# Solve the no connection puzzle

(Redirected from No connection puzzle)
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.

Produce and show here   one   solution to the puzzle.

## 11l

Translation of: Python
```V 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)]

F ok(conn, perm)
R abs(perm[conn[0]] - perm[conn[1]]) != 1

F solve()
[[Int]] r
V perm = Array(1..8)
L
I all(:connections.map(conn -> ok(conn, @perm)))
r [+]= copy(perm)
I !perm.next_permutation()
L.break
R r

V solutions = solve()
print(‘A, B, C, D, E, F, G, H = ’solutions[0].join(‘, ’))```
Output:
```A, B, C, D, E, F, G, H = 3, 4, 7, 1, 8, 2, 5, 6
```

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.

```With
Connection_Types,
Connection_Combinations;

procedure main is
Result : Connection_Types.Partial_Board renames Connection_Combinations;
begin
end;
```
```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;
```
```Pragma Ada_2012;

with Connection_Types;
use  Connection_Types;

Function Connection_Combinations return Partial_Board;
```
```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;
```
```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.
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;

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;

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

## 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) ⍝ node i connects with nodes i⊃links
tries←8 perms 8
fails←{1∊{1∊⍵∊¯1 0 1}¨|⍺-¨⍺∘{⍺[⍵]}¨⍵}
⍝ 16
tries[''⍴solns;]
}
```

## ARM Assembly

Works with: as version Raspberry Pi
```/* ARM assembly Raspberry PI  */
/*  program noconnpuzzle.s   */

/************************************/
/* Constantes                       */
/************************************/
.equ STDOUT, 1     @ Linux output console
.equ EXIT,   1     @ Linux syscall
.equ WRITE,  4     @ Linux syscall

.equ NBBOX,  8
.equ POSA,   5

/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessDeb:           .ascii "a="
sMessValeur_a:     .fill 11, 1, ' '            @ size => 11
.ascii "b="
sMessValeur_b:     .fill 11, 1, ' '            @ size => 11
.ascii "c="
sMessValeur_c:     .fill 11, 1, ' '            @ size => 11
.ascii "d="
sMessValeur_d:     .fill 11, 1, ' '            @ size => 11
.ascii "\n"
.ascii "e="
sMessValeur_e:     .fill 11, 1, ' '            @ size => 11
.ascii "f="
sMessValeur_f:     .fill 11, 1, ' '            @ size => 11
.ascii "g="
sMessValeur_g:     .fill 11, 1, ' '            @ size => 11
.ascii "h="
sMessValeur_h:     .fill 11, 1, ' '            @ size => 11

szCarriageReturn:   .asciz "\n************************\n"

szMessLine1:            .asciz "               \n"
szMessLine2:            .asciz "    /|\\ /|\\  \n"
szMessLine3:            .asciz "   / | X | \\     \n"
szMessLine4:            .asciz "  /  |/ \\|  \\     \n"
szMessLine5:            .asciz "   -   - | -     \n"
szMessLine6:            .asciz "  \\  |\\ /|  /  \n"
szMessLine7:            .asciz "   \\ | X | /  \n"
szMessLine8:            .asciz "    \\|/ \\|/   \n"
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
.align 4
iValues_a:                .skip 4 * NBBOX
iValues_b:                .skip 4 * NBBOX - 1
iValues_c:                .skip 4 * NBBOX - 2
iValues_d:                .skip 4 * NBBOX - 3
iValues_e:                .skip 4 * NBBOX - 4
iValues_f:                .skip 4 * NBBOX - 5
iValues_g:                .skip 4 * NBBOX - 6
iValues_h:                .skip 4 * NBBOX - 7
sConvValue:               .skip 12
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main
main:                                             @ entry of program
mov r0,#1
mov r1,#8
bl searchPb

100:                                              @ standard end of the program
mov r0, #0                                    @ return code
mov r7, #EXIT                                 @ request to exit program
svc #0                                        @ perform the system call

/******************************************************************/
/*     search problem  unique solution                            */
/******************************************************************/
/* r0 contains start digit */
/* r1 contains end digit */
searchPb:
push {r0-r12,lr}                                  @ save  registers
@ init
ldr r3,iAdriValues_a                              @ area value a
mov r4,#0
1:                                                    @ loop init value a
str r0,[r3,r4,lsl #2]
cmp r0,r1
ble 1b

mov r12,#-1
2:
add r12,#1                                        @ increment indice a
cmp r12,#NBBOX-1
bgt 90f
ldr r0,iAdriValues_a                              @ area value a
ldr r1,iAdriValues_b                              @ area value b
mov r2,r12                                        @ indice  a
mov r3,#NBBOX                                     @ number of origin values
bl prepValues
mov r11,#-1
3:
add r11,#1                                        @ increment indice b
cmp r11,#NBBOX - 2
bgt 2b
ldr r0,iAdriValues_b                              @ area value b
ldr r1,iAdriValues_c                              @ area value c
mov r2,r11                                        @ indice b
mov r3,#NBBOX -1                                  @ number of origin values
bl prepValues
mov r10,#-1
4:
cmp r10,#NBBOX - 3
bgt 3b
ldr r0,[r0,r12,lsl #2]
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 4b
mov r2,r10
mov r3,#NBBOX - 2
bl prepValues
mov r9,#-1
5:
cmp r9,#NBBOX - 4
bgt 4b
@ control d   / a b c
ldr r0,[r0,r9,lsl #2]
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 5b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 5b
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 5b

mov r2,r9
mov r3,#NBBOX - 3
bl prepValues
mov r8,#-1
6:
cmp r8,#NBBOX - 5
bgt 5b
@ control e   / a b d
ldr r0,[r0,r8,lsl #2]
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 6b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 6b
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 6b

mov r2,r8
mov r3,#NBBOX - 4
bl prepValues
mov r7,#-1
7:
cmp r7,#NBBOX - 6
bgt 6b
@ control f   / b e
ldr r0,[r0,r7,lsl #2]
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 7b
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 7b

mov r2,r7
mov r3,#NBBOX - 5
bl prepValues
mov r6,#-1
8:
cmp r6,#NBBOX - 7
bgt 7b
@ control g   / c d e
ldr r0,[r0,r6,lsl #2]
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 8b
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 8b
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 8b
mov r2,r6
mov r3,#NBBOX - 6
bl prepValues
mov r5,#-1
9:
cmp r5,#NBBOX - 8
bgt 8b
@ control h   / d e f
ldr r0,[r0,r5,lsl #2]
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 9b
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 9b
ldr r1,[r1,r7,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
cmp r2,#1
beq 9b
@ solution ok   display text
ldr r0,[r0,r12,lsl #2]
bl conversion10
ldr r0,[r0,r11,lsl #2]
bl conversion10
ldr r0,[r0,r10,lsl #2]
bl conversion10
ldr r0,[r0,r9,lsl #2]
bl conversion10
ldr r0,[r0,r8,lsl #2]
bl conversion10
ldr r0,[r0,r7,lsl #2]
bl conversion10
ldr r0,[r0,r6,lsl #2]
bl conversion10
ldr r0,[r0,r5,lsl #2]
bl conversion10
bl affichageMess

@ display design
ldr r0,[r0,r12,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA]
ldr r0,[r0,r11,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA+4]
bl affichageMess
bl affichageMess
bl affichageMess
bl affichageMess
ldr r0,[r0,r10,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA-4]
ldr r0,[r0,r9,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA]
ldr r0,[r0,r8,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA+4]
ldr r0,[r0,r7,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA+8]
bl affichageMess
bl affichageMess
bl affichageMess
bl affichageMess
ldr r0,[r0,r6,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA]
ldr r0,[r0,r5,lsl #2]
bl conversion10
ldrb r2,[r1]
strb r2,[r0,#POSA+4]
bl affichageMess

