Solve the no connection puzzle: Difference between revisions

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⍝ ⍴⍸~tries{1∊{1∊⍵∊¯1 0 1}¨|(⍺[⍳8])-¨⍺∘{⍺[⍵]}¨⍵}⍤1⊢links
⍝ ⍴⍸~tries{1∊{1∊⍵∊¯1 0 1}¨|(⍺[⍳8])-¨⍺∘{⍺[⍵]}¨⍵}⍤1⊢links
⍝ 16
⍝ 16
solns←⍸~tries{1∊{1∊⍵∊¯1 0 1}¨|(⍺[⍳8])-¨⍺∘{⍺[⍵]}¨⍵}⍤1⊢links
tries[''⍴solns;]
tries[''⍴solns;]
}
}
</lang>
</lang>

Revision as of 04:16, 17 September 2018

Task
Solve the no connection puzzle
You are encouraged to solve this task according to the task description, using any language you may know.

You are given a box with eight holes labelled   A-to-H,   connected by fifteen straight lines in the pattern as shown below: 

             A   B
            /│\ /│\
           / │ X │ \
          /  │/ \│  \
         C ─ D ─ E ─ F
          \  │\ /│  /
           \ │ X │ /
            \│/ \│/
             G   H

You are also given eight pegs numbered   1-to-8.


Objective

Place the eight pegs in the holes so that the (absolute) difference between any two numbers connected by any line is greater than one.


Example

In this attempt: 

             4   7
            /│\ /│\
           / │ X │ \
          /  │/ \│  \
         8 ─ 1 ─ 6 ─ 2
          \  │\ /│  /
           \ │ X │ /
            \│/ \│/
             3   5

Note that   7   and   6   are connected and have a difference of   1,   so it is   not   a solution.


Task

Produce and show here   one   solution to the puzzle.


Related tasks


See also

No Connection Puzzle (youtube).

Ada

This solution is a bit longer than it actually needs to be; however, it uses tasks to find the solution and the used types and solution-generating functions are well-separated, making it more amenable to other solutions or altering it to display all solutions. <lang Ada> With Ada.Text_IO, Connection_Types, Connection_Combinations;

procedure main is 

  Result : Connection_Types.Partial_Board renames Connection_Combinations;

begin 

  Ada.Text_IO.Put_Line( Connection_Types.Image(Result) );

end;</lang> <lang Ada>Pragma Ada_2012;

Package Connection_Types with Pure is 

  -- Name of the nodes.
  Type Node is (A, B, C, D, E, F, G, H);

  -- Type for indicating if a node is connected.
  Type Connection_List is array(Node) of Boolean
    with Size => 8, Object_Size => 8, Pack;

  Function "&"( Left : Connection_List; Right : Node ) return Connection_List;
  
  -- The actual map of the network connections.
  Network : Constant Array (Node) of Connection_List:=
    (
     A => (C|D|E	=> True, others => False),
     B => (D|E|F	=> True, others => False),
     C => (A|D|G	=> True, others => False),
     D => (C|A|B|E|H|G	=> True, others => False),
     E => (D|A|B|F|H|G => True, others => False),
     F => (B|E|H	=> True, others => False),
     G => (C|D|E	=> True, others => False),
     H => (D|E|F	=> True, others => False)
    );

  -- Values of the nodes.
  Type Peg is range 1..8;

  -- Indicator for which values have been assigned.
  Type Used_Peg is array(Peg) of Boolean
    with Size => 8, Object_Size => 8, Pack;

  Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg;



  -- Type describing the layout of the network.
  Type Partial_Board is array(Node range <>) of Peg;
  Subtype Board is Partial_Board(Node);

  -- Determines if the given board is a solution or partial-solution.
  Function Is_Solution	( Input : Partial_Board ) return Boolean;

  -- Displays the board as text.
  Function Image	( Input : Partial_Board ) Return String;

End Connection_Types;</lang> <lang Ada> Pragma Ada_2012;

with Connection_Types; use Connection_Types;

Function Connection_Combinations return Partial_Board; </lang> <lang Ada>Pragma Ada_2012;

Package Body Connection_Types is 

  New_Line : Constant String := ASCII.CR & ASCII.LF;

  ---------------------
  --  Solution Test  --
  ---------------------
  
  Function Is_Solution( Input : Partial_Board ) return Boolean is
    (for all Index in Input'Range =>
       (for all Connection in Node'Range =>
            (if Network(Index)(Connection) and Connection in Input'Range
             then abs (Input(Index) - Input(Connection)) > 1
            )
       )
    );
  
  ------------------------
  --  Concat Operators  --
  ------------------------
  
  Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg is
  begin
     return Result : Used_Peg := Left do
        Result(Right):= True;
     end return;
  end "&";

  Function "&"(Left : Connection_List; Right : Node) return Connection_List is
  begin
     Return Result : Connection_List := Left do
        Result(Right):= True;
     end return;        
  end "&";   

  -----------------------
  --  IMAGE FUNCTIONS  --
  -----------------------

  Function Image(Input : Peg) Return Character is
    ( Peg'Image(Input)(2) );

  Function Image(Input : Peg) Return String is
    ( 1 => Image(Input) );

  Function Image(Input : Partial_Board; Item : Node) Return String is
    ( 1 => (if Item not in Input'Range then '*' else Image(Input(Item)) ));

  Function Image( Input : Partial_Board ) Return String is
     A : String renames Image(Input, Connection_Types.A);
     B : String renames Image(Input, Connection_Types.B);
     C : String renames Image(Input, Connection_Types.C);
     D : String renames Image(Input, Connection_Types.D);
     E : String renames Image(Input, Connection_Types.E);
     F : String renames Image(Input, Connection_Types.F);
     G : String renames Image(Input, Connection_Types.G);
     H : String renames Image(Input, Connection_Types.H);
  begin
     return

" "&A&" "&B & New_Line & " /|\ /|\" & New_Line & " / | X | \" & New_Line & " / |/ \| \" & New_Line & " "&C&" - "&D&" - "&E&" - "&F & New_Line & " \ |\ /| /" & New_Line & " \ | X | /" & New_Line & " \|/ \|/" & New_Line & " "&G&" "&H & New_Line; 

  end Image;

End Connection_Types;</lang> <lang Ada>Function Connection_Combinations return Partial_Board is

begin 

  Return Result : Board do
     declare
        
        -- The Generate task takes two parameters
        --   (1) a list of pegs already in use, and
        --   (2) a partial-board
        -- and, if the state given is a viable yet incomplete solution, it
        -- takes a peg and adds it to the state creating a new task with
        -- that peg in its used list.
        --
        -- When a complete solution is found it is copied into result.
        task type Generate(
                           Pegs	: not null access Used_Peg:= new Used_Peg'(others => False);
                           State	: not null access Partial_Board:= new Partial_Board'(Node'Last..Node'First => <>)
                          ) is
        end Generate;

        -- An access to Generate and array thereof, for creating the
        -- children tasks.
        type Generator  is access all Generate;
        type Generators is array(Peg range <>) of Generator;
        
        -- Gen handles the actual creation of a new task and state.
        Function Gen(P : Peg; G : not null access Generate) return Generator is
        begin
           return (if G.Pegs(P) then null
                   else new Generate(
                     Pegs     => new Used_Peg'(G.Pegs.all & P),
                     State    => New Partial_Board'(G.All.State.All & P)
                    )
                  );
        end;

        task body Generate is
        begin
           if Is_Solution(State.All) then
              -- If the state is a partial board, we make children to
              -- complete the calculations.
              if State'Length <= Node'Pos(Node'Last) then
                 declare
                    Subtasks : Constant Generators:=
                      (
                       Gen(1, Generate'Access),
                       Gen(2, Generate'Access),
                       Gen(3, Generate'Access),
                       Gen(4, Generate'Access),
                       Gen(5, Generate'Access),
                       Gen(6, Generate'Access),
                       Gen(7, Generate'Access),
                       Gen(8, Generate'Access)
                      );
                 begin
                    null;
                 end;
              else
                 Result:= State.All;
              end if;
           else
              -- The current state is not a solution, so we do not continue it.
              Null;
           end if;
           
        end Generate;
        
        Master : Generate;
     begin
        null;
     end;
  End return;

End Connection_Combinations; </lang>

Output:
        4   5
       /|\ /|\
      / | X | \
     /  |/ \|  \
    7 - 1 - 8 - 2
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        3   6

