Pancake numbers: Difference between revisions

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Adrian Monk has problems and an assistant, Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.

The task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.

Sorting_algorithms/Pancake_sort actually performs the sort some giving the number of flips used. How do these compare with p(n)?

Few people know p(20), generously I shall award an extra credit for anyone doing more than p(16).

References

F#

<lang fsharp> // Pancake numbers. Nigel Galloway: October 28th., 2020 let pKake z=let n=[for n in 1..z-1->Array.ofList([n.. -1..0]@[n+1..z-1])]

           let e=let rec fG n g=match g with 0->n |_->fG (n*g) (g-1) in fG 1 z
           let rec fN i g l=match (Set.count g)-e with 0->(i,List.last l)
                                                      |_->let l=l|>List.collect(fun g->[for n in n->List.permute(fun g->n.[g]) g])|>Set.ofList
                                                          fN (i+1) (Set.union g l) (Set.difference l g|>Set.toList)
           fN 0 (set1..z) 1..z

[1..9]|>List.iter(fun n->let i,g=pKake n in printfn "Maximum number of flips to sort %d elements is %d. e.g %A->%A" n i g [1..n]) </lang>

Output:

Maximum number of flips to sort 1 elements is 0. e.g [1]->[1]
Maximum number of flips to sort 2 elements is 1. e.g [2; 1]->[1; 2]
Maximum number of flips to sort 3 elements is 3. e.g [1; 3; 2]->[1; 2; 3]
Maximum number of flips to sort 4 elements is 4. e.g [4; 2; 3; 1]->[1; 2; 3; 4]
Maximum number of flips to sort 5 elements is 5. e.g [5; 3; 1; 4; 2]->[1; 2; 3; 4; 5]
Maximum number of flips to sort 6 elements is 7. e.g [5; 3; 6; 1; 4; 2]->[1; 2; 3; 4; 5; 6]
Maximum number of flips to sort 7 elements is 8. e.g [7; 3; 1; 5; 2; 6; 4]->[1; 2; 3; 4; 5; 6; 7]
Maximum number of flips to sort 8 elements is 9. e.g [8; 6; 2; 4; 7; 3; 5; 1]->[1; 2; 3; 4; 5; 6; 7; 8]
Maximum number of flips to sort 9 elements is 10. e.g [9; 7; 5; 2; 8; 1; 4; 6; 3]->[1; 2; 3; 4; 5; 6; 7; 8; 9]

Go

Translation of: Phix

<lang go>package main

import "fmt"

func pancake(n int) int {

   gap, sum, adj := 2, 2, -1
   for sum < n {
       adj++
       gap = gap*2 - 1
       sum += gap
   }
   return n + adj

}

func main() {

   for i := 0; i < 4; i++ {
       for j := 1; j < 6; j++ {
           n := i*5 + j
           fmt.Printf("p(%2d) = %2d  ", n, pancake(n))
       }
       fmt.Println()
   }

}</lang>

Output:

p( 1) =  0  p( 2) =  1  p( 3) =  3  p( 4) =  4  p( 5) =  5  
p( 6) =  7  p( 7) =  8  p( 8) =  9  p( 9) = 10  p(10) = 11  
p(11) = 13  p(12) = 14  p(13) = 15  p(14) = 16  p(15) = 17  
p(16) = 18  p(17) = 19  p(18) = 20  p(19) = 21  p(20) = 23

Julia

Translation of: Phix

<lang julia>function pancake(len)

   gap, gapsum, adj = 2, 2, -1
   while gapsum < len
       adj += 1
       gap = gap * 2 - 1
       gapsum += gap
   end
   return len + adj

end

for i in 1:25

   print("pancake(", lpad(i, 2), ") = ", rpad(pancake(i), 5)) 
   i % 5 == 0 && println()

end

</lang>

Output:

pancake( 1) = 0    pancake( 2) = 1    pancake( 3) = 3    pancake( 4) = 4    pancake( 5) = 5    
pancake( 6) = 7    pancake( 7) = 8    pancake( 8) = 9    pancake( 9) = 10   pancake(10) = 11
pancake(11) = 13   pancake(12) = 14   pancake(13) = 15   pancake(14) = 16   pancake(15) = 17
pancake(16) = 18   pancake(17) = 19   pancake(18) = 20   pancake(19) = 21   pancake(20) = 23
pancake(21) = 24   pancake(22) = 25   pancake(23) = 26   pancake(24) = 27   pancake(25) = 28