//b 9b                   @ loop for other solution
90:

100:
pop {r0-r12,lr}                               @ restaur registers
bx lr                                         @return

/******************************************************************/
/*     copy value area  and substract value of indice             */
/******************************************************************/
/* r0 contains the address of values origin */
/* r1 contains the address of values destination */
/* r2 contains value indice to substract     */
/* r3 contains origin values number          */
prepValues:
push {r1-r6,lr}                                @ save  registres
mov r4,#0                                      @ indice origin value
mov r5,#0                                      @ indice destination value
1:
cmp r4,r2                                      @ substract indice ?
beq 2f                                         @ yes -> jump
ldr r6,[r0,r4,lsl #2]                          @ no -> copy value
str r6,[r1,r5,lsl #2]
add r5,#1                                      @ increment destination indice
2:
add r4,#1                                       @ increment origin indice
cmp r4,r3                                       @ end ?
blt 1b
100:
pop {r1-r6,lr}                                 @ restaur registres
bx lr                                          @return
/******************************************************************/
/*     display text with size calculation                         */
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
push {r0,r1,r2,r7,lr}                          @ save  registres
mov r2,#0                                      @ counter length
1:                                                 @ loop length calculation
ldrb r1,[r0,r2]                                @ read octet start position + index
cmp r1,#0                                      @ if 0 its over
bne 1b                                         @ and loop
@ so here r2 contains the length of the message
mov r1,r0                                      @ address message in r1
mov r0,#STDOUT                                 @ code to write to the standard output Linux
mov r7, #WRITE                                 @ code call system "write"
svc #0                                         @ call systeme
pop {r0,r1,r2,r7,lr}                           @ restaur des  2 registres */
bx lr                                          @ return
/******************************************************************/
/*     Converting a register to a decimal unsigned                */
/******************************************************************/
/* r0 contains value and r1 address area   */
/* r0 return size of result (no zero final in area) */
/* area size => 11 bytes          */
.equ LGZONECAL,   10
conversion10:
push {r1-r4,lr}                                 @ save registers
mov r3,r1
mov r2,#LGZONECAL
1:                                                  @ start loop
bl divisionpar10U                               @ unsigned  r0 <- dividende. quotient ->r0 reste -> r1
strb r1,[r3,r2]                                 @ store digit on area
cmp r0,#0                                       @ stop if quotient = 0
subne r2,#1                                     @ else previous position
bne 1b                                          @ and loop
@ and move digit from left of area
mov r4,#0
2:
ldrb r1,[r3,r2]
strb r1,[r3,r4]
cmp r2,#LGZONECAL
ble 2b
@ and move spaces in end on area
mov r0,r4                                         @ result length
mov r1,#' '                                       @ space
3:
strb r1,[r3,r4]                                   @ store space in area
cmp r4,#LGZONECAL
ble 3b                                            @ loop if r4 <= area size

100:
pop {r1-r4,lr}                                    @ restaur registres
bx lr                                             @return

/***************************************************/
/*   division par 10   unsigned                    */
/***************************************************/
/* r0 dividende   */
/* r0 quotient    */
/* r1 remainder   */
divisionpar10U:
push {r2,r3,r4, lr}
mov r4,r0                                          @ save value
ldr r3,iMagicNumber                                @ r3 <- magic_number    raspberry 1 2
umull r1, r2, r3, r0                               @ r1<- Lower32Bits(r1*r0) r2<- Upper32Bits(r1*r0)
mov r0, r2, LSR #3                                 @ r2 <- r2 >> shift 3
add r2,r0,r0, lsl #2                               @ r2 <- r0 * 5
sub r1,r4,r2, lsl #1                               @ r1 <- r4 - (r2 * 2)  = r4 - (r0 * 10)
pop {r2,r3,r4,lr}
bx lr                                              @ leave function
iMagicNumber:  	.int 0xCCCCCCCD```
```a=3          b=4          c=7          d=1
e=8          f=2          g=5          h=6
************************
3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6

```

## 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
}
}
}
;-----------------------------------------------------------------------
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, ",")
}
```

Outputs:

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

## C

Translation of: Go
```#include <stdbool.h>
#include <stdio.h>
#include <math.h>

int connections[15][2] = {
{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
};

int pegs[8];
int num = 0;

bool valid() {
int i;
for (i = 0; i < 15; i++) {
if (abs(pegs[connections[i][0]] - pegs[connections[i][1]]) == 1) {
return false;
}
}
return true;
}

void swap(int *a, int *b) {
int t = *a;
*a = *b;
*b = t;
}

void printSolution() {
printf("----- %d -----\n", num++);
printf("  %d %d\n", /*        */ pegs[0], pegs[1]);
printf("%d %d %d %d\n", pegs[2], pegs[3], pegs[4], pegs[5]);
printf("  %d %d\n", /*        */ pegs[6], pegs[7]);
printf("\n");
}

void solution(int le, int ri) {
if (le == ri) {
if (valid()) {
printSolution();
}
} else {
int i;
for (i = le; i <= ri; i++) {
swap(pegs + le, pegs + i);
solution(le + 1, ri);
swap(pegs + le, pegs + i);
}
}
}

int main() {
int i;
for (i = 0; i < 8; i++) {
pegs[i] = i + 1;
}

solution(0, 8 - 1);
return 0;
}
```
Output:
```----- 0 -----
3 4
7 1 8 2
5 6

----- 1 -----
3 5
7 1 8 2
4 6

----- 2 -----
3 6
7 1 8 2
4 5

----- 3 -----
3 6
7 1 8 2
5 4

----- 4 -----
4 3
2 8 1 7
6 5

----- 5 -----
4 5
2 8 1 7
6 3

----- 6 -----
4 5
7 1 8 2
3 6

----- 7 -----
4 6
7 1 8 2
3 5

----- 8 -----
5 3
2 8 1 7
6 4

----- 9 -----
5 4
2 8 1 7
6 3

----- 10 -----
5 4
7 1 8 2
3 6

----- 11 -----
5 6
7 1 8 2
3 4

----- 12 -----
6 3
2 8 1 7
5 4

----- 13 -----
6 3
2 8 1 7
4 5

----- 14 -----
6 4
2 8 1 7
5 3

----- 15 -----
6 5
2 8 1 7
4 3```

## C#

Translation of: Java
```using System;
using System.Collections.Generic;
using System.Linq;

public class NoConnection
{
static int[][] links = new int[][] {
new int[] {2, 3, 4}, // A to C,D,E
new int[] {3, 4, 5}, // B to D,E,F
new int[] {2, 4},    // D to C, E
new int[] {5},       // E to F
new int[] {2, 3, 4}, // G to C,D,E
new int[] {3, 4, 5}, // H to D,E,F
};

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

public static void Main(string[] args)
{
List<int> vals = Enumerable.Range(1, 8).ToList();
Random rng = new Random();

do
{
vals = vals.OrderBy(a => rng.Next()).ToList();
for (int i = 0; i < pegs.Length; i++)
pegs[i] = vals[i];

} while (!Solved());

PrintResult();
}

static bool Solved()
{
for (int i = 0; i < links.Length; i++)
if (Math.Abs(pegs[i] - peg) == 1)
return false;
return true;
}

static void PrintResult()
{
Console.WriteLine(\$"  {pegs[0]} {pegs[1]}");
Console.WriteLine(\$"{pegs[2]} {pegs[3]} {pegs[4]} {pegs[5]}");
Console.WriteLine(\$"  {pegs[6]} {pegs[7]}");
}
}
```
Output:
```  6 1
4 3 8 7
2 5