APL

Works with: Dyalog APL 17.0 Unicode

<lang APL> 


    perms←{
    ⍝∇ 20100513/20140818 ra⌈ --()--
       1=⍴⍴⍵:⍵[∇ ⍬⍴⍴⍵]
      ↑{0∊⍴⍵:⍵ ⋄ (⍺[1]⌷⍵),(1↓⍺)∇ ⍵~⍺[1]⌷⍵}∘(⍳⍵)¨↓⍉1+(⌽⍳⍵)⊤¯1+⍳!⍵
  }

solution←{ 

   links←  (3 4 5) (4 5 6) (1 4 7) (1 2 3 5 7 8) (1 2 4 6 7 8) (2 5 8) (3 4 5) (4 5 6) 
   tries←8 perms 8            
 ⍝    ⍴⍸~tries{1∊{1∊⍵∊¯1 0 1}¨|(⍺[⍳8])-¨⍺∘{⍺[⍵]}¨⍵}⍤1⊢links
 ⍝ 16
  solns←⍸~tries{1∊{1∊⍵∊¯1 0 1}¨|(⍺[⍳8])-¨⍺∘{⍺[⍵]}¨⍵}⍤1⊢links
  tries[⍴solns;]
  }
</lang>

AutoHotkey

<lang AutoHotkey>oGrid := [[ "", "X", "X"] ; setup oGrid ,[ "X", "X", "X", "X"] ,[ "", "X", "X"]]

oNeighbor := [], oCell := [], oRoute := [] , oVisited := [] ; initialize objects

for row, oRow in oGrid for col, val in oRow if val ; for each valid cell in oGrid oNeighbor[row, col] := Neighbors(row, col, oGrid) ; list valid no-connection neighbors

Solve: for row, oRow in oGrid for col , val in oRow if val ; for each valid cell in oGrid if (oSolution := SolveNoConnect(row, col, 1)).8 ; solve for this cell break, Solve ; if solution found stop

show solution

for i , val in oSolution oCell[StrSplit(val, ":").1 , StrSplit(val, ":").2] := i

A := oCell[1, 2] , B := oCell[1, 3] C := oCell[2, 1], D := oCell[2, 2] , E := oCell[2, 3], F := oCell[2, 4] G := oCell[3, 2] , H := oCell[3, 3] sol = ( 

   %A%   %B%
  /|\ /|\
 / | X | \
/  |/ \|  \

%C% - %D% - %E% - %F% 

\  |\ /|  /
 \ | X | /
  \|/ \|/
   %G%   %H%

) MsgBox % sol return

-----------------------------------------------------------------------

SolveNoConnect(row, col, val){ global oRoute.push(row ":" col) ; save route oVisited[row, col] := true ; mark this cell visited

if oRoute[8] ; if solution found return true ; end recursion

for each, nn in StrSplit(oNeighbor[row, col], ",") ; for each no-connection neighbor of cell { rowX := StrSplit(nn, ":").1 , colX := StrSplit(nn, ":").2 ; get coords of this neighbor if !oVisited[rowX, colX] ; if not previously visited { oVisited[rowX, colX] := true ; mark this cell visited val++ ; increment if (SolveNoConnect(rowX, colX, val)) ; recurse return oRoute ; if solution found return route } } oRoute.pop() ; Solution not found, backtrack oRoute oVisited[row, col] := false ; Solution not found, remove mark }

-----------------------------------------------------------------------

Neighbors(row, col, oGrid){ ; return distant neighbors of oGrid[row,col] for r , oRow in oGrid for c, v in oRow if (v="X") && (abs(row-r) > 1 || abs(col-c) > 1) list .= r ":"c "," if (row<>2) && oGrid[row, col] list .= oGrid[row, col+1] ? row ":" col+1 "," : oGrid[row, col-1] ? row ":" col-1 "," : "" return Trim(list, ",") }</lang>

Outputs:

    3   5
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    4   6

Chapel

<lang chapel>type hole = int; param A : hole = 1; param B : hole = A+1; param C : hole = B+1; param D : hole = C+1; param E : hole = D+1; param F : hole = E+1; param G : hole = F+1; param H : hole = G+1; param starting : int = 0; const holes : domain(hole) = { A,B,C,D,E,F,G,H }; const graph : [holes] domain(hole) = [ A => { C,D,E }, 

                                       B => { D,E,F },
                                       C => { A,D,G },
                                       D => { A,B,C,E,G,H },
                                       E => { A,B,D,F,G,H },
                                       F => { B,E,H },
                                       G => { C,D,E },
                                       H => { D,E,F } 
                                     ];

proc check( configuration : [] int, idx : hole ) : bool { 

 var good = true;
 for adj in graph[idx] {
   if adj >= idx then continue;
   if abs( configuration[idx] - configuration[adj] ) <= 1 {
     good = false;
     break;
   }
 }
 
 return good;

}

proc solve( configuration : [] int, pegs : domain(int), idx : hole = A ) : bool { 

 for value in pegs {
   configuration[idx] = value;
   if check( configuration, idx ) {
     if idx < holes.size {
       var prePegs = pegs;
       if solve( configuration, prePegs - value, idx + 1 ){
         return true;  
       }
     } else {
       return true;
     }
   }
 }
 configuration[idx] = starting;
 return false;

}

proc printBoard( configuration : [] int ){ return "\n " + configuration[A] + " " + configuration[B]+ "\n" + " /|\\ /|\\ \n"+ " / | X | \\ \n"+ " / |/ \\| \\ \n"+ " " + configuration[C] +" - " + configuration[D] + " - " + configuration[E] + " - " + configuration[F] + " \n"+ " \\ |\\ /| / \n"+ " \\ | X | / \n"+ " \\|/ \\|/ \n"+ " " + configuration[G] + " " + configuration[H]+ "\n";

}


proc main(){ 

 var configuration : [holes] int;
 for idx in holes do configuration[idx] = starting;
 
 var pegs : domain(int) = {1,2,3,4,5,6,7,8};
 solve( configuration, pegs );

 writeln( printBoard( configuration ) );

} </lang> 

       4   5
      /|\ /|\ 
     / | X | \ 
    /  |/ \|  \ 
   7 - 1 - 8 - 2 
    \  |\ /|  / 
     \ | X | / 
      \|/ \|/ 
       3   6

D

<lang d>void main() @safe { 

   import std.stdio, std.math, std.algorithm, std.traits, std.string;

   enum Peg { A, B, C, D, E, F, G, H }
   immutable Peg[2][15] connections =
           [[Peg.A, Peg.C], [Peg.A, Peg.D], [Peg.A, Peg.E],
            [Peg.B, Peg.D], [Peg.B, Peg.E], [Peg.B, Peg.F],
            [Peg.C, Peg.D], [Peg.D, Peg.E], [Peg.E, Peg.F],
            [Peg.G, Peg.C], [Peg.G, Peg.D], [Peg.G, Peg.E],
            [Peg.H, Peg.D], [Peg.H, Peg.E], [Peg.H, Peg.F]];

   immutable board = r"
       A   B
      /|\ /|\
     / | X | \
    /  |/ \|  \
   C - D - E - F
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       G   H";

   Peg[EnumMembers!Peg.length] perm = [EnumMembers!Peg];
   do if (connections[].all!(con => abs(perm[con[0]] - perm[con[1]]) > 1))
       return board.tr("ABCDEFGH", "%(%d%)".format(perm)).writeln;
   while (perm[].nextPermutation);

}</lang>

Output:
        2   3
       /|\ /|\
      / | X | \
     /  |/ \|  \
    6 - 0 - 7 - 1
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        4   5

Alternative version

Using a simple backtracking.

Translation of: Go

<lang d>import std.stdio, std.algorithm, std.conv, std.string, std.typecons;

// Holes A=0, B=1, ..., H=7 // With connections: const board = r" 

      A   B
     /|\ /|\
    / | X | \
   /  |/ \|  \
  C - D - E - F
   \  |\ /|  /
    \ | X | /
     \|/ \|/
      G   H";

struct Connection { uint a, b; }

immutable Connection[] connections = [ 

   {0, 2}, {0, 3}, {0, 4}, // A to C,D,E
   {1, 3}, {1, 4}, {1, 5}, // B to D,E,F
   {6, 2}, {6, 3}, {6, 4}, // G to C,D,E
   {7, 3}, {7, 4}, {7, 5}, // H to D,E,F
   {2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F

];

alias Pegs = uint[8];

int absDiff(in uint a, in uint b) pure nothrow @safe @nogc { 

   return (a > b) ? (a - b) : (b - a);

}

/** Solution is a simple recursive brute force solver, it stops at the first found solution. It returns the solution, the number of positions tested, and the number of pegs swapped. */ Tuple!(Pegs,"p", uint,"tests", uint,"swaps") solve() pure nothrow @safe @nogc { 

   uint tests = 0, swaps = 0;
   Pegs p = [1, 2, 3, 4, 5, 6, 7, 8];

   bool recurse(in uint i) nothrow @safe @nogc {
       if (i >= p.length.signed - 1) {
           tests++;
           return connections.all!(c => absDiff(p[c.a], p[c.b]) > 1);
       }

       // Try each remain peg from.
       foreach (immutable j;  i .. p.length) {
           swaps++;
           swap(p[i], p[j]);
           if (recurse(i + 1))
               return true;
           swap(p[i], p[j]);
       }
       return false;
   }

   recurse(0);
   return typeof(return)(p, tests, swaps);

}

void main() { 

   immutable sol = solve();
   board.tr("ABCDEFGH", "%(%d%)".format(sol.p)).writeln;
   writeln("Tested ", sol.tests, " positions and did ", sol.swaps, " swaps.");

}</lang>

Output:
       3   4
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       5   6
Tested 12094 positions and did 20782 swaps.