Exhaustive search, breadth first method. <lang julia>function pancake(len)

   goalstack = collect(1:len)
   stacks, numstacks = Dict(goalstack => 0), 1
   newstacks = deepcopy(stacks)
   for i in 1:1000
       nextstacks = Dict()
       for (arr, steps) in newstacks, pos in 2:len
           newstack = vcat(reverse(arr[1:pos]), arr[pos+1:end])
           haskey(stacks, newstack) || (nextstacks[newstack] = i)
       end
       newstacks = nextstacks
       stacks = merge(stacks, newstacks)
       perms = length(stacks)
       perms == numstacks && return i - 1
       numstacks = perms
   end

end

for i in 1:10

   print("pancake(", lpad(i, 2), ") = ", rpad(pancake(i), 5)) 
   i % 5 == 0 && println()

end

</lang>

Output:

pancake( 1) = 0    pancake( 2) = 1    pancake( 3) = 3    pancake( 4) = 4    pancake( 5) = 5
pancake( 6) = 7    pancake( 7) = 8    pancake( 8) = 9    pancake( 9) = 10   pancake(10) = 11

Phix

Extra credit to anyone who can prove that this is in any way wrong?
The algorithm is freshly made up today, from scratch, by yours truly.
It agrees with https://oeis.org/A058986/b058986.txt which would put p(20) as either 22 or 23.
Note that several other references/links disagree on p(17) and up. <lang Phix>function pancake(integer n)

   integer gap = 2, sum_gaps = gap, adj = -1
   while sum_gaps<n do
       adj += 1
       gap = gap*2-1
       sum_gaps += gap
   end while
   n += adj
   return n

end function sequence t = tagset(20),

        r = columnize({t,apply(t,pancake)}),
        p = apply(true,sprintf,{{"p(%2d) = %2d"},r})

printf(1,"%s\n",join_by(p,1,5))</lang>

Output:

p( 1) =  0   p( 2) =  1   p( 3) =  3   p( 4) =  4   p( 5) =  5
p( 6) =  7   p( 7) =  8   p( 8) =  9   p( 9) = 10   p(10) = 11
p(11) = 13   p(12) = 14   p(13) = 15   p(14) = 16   p(15) = 17
p(16) = 18   p(17) = 19   p(18) = 20   p(19) = 21   p(20) = 23

vs. max() of ten runs each of pancake_sort(shuffle(tagset(n))), modified to return the number of flips it made:

p( 1) =  0   p( 2) =  1   p( 3) =  3   p( 4) =  5   p( 5) =  6
p( 6) =  9   p( 7) = 10   p( 8) = 11   p( 9) = 12   p(10) = 15
p(11) = 16   p(12) = 17   p(13) = 20   p(14) = 22   p(15) = 25
p(16) = 28   p(17) = 28   p(18) = 31   p(19) = 33   p(20) = 34

Obviously the sort focuses on getting one pancake at a time into place, and therefore runs closer to 2n flips.

