⚠️ Warning: This is a draft ⚠️

This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.

If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.

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

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

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

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

;Example: In this attempt:

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

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

;Task Produce and show here ''one'' solution to the puzzle.

;Related tasks: :* [[A* search algorithm]] :* [[Solve a Holy Knight's tour]] :* [[Knight's tour]] :* [[N-queens problem]] :* [[Solve a Hidato puzzle]] :* [[Solve a Holy Knight's tour]] :* [[Solve a Hopido puzzle]] :* [[Solve a Numbrix puzzle]] :* [[4-rings or 4-squares puzzle]]

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;
```
```

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;

```

{{out}}

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

```

## APL

{{works with|Dyalog APL 17.0 Unicode}}

```

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|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
```

## 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);
}
```

{{out}}

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

```

### Alternative version

Using a simple backtracking. {{trans|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.");
}
```

{{out}}

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

## Elixir

{{trans|Ruby}} This solution uses HLPsolver from [[Solve_a_Hidato_puzzle#Elixir | 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
```

{{out}}

```
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
```

{{out}}

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

```

## 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
}
```

{{out}}

```

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

```

```import Data.List (intercalate, permutations)

solution :: [Int]
solution@(a:b:c:d:e:f:g:h:_) = head \$ filter isSolution (permutations [1 .. 8])
where
isSolution :: [Int] -> Bool
isSolution (a:b:c:d:e:f:g:h:_) =
all ((> 1) . abs) \$
zipWith
(-)
[a, c, g, e, a, c, g, e, b, d, h, f, b, d, h, f]
[d, d, d, d, c, g, e, a, e, e, e, e, d, h, f, b]

main :: IO ()
main =
(putStrLn . unlines) \$
let rightShift s
| length s > 3 = s
| otherwise = "  " ++ s
in intercalate
"\n"
(zipWith (\x y -> x : (" = " ++ show y)) ['A' .. 'H'] solution) :
((rightShift . unwords . fmap show) <\$> [[], [a, b], [c, d, e, f], [g, h]])
```

{{Out}}

```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:

```J
holes=:;:'A B C D E F G H'

connections=:".;._2]0 :0
holes e.;:'C D E'          NB. A
holes e.;:'D E F'          NB. B
holes e.;:'A D G'          NB. C
holes e.;:'A B C E G H'    NB. D
holes e.;:'A B D F G H'    NB. E
holes e.;:'B E H'          NB. F
holes e.;:'C D E'          NB. G
holes e.;:'D E F'          NB. H
)
assert (-:|:) connections NB. catch typos

pegs=: 1+(A.&i.~ !)8

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

box=:0 :0
A   B
/|\ /|\
/ | X | \
/  |/ \|  \
C - D - E - F
\  |\ /|  /
\ | X | /
\|/ \|/
G   H
)

disp=:verb define
rplc&(,holes;&":&>y) box
)
```

Intermezzo:

```J
(#~ 1 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)

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

## JavaScript

### ES6

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

{{Out}}

```txt
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|1.4}}
We present a generate-and-test solver for a slightly more general version of the problem, in which there are N pegs and holes, and in which the connectedness of holes is defined by an array such that holes i and j are connected if and only if [i,j] is a member of the
array.

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

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

'''Part 1: Generic functions'''

```jq
# Short-circuit determination of whether (a|condition)
# is true for all a in array:
def forall(array; condition):
def check:
. as \$ix
| if \$ix == (array|length) then true
elif (array[\$ix] | condition) then (\$ix + 1) | check
else false
end;
0 | check;

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

# 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'''

```jq
# 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
| forall( connections;
((\$p[.[0]] - \$p[.[1]])|abs) != 1 );

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

'''Part 3: The 8-peg no-connection puzzle'''

```jq
# 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''':

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

# To count the number of solutions:
# count(solve)

# jq 1.4 lacks facilities for harnessing generators,
# but the following will suffice here:
def one(f): reduce f as \$s
(null; if . == null then \$s else . end);

one(solve) | pp

```

{{out}}

```sh
\$ jq -n -r -f no_connection.jq

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

## Julia

```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}}
```txt

Found 16 solutions.
3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6

```

## Kotlin

{{trans|Go}}

```scala
// version 1.2.0

import kotlin.math.abs

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

val connections = listOf(
0 to 2, 0 to 3, 0 to 4,   // A to C, D, E
1 to 3, 1 to 4, 1 to 5,   // B to D, E, F
6 to 2, 6 to 3, 6 to 4,   // G to C, D, E
7 to 3, 7 to 4, 7 to 5,   // H to D, E, F
2 to 3, 3 to 4, 4 to 5    // C-D, D-E, E-F
)