Elixir

Translation of: Ruby

This solution uses HLPsolver from here <lang elixir># It solved if connected A and B, connected G and H (according to the video).

  1. require HLPsolver

adjacent = for i <- -2..2, j <- -2..2, not(i in -1..1 and j in -1..1), do: {i,j} layout = ~S""" 

      A - B
     /|\ /|\ 
    / | X | \ 
   /  |/ \|  \ 
  C - D - E - F
   \  |\ /|  /
    \ | X | /
     \|/ \|/
      G - H

""" board = """ 

 . 0 0 .
 0 1 0 0 
 . 0 0 .

""" HLPsolver.solve(board, adjacent, false) |> Enum.sort |> Enum.map(fn {_,cell} -> cell.value end) |> Enum.zip(~w[A B C D E F G H]) |> Enum.reduce(layout, fn {n,c},acc -> String.replace(acc, c, to_string(n)) end) |> IO.puts</lang>

Output:
       4 - 6
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       3 - 5

Factor

<lang factor>USING: assocs interpolate io kernel math math.combinatorics math.ranges math.parser multiline pair-rocket sequences sequences.generalizations ;

STRING: diagram 

   ${}   ${}
  /|\ /|\
 / | X | \
/  |/ \|  \

${} - ${} - ${} - ${} 

\  |\ /|  /
 \ | X | /
  \|/ \|/
   ${}   ${}

CONSTANT: adjacency H{ 

   0 => { 2 3 4 }
   1 => { 3 4 5 }
   2 => { 0 3 6 }
   3 => { 0 1 2 4 6 7 }
   4 => { 0 1 3 5 6 7 }
   5 => { 1 4 7 }
   6 => { 2 3 4 }
   7 => { 3 4 5 }

}

any-consecutive? ( seq n -- ? ) [ - abs 1 = ] curry any? ;
neighbors ( elt seq i -- seq elt )
   adjacency at swap nths swap ;
solution? ( permutation-seq -- ? )
   dup [ neighbors any-consecutive? ] with find-index nip not ;
find-solution ( -- seq )
   8 [1,b] [ solution? ] find-permutation ;
display-solution ( seq -- )
   [ number>string ] map 8 firstn diagram interpolate>string
   print ;
main ( -- ) find-solution display-solution ;

MAIN: main</lang>

Output:
    3   4
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    5   6

Go

A simple recursive brute force solution. <lang go>package main

import ( "fmt" "strings" )

func main() { p, tests, swaps := Solution() fmt.Println(p) fmt.Println("Tested", tests, "positions and did", swaps, "swaps.") }

// Holes A=0, B=1, …, H=7 // With connections: const conn = ` 

      A   B
     /|\ /|\
    / | X | \
   /  |/ \|  \
  C - D - E - F
   \  |\ /|  /
    \ | X | /
     \|/ \|/
      G   H`

var connections = []struct{ a, b int }{ {0, 2}, {0, 3}, {0, 4}, // A to C,D,E {1, 3}, {1, 4}, {1, 5}, // B to D,E,F {6, 2}, {6, 3}, {6, 4}, // G to C,D,E {7, 3}, {7, 4}, {7, 5}, // H to D,E,F {2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F }

type pegs [8]int

// Valid checks if the pegs are a valid solution. // If the absolute difference between any pair of connected pegs is // greater than one it is a valid solution. func (p *pegs) Valid() bool { for _, c := range connections { if absdiff(p[c.a], p[c.b]) <= 1 { return false } } return true }

// Solution is a simple recursive brute force solver, // it stops at the first found solution. // It returns the solution, the number of positions tested, // and the number of pegs swapped. func Solution() (p *pegs, tests, swaps int) { var recurse func(int) bool recurse = func(i int) bool { if i >= len(p)-1 { tests++ return p.Valid() } // Try each remain peg from p[i:] in p[i] for j := i; j < len(p); j++ { swaps++ p[i], p[j] = p[j], p[i] if recurse(i + 1) { return true } p[i], p[j] = p[j], p[i] } return false } p = &pegs{1, 2, 3, 4, 5, 6, 7, 8} recurse(0) return }

func (p *pegs) String() string { return strings.Map(func(r rune) rune { if 'A' <= r && r <= 'H' { return rune(p[r-'A'] + '0') } return r }, conn) }

func absdiff(a, b int) int { if a > b { return a - b } return b - a }</lang>

Output:

       3   4
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       5   6
Tested 12094 positions and did 20782 swaps.

Haskell

<lang haskell>import Data.List (intercalate, permutations)

solution :: [Int] solution@(a:b:c:d:e:f:g:h:_) = head $ filter isSolution (permutations [1 .. 8]) 

 where
   isSolution :: [Int] -> Bool
   isSolution (a:b:c:d:e:f:g:h:_) =
     all ((> 1) . abs) $
     zipWith
       (-)
       [a, c, g, e, a, c, g, e, b, d, h, f, b, d, h, f]
       [d, d, d, d, c, g, e, a, e, e, e, e, d, h, f, b]

main :: IO () main = 

 (putStrLn . unlines) $
 let rightShift s
       | length s > 3 = s
       | otherwise = "  " ++ s
 in intercalate
      "\n"
      (zipWith (\x y -> x : (" = " ++ show y)) ['A' .. 'H'] solution) :
    ((rightShift . unwords . fmap show) <$> [[], [a, b], [c, d, e, f], [g, h]])</lang>
Output:
A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6

  3 4
7 1 8 2
  5 6

J

Supporting code:

<lang J>holes=:;:'A B C D E F G H'

connections=:".;._2]0 :0 

holes e.;:'C D E'          NB. A
holes e.;:'D E F'          NB. B
holes e.;:'A D G'          NB. C
holes e.;:'A B C E G H'    NB. D
holes e.;:'A B D F G H'    NB. E
holes e.;:'B E H'          NB. F
holes e.;:'C D E'          NB. G
holes e.;:'D E F'          NB. H

) assert (-:|:) connections NB. catch typos

pegs=: 1+(A.&i.~ !)8

attempt=: [: <./@(-.&0)@,@:| connections * -/~


box=:0 :0 

       A   B
      /|\ /|\
     / | X | \
    /  |/ \|  \
   C - D - E - F
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       G   H

)

disp=:verb define 

 rplc&(,holes;&":&>y) box

)</lang>

Intermezzo:

<lang J> (#~ 1<attempt"1) pegs 3 4 7 1 8 2 5 6 3 5 7 1 8 2 4 6 3 6 7 1 8 2 4 5 3 6 7 1 8 2 5 4 4 3 2 8 1 7 6 5 4 5 2 8 1 7 6 3 4 5 7 1 8 2 3 6 4 6 7 1 8 2 3 5 5 3 2 8 1 7 6 4 5 4 2 8 1 7 6 3 5 4 7 1 8 2 3 6 5 6 7 1 8 2 3 4 6 3 2 8 1 7 4 5 6 3 2 8 1 7 5 4 6 4 2 8 1 7 5 3 6 5 2 8 1 7 4 3</lang>

Since there's more than one arrangement where the pegs satisfy the task constraints, and since the task calls for one solution, we will need to pick one of them. We can use the "first" function to satisfy this important constraint.

<lang J> disp {. (#~ 1<attempt"1) pegs 

       3   4
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       5   6

</lang>

Video

If we follow the video and also connect A and B as well as G and H, we get only four solutions (which we can see are reflections / rotations of each other):

<lang J> (#~ 1<attempt"1) pegs 3 5 7 1 8 2 4 6 4 6 7 1 8 2 3 5 5 3 2 8 1 7 6 4 6 4 2 8 1 7 5 3</lang>

The first of these looks like this:

<lang J> disp {. (#~ 1<attempt"1) pegs 

       3 - 5
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       4 - 6

</lang>

For this puzzle, we can also see that the solution can be described as: put the starting and ending numbers in the middle - everything else follows from there. It's perhaps interesting that we get this solution even if we do not explicitly put that logic into our code - it's built into the puzzle itself and is still the only solution no matter how we arrive there.

Java

The backtracking is getting tiresome, we'll try a stochastic solution for a change.

