int{undot(fbn51s[0]), undot(fbn51s[1])}
   perms = mapToPerms(factNums)
   shoe := []rune("A♠K♠Q♠J♠T♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥T♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦T♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣T♣9♣8♣7♣6♣5♣4♣3♣2♣")
   cards := make([]string, 52)
   for i := 0; i < 52; i++ {
       cards[i] = string(shoe[2*i : 2*i+2])
       if cards[i][0] == 'T' {
           cards[i] = "10" + cards[i][1:]
       }
   }
   for i, fbn51 := range fbn51s {
       fmt.Println(fbn51)
       for _, d := range perms[i] {
           fmt.Print(cards[d])
       }
       fmt.Println("\n")
   }

   // Create a random 51 digit factorial base number and produce a shuffle from that.
   fbn51 := make([]int, 51)
   for i := 0; i < 51; i++ {
       fbn51[i] = rand.Intn(52 - i)
   }
   fmt.Println(join(fbn51, "."))
   perms = mapToPerms([][]int{fbn51})
   for _, d := range perms[0] {
       fmt.Print(cards[d])
   }
   fmt.Println()

}</lang>

Output:

Random for Part 4: 

0.0.0 -> 0123
0.0.1 -> 0132
0.1.0 -> 0213
0.1.1 -> 0231
0.2.0 -> 0312
0.2.1 -> 0321
1.0.0 -> 1023
1.0.1 -> 1032
1.1.0 -> 1203
1.1.1 -> 1230
1.2.0 -> 1302
1.2.1 -> 1320
2.0.0 -> 2013
2.0.1 -> 2031
2.1.0 -> 2103
2.1.1 -> 2130
2.2.0 -> 2301
2.2.1 -> 2310
3.0.0 -> 3012
3.0.1 -> 3021
3.1.0 -> 3102
3.1.1 -> 3120
3.2.0 -> 3201
3.2.1 -> 3210

Permutations generated = 39916800
compared to 11! which  = 39916800

39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
A♣3♣7♠4♣10♦8♦Q♠K♥2♠10♠4♦7♣J♣5♥10♥10♣K♣2♣3♥5♦J♠6♠Q♣5♠K♠A♦3♦Q♥8♣6♦9♠8♠4♠9♥A♠6♥5♣2♦7♥8♥9♣6♣7♦A♥J♦Q♦9♦2♥3♠J♥4♥K♦

51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
2♣5♣J♥4♥J♠A♠5♥A♣6♦Q♠9♣3♦Q♥J♣10♥K♣10♣5♦7♥10♦3♠8♥10♠7♠6♥5♠K♥4♦A♥4♣2♥9♦Q♣8♣7♦6♣3♥6♠7♣2♦J♦9♥A♦Q♦8♦4♠K♦K♠3♣2♠8♠9♠

18.14.25.48.18.9.1.16.15.11.41.8.26.19.36.11.8.21.20.15.15.14.27.10.5.24.0.11.18.12.6.8.5.14.16.10.13.13.9.7.11.1.1.7.0.2.5.0.3.0.0
9♥K♥K♦2♣7♥5♠K♠6♥8♥A♥3♣4♠4♦J♦5♣J♥3♠6♦7♦A♦Q♦2♥7♣10♥8♠8♣A♠10♦Q♣8♦2♠4♥6♠J♣6♣3♦10♣9♣5♦3♥4♣J♠10♠A♣Q♠Q♥K♣9♠2♦7♠5♥9♦

Perl 6

Works with: Rakudo version 2018.11

Using my interpretation of the task instructions as shown on the discussion page.