```

## C++

Translation of: C
```#include <array>
#include <iostream>
#include <vector>

std::vector<std::pair<int, int>> 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
};
std::array<int, 8> pegs;
int num = 0;

void printSolution() {
std::cout << "----- " << num++ << " -----\n";
std::cout << "  " /*     */ << pegs[0] << ' ' << pegs[1] << '\n';
std::cout << pegs[2] << ' ' << pegs[3] << ' ' << pegs[4] << ' ' << pegs[5] << '\n';
std::cout << "  " /*     */ << pegs[6] << ' ' << pegs[7] << '\n';
std::cout << '\n';
}

bool valid() {
for (size_t i = 0; i < connections.size(); i++) {
if (abs(pegs[connections[i].first] - pegs[connections[i].second]) == 1) {
return false;
}
}
return true;
}

void solution(int le, int ri) {
if (le == ri) {
if (valid()) {
printSolution();
}
} else {
for (size_t i = le; i <= ri; i++) {
std::swap(pegs[le], pegs[i]);
solution(le + 1, ri);
std::swap(pegs[le], pegs[i]);
}
}
}

int main() {
pegs = { 1, 2, 3, 4, 5, 6, 7, 8 };
solution(0, pegs.size() - 1);
return 0;
}
```
Output:
```----- 0 -----
3 4
7 1 8 2
5 6

----- 1 -----
3 5
7 1 8 2
4 6

----- 2 -----
3 6
7 1 8 2
4 5

----- 3 -----
3 6
7 1 8 2
5 4

----- 4 -----
4 3
2 8 1 7
6 5

----- 5 -----
4 5
2 8 1 7
6 3

----- 6 -----
4 5
7 1 8 2
3 6

----- 7 -----
4 6
7 1 8 2
3 5

----- 8 -----
5 3
2 8 1 7
6 4

----- 9 -----
5 4
2 8 1 7
6 3

----- 10 -----
5 4
7 1 8 2
3 6

----- 11 -----
5 6
7 1 8 2
3 4

----- 12 -----
6 3
2 8 1 7
5 4

----- 13 -----
6 3
2 8 1 7
4 5

----- 14 -----
6 4
2 8 1 7
5 3

----- 15 -----
6 5
2 8 1 7
4 3```

## 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;
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 ) );

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

## 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);
}
```
Output:
```        2   3
/|\ /|\
/ | X | \
/  |/ \|  \
6 - 0 - 7 - 1
\  |\ /|  /
\ | X | /
\|/ \|/
4   5
```

### Alternative version

Using a simple backtracking.

Translation of: Go
```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.");
}
```
Output:
```       3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6
Tested 12094 positions and did 20782 swaps.```

## Delphi

Works with: Delphi version 6.0

```{ This item would normally be in a separate library. It is presented here for clarity}

{Permutator object steps through all permutation of array items}
{Zero-Based = True = 0..Permutions-1 False = 1..Permutaions}
{Permutation set on "Create(Size)" or by "Permutations" property}
{Permutation are contained in the array "Indices"}

type TPermutator = class(TObject)
private
FZeroBased: boolean;
FBase: integer;
FPermutations: integer;
procedure SetZeroBased(const Value: boolean);
procedure SetPermutations(const Value: integer);
protected
FMax: integer;
public
Indices: TIntegerDynArray;
constructor Create(Size: integer);
procedure Reset;
function Next: boolean;
property ZeroBased: boolean read FZeroBased write SetZeroBased;
property Permutations: integer read FPermutations write SetPermutations;
end;

procedure TPermutator.Reset;
var I: integer;
begin
FMax:=High(Indices);
for I:= 0 to High(Indices) do Indices[I]:= I+FBase;
end;

procedure TPermutator.SetPermutations(const Value: integer);
begin
if FPermutations<>Value then
begin
FPermutations := Value;
SetLength(Indices,Value);
Reset;
end;
end;

constructor TPermutator.Create(Size: integer);
begin
ZeroBased:=True;
Permutations:=Size;
Reset;
end;

function TPermutator.Next: boolean;
{Returns true when sequence completed}
var I,T: integer;
begin
while true do
begin
T:= Indices[0];
for I:=0 to FMax-1 do Indices[I]:= Indices[I+1];
Indices[FMax]:= T;
if T<>(FMax+FBase) then
begin
FMax:=High(Indices);
break;
end
else FMax:= FMax-1;
if FMax<0 then break;
end;
Result:=FMax<1;
if Result then Reset;
end;

procedure TPermutator.SetZeroBased(const Value: boolean);
begin
if FZeroBased<>Value then
begin
FZeroBased := Value;
if Value then FBase:=0
else FBase:=1;
Reset;
end;
end;

{------------------------------------------------------------------------------}

{Network structures}
{Puzzle node}

type TPuzNode = record
Name: string;
Value: integer;
end;
type PPuzNode = ^TPuzNode;

{Edges connecting nodes}

type TPuzEdge = record
N1,N2: ^TPuzNode;
end;

{All edges in puzzle}

var Edges: array [0..14] of TPuzEdge;

{All nodes in puzzle}

var A: TPuzNode = (Name: 'A'; Value: 1);
var B: TPuzNode = (Name: 'B'; Value: 2);
var C: TPuzNode = (Name: 'C'; Value: 3);
var D: TPuzNode = (Name: 'D'; Value: 4);
var E: TPuzNode = (Name: 'E'; Value: 5);
var F: TPuzNode = (Name: 'F'; Value: 6);
var G: TPuzNode = (Name: 'G'; Value: 7);
var H: TPuzNode = (Name: 'H'; Value: 8);

{Array of pointers to puzzle nodes }

var PuzNodes: array [0..7] of Pointer;

procedure BuildNetwork;
{Build puzzle net work}
begin
{Put pointers to nodes in array}
PuzNodes[0]:=@A;
PuzNodes[1]:=@B;
PuzNodes[2]:=@C;
PuzNodes[3]:=@D;
PuzNodes[4]:=@E;
PuzNodes[5]:=@F;
PuzNodes[6]:=@G;
PuzNodes[7]:=@H;
{Set up all edges}
Edges[0].N1:=@A; Edges[0].N2:=@C;
Edges[1].N1:=@A; Edges[1].N2:=@D;
Edges[2].N1:=@A; Edges[2].N2:=@E;
Edges[3].N1:=@B; Edges[3].N2:=@D;
Edges[4].N1:=@B; Edges[4].N2:=@E;
Edges[5].N1:=@B; Edges[5].N2:=@F;
Edges[6].N1:=@G; Edges[6].N2:=@C;
Edges[7].N1:=@G; Edges[7].N2:=@D;
Edges[8].N1:=@G; Edges[8].N2:=@E;
Edges[9].N1:=@H; Edges[9].N2:=@D;
Edges[10].N1:=@H; Edges[10].N2:=@E;
Edges[11].N1:=@H; Edges[11].N2:=@F;
Edges[12].N1:=@C; Edges[12].N2:=@D;
Edges[13].N1:=@D; Edges[13].N2:=@E;
Edges[14].N1:=@E; Edges[14].N2:=@F;
end;

function ValidPattern: boolean;
{Test if pattern of node values is valid}
{i.e., edges values are greater than 1}
var I: integer;
begin
Result:=False;
for I:=0 to High(Edges) do
if abs(Edges[I].N2.Value-Edges[I].N1.Value)<2 then exit;
Result:=True;
end;

function Permutate: boolean;
{Use permutator object to iterate through all combinations}
var PM: TPermutator;
var I: integer;
begin
{Create with 8 items}
PM:=TPermutator.Create(8);
try
{Set to make it 1..8}
PM.ZeroBased:=False;
Result:=True;
{Iterate through all permutation}
while not PM.Next do
begin
{Copy permutation into network}
for I:=0 to High(PM.Indices) do
PPuzNode(PuzNodes[I])^.Value:=PM.Indices[I];
{If permutation is valid exit}
if ValidPattern then exit;
end;
{No valid permutation found}
Result:=False;
finally PM.Free; end;
end;