Works with: Java version 8

<lang java>import static java.lang.Math.abs; import java.util.*; import static java.util.stream.Collectors.toList; import static java.util.stream.IntStream.range;

public class NoConnection { 

   // adopted from Go
   static int[][] links = {
       {2, 3, 4}, // A to C,D,E
       {3, 4, 5}, // B to D,E,F
       {2, 4},    // D to C, E
       {5},       // E to F
       {2, 3, 4}, // G to C,D,E
       {3, 4, 5}, // H to D,E,F
   };

   static int[] pegs = new int[8];

   public static void main(String[] args) {

       List<Integer> vals = range(1, 9).mapToObj(i -> i).collect(toList());
       do {
           Collections.shuffle(vals);
           for (int i = 0; i < pegs.length; i++)
               pegs[i] = vals.get(i);

       } while (!solved());

       printResult();
   }

   static boolean solved() {
       for (int i = 0; i < links.length; i++)
           for (int peg : links[i])
               if (abs(pegs[i] - peg) == 1)
                   return false;
       return true;
   }

   static void printResult() {
       System.out.printf("  %s %s%n", pegs[0], pegs[1]);
       System.out.printf("%s %s %s %s%n", pegs[2], pegs[3], pegs[4], pegs[5]);
       System.out.printf("  %s %s%n", pegs[6], pegs[7]);
   }

}</lang> (takes about 500 shuffles on average) 

       4  5       
    2  8  1  7    
       6  3

JavaScript

ES6

Translation of: Haskell

<lang JavaScript>(() => { 

   'use strict';

   // GENERIC FUNCTIONS ------------------------------------------------------

   // abs :: Num a => a -> a
   const abs = Math.abs;

   // all :: (a -> Bool) -> [a] -> Bool
   const all = (f, xs) => xs.every(f);

   // concatMap :: (a -> [b]) -> [a] -> [b]
   const concatMap = (f, xs) => [].concat.apply([], xs.map(f));

   // delete_ :: Eq a => a -> [a] -> [a]
   const delete_ = (x, xs) =>
       deleteBy((a, b) => a === b, x, xs);

   // deleteBy :: (a -> a -> Bool) -> a -> [a] -> [a]
   const deleteBy = (f, x, xs) =>
       xs.length > 0 ? (
           f(x, xs[0]) ? (
               xs.slice(1)
           ) : [xs[0]].concat(deleteBy(f, x, xs.slice(1)))
       ) : [];

   // enumFromTo :: Enum a => a -> a -> [a]
   const enumFromTo = (m, n) => {
       const [tm, tn] = [typeof m, typeof n];
       return tm !== tn ? undefined : (() => {
           const
               blnS = (tm === 'string'),
               [base, end] = [m, n].map(blnS ? (s => s.codePointAt(0)) : id);
           return Array.from({
               length: Math.floor(end - base) + 1
           }, (_, i) => blnS ? String.fromCodePoint(base + i) : m + i);
       })();
   };

   // id :: a -> a
   const id = x => x;

   // justifyRight :: Int -> Char -> Text -> Text
   const justifyRight = (n, cFiller, strText) =>
       n > strText.length ? (
           (cFiller.repeat(n) + strText)
           .slice(-n)
       ) : strText;

   // permutations :: [a] -> a
   const permutations = xs =>
       xs.length ? concatMap(x => concatMap(ys => [
               [x].concat(ys)
           ],
           permutations(delete_(x, xs))), xs) : [
           []
       ];

   // show :: a -> String
   const show = x => JSON.stringify(x);

   // unlines :: [String] -> String
   const unlines = xs => xs.join('\n');

   // until :: (a -> Bool) -> (a -> a) -> a -> a
   const until = (p, f, x) => {
       let v = x;
       while (!p(v)) v = f(v);
       return v;
   };

   // unwords :: [String] -> String
   const unwords = xs => xs.join(' ');

   // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
   const zipWith = (f, xs, ys) => {
       const ny = ys.length;
       return (xs.length <= ny ? xs : xs.slice(0, ny))
           .map((x, i) => f(x, ys[i]));
   };



   // CONNECTION PUZZLE ------------------------------------------------------

   // universe :: Int
   const universe = permutations(enumFromTo(1, 8));

   // isSolution :: [Int] -> Bool
   const isSolution = ([a, b, c, d, e, f, g, h]) =>
       all(x => abs(x) > 1, [a - d, c - d, g - d, e - d, a - c, c - g, g - e,
           e - a, b - e, d - e, h - e, f - e, b - d, d - h, h - f, f - b
       ]);

   // firstSolution :: [Int]
   const firstSolution = universe[until(
       i => isSolution(universe[i]),
       i => i + 1,
       0
   )];

   // TEST -------------------------------------------------------------------

   // [Int]
   const [a, b, c, d, e, f, g, h] = firstSolution;

   return unlines(
       zipWith(
           (a, n) => a + ' = ' + n.toString(),
           enumFromTo('A', 'H'),
           firstSolution
       )
       .concat(
           [
               [],
               [a, b],
               [c, d, e, f],
               [g, h]
           ].map(xs => justifyRight(5, ' ', unwords(xs.map(show))))
       )
   );

})();</lang>

Output:
A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
     
  3 4
7 1 8 2
  5 6

jq

Works with: jq version 1.4

We present a generate-and-test solver for a slightly more general version of the problem, in which there are N pegs and holes, and in which the connectedness of holes is defined by an array such that holes i and j are connected if and only if [i,j] is a member of the array.

The jq index origin is 0, and so in the following, the pegs and holes are internally numbered from 0 to (N-1) inclusive. That is, we interpret a permutation, p, of 0 .. (N-1) as meaning that the i-th peg is numbered p[i], for i in 0 .. (N-1).

However the pretty-print function shows solutions using the 1-to-8 numbering scheme for pegs, and the A-to-H lettering scheme for holes.

Part 1: Generic functions <lang jq># Short-circuit determination of whether (a|condition)

  1. is true for all a in array:

def forall(array; condition): 

 def check:
   . as $ix
   | if $ix == (array|length) then true
     elif (array[$ix] | condition) then ($ix + 1) | check
     else false
     end;
 0 | check;
  1. permutations of 0 .. (n-1)

def permutations(n): 

 # Given a single array, generate a stream by inserting n at different positions:
 def insert(m;n):
    if m >= 0 then (.[0:m] + [n] + .[m:]), insert(m-1;n) else empty end;
 if n==0 then []
 elif n == 1 then [1]
 else
   permutations(n-1) | insert(n-1; n)
 end;
  1. Count the number of items in a stream

def count(f): reduce f as $_ (0; .+1);</lang>

Part 2: The no-connections puzzle for N pegs and holes <lang jq># Generate a stream of solutions.

  1. Input should be the connections array, i.e. an array of [i,j] pairs;
  2. N is the number of pegs and holds.

def solutions(N): 

 def abs: if . < 0 then -. else . end;

 # Is the proposed permutation (the input) ok?
 def ok(connections):
   . as $p
   | forall( connections; 
             (($p[.[0]] - $p[.[1]])|abs) != 1 );

  . as $connections | permutations(N) | select(ok($connections);</lang>

Part 3: The 8-peg no-connection puzzle <lang jq># The connectedness matrix:

  1. In this table, 0 represents "A", etc, and an entry [i,j]
  2. signifies that the holes with indices i and j are connected.

def connections: 

 [[0, 2], [0, 3], [0, 4],
  [1, 3], [1, 4], [1, 5],
  [6, 2], [6, 3], [6, 4],
  [7, 3], [7, 4], [7, 5],
  [2, 3], [3, 4], [4, 5]]

def solve: 

 connections | solutions(8);
  1. pretty-print a solution for the 8-peg puzzle

def pp: 

 def pegs: ["A", "B", "C", "D", "E", "F", "G", "H"];
 . as $in
 | ("
        A   B
       /|\\ /|\\
      / | X | \\
     /  |/ \\|  \\
    C - D - E - F
     \\  |\\ /|  /
      \\ | X | /
       \\|/ \\|/
        G   H

" | explode) as $board 

   | (pegs | map(explode)) as $letters
   | $letters
   | reduce range(0;length) as $i ($board; index($letters[$i]) as $ix | .[$ix] = $in[$i] + 48)
   | implode;</lang>

Examples: <lang jq># To print all the solutions:

  1. solve | pp
  1. To count the number of solutions:
  2. count(solve)
  1. jq 1.4 lacks facilities for harnessing generators,
  2. but the following will suffice here:

def one(f): reduce f as $s 

 (null; if . == null then $s else . end);

one(solve) | pp </lang>

Output:

<lang sh>$ jq -n -r -f no_connection.jq 

        5   6
       /|\ /|\
      / | X | \
     /  |/ \|  \
    7 - 1 - 8 - 2
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        3   4</lang>