exhaustive search

Translation of: Julia

<lang Phix>function visitor(sequence arr, integer steps, sequence userdata)

   integer {len,stacks,i,nextstacks} = userdata
   for pos=2 to len do
       sequence newstack = reverse(arr[1..pos])&arr[pos+1..$]
       if getd_index(newstack,stacks)=NULL then
           setd(newstack,i,nextstacks)
       end if
   end for
   return 1

end function

function merge(sequence arr, integer steps, integer stacks)

   setd(arr,steps,stacks)
   return 1

end function

function pancake(integer len)

   sequence goalstack = tagset(len)
   integer stacks = new_dict(Template:Goalstack,0),
           numstacks = 1,
           newstacks = new_dict(stacks)
   for i=1 to 1000 do
       integer nextstacks = new_dict()
       traverse_dict(visitor,{len,stacks,i,nextstacks},newstacks)
       destroy_dict(newstacks)
       newstacks = nextstacks
       traverse_dict(merge,stacks,newstacks)
       integer perms = dict_size(stacks)
       if perms == numstacks then return i-1 end if
       numstacks = perms
   end for

end function

atom t0 = time() for n=1 to 8 do -- (for <3s)

   integer pn = pancake(n)
   printf(1,"p(%d) = %d (%s)\n",{n,pn,elapsed(time()-t0)})

end for</lang>

Output:

p(1) = 0 (0s)
p(2) = 1 (0.1s)
p(3) = 3 (0.1s)
p(4) = 4 (0.1s)
p(5) = 5 (0.1s)
p(6) = 7 (0.1s)
p(7) = 8 (0.3s)
p(8) = 9 (2.2s)
p(9) = 10 (25.5s)
p(10) = 11 (5 minutes and 35s)

After p(7), each subsequent p(n) takes about n times as long to complete.

REXX

Translation of: Go

Translation of: Phix

<lang rexx>/*REXX program calculates/displays ten pancake numbers (from 1 ──► 20, inclusive). */

    pad= center(        , 10)                                         /*indentation. */

say pad center('pancakes', 10 ) center('pancake flips', 15 ) /*show the hdr.*/ say pad center( , 10, "─") center(, 15, "─") /* " " sep.*/

        do #=1  to 20;  say pad  center(#, 10)  center( pancake(#), 15) /*index, flips.*/
        end   /*#*/

exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ pancake: procedure; parse arg n; adj= -1 /*obtain N; initialize the ADJustment.*/

        gap= 2;  sum= 2                         /*initialize the  GAP  and the  SUM.   */
                             do while sum<n     /*perform while  SUM is less than  N.  */
                             adj= adj + 1       /*bump the value of the ADJustment.    */
                             gap= gap*2 - 1     /*calculate the GAP.                   */
                             sum= sum + gap     /*add the  GAP  to the  SUM.           */
                             end   /*while*/
        return n + adj</lang>

output when using the default inputs:

            pancakes   pancake flips
           ────────── ───────────────
               1             0
               2             1
               3             3
               4             4
               5             5
               6             7
               7             8
               8             9
               9            10
               10           11
               11           13
               12           14
               13           15
               14           16
               15           17
               16           18
               17           19
               18           20
               19           21
               20           23

Wren

Translation of: Phix

Library: Wren-fmt

Well, it's hard to believe it can be as simple as this but Pete's algorithm does at least give the same answers as the OEIS sequence for n <= 19 which is usually taken as the authority on these matters.

Clearly, for non-trivial 'n', the number of flips required for the pancake sorting task will generally be more as no attempt is being made there to minimize the number of flips, just to get the data into sorted order. <lang ecmascript>import "/fmt" for Fmt

var pancake = Fn.new { |n|

   var gap = 2
   var sum = 2
   var adj = -1
   while (sum < n) {
       adj = adj + 1
       gap = gap*2 - 1
       sum = sum + gap
   }
   return n + adj

}

for (i in 0..3) {

   for (j in 1..5) {
       var n = i*5 + j
       Fmt.write("p($2d) = $2d  ", n, pancake.call(n))
   }
   System.print()

}</lang>

Output:

p( 1) =  0  p( 2) =  1  p( 3) =  3  p( 4) =  4  p( 5) =  5  
p( 6) =  7  p( 7) =  8  p( 8) =  9  p( 9) = 10  p(10) = 11  
p(11) = 13  p(12) = 14  p(13) = 15  p(14) = 16  p(15) = 17  
p(16) = 18  p(17) = 19  p(18) = 20  p(19) = 21  p(20) = 23