{String to display game board}

var GameBoard: string =
'       A   B'+CRLF+
'      /|\ /|\'+CRLF+
'     / | X | \'+CRLF+
'    /  |/ \|  \'+CRLF+
'   C - D - E - F'+CRLF+
'    \  |\ /|  /'+CRLF+
'     \ | X | /'+CRLF+
'      \|/ \|/'+CRLF+
'       G   H'+CRLF;

procedure ShowPuzzle(Memo: TMemo);
{Display game board with correct answer inserted}
var I,Inx: integer;
var S: string;
var PN: TPuzNode;
begin
S:=GameBoard;
{Search for Letters A..H}
for I:=1 to Length(S) do
if S[I] in ['A'..'H'] then
begin
{Convert A..H to index}
Inx:=byte(S[I]) - \$41;
{Get node A..H}
PN:=PPuzNode(PuzNodes[Inx])^;
{Store value in corresponding node}
S[I]:=char(PN.Value+\$30);
end;
{Display board}
end;

procedure ConnectionPuzzle(Memo: TMemo);
{Solve connection puzzle}
var S: string;
var I: integer;
var PN: TPuzNode;
begin
BuildNetwork;
Permutate;
{Display result}
S:='';
for I:=0 to High(PuzNodes) do
begin
PN:=PPuzNode(PuzNodes[I])^;
S:=S+PN.Name+'='+IntToStr(PN.Value)+' ';
end;
{Show puzzle with values inserted}
ShowPuzzle(Memo);
end;
```
Output:
```A=5 B=6 C=7 D=1 E=8 F=2 G=3 H=4
5   6
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
3   4

Elapsed Time: 2.092 ms.

```

## Elixir

Translation of: Ruby

This solution uses HLPsolver from here

```# It solved if connected A and B, connected G and H (according to the video).

# 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 .
"""
|> 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
```
Output:
```       4 - 6
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
3 - 5
```

## Factor

```USING: assocs interpolate io kernel math math.combinatorics
math.ranges math.parser multiline pair-rocket sequences
sequences.generalizations ;

STRING: diagram
\${}   \${}
/|\ /|\
/ | X | \
/  |/ \|  \
\${} - \${} - \${} - \${}
\  |\ /|  /
\ | X | /
\|/ \|/
\${}   \${}
;

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

## Fortran

Works with: gfortran version 11.2.1
```! This is free and unencumbered software released into the public domain,

program no_connection_puzzle

implicit none

! The names of the holes.
integer, parameter :: a = 1
integer, parameter :: b = 2
integer, parameter :: c = 3
integer, parameter :: d = 4
integer, parameter :: e = 5
integer, parameter :: f = 6
integer, parameter :: g = 7
integer, parameter :: h = 8

integer :: holes(a:h)

call find_solutions (holes, a)

contains

recursive subroutine find_solutions (holes, current_hole_index)
integer, intent(inout) :: holes(a:h)
integer, intent(in) :: current_hole_index

integer :: peg_number

! Recursively construct and print possible solutions, quitting
! any partial solution that does not satisfy constraints.
do peg_number = 1, 8
holes(current_hole_index) = peg_number
if (satisfies_the_constraints (holes, current_hole_index)) then
if (current_hole_index == h) then
call print_solution (holes)
write (*, '()')
else
call find_solutions (holes, current_hole_index + 1)
end if
end if
end do
end subroutine find_solutions

function satisfies_the_constraints (holes, i) result (satisfies)
integer, intent(inout) :: holes(a:h)
integer, intent(in) :: i    ! Where the new peg goes.
logical :: satisfies

integer :: j

! The most recently inserted peg must not be a duplicate of one
satisfies = all (holes(a : i - 1) /= holes(i))

if (satisfies) then
! ‘Fill’ the unfilled holes with fake pegs that cause
! differences with them always to be larger than 1.
do j = i + 1, h
holes(j) = 100 * j
end do

! Check that the differences are satisfactory.
satisfies = 1 < abs (holes(a) - holes(c)) .and.     &
&      1 < abs (holes(a) - holes(d)) .and.     &
&      1 < abs (holes(a) - holes(e)) .and.     &
&      1 < abs (holes(c) - holes(g)) .and.     &
&      1 < abs (holes(d) - holes(g)) .and.     &
&      1 < abs (holes(e) - holes(g)) .and.     &
&      1 < abs (holes(b) - holes(d)) .and.     &
&      1 < abs (holes(b) - holes(e)) .and.     &
&      1 < abs (holes(b) - holes(f)) .and.     &
&      1 < abs (holes(d) - holes(h)) .and.     &
&      1 < abs (holes(e) - holes(h)) .and.     &
&      1 < abs (holes(f) - holes(h)) .and.     &
&      1 < abs (holes(c) - holes(d)) .and.     &
&      1 < abs (holes(d) - holes(e)) .and.     &
&      1 < abs (holes(e) - holes(f))
end if
end function satisfies_the_constraints

subroutine print_solution (holes)
integer, intent(in) :: holes(a:h)

write (*, '(I5, I4)') holes(a), holes(b)
write (*, '("   /│\ /│\")')
write (*, '("  / │ X │ \")')
write (*, '(" /  │/ \│  \")')
write (*, '(3(I1, "───"), I1)') holes(c), holes(d), holes(e), holes(f)
write (*, '(" \  │\ /│  /")')
write (*, '("  \ │ X │ /")')
write (*, '("   \│/ \│/")')
write (*, '(I5, I4)') holes(g), holes(h)
end subroutine print_solution

end program no_connection_puzzle
```

The first solution printed:

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

## FreeBASIC

Translation of: Phix
```Dim As String txt = "" _
"        A   B" & Chr(10) & _
"       /|\ /|\" & Chr(10) & _
"      / | X | \" & Chr(10) & _
"     /  |/ \|  \" & Chr(10) & _
"    C - D - E - F" & Chr(10) & _
"     \  |\ /|  /" & Chr(10) & _
"      \ | X | /" & Chr(10) & _
"       \|/ \|/" & Chr(10) & _
"        G   H"

Dim Shared As String links(1 To 8)

Sub ReplaceString(Byref cad As String, oldChar As String, newChar As String)
Dim As Integer posic
Do
If posic = 0 Then Exit Do
Loop
End Sub

Function solve(s As String, idx As Integer, part As String) As String
Dim As Integer v, p, i, j
Dim As String res

For i = 1 To Len(s)
v = Val(Mid(s, i, 1))
For j = 1 To Len(links(idx))
p = Asc(Mid(links(idx), j, 1)) - 64
If Abs(v - Val(Mid(part, p, 1))) < 2 Then v = 0: Exit For
Next j
If v <> 0 Then
If Len(s) = 1 Then Return part + Chr(48 + v)
res = solve(Left(s, i - 1) & Mid(s, i + 1), idx + 1, part & Chr(48 + v))
If Len(res) > 0 Then Return res
End If
Next i

Return ""
End Function

Dim As String result = solve("12345678", 1, "")
For i As Integer = 1 To Len(result)
ReplaceString(txt, Chr(64 + i), Mid(result, i, 1))
Next i
Print txt

Sleep
```
Output:
`Same as Phix entry.`

## Go

A simple recursive brute force solution.