Kotlin

Translation of: Go

<lang scala>// version 1.2.0

import kotlin.math.abs

// Holes A=0, B=1, …, H=7 // With connections: const val conn = """ 

      A   B
     /|\ /|\
    / | X | \
   /  |/ \|  \
  C - D - E - F
   \  |\ /|  /
    \ | X | /
     \|/ \|/
      G   H

"""

val connections = listOf( 

   0 to 2, 0 to 3, 0 to 4,   // A to C, D, E
   1 to 3, 1 to 4, 1 to 5,   // B to D, E, F
   6 to 2, 6 to 3, 6 to 4,   // G to C, D, E
   7 to 3, 7 to 4, 7 to 5,   // H to D, E, F
   2 to 3, 3 to 4, 4 to 5    // C-D, D-E, E-F

)

// 'isValid' checks if the pegs are a valid solution. // If the absolute difference between any pair of connected pegs is // greater than one it is a valid solution. fun isValid(pegs: IntArray): Boolean { 

  for ((a, b) in connections) {
      if (abs(pegs[a] - pegs[b]) <= 1) return false
  }
  return true

}

fun swap(pegs: IntArray, i: Int, j: Int) { 

   val tmp = pegs[i]
   pegs[i] = pegs[j]
   pegs[j] = tmp

}

// 'solve' is a simple recursive brute force solver, // it stops at the first found solution. // It returns the solution, the number of positions tested, // and the number of pegs swapped.

fun solve(): Triple<IntArray, Int, Int> { 

   val pegs = IntArray(8) { it + 1 }
   var tests = 0
   var swaps = 0

   fun recurse(i: Int): Boolean {
       if (i >= pegs.size - 1) {
           tests++
           return isValid(pegs)
       }
       // Try each remaining peg from pegs[i] onwards
       for (j in i until pegs.size) {
           swaps++
           swap(pegs, i, j)
           if (recurse(i + 1)) return true
           swap(pegs, i, j)
       }
       return false
   }

   recurse(0)
   return Triple(pegs, tests, swaps)

}

fun pegsAsString(pegs: IntArray): String { 

   val ca = conn.toCharArray()
   for ((i, c) in ca.withIndex()) {
       if (c in 'A'..'H') ca[i] = '0' + pegs[c - 'A']
   }
   return String(ca)

}

fun main(args: Array<String>) { 

   val (p, tests, swaps) = solve()
   println(pegsAsString(p))
   println("Tested $tests positions and did $swaps swaps.")

}</lang>

Output:
       3   4
      /|\ /|\
     / | X | \
    /  |/ \|  \
   7 - 1 - 8 - 2
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       5   6

Tested 12094 positions and did 20782 swaps.

M2000 Interpreter

Final Version, print all solutions (16 from 40320 permutations)

Press space bar to see solutions so far. <lang M2000 Interpreter> Module no_connection_puzzle { 

     \\ Holes
     Inventory Connections="A":="CDE","B":="DEF","C":="ADG", "D":="ABCEGH"
     Append Connections, "E":="ABDFGH","F":="HEB", "G":="CDE","H":="DEF"
     Inventory ToDelete, Solutions
     \\ eliminate double connnections
     con=each(Connections)
     While con {
           m$=eval$(con, con^)
           c$=eval$(con)
           If c$="*" Then continue
           For i=1 to len(C$) {
                d$=mid$(c$,i,1)
                r$=Filter$(Connections$(d$), m$)
                If r$<>"" Then  {
                        Return connections, d$:=r$
                 }  else   {
                       If m$=connections$(d$) Then {
                             Return connections, d$:="*"  : If not exist(todelete, d$)  Then  Append todelete, d$
                       }
                 }
           }
     }
     con=each(todelete)
     While con {
           Delete Connections, eval$(con)
     }
     Inventory Holes
     For i=0 to 7 : Append Holes, Chr$(65+i):=i : Next i
     CheckValid=lambda Holes, Connections (a$, arr) -> {
           val=Array(arr, Holes(a$))
           con$=Connections$(a$)
           res=True
           For i=1 to Len(con$) {
                If Abs(Array(Arr, Holes(mid$(con$,i,1)))-val)<2 Then res=False: Exit
           }
           =res
     }
     a=(1,2,3,4,5,6,7,8)
     h=(,)
     solution=(,)
     done=False
     counter=0
     Print "Wait..."
     P(h, a)
     sol=Each(Solutions)
     While sol {
           Print "Solution:";sol^+1
           Disp(Eval(Solutions))
           aa$=Key$
     }
     Sub P(h, a)
     If len(a)<=1 Then process(cons(h, a)) : Exit Sub
     local b=cons(a)
     For i=1 to len(b) {
                 b=cons(cdr(b),car(b))
                 P(cons(h,car(b)), cdr(b))
     }
     End Sub
     Sub Process(a)
           counter++
           Print counter
           If keypress(32) Then {
           local  sol=Each(Solutions)
                       aa$=Key$
                       While sol {
                                   Print "Solution:";sol^+1
                                   Disp(Eval(Solutions))
                                   aa$=Key$
                       }  
           }
           hole=each(Connections)
           done=True
           While hole {
                 If not CheckValid(Eval$(hole, hole^), a) Then done=False : Exit
           }
           If done Then Append Solutions, Len(Solutions):=a : Print a
     End Sub
     Sub Disp(a)
           Print format$("    {0}   {1}", array(a), array(a,1))
           Print "   /|\ /|\"
           Print "  / | X | \"
           Print " /  |/ \|  \"
           Print Format$("{0} - {1} - {2} - {3}", array(a,2),array(a,3), array(a,4), array(a,5))
           Print " \  |\ /|  /"
           Print "  \ | X | /"
           Print "   \|/ \|/"
           Print Format$("    {0}   {1}", array(a,6), array(a,7))
     End Sub

} no_connection_puzzle

</lang>

Output:
    3   5
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    4   6

Mathematica

This one simply takes all permutations of the pegs and filters out invalid solutions. <lang Mathematica>sol = Fold[ 

   Select[#, 
     Function[perm, Abs[perm[[#21]] - perm[[#22]]] > 1]] &, 
   Permutations[
    Range[8]], {{1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {2, 6}, {3, 
     4}, {3, 7}, {4, 5}, {4, 7}, {4, 8}, {5, 6}, {5, 7}, {5, 8}, {6, 
     8}}]1;

Print[StringForm[ 

  "    ``   ``\n   /|\\ /|\\\n  / | X | \\\n /  |/ \\|  \\\n`` - `` \

- `` - ``\n \\ |\\ /| /\n \\ | X | /\n \\|/ \\|/\n `` ``", 

  Sequence @@ sol]];</lang>
Output:
    3   4
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    5   6

Perl 6

This uses a Warnsdorff solver, which cuts down the number of tries by more than a factor of six over the brute force approach. This same solver is used in:

The idiosyncratic adjacency diagram is dealt with by the simple expedient of bending the two vertical lines || into two bows )(, such that adjacency can be calculated simply as a distance of 2 or less. <lang perl6>my @adjacent = gather -> $y, $x { 

   take [$y,$x] if abs($x|$y) > 2;

} for flat -5 .. 5 X -5 .. 5;

solveboard q:to/END/; 

   . _ . . _ .
   . . . . . .
   _ . _ 1 . _
   . . . . . .
   . _ . . _ .
   END

sub solveboard($board) { 

   my $max = +$board.comb(/\w+/);
   my $width = $max.chars;

   my @grid;
   my @known;
   my @neigh;
   my @degree;

   @grid = $board.lines.map: -> $line {
       [ $line.words.map: { /^_/ ?? 0 !! /^\./ ?? Rat !! $_ } ]
   }

   sub neighbors($y,$x --> List) {
       eager gather for @adjacent {
           my $y1 = $y + .[0];
           my $x1 = $x + .[1];
           take [$y1,$x1] if defined @grid[$y1][$x1];
       }
   }

   for ^@grid -> $y {
       for ^@grid[$y] -> $x {
           if @grid[$y][$x] -> $v {
               @known[$v] = [$y,$x];
           }
           if @grid[$y][$x].defined {
               @neigh[$y][$x] = neighbors($y,$x);
               @degree[$y][$x] = +@neigh[$y][$x];
           }
       }
   }
   print "\e[0H\e[0J";

   my $tries = 0;

   try_fill 1, @known[1];

   sub try_fill($v, $coord [$y,$x] --> Bool) {
       return True if $v > $max;
       $tries++;

       my $old = @grid[$y][$x];

       return False if +$old and $old != $v;
       return False if @known[$v] and @known[$v] !eqv $coord;