<lang perl6>sub postfix:<!> (Int $n) { (flat 1, [\*] 1..*)[$n] }

multi base (Int $n is copy, 'F', $length? is copy) { 

   constant @fact = [\*] 1 .. *;
   my $i = $length // @fact.first: * > $n, :k;
   my $f;
   [ @fact[^$i].reverse.map: { ($n, $f) = $n.polymod($_); $f } ]

}

sub fpermute (@a is copy, *@f) { (^@f).map: { @a[$_ .. $_ + @f[$_]].=rotate(-1) }; @a }

put "Part 1: Generate table"; put $_.&base('F', 3).join('.') ~ ' -> ' ~ [0,1,2,3].&fpermute($_.&base('F', 3)).join for ^24;

put "\nPart 2: Compare 11! to 11! " ~ '¯\_(ツ)_/¯';

  1. This is kind of a weird request. Since we don't actually need to _generate_
  2. the permutations, only _count_ them: compare count of 11! vs count of 11!

put "11! === 11! : {11! === 11!}";

put "\nPart 3: Generate the given task shuffles"; my \Ω = <A♠ K♠ Q♠ J♠ 10♠ 9♠ 8♠ 7♠ 6♠ 5♠ 4♠ 3♠ 2♠ A♥ K♥ Q♥ J♥ 10♥ 9♥ 8♥ 7♥ 6♥ 5♥ 4♥ 3♥ 2♥ 

        A♦ K♦ Q♦ J♦ 10♦ 9♦ 8♦ 7♦ 6♦ 5♦ 4♦ 3♦ 2♦ A♣ K♣ Q♣ J♣ 10♣ 9♣ 8♣ 7♣ 6♣ 5♣ 4♣ 3♣ 2♣

>;

my @books = < 

   39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
   51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1

>;

put "Original deck:"; put Ω.join;

put "\n$_\n" ~ Ω[(^Ω).&fpermute($_.split: '.')].join for @books;

put "\nPart 4: Generate a random shuffle"; my @shoe = (+Ω … 2).map: { (^$_).pick }; put @shoe.join('.'); put Ω[(^Ω).&fpermute(@shoe)].join;

put "\nSeems to me it would be easier to just say: Ω.pick(*).join"; put Ω.pick(*).join;</lang>

Output:
Part 1: Generate table
0.0.0 -> 0123
0.0.1 -> 0132
0.1.0 -> 0213
0.1.1 -> 0231
0.2.0 -> 0312
0.2.1 -> 0321
1.0.0 -> 1023
1.0.1 -> 1032
1.1.0 -> 1203
1.1.1 -> 1230
1.2.0 -> 1302
1.2.1 -> 1320
2.0.0 -> 2013
2.0.1 -> 2031
2.1.0 -> 2103
2.1.1 -> 2130
2.2.0 -> 2301
2.2.1 -> 2310
3.0.0 -> 3012
3.0.1 -> 3021
3.1.0 -> 3102
3.1.1 -> 3120
3.2.0 -> 3201
3.2.1 -> 3210

Part 2: Compare 11! to 11! ¯\_(ツ)_/¯
11! === 11! : True

Part 3: Generate the given task shuffles
Original deck:
A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣

39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
A♣3♣7♠4♣10♦8♦Q♠K♥2♠10♠4♦7♣J♣5♥10♥10♣K♣2♣3♥5♦J♠6♠Q♣5♠K♠A♦3♦Q♥8♣6♦9♠8♠4♠9♥A♠6♥5♣2♦7♥8♥9♣6♣7♦A♥J♦Q♦9♦2♥3♠J♥4♥K♦

51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
2♣5♣J♥4♥J♠A♠5♥A♣6♦Q♠9♣3♦Q♥J♣10♥K♣10♣5♦7♥10♦3♠8♥10♠7♠6♥5♠K♥4♦A♥4♣2♥9♦Q♣8♣7♦6♣3♥6♠7♣2♦J♦9♥A♦Q♦8♦4♠K♦K♠3♣2♠8♠9♠

Part 4: Generate a random shuffle
47.9.46.16.28.8.36.27.29.1.9.27.1.16.21.22.28.34.30.8.19.27.18.22.3.25.15.20.12.14.8.9.11.1.4.0.3.5.4.2.2.10.8.1.6.1.2.4.1.2.1
6♣5♠5♣10♥10♦6♠K♣9♦6♦K♠2♠5♦Q♠5♥Q♦8♦J♣2♣8♣A♥K♦9♣A♦2♦9♠4♣3♥A♣7♥2♥Q♥9♥4♥J♠4♠A♠3♠8♥J♥7♠K♥3♣10♣8♠Q♣6♥7♦7♣J♦3♦4♦10♠