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

## Groovy

Translation of: Java
```import java.util.stream.Collectors
import java.util.stream.IntStream

class NoConnection {
[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]

static void main(String[] args) {
List<Integer> vals = IntStream.range(1, 9)
.mapToObj({ it })
.collect(Collectors.toList())

while (true) {
Collections.shuffle(vals)
for (int i = 0; i < pegs.length; i++) {
pegs[i] = vals.get(i)
}
if (solved()) {
break
}
}

printResult()
}

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

static void printResult() {
println("  \${pegs[0]} \${pegs[1]}")
println("\${pegs[2]} \${pegs[3]} \${pegs[4]} \${pegs[5]}")
println("  \${pegs[6]} \${pegs[7]}")
}
}
```
Output:
```  6 7
2 3 8 1
4 5```

```import Data.List (permutations)

solution :: [Int]
solution@(a : b : c : d : e : f : g : h : _) =
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) \$
unlines
( zipWith
(\x y -> x : (" = " <> show y))
['A' .. 'H']
solution
) :
( rightShift . unwords . fmap show
<\$> [[], [a, b], [c, d, e, f], [g, h]]
)
where
rightShift s
| length s > 3 = s
| otherwise = "  " <> s
```
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:

```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
)
```

Intermezzo:

```   (#~ 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
```

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.

```   disp {. (#~ 1<attempt"1) pegs
3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6
```

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):

```   (#~ 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
```

The first of these looks like this:

```   disp {. (#~ 1<attempt"1) pegs
3 - 5
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
4 - 6
```

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
```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 {

{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++)
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]);
}
}
```

(takes about 500 shuffles on average)

```       4  5
2  8  1  7
6  3     ```

## JavaScript

### ES6

```(() => {
'use strict';

// -------------- NO CONNECTION PUZZLE ---------------

// solvedPuzzle :: () -> [Int]
const solvedPuzzle = () => {

// 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
]);

return universe[
until(i => isSolution(universe[i]))(
succ
)(0)
];
}

// ---------------------- TEST -----------------------
const main = () => {
const
firstSolution = solvedPuzzle(),
[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 => unwords(xs.map(show))
))
);
}

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

// abs :: Num -> Num
const abs =
// Absolute value of a given number - without the sign.
Math.abs;

// all :: (a -> Bool) -> [a] -> Bool
const all = p =>
// True if p(x) holds for every x in xs.
xs => [...xs].every(p);

// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
// A function defined by the right-to-left
// composition of all the functions in fs.
fs.reduce(
(f, g) => x => f(g(x)),
x => x
);

// enumFromTo :: Enum a => a -> a -> [a]
const enumFromTo = m => n => {
const [x, y] = [m, n].map(fromEnum),
b = x + (isNaN(m) ? 0 : m - x);
return Array.from({
length: 1 + (y - x)
}, (_, i) => toEnum(m)(b + i));
};

// fromEnum :: Enum a => a -> Int
typeof x !== 'string' ? (
x.constructor === Object ? (
x.value
) : parseInt(Number(x))
) : x.codePointAt(0);

// length :: [a] -> Int
const length = xs =>
// Returns Infinity over objects without finite
// length. This enables zip and zipWith to choose
// the shorter argument when one is non-finite,
// like cycle, repeat etc
'GeneratorFunction' !== xs.constructor
.constructor.name ? (
xs.length
) : Infinity;

// list :: StringOrArrayLike b => b -> [a]
const list = xs =>
// xs itself, if it is an Array,
// or an Array derived from xs.
Array.isArray(xs) ? (
xs
) : Array.from(xs || []);

// permutations :: [a] -> [[a]]
const permutations = xs => (
ys => ys.reduceRight(
(a, y) => a.flatMap(
ys => Array.from({
length: 1 + ys.length
}, (_, i) => i)
.map(n => ys.slice(0, n)
.concat(y)
.concat(ys.slice(n))
)
), [
[]
]
)
)(list(xs));

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

// succ :: Enum a => a -> a
const succ = x =>
1 + x;

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => 'GeneratorFunction' !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// toEnum :: a -> Int -> a
const toEnum = e =>
// The first argument is a sample of the type
// allowing the function to make the right mapping
x => ({
'number': Number,
'string': String.fromCodePoint,
'boolean': Boolean,
'object': v => e.min + v
} [typeof e])(x);

// unlines :: [String] -> String
const unlines = xs =>
// A single string formed by the intercalation
// of a list of strings with the newline character.
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 =>
// A space-separated string derived
// from a list of words.
xs.join(' ');

// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = f =>
// Use of `take` and `length` here allows
// zipping with non-finite lists
// i.e. generators like cycle, repeat, iterate.
xs => ys => {
const n = Math.min(length(xs), length(ys));
return Infinity > n ? (
(([as, bs]) => Array.from({
length: n
}, (_, i) => f(as[i])(
bs[i]
)))([xs, ys].map(
compose(take(n), list)
))
) : zipWithGen(f)(xs)(ys);
};

return main();
})();
```
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 1.6

Also works with gojq, the Go implementation of jq

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

```# permutations of 0 .. (n-1) inclusive
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;

# Count the number of items in a stream
def count(f): reduce f as \$_ (0; .+1);```

Part 2: The no-connections puzzle for N pegs and holes

```# Generate a stream of solutions.
# Input should be the connections array, i.e. an array of [i,j] pairs;
# 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
| all( connections[];
((\$p[.[0]] - \$p[.[1]])|abs) != 1 );

. as \$connections | permutations(N) | select(ok(\$connections));```

Part 3: The 8-peg no-connection puzzle

```# The connectedness matrix
# In this table, 0 represents "A", etc, and an entry [i,j]
# 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);

# 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;```

Examples:

```# To print all the solutions:
# solve | pp

# To count the number of solutions:
# count(solve)
# => 16

limit(1; solve) | pp```
Output:

Invocation: jq -n -r -f no_connection.jq

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

## Julia

```using Combinatorics

const HOLES = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H']
const PEGS = [1, 2, 3, 4, 5, 6, 7, 8]
const EDGES = [('A', 'C'), ('A', 'D'), ('A', 'E'),
('B', 'D'), ('B', 'E'), ('B', 'F'),
('C', 'G'), ('C', 'D'), ('D', 'G'),
('D', 'E'), ('D', 'H'), ('E', 'F'),
('E', 'G'), ('E', 'H'), ('F', 'H')]

goodperm(p) = all(e->abs(p[e[1]-'A'+1] - p[e[2]-'A'+1]) > 1, EDGES)

goodplacements() = [p for p in permutations(PEGS) if goodperm(p)]

const BOARD = raw"""
A   B
/|\ /|\
/ | X | \
/  |/ \|  \
C - D - E - F
\  |\ /|  /
\ | X | /
\|/ \|/
G   H
"""

function printsolutions()
solutions = goodplacements()
println("Found \$(length(solutions)) solutions.")
for soln in solutions
board = BOARD
for (i, n) in enumerate(soln)
board = replace(board, string('A' + i - 1) => string(n))
end
println(board); exit(1) # remove this exit for all solutions
end
end

printsolutions()
```
Output:
```
Found 16 solutions.

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

```

## Kotlin

Translation of: Go
```// 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.")
}
```
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.

```Module no_connection_puzzle {
\\ Holes
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```
Output:
```    3   5
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
4   6
```

## m4

Translation of: Fortran

Unlike the Fortran from which it was migrated, this m4 program stops at the first solution. The holes are represented by positions in a string; you can regard the string as a variable-size array. (m4 is, of course, a string-manipulation language.)

The program ought to work with any POSIX-compliant m4. The display has been changed to use only ASCII characters, because very old m4 cannot handle UTF-8.