       @grid[$y][$x] = $v;               # conjecture grid value

       print "\e[0H";                    # show conjectured board
       for @grid -> $r {
           say do for @$r {
               when Rat { ' ' x $width }
               when 0   { '_' x $width }
               default  { .fmt("%{$width}d") }
           }
       }



       my @neighbors = @neigh[$y][$x][];

       my @degrees;
       for @neighbors -> \n [$yy,$xx] {
           my $d = --@degree[$yy][$xx];  # conjecture new degrees
           push @degrees[$d], n;         # and categorize by degree
       }

       for @degrees.grep(*.defined) -> @ties {
           for @ties.reverse {           # reverse works better for this hidato anyway
               return True if try_fill $v + 1, $_;
           }
       }

       for @neighbors -> [$yy,$xx] {
           ++@degree[$yy][$xx];          # undo degree conjectures
       }

       @grid[$y][$x] = $old;             # undo grid value conjecture
       return False;
   }
    
   say "$tries tries";

}</lang>

Output:
  4     3  
           
2   8 1   7
           
  6     5  
18 tries

Phix

Brute force solution. I ordered the links highest letter first, then grouped by start letter to eliminate things asap. Nothing to eliminate when placing A and B, when placing C, check that CA>1, when placing D, check that DA,DB,DC are all >1, etc. <lang Phix> constant txt = """ 

       A   B
      /|\ /|\
     / | X | \
    /  |/ \|  \
   C - D - E - F
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       G   H"""

--constant links = "CA DA DB DC EA EB ED FB FE GC GD GE HD HE HF" constant links = {"","","A","ABC","ABD","BE","CDE","DEF"}

function solve(sequence s, integer idx, sequence part) object res integer v, p 

   for i=1 to length(s) do
       v = s[i]
       for j=1 to length(links[idx]) do
           p = links[idx][j]-'@'
           if abs(v-part[p])<2 then v=0 exit end if
       end for
       if v then
           if length(s)=1 then return part&v end if
           res = solve(s[1..i-1]&s[i+1..$],idx+1,part&v)
           if sequence(res) then return res end if
       end if
   end for
   return 0

end function

printf(1,substitute_all(txt,"ABCDEFGH",solve("12345678",1,"")))</lang>

Output:
        3   4
       /|\ /|\
      / | X | \
     /  |/ \|  \
    7 - 1 - 8 - 2
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        5   6

Prolog

Works with SWi-Prolog with module clpfd written by Markus Triska

We first compute a list of nodes, with sort this list, and we attribute a value at the nodes. <lang Prolog>:- use_module(library(clpfd)).

edge(a, c). edge(a, d). edge(a, e). edge(b, d). edge(b, e). edge(b, f). edge(c, d). edge(c, g). edge(d, e). edge(d, g). edge(d, h). edge(e, f). edge(e, g). edge(e, h). edge(f, h).

connected(A, B) :- ( edge(A,B); edge(B, A)).

no_connection_puzzle(Vs) :- % construct the arranged list of the nodes bagof(A, B^(edge(A,B); edge(B, A)), Lst), sort(Lst, L), length(L, Len),

% construct the list of the values length(Vs, Len), Vs ins 1..Len, all_distinct(Vs),

% two connected nodes must have values different for more than 1 set_constraints(L, Vs), label(Vs).


set_constraints([], []).

set_constraints([H | T], [VH | VT]) :- set_constraint(H, T, VH, VT), set_constraints(T, VT).


set_constraint(_, [], _, []). set_constraint(H, [H1 | T1], V, [VH | VT]) :- connected(H, H1), ( V - VH #> 1; VH - V #> 1), set_constraint(H, T1, V, VT).

set_constraint(H, [H1 | T1], V, [_VH | VT]) :- \+connected(H, H1), set_constraint(H, T1, V, VT).

</lang> Output : 

 ?- no_connection_puzzle(Vs).
Vs = [4, 3, 2, 8, 1, 7, 6, 5] .

 27 ?- setof(Vs, no_connection_puzzle(Vs), R), length(R, Len).
R = [[3, 4, 7, 1, 8, 2, 5, 6], [3, 5, 7, 1, 8, 2, 4|...], [3, 6, 7, 1, 8, 2|...], [3, 6, 7, 1, 8|...], [4, 3, 2, 8|...], [4, 5, 2|...], [4, 5|...], [4|...], [...|...]|...],
Len = 16.

Python

A brute force search solution. <lang python>from __future__ import print_function from itertools import permutations from enum import Enum

A, B, C, D, E, F, G, H = Enum('Peg', 'A, B, C, D, E, F, G, H')

connections = ((A, C), (A, D), (A, E), 

              (B, D), (B, E), (B, F),
              (G, C), (G, D), (G, E),
              (H, D), (H, E), (H, F),
              (C, D), (D, E), (E, F))


def ok(conn, perm): 

   """Connected numbers ok?"""
   this, that = (c.value - 1 for c in conn)
   return abs(perm[this] - perm[that]) != 1


def solve(): 

   return [perm for perm in permutations(range(1, 9))
           if all(ok(conn, perm) for conn in connections)]


if __name__ == '__main__': 

   solutions = solve()
   print("A, B, C, D, E, F, G, H =", ', '.join(str(i) for i in solutions[0]))</lang>
Output:
A, B, C, D, E, F, G, H = 3, 4, 7, 1, 8, 2, 5, 6


All solutions pretty printed

Add the following code after that above: <lang python>def pp(solution): 

   """Prettyprint a solution"""
   boardformat = r"""
        A   B
       /|\ /|\
      / | X | \
     /  |/ \|  \
    C - D - E - F
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        G   H"""
   for letter, number in zip("ABCDEFGH", solution):
       boardformat = boardformat.replace(letter, str(number))
   print(boardformat)


if __name__ == '__main__': 

   for i, s in enumerate(solutions, 1):
       print("\nSolution", i, end=)
       pp(s)</lang>
Extra output
Solution 1
         3   4
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         5   6

Solution 2
         3   5
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         4   6

Solution 3
         3   6
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         4   5

Solution 4
         3   6
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         5   4

Solution 5
         4   3
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         6   5

Solution 6
         4   5
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         6   3

Solution 7
         4   5
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         3   6

Solution 8
         4   6
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         3   5

Solution 9
         5   3
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         6   4

Solution 10
         5   4
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         6   3

Solution 11
         5   4
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         3   6

Solution 12
         5   6
        /|\ /|\
       / | X | \
      /  |/ \|  \
     7 - 1 - 8 - 2
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         3   4

Solution 13
         6   3
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         4   5

Solution 14
         6   3
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         5   4

Solution 15
         6   4
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         5   3

Solution 16
         6   5
        /|\ /|\
       / | X | \
      /  |/ \|  \
     2 - 8 - 1 - 7
      \  |\ /|  /
       \ | X | /
        \|/ \|/
         4   3

Racket

<lang racket>#lang racket

Solve the no connection puzzle. Tim Brown Oct. 2014
absolute difference of a and b if they are both true

(define (and- a b) (and a b (abs (- a b))))

Finds the differences of all established connections in the network

(define (network-diffs (A #f) (B #f) (C #f) (D #f) (E #f) (F #f) (G #f) (H #f)) 

 (list (and- A C) (and- A D) (and- A E)
       (and- B D) (and- B E) (and- B F)
       (and- C D) (and- C G)
       (and- D E) (and- D G) (and- D H)
       (and- E F) (and- E G) (and- E H)
       (and- F G)))
Make sure there is “no connection” in the network N; return N if good

(define (good-network? N) 

 (and (for/and ((d (filter values (apply network-diffs N)))) (> d 1)) N))
possible optimisation is to reverse the arguments to network-diffs, reverse the return value from
this function and make this a cons but we're pretty quick here as it is.