Seems to me it would be easier to just say: Ω.pick(*).join
5♦3♠8♦10♦2♥7♠7♦Q♦A♠5♣8♣Q♠4♠2♦K♦5♠Q♥7♣10♠2♠K♠J♣9♣3♣4♥3♥4♦3♦Q♣2♣4♣J♦9♠A♣J♠10♣6♣9♦6♠10♥6♥9♥J♥7♥K♥A♦8♠A♥5♥8♥K♣6♦

zkl

<lang zkl>fcn fpermute(omega,num){ // eg (0,1,2,3), (0,0,0)..(3,2,1) 

  omega=omega.copy(); 	  // omega gonna be mutated
  foreach m,g in ([0..].zip(num)){ if(g) omega.insert(m,omega.pop(m+g)) }
  omega

}</lang>

Part 1, Generate permutation table:

<lang zkl>foreach a,b,c in (4,3,2){ 

  println("%d.%d.%d --> %s".fmt(a,b,c, fpermute(T(0,1,2,3),T(a,b,c)).concat()));

}</lang>

Output:
0.0.0 --> 0123
0.0.1 --> 0132
0.1.0 --> 0213
0.1.1 --> 0231
0.2.0 --> 0312
0.2.1 --> 0321
1.0.0 --> 1023
1.0.1 --> 1032
1.1.0 --> 1203
1.1.1 --> 1230
1.2.0 --> 1302
1.2.1 --> 1320
2.0.0 --> 2013
2.0.1 --> 2031
2.1.0 --> 2103
2.1.1 --> 2130
2.2.0 --> 2301
2.2.1 --> 2310
3.0.0 --> 3012
3.0.1 --> 3021
3.1.0 --> 3102
3.1.1 --> 3120
3.2.0 --> 3201
3.2.1 --> 3210
Part 3, Generate the given task shuffles:

<lang zkl>deck:=List(); foreach s,c in ("\u2660 \u2665 \u2666 \u2663".split(), 

               "A K Q J 10 9 8 7 6 5 4 3 2".split()){ deck.append(c+s) }

books:=List( 

  "39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0",
  "51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1")
  .apply(fcn(s){ s.split(".").apply("toInt") });

foreach book in (books){ println(fpermute(deck,book).concat("")); }</lang>

Output:
A♣3♣7♠4♣10♦8♦Q♠K♥2♠10♠4♦7♣J♣5♥10♥10♣K♣2♣3♥5♦J♠6♠Q♣5♠K♠A♦3♦Q♥8♣6♦9♠8♠4♠9♥A♠6♥5♣2♦7♥8♥9♣6♣7♦A♥J♦Q♦9♦2♥3♠J♥4♥K♦
2♣5♣J♥4♥J♠A♠5♥A♣6♦Q♠9♣3♦Q♥J♣10♥K♣10♣5♦7♥10♦3♠8♥10♠7♠6♥5♠K♥4♦A♥4♣2♥9♦Q♣8♣7♦6♣3♥6♠7♣2♦J♦9♥A♦Q♦8♦4♠K♦K♠3♣2♠8♠9♠
Part 4, Generate a random shuffle:

<lang zkl>r:=[52..2,-1].pump(List,(0).random); println(r.concat("."),"\n",fpermute(deck,r).concat(""));</lang>

Output:
36.21.48.31.19.37.16.39.43.1.27.23.30.22.14.32.31.2.27.11.5.24.28.20.23.20.17.19.23.13.11.12.3.12.1.0.11.1.8.10.6.2.8.3.7.1.1.4.2.2.1
4♦6♥3♣8♦8♥Q♣J♥8♣2♣K♠9♦K♦2♦A♦Q♥9♣10♣J♠A♣A♥7♠3♦5♣10♦K♣7♦2♥6♦4♣7♥10♥5♥9♠3♥Q♠A♠J♦8♠4♥J♣K♥5♠7♣3♠6♣6♠4♠5♦9♥Q♦2♠10♠
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