```divert(-1)

define(`abs',`eval(((( \$1 ) < 0) * (-( \$1 ))) + ((0 <= ( \$1 )) * ( \$1 )))')

define(`display_solution',
`    substr(\$1,0,1)   substr(\$1,1,1)
/|\ /|\
/ | X | \
/  |/ \|  \
substr(\$1,2,1)`---'substr(\$1,3,1)`---'substr(\$1,4,1)`---'substr(\$1,5,1)
\  |\ /|  /
\ | X | /
\|/ \|/
substr(\$1,6,1)   substr(\$1,7,1)')

define(`satisfies_constraints',
`eval(satisfies_no_duplicates_constraint(\$1) && satisfies_difference_constraints(\$1))')

define(`satisfies_no_duplicates_constraint',
`eval(index(all_but_last(\$1),last(\$1)) == -1)')

define(`satisfies_difference_constraints',
`pushdef(`A',ifelse(eval(1 <= len(\$1)),1,substr(\$1,0,1),100))`'dnl
pushdef(`B',ifelse(eval(2 <= len(\$1)),1,substr(\$1,1,1),200))`'dnl
pushdef(`C',ifelse(eval(3 <= len(\$1)),1,substr(\$1,2,1),300))`'dnl
pushdef(`D',ifelse(eval(4 <= len(\$1)),1,substr(\$1,3,1),400))`'dnl
pushdef(`E',ifelse(eval(5 <= len(\$1)),1,substr(\$1,4,1),500))`'dnl
pushdef(`F',ifelse(eval(6 <= len(\$1)),1,substr(\$1,5,1),600))`'dnl
pushdef(`G',ifelse(eval(7 <= len(\$1)),1,substr(\$1,6,1),700))`'dnl
pushdef(`H',ifelse(eval(8 <= len(\$1)),1,substr(\$1,7,1),800))`'dnl
eval(1 < abs((A) - (C)) &&
1 < abs((A) - (D)) &&
1 < abs((A) - (E)) &&
1 < abs((C) - (G)) &&
1 < abs((D) - (G)) &&
1 < abs((E) - (G)) &&
1 < abs((B) - (D)) &&
1 < abs((B) - (E)) &&
1 < abs((B) - (F)) &&
1 < abs((D) - (H)) &&
1 < abs((E) - (H)) &&
1 < abs((F) - (H)) &&
1 < abs((C) - (D)) &&
1 < abs((D) - (E)) &&
1 < abs((E) - (F)))'`dnl
popdef(`A',`B',`C',`D',`E',`F',`G',`H')')

define(`all_but_last',`substr(\$1,0,decr(len(\$1)))')
define(`last',`substr(\$1,decr(len(\$1)))')

define(`last_is_eight',`eval((last(\$1)) == 8)')
define(`strip_eights',`ifelse(last_is_eight(\$1),1,`\$0(all_but_last(\$1))',`\$1')')

define(`incr_last',`all_but_last(\$1)`'incr(last(\$1))')

define(`solve_puzzle',`_\$0(1)')
define(`_solve_puzzle',
`ifelse(eval(len(\$1) == 8 && satisfies_constraints(\$1)),1,`display_solution(\$1)',
satisfies_constraints(\$1),1,`\$0(\$1`'1)',
last_is_eight(\$1),1,`\$0(incr_last(strip_eights(\$1)))',
`\$0(incr_last(\$1))')')

divert`'dnl
dnl
solve_puzzle```
Output:
```    3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7---1---8---2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6```

## Mathematica /Wolfram Language

This one simply takes all permutations of the pegs and filters out invalid solutions.

```sol = Fold[
Select[#,
Function[perm, Abs[perm[[#2[[1]]]] - perm[[#2[[2]]]]] > 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]];
```
Output:
```    3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6```

## Nim

Translation of: C++

I choose to use one-based indexing for the array of pegs. It seems more logical here and Nim allows to choose any starting index for static arrays.

```import strformat

const Connections = [(1, 3), (1, 4), (1, 5),  # A to C, D, E
(2, 4), (2, 5), (2, 6),  # B to D, E, F
(7, 3), (7, 4), (7, 5),  # G to C, D, E
(8, 4), (8, 5), (8, 6),  # H to D, E, F
(3, 4), (4, 5), (5, 6)]  # C-D, D-E, E-F

type
Peg = 1..8
Pegs = array[1..8, Peg]

func valid(pegs: Pegs): bool =
for (src, dst) in Connections:
if abs(pegs[src] - pegs[dst]) == 1:
return false
result = true

proc print(pegs: Pegs; num: Positive) =
echo &"----- {num} -----"
echo &"  {pegs[1]} {pegs[2]}"
echo &"{pegs[3]} {pegs[4]} {pegs[5]} {pegs[6]}"
echo &"  {pegs[7]} {pegs[8]}"
echo()

proc findSolution(pegs: var Pegs; left, right: Natural; solCount = 0): Natural =
var solCount = solCount
if left == right:
if pegs.valid():
inc solCount
pegs.print(solCount)
else:
for i in left..right:
swap pegs[left], pegs[i]
solCount = pegs.findSolution(left + 1, right, solCount)
swap pegs[left], pegs[i]
result = solCount

when isMainModule:

var pegs = [Peg 1, 2, 3, 4, 5, 6, 7, 8]
```
Output:
```----- 1 -----
3 4
7 1 8 2
5 6

----- 2 -----
3 5
7 1 8 2
4 6

----- 3 -----
3 6
7 1 8 2
4 5

----- 4 -----
3 6
7 1 8 2
5 4

----- 5 -----
4 3
2 8 1 7
6 5

----- 6 -----
4 5
2 8 1 7
6 3

----- 7 -----
4 5
7 1 8 2
3 6

----- 8 -----
4 6
7 1 8 2
3 5

----- 9 -----
5 3
2 8 1 7
6 4

----- 10 -----
5 4
2 8 1 7
6 3

----- 11 -----
5 4
7 1 8 2
3 6

----- 12 -----
5 6
7 1 8 2
3 4

----- 13 -----
6 3
2 8 1 7
5 4

----- 14 -----
6 3
2 8 1 7
4 5

----- 15 -----
6 4
2 8 1 7
5 3

----- 16 -----
6 5
2 8 1 7
4 3```

## Perl

```#!/usr/bin/perl

use strict;
use warnings;

my \$gap = qr/.{3}/s;

find( <<terminator );
-AB-
CDEF
-GH-
terminator

sub find
{
my \$p = shift;
\$p =~ /(\d)\$gap.{0,2}(\d)(??{abs \$1 - \$2 <= 1 ? '' : '(*F)'})/ ||
\$p =~ /^.*\n.*(\d)(\d)(??{abs \$1 - \$2 <= 1 ? '' : '(*F)'})/ and return;
if( \$p =~ /[A-H]/ )
{
find( \$p =~ s/[A-H]/\$_/r ) for grep \$p !~ \$_, 1 .. 8;
}
else
{
print \$p =~ tr/-/ /r;
exit;
}
}
```
Output:
``` 34
7182
56
```

## 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.