(define (find-good-network pegs (n/w null)) 

 (if (null? pegs) n/w
     (for*/or ((p pegs))
       (define n/w+ (append n/w (list p)))
       (and (good-network? n/w+)
            (find-good-network (remove p pegs =) n/w+)))))

(define (render-puzzle pzl) 

 (apply printf (regexp-replace* "O" #<<EOS
   O   O
  /|\ /|\
 / | X | \
/  |/ \|  \

O - O - O - O 

\  |\ /|  /
 \ | X | /
  \|/ \|/
   O   O~%

EOS 

                                "~a") pzl))

(render-puzzle (find-good-network '(1 2 3 4 5 6 7 8)))</lang>

Output:
    3   4
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    5   6

REXX

unannotated solutions

<lang rexx>/*REXX program solves the "no-connection" puzzle (the puzzle has eight pegs). */ parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */ if limit== | limit=="." then limit=1 /* ║ A B ║ */ 

                                                      /* ║         /│\  /│\          ║ */

@. = /* ║ / │ \/ │ \ ║ */ @.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */ @.2 = 'B D E F' /* ║ / │/ \│ \ ║ */ @.3 = 'C A D G' /* ║ C────D────E────F ║ */ @.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */ @.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */ @.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */ @.7 = 'G C D E' /* ║ \│/ \│/ ║ */ @.8 = 'H D E F' /* ║ G H ║ */ cnt=0 /* ╚═══════════════════════════╝ */ 

                 do pegs=1  while  @.pegs\==;    _=word(@.pegs,1)
                 subs=0
                            do #=1  for  words(@.pegs) -1  /*create list of node paths.*/
                            __=word(@.pegs, # + 1);    if __>_  then iterate
                            subs=subs + 1;             !._.subs=__
                            end  /*#*/
                 !._.0=subs                     /*assign the number of the node paths. */
                 end   /*pegs*/

pegs=pegs-1 /*the number of pegs to be seated. */ _=' ' /*_ is used for indenting the output.*/ 

       do        a=1  for pegs;     if ?('A')  then iterate
        do       b=1  for pegs;     if ?('B')  then iterate
         do      c=1  for pegs;     if ?('C')  then iterate
          do     d=1  for pegs;     if ?('D')  then iterate
           do    e=1  for pegs;     if ?('E')  then iterate
            do   f=1  for pegs;     if ?('F')  then iterate
             do  g=1  for pegs;     if ?('G')  then iterate
              do h=1  for pegs;     if ?('H')  then iterate
              say _ 'a='a _  'b='||b _  'c='c _  'd='d _  'e='e _  'f='f _  'g='g _ 'h='h
              cnt=cnt+1;        if cnt==limit  then leave a
              end   /*h*/
             end    /*g*/
            end     /*f*/
           end      /*e*/
          end       /*d*/
         end        /*c*/
        end         /*b*/
       end          /*a*/

say /*display a blank line to the terminal.*/ s= left('s', cnt\==1) /*handle the case of plurals (or not).*/ say 'found ' cnt " solution"s'.' /*display the number of solutions found*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ ?: parse arg node; nn=value(node) 

  nH=nn+1
            do cn=c2d('A')  to c2d(node) - 1;    if value( d2c(cn) )==nn  then return 1
            end   /*cn*/                       /* [↑]  see if there any are duplicates.*/
  nL=nn-1
            do ch=1  for !.node.0              /* [↓]  see if there any  ¬= ±1  values.*/
            $=!.node.ch;        fn=value($)    /*the node name  and  its current peg #.*/
            if nL==fn | nH==fn  then return 1  /*if ≡ ±1,  then the node can't be used.*/
            end   /*ch*/                       /* [↑]  looking for suitable number.    */
  return 0                                     /*the subroutine arg value passed is OK.*/</lang>

output   when using the default input: 

     a=3      b=4      c=7      d=1      e=8      f=2      g=5      h=6

found  1  solution.

output   when using the input of:   999 

     a=3      b=4      c=7      d=1      e=8      f=2      g=5      h=6
     a=3      b=5      c=7      d=1      e=8      f=2      g=4      h=6
     a=3      b=6      c=7      d=1      e=8      f=2      g=4      h=5
     a=3      b=6      c=7      d=1      e=8      f=2      g=5      h=4
     a=4      b=3      c=2      d=8      e=1      f=7      g=6      h=5
     a=4      b=5      c=2      d=8      e=1      f=7      g=6      h=3
     a=4      b=5      c=7      d=1      e=8      f=2      g=3      h=6
     a=4      b=6      c=7      d=1      e=8      f=2      g=3      h=5
     a=5      b=3      c=2      d=8      e=1      f=7      g=6      h=4
     a=5      b=4      c=2      d=8      e=1      f=7      g=6      h=3
     a=5      b=4      c=7      d=1      e=8      f=2      g=3      h=6
     a=5      b=6      c=7      d=1      e=8      f=2      g=3      h=4
     a=6      b=3      c=2      d=8      e=1      f=7      g=4      h=5
     a=6      b=3      c=2      d=8      e=1      f=7      g=5      h=4
     a=6      b=4      c=2      d=8      e=1      f=7      g=5      h=3
     a=6      b=5      c=2      d=8      e=1      f=7      g=4      h=3

found  16  solutions.

annotated solutions

Usage note:   if the   limit   (the 1st argument)   is negative, a diagram (node graph) is shown. <lang rexx>/*REXX program solves the "no-connection" puzzle (the puzzle has eight pegs). */ @abc='ABCDEFGHIJKLMNOPQRSTUVWXYZ' parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */ if limit== | limit=="." then limit=1 /* ║ A B ║ */ oLimit=limit; limit=abs(limit) /* ║ /│\ /│\ ║ */ @. = /* ║ / │ \/ │ \ ║ */ @.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */ @.2 = 'B D E F' /* ║ / │/ \│ \ ║ */ @.3 = 'C A D G' /* ║ C────D────E────F ║ */ @.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */ @.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */ @.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */ @.7 = 'G C D E' /* ║ \│/ \│/ ║ */ @.8 = 'H D E F' /* ║ G H ║ */ cnt=0 /* ╚═══════════════════════════╝ */ 

                 do pegs=1  while  @.pegs\==;    _=word(@.pegs, 1)
                 subs=0
                            do #=1  for  words(@.pegs) -1  /*create list of node paths.*/
                            __=word(@.pegs, #+1);      if __>_  then iterate
                            subs=subs + 1;             !._.subs=__
                            end  /*#*/
                 !._.0=subs                    /*assign the number of the node paths.  */
                 end   /*pegs*/

pegs=pegs - 1 /*the number of pegs to be seated. */ _=' ' /*_ is used for indenting the output. */ 

       do        a=1  for pegs;     if ?('A')  then iterate
        do       b=1  for pegs;     if ?('B')  then iterate
         do      c=1  for pegs;     if ?('C')  then iterate
          do     d=1  for pegs;     if ?('D')  then iterate
           do    e=1  for pegs;     if ?('E')  then iterate
            do   f=1  for pegs;     if ?('F')  then iterate
             do  g=1  for pegs;     if ?('G')  then iterate
              do h=1  for pegs;     if ?('H')  then iterate
              call showNodes
              cnt=cnt+1;        if cnt==limit  then leave a
              end   /*h*/
             end    /*g*/
            end     /*f*/
           end      /*e*/
          end       /*d*/
         end        /*c*/
        end         /*b*/
       end          /*a*/

say /*display a blank line to the terminal.*/ s=left('s', cnt\==1) /*handle the case of plurals (or not).*/ say 'found ' cnt " solution"s'.' /*display the number of solutions found*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ ?: parse arg node; nn=value(node) 

  nH=nn+1
            do cn=c2d('A')  to c2d(node)-1;  if value( d2c(cn) )==nn  then return 1
            end   /*cn*/                        /* [↑]  see if there're any duplicates.*/
  nL=nn-1
            do ch=1  for !.node.0               /* [↓]  see if there any ¬= ±1  values.*/
            $=!.node.ch;        fn=value($)     /*the node name  and its current peg #.*/
            if nL==fn | nH==fn  then return 1   /*if ≡ ±1, then the node can't be used.*/
            end   /*ch*/                        /* [↑]  looking for suitable number.   */
  return 0                                      /*the subroutine arg value passed is OK*/

/*──────────────────────────────────────────────────────────────────────────────────────*/ showNodes: _=left(, 5) /*_ is used for padding the output. */ show=0 /*indicates no graph has been found yet*/ 

     do box=1  for sourceline()  while oLimit<0 /*Negative?  Then display the diagram. */
     xw=sourceline(box)                         /*get a source line of this program.   */
     p2=lastpos('*', xw)                        /*the position of    last     asterisk.*/
     p1=lastpos('*', xw, max(1, p2-1) )         /* "      "     " penultimate     "    */
     if pos('╔', xw)\==0  then show=1           /*Have found the top-left box corner ? */
     if \show             then iterate          /*Not found?  Then skip this line.     */
     xb=substr(xw, p1+1, p2-p1-2)               /*extract the  "box"  part of line.    */
     xt=xb                                      /*get a working copy of the box.       */
                    do jx=1  for pegs           /*do a substitution for all the pegs.  */
                    @=substr(@abc, jx, 1)       /*get the name of the peg  (A ──► Z).  */
                    xt=translate(xt,value(@),@) /*substitute the peg name with a value.*/
                    end   /*jx*/                /* [↑]    graph is limited to 26 nodes.*/
     say _ xb _ _ xt                            /*display one line of the graph.       */
     if pos('╝', xw)\==0  then return           /*Is this last line of graph? Then stop*/
     end   /*box*/

say _ 'a='a _ 'b='||b _ 'c='c _ 'd='d _ ' e='e _ 'f='f _ 'g='g _ 'h='h return</lang> output when using the input of:   -3 