```with javascript_semantics
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"

function solve(sequence s, integer idx, sequence part)
object res
integer v, p
for i=1 to length(s) do
v = s[i]
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,"")))
```
Output:
```        3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6
```

## Picat

```import cp.

no_connection_puzzle(X) =>
N = 8,
X = new_list(N),
X :: 1..N,
Graph =
{{1,3}, {1,4}, {1,5},
{2,4}, {2,5}, {2,6},
{3,1}, {3,4}, {3,7},
{4,1}, {4,2}, {4,3}, {4,5}, {4,7}, {4,8},
{5,1}, {5,2}, {5,4}, {5,6}, {5,7}, {5,8},
{6,2}, {6,5}, {6,8},
{7,3}, {7,4}, {7,5},
{8,4}, {8,5}, {8,6}},

all_distinct(X),
foreach(I in 1..Graph.length)
abs(X[Graph[I,1]]-X[Graph[I,2]]) #> 1
end,

% symmetry breaking
X[1] #< X[N],

solve(X),
println(X),
nl,
[A,B,C,D,E,F,G,H] = X,
Solution = to_fstring(
"    %d   %d       \n"++
"   /|\\ /|\\      \n"++
"  / | X | \\      \n"++
" /  |/ \\|  \\    \n"++
"%d - %d - %d - %d \n"++
" \\  |\\ /|  /    \n"++
"  \\ | X | /      \n"++
"   \\|/ \\|/      \n"++
"    %d   %d       \n",
A,B,C,D,E,F,G,H),
println(Solution).```
Output:
```Picat> no_connection_puzzle(_X)
[3,4,7,1,8,2,5,6]

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.

```:- 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).
```

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.

```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]))
```
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:

```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)
```
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```

## Quackery

`rank->perm` and `!` are defined at Permutations/Rank of a permutation#Quackery.

```  [ ' [ [ 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 ] ] ] is connections (     --> [ )

[ dip dup do
unrot peek dip peek - abs 1 = ] is invalid     ( [ [ --> b )

[ true swap
connections witheach
[ dip dup invalid if
[ dip not conclude ] ]
drop ]                          is allvalid    (   [ --> b )

say "  A B C D E F G H" cr
8 ! times
[ i^ 8 rank->perm
allvalid if
[ sp
i^ 8 rank->perm
witheach
[ sp 1+ echo ]
cr conclude ] ]```
Output:
```  A B C D E F G H
3 4 7 1 8 2 5 6
```

As noted in the talk for this page, "According to the video, A and B, G and H are also connected".

Adding these to the list of connections (i.e. `[ 0 1 ] [ 6 7 ]`) and generating all the solutions by removing the word `conclude` from the last line of the code gives:

Output:
```  A B C D E F G H
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
```

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

## Raku

(formerly 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.

```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) {
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";
}
```
Output:
```  4     3

2   8 1   7

6     5
18 tries```

## Red

### Basic version

```Red ["Solve the no connection puzzle"]

points: [a b c d e f g h]
; 'links' series will be scanned by pairs: [a c], [a d] etc.
links: [a c a d a e b d b e b f c d c g d e d g d h e f e g e h f h]
allpegs: [1 2 3 4 5 6 7 8]

; check if two points are connected (then game is lost) i.e.
; both are have a value (not zero) and absolute difference is 1
connected: func [x y] [all [
x * y <> 0
1 = absolute (x - y)
]]
; a list of points is valid if no connexion is found
isvalid: function [pegs [block!]] [
; assign pegs values to points, or 0 for remaining points
set points append/dup copy pegs 0 8
foreach [x y] links [if connected get x get y [return false]]
true
]
; recursively build a list of up to 8 valid points
check: function [pegs [block!]] [
if isvalid pegs [
rest: difference allpegs pegs
either empty? rest [
print rejoin ["Here is a solution: " pegs]
halt ; comment this line to get all solutions
][
foreach peg rest [check append copy pegs peg]
]
]
]
check []
```
Output:
```Here is a solution: 3 4 7 1 8 2 5 6
(halted)
```

### With graphics

```Red [Needs: 'View]

points: [a b c d e f g h]
; 'links' series will be scanned by pairs: [a c], [a d] etc.
links: [a c a d a e b d b e b f c d c g d e d g d h e f e g e h f h]
allpegs: [1 2 3 4 5 6 7 8]

; check if two points are connected (then game is lost) i.e.
; both are have a value (not zero) and absolute difference is 1
connected: func [x y] [all [
x * y <> 0
1 = absolute (x - y)
]]
; a list of points is valid if no connexion is found
isvalid: function [pegs [block!]] [
; assign pegs values to points, or 0 for remaining points
set points append/dup copy pegs 0 8
foreach [x y] links [if connected get x get y [return false]]
true
]
; recursively build a list of up to 8 valid points
check: function [pegs [block!]] [
if isvalid pegs [
rest: difference allpegs pegs
either empty? rest [
vis points
][
foreach peg rest [check append copy pegs peg]
]
]
]
; view solution found
vis: function [points] [
pos: [100x0 200x0 0x100 100x100 200x100 300x100 100x200 200x200]
offs: 30x30
pos-of: function [x] [pick pos index? find points x]
val-at: function [p] [get pick points index? find pos p]
visu: layout [img: image 362x262 draw []]
foreach [x y] links [append img/draw reduce [
'line offs + pos-of x offs + pos-of y]]
append img/draw [fill-pen snow]
foreach p pos [append img/draw reduce [
'circle offs + p 15 'text 21x15 + p form val-at p]]
view/options visu [text: "Solution to the no-connection puzzle"]
]
check []
```
Output:

## REXX

### unannotated solutions

```/*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.*/
```
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 default 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.

```/*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 ? */
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
```
output   when using the default inputs of:     -1
```       ╔═══════════════════════════╗              ╔═══════════════════════════╗
║          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:

```#  Solve No Connection Puzzle
#
#  Nigel_Galloway
#  October 6th., 2014
require 'HLPSolver'
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.solve
```
Output:
```Problem:
0  0
0  0  1  0
0  0

Solution:
5  3
2  8  1  7
6  4
```

## Scala

Output:

Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).

```object NoConnection extends App {

Seq(2, 3, 4), // A to C,D,E
Seq(3, 4, 5), // B to D,E,F
Seq(2, 4), // D to C, E
Seq(5), // E to F
Seq(2, 3, 4), // G to C,D,E
Seq(3, 4, 5)) // H to D,E,F

private def genRandom: LazyList[Seq[Int]] = util.Random.shuffle((1 to 8).toList) #:: genRandom

private def notSolved(links: Seq[Seq[Int]], pegs: Seq[Int]): Boolean =
i => !links(i).forall(peg => math.abs(pegs(i) - peg) == 1))

private def printResult(pegs: Seq[Int]) = {
println(f"\${pegs(0)}%3d\${pegs(1)}%2d")
println(f"\${pegs(2)}%1d\${pegs(3)}%2d\${pegs(4)}%2d\${pegs(5)}%2d")
println(f"\${pegs(6)}%3d\${pegs(7)}%2d")
}

}
```

## Tcl

Library: Tcllib (Package: struct::list)
```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} {
puts [string map [join [
lmap name {A B C D E F G H} val \$p {list \$name \$val}
]] \$layout]
break
}
}
```
Output:
```        3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6```

## Wren

Translation of: Kotlin
Library: Wren-dynamic
```import "./dynamic" for Tuple

var Solution = Tuple.create("Solution", ["p", "tests", "swaps"])

// Holes A=0, B=1, …, H=7
// With connections:
var conn = "
A   B
/|\\ /|\\
/ | X | \\
/  |/ \\|  \\
C - D - E - F
\\  |\\ /|  /
\\ | X | /
\\|/ \\|/
G   H
"

var 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
]

// '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.
var isValid = Fn.new { |pegs|
for (c in connections) {
if ((pegs[c[0]] - pegs[c[1]]).abs <= 1) return false
}
return true
}

var swap = Fn.new { |pegs, i, j|
var 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.
var solve
solve = Fn.new {
var pegs = List.filled(8, 0)
for (i in 0..7) pegs[i] = i + 1
var tests = 0
var swaps = 0

var recurse // recursive closure
recurse = Fn.new { |i|
if (i >= pegs.count - 1) {
tests = tests + 1
return isValid.call(pegs)
}
// Try each remaining peg from pegs[i] onwards
for (j in i...pegs.count) {
swaps = swaps + 1
swap.call(pegs, i, j)
if (recurse.call(i + 1)) return true
swap.call(pegs, i, j)
}
return false
}

recurse.call(0)
return Solution.new(pegs, tests, swaps)
}

var pegsAsString = Fn.new { |pegs|
var ca = conn.toList
var i = 0
for (c in ca) {
if ("ABCDEFGH".contains(c)) ca[i] = String.fromByte(48 + pegs[c.bytes[0] - 65])
i = i + 1
}
return ca.join()
}

var s = solve.call()
System.print(pegsAsString.call(s.p))
System.print("Tested %(s.tests) positions and did %(s.swaps) swaps.")
```
Output:
```       3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6

Tested 12094 positions and did 20782 swaps.
```

## 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 | \
/  |/ \|  \
# - # - # - #
\  |\ /|  /
\ | 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);
]```
Output:
```    5   6
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
3   4
```

## Zig

```const std = @import("std");
const print = std.debug.print;

const PA = 0;
const PB = 1;
const PC = 2;
const PD = 3;
const PE = 4;
const PF = 5;
const PG = 6;
const PH = 7;

const Pair = struct { u8, u8 };
const Pairs = [_]Pair{ .{ PA, PC }, .{ PA, PD }, .{ PA, PE }, .{ PB, PD }, .{ PB, PE }, .{ PB, PF }, .{ PC, PG }, .{ PC, PD }, .{ PD, PE }, .{ PD, PG }, .{ PD, PH }, .{ PE, PG }, .{ PE, PH }, .{ PF, PH }, .{ PE, PF } };
var t: [8]u32 = .{0} ** 8;

inline fn abs(p: Pair) u32 {
if (t[p.@"0"] < t[p.@"1"]) return t[p.@"1"] - t[p.@"0"];
return t[p.@"0"] - t[p.@"1"];
}

fn check() bool {
var r: bool = true;
for (Pairs) |p| {
r = r and abs(p) > 1;
if (!r) break;
}

return r;
}

fn has(a: u32) bool {
for (t) |v| {
if (v == a) return true;
}
return false;
}

fn solve(lvl: u32) bool {
if (lvl == 8) return check();
for (1..9) |v| {
if (has(v)) continue;
t[lvl] = v;
if (solve(lvl + 1)) return true;
}
t[lvl] = 0;
return false;
}

pub fn main() void {
_ = solve(0);
print("{{ A, B, C, D, E, F, G, H }} = {any}", .{t});
}
```

Output:

```{ A, B, C, D, E, F, G, H } = { 3, 4, 7, 1, 8, 2, 5, 6 }
```
```const std = @import("std");
const print = std.debug.print;

const PA = 0;
const PB = 1;
const PC = 2;
const PD = 3;
const PE = 4;
const PF = 5;
const PG = 6;
const PH = 7;

const Pair = struct { u8, u8 };
const Pairs = [_]Pair{ .{ PA, PC }, .{ PA, PD }, .{ PA, PE }, .{ PB, PD }, .{ PB, PE }, .{ PB, PF }, .{ PC, PG }, .{ PC, PD }, .{ PD, PE }, .{ PD, PG }, .{ PD, PH }, .{ PE, PG }, .{ PE, PH }, .{ PF, PH }, .{ PE, PF } };
var t: [8]u32 = .{0} ** 8;

inline fn abs(p: Pair) u32 {
if (t[p.@"0"] < t[p.@"1"]) return t[p.@"1"] - t[p.@"0"];
return t[p.@"0"] - t[p.@"1"];
}

fn check() bool {
var r: bool = true;
for (Pairs) |p| {
r = r and abs(p) > 1;
if (!r) break;
}

return r;
}

fn has(a: u32) bool {
for (t) |v| {
if (v == a) return true;
}
return false;
}

fn solve(lvl: u32) bool {
if (lvl == 8) return check();
for (1..9) |v| {
if (has(v)) continue;
t[lvl] = v;
if (solve(lvl + 1)) {
print("{{ A, B, C, D, E, F, G, H }} = {any}\n", .{t});
}
}
t[lvl] = 0;
return false;
}

pub fn main() void {
_ = solve(0);
}
```

Output:

```{ A, B, C, D, E, F, G, H } = { 3, 4, 7, 1, 8, 2, 5, 6 }
{ A, B, C, D, E, F, G, H } = { 3, 5, 7, 1, 8, 2, 4, 6 }
{ A, B, C, D, E, F, G, H } = { 3, 6, 7, 1, 8, 2, 4, 5 }
{ A, B, C, D, E, F, G, H } = { 3, 6, 7, 1, 8, 2, 5, 4 }
{ A, B, C, D, E, F, G, H } = { 4, 3, 2, 8, 1, 7, 6, 5 }
{ A, B, C, D, E, F, G, H } = { 4, 5, 2, 8, 1, 7, 6, 3 }
{ A, B, C, D, E, F, G, H } = { 4, 5, 7, 1, 8, 2, 3, 6 }
{ A, B, C, D, E, F, G, H } = { 4, 6, 7, 1, 8, 2, 3, 5 }
{ A, B, C, D, E, F, G, H } = { 5, 3, 2, 8, 1, 7, 6, 4 }
{ A, B, C, D, E, F, G, H } = { 5, 4, 2, 8, 1, 7, 6, 3 }
{ A, B, C, D, E, F, G, H } = { 5, 4, 7, 1, 8, 2, 3, 6 }
{ A, B, C, D, E, F, G, H } = { 5, 6, 7, 1, 8, 2, 3, 4 }
{ A, B, C, D, E, F, G, H } = { 6, 3, 2, 8, 1, 7, 4, 5 }
{ A, B, C, D, E, F, G, H } = { 6, 3, 2, 8, 1, 7, 5, 4 }
{ A, B, C, D, E, F, G, H } = { 6, 4, 2, 8, 1, 7, 5, 3 }
{ A, B, C, D, E, F, G, H } = { 6, 5, 2, 8, 1, 7, 4, 3 }
```

## zkl

Translation of: D
```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();
#<<<   // 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\$;
#<<<	// 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
}```

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
```