       ╔═══════════════════════════╗              ╔═══════════════════════════╗
       ║          A    B           ║              ║          3    4           ║
       ║         /│\  /│\          ║              ║         /│\  /│\          ║
       ║        / │ \/ │ \         ║              ║        / │ \/ │ \         ║
       ║       /  │ /\ │  \        ║              ║       /  │ /\ │  \        ║
       ║      /   │/  \│   \       ║              ║      /   │/  \│   \       ║
       ║     C────D────E────F      ║              ║     7────1────8────2      ║
       ║      \   │\  /│   /       ║              ║      \   │\  /│   /       ║
       ║       \  │ \/ │  /        ║              ║       \  │ \/ │  /        ║
       ║        \ │ /\ │ /         ║              ║        \ │ /\ │ /         ║
       ║         \│/  \│/          ║              ║         \│/  \│/          ║
       ║          G    H           ║              ║          5    6           ║
       ╚═══════════════════════════╝              ╚═══════════════════════════╝
       ╔═══════════════════════════╗              ╔═══════════════════════════╗
       ║          A    B           ║              ║          3    5           ║
       ║         /│\  /│\          ║              ║         /│\  /│\          ║
       ║        / │ \/ │ \         ║              ║        / │ \/ │ \         ║
       ║       /  │ /\ │  \        ║              ║       /  │ /\ │  \        ║
       ║      /   │/  \│   \       ║              ║      /   │/  \│   \       ║
       ║     C────D────E────F      ║              ║     7────1────8────2      ║
       ║      \   │\  /│   /       ║              ║      \   │\  /│   /       ║
       ║       \  │ \/ │  /        ║              ║       \  │ \/ │  /        ║
       ║        \ │ /\ │ /         ║              ║        \ │ /\ │ /         ║
       ║         \│/  \│/          ║              ║         \│/  \│/          ║
       ║          G    H           ║              ║          4    6           ║
       ╚═══════════════════════════╝              ╚═══════════════════════════╝
       ╔═══════════════════════════╗              ╔═══════════════════════════╗
       ║          A    B           ║              ║          3    6           ║
       ║         /│\  /│\          ║              ║         /│\  /│\          ║
       ║        / │ \/ │ \         ║              ║        / │ \/ │ \         ║
       ║       /  │ /\ │  \        ║              ║       /  │ /\ │  \        ║
       ║      /   │/  \│   \       ║              ║      /   │/  \│   \       ║
       ║     C────D────E────F      ║              ║     7────1────8────2      ║
       ║      \   │\  /│   /       ║              ║      \   │\  /│   /       ║
       ║       \  │ \/ │  /        ║              ║       \  │ \/ │  /        ║
       ║        \ │ /\ │ /         ║              ║        \ │ /\ │ /         ║
       ║         \│/  \│/          ║              ║         \│/  \│/          ║
       ║          G    H           ║              ║          4    5           ║
       ╚═══════════════════════════╝              ╚═══════════════════════════╝

found  3  solutions.

Ruby

Be it Golden Frogs jumping on trancendental lilly pads, or a Knight on a board, or square pegs into round holes this is essentially a Hidato Like Problem, so I use HLPSolver: <lang ruby>

  1. Solve No Connection Puzzle
  3. Nigel_Galloway
  4. October 6th., 2014

require 'HLPSolver' ADJACENT = 0,0 A,B,C,D,E,F,G,H = [0,1],[0,2],[1,0],[1,1],[1,2],[1,3],[2,1],[2,2]

board1 = <<EOS 

 . 0 0 .
 0 0 1 0 
 . 0 0 .

EOS g = HLPsolver.new(board1) g.board[A[0]][A[1]].adj = [B,G,H,F] g.board[B[0]][B[1]].adj = [A,C,G,H] g.board[C[0]][C[1]].adj = [B,E,F,H] g.board[D[0]][D[1]].adj = [F] g.board[E[0]][E[1]].adj = [C] g.board[F[0]][F[1]].adj = [A,C,D,G] g.board[G[0]][G[1]].adj = [A,B,F,H] g.board[H[0]][H[1]].adj = [A,B,C,G] g.solve </lang>

Output:
Problem:
     0  0
  0  0  1  0
     0  0

Solution:
     5  3
  2  8  1  7
     6  4

Tcl

Library: Tcllib (Package: struct::list)

<lang tcl>package require Tcl 8.6 package require struct::list

proc haveAdjacent {a b c d e f g h} { 

   expr {

[edge $a $c] || [edge $a $d] || [edge $a $e] || [edge $b $d] || [edge $b $e] || [edge $b $f] || [edge $c $d] || [edge $c $g] || [edge $d $e] || [edge $d $g] || [edge $d $h] || [edge $e $f] || [edge $e $g] || [edge $e $h] || [edge $f $h] 

   }

} proc edge {x y} { 

   expr {abs($x-$y) == 1}

}

set layout [string trim { 

       A   B
      /|\ /|\ 
     / | X | \ 
    /  |/ \|  \ 
   C - D - E - F
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       G   H

} \n] struct::list foreachperm p {1 2 3 4 5 6 7 8} { 

   if {![haveAdjacent {*}$p]} {

puts [string map [join [ lmap name {A B C D E F G H} val $p {list $name $val} ]] $layout] break 

   }

}</lang>

Output:
        3   4
       /|\ /|\ 
      / | X | \ 
     /  |/ \|  \ 
    7 - 1 - 8 - 2
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        5   6

XPL0

<lang XPL0>include c:\cxpl\codes;

int Hole, Max, I; char Box(8), Str; def A, B, C, D, E, F, G, H; [for Hole:= 0 to 7 do Box(Hole):= Hole+1; Max:= 7; while abs(Box(D)-Box(A)) < 2 or abs(Box(D)-Box(C)) < 2 or 

       abs(Box(D)-Box(G)) < 2  or  abs(Box(D)-Box(E)) < 2  or
       abs(Box(A)-Box(C)) < 2  or  abs(Box(C)-Box(G)) < 2  or
       abs(Box(G)-Box(E)) < 2  or  abs(Box(E)-Box(A)) < 2  or
       abs(Box(E)-Box(B)) < 2  or  abs(Box(E)-Box(H)) < 2  or
       abs(Box(E)-Box(F)) < 2  or  abs(Box(B)-Box(D)) < 2  or
       abs(Box(D)-Box(H)) < 2  or  abs(Box(H)-Box(F)) < 2  or
       abs(Box(F)-Box(B)) < 2  do
          loop [I:= Box(0);                    \next permutation
               for Hole:= 0 to Max-1 do Box(Hole):= Box(Hole+1);
               Box(Max):= I;
               if I # Max+1 then [Max:= 7;  quit]
               else Max:= Max-1];

Str:= " 

   #   #
  /|\ /|\
 / | X | \
/  |/ \|  \
  1. - # - # - #
\  |\ /|  /
 \ | X | /
  \|/ \|/
   #   #

"; Hole:= 0; I:= 0; repeat if Str(I)=^# then [Str(I):= Box(Hole)+^0; Hole:= Hole+1]; 

       I:= I+1;

until Hole = 8; Text(0, Str); ]</lang>

Output:
    5   6
   /|\ /|\
  / | X | \
 /  |/ \|  \
7 - 1 - 8 - 2
 \  |\ /|  /
  \ | X | /
   \|/ \|/
    3   4

zkl

Translation of: D

<lang zkl>const PegA=0, PegB=1, PegC=2, PegD=3, PegE=4, PegF=5, PegG=6, PegH=7; connections:=T( 

  T(PegA, PegC), T(PegA, PegD), T(PegA, PegE),
  T(PegB, PegD), T(PegB, PegE), T(PegB, PegF),
  T(PegC, PegD), T(PegD, PegE), T(PegE, PegF),
  T(PegG, PegC), T(PegG, PegD), T(PegG, PegE),
  T(PegH, PegD), T(PegH, PegE), T(PegH, PegF) );

CZ:=connections.len();

  1. <<< // Use "raw" string in a "here doc" so \ isn't a quote char

board:= 0'$ A B 

      /|\ /|\
     / | X | \
    /  |/ \|  \
   C - D - E - F
    \  |\ /|  /
     \ | X | /
      \|/ \|/
       G   H$;
  1. <<< // end "here doc"

perm:=T(PegA,PegB,PegC,PegD,PegE,PegF,PegG,PegH); // Peg[8] foreach p in (Utils.Helpers.permuteW(perm)){ // permutation iterator 

  if(connections.filter1('wrap([(a,b)]){ (p[a] - p[b]).abs()<=1 })) continue;
  board.translate("ABCDEFGH",p.apply('+(1)).concat()).println(); 
  break;  // comment out to see all 16 solutions

}</lang> The filter1 method stops on the first True, so it acts like a conditional or.

Output:
        5   6
       /|\ /|\
      / | X | \
     /  |/ \|  \
    7 - 1 - 8 - 2
     \  |\ /|  /
      \ | X | /
       \|/ \|/
        3   4
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