// 'isValid' checks if the pegs are a valid solution.
// If the absolute difference between any pair of connected pegs is
// greater than one it is a valid solution.
fun isValid(pegs: IntArray): Boolean {
for ((a, b) in connections) {
if (abs(pegs[a] - pegs[b]) <= 1) return false
}
return true
}

fun swap(pegs: IntArray, i: Int, j: Int) {
val tmp = pegs[i]
pegs[i] = pegs[j]
pegs[j] = tmp
}

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

fun solve(): Triple {
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) {
val (p, tests, swaps) = solve()
println(pegsAsString(p))
println("Tested \$tests positions and did \$swaps swaps.")
}
```

{{out}}

```txt

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.

```M2000 Interpreter

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

```

{{out}}

```txt

3   5
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
4   6
```
## Mathematica This one simply takes all permutations of the pegs and filters out invalid solutions. ```Mathematica 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]]; ``` {{out}} ```txt 3 4 /|\ /|\ / | X | \ / |/ \| \ 7 - 1 - 8 - 2 \ |\ /| / \ | X | / \|/ \|/ 5 6 ``` ## Perl ```perl #!/usr/bin/perl use strict; use warnings; my \$gap = qr/.{3}/s; find( <|| into two bows )(, such that adjacency can be calculated simply as a distance of 2 or less. ```perl6 my @adjacent = gather -> \$y, \$x { take [\$y,\$x] if abs(\$x|\$y) > 2; } for flat -5 .. 5 X -5 .. 5; solveboard q:to/END/; . _ . . _ . . . . . . . _ . _ 1 . _ . . . . . . . _ . . _ . END sub solveboard(\$board) { my \$max = +\$board.comb(/\w+/); my \$width = \$max.chars; my @grid; my @known; my @neigh; my @degree; @grid = \$board.lines.map: -> \$line { [ \$line.words.map: { /^_/ ?? 0 !! /^\./ ?? Rat !! \$_ } ] } sub neighbors(\$y,\$x --> List) { eager gather for @adjacent { my \$y1 = \$y + .[0]; my \$x1 = \$x + .[1]; take [\$y1,\$x1] if defined @grid[\$y1][\$x1]; } } for ^@grid -> \$y { for ^@grid[\$y] -> \$x { if @grid[\$y][\$x] -> \$v { @known[\$v] = [\$y,\$x]; } if @grid[\$y][\$x].defined { @neigh[\$y][\$x] = neighbors(\$y,\$x); @degree[\$y][\$x] = +@neigh[\$y][\$x]; } } } print "\e[0H\e[0J"; my \$tries = 0; try_fill 1, @known[1]; sub try_fill(\$v, \$coord [\$y,\$x] --> Bool) { return True if \$v > \$max; \$tries++; my \$old = @grid[\$y][\$x]; return False if +\$old and \$old != \$v; return False if @known[\$v] and @known[\$v] !eqv \$coord; @grid[\$y][\$x] = \$v; # conjecture grid value print "\e[0H"; # show conjectured board for @grid -> \$r { say do for @\$r { when Rat { ' ' x \$width } when 0 { '_' x \$width } default { .fmt("%{\$width}d") } } } my @neighbors = @neigh[\$y][\$x][]; my @degrees; for @neighbors -> \n [\$yy,\$xx] { my \$d = --@degree[\$yy][\$xx]; # conjecture new degrees push @degrees[\$d], n; # and categorize by degree } for @degrees.grep(*.defined) -> @ties { for @ties.reverse { # reverse works better for this hidato anyway return True if try_fill \$v + 1, \$_; } } for @neighbors -> [\$yy,\$xx] { ++@degree[\$yy][\$xx]; # undo degree conjectures } @grid[\$y][\$x] = \$old; # undo grid value conjecture return False; } say "\$tries tries"; } ``` {{out}} ```txt 4 3 2 8 1 7 6 5 18 tries ``` ## Phix Brute force solution. I ordered the links highest letter first, then grouped by start letter to eliminate things asap. Nothing to eliminate when placing A and B, when placing C, check that CA>1, when placing D, check that DA,DB,DC are all >1, etc. ```Phix constant txt = """ A B /|\ /|\ / | X | \ / |/ \| \ C - D - E - F \ |\ /| / \ | X | / \|/ \|/ G H""" --constant links = "CA DA DB DC EA EB ED FB FE GC GD GE HD HE HF" constant links = {"","","A","ABC","ABD","BE","CDE","DEF"} function solve(sequence s, integer idx, sequence part) object res integer v, p for i=1 to length(s) do v = s[i] for j=1 to length(links[idx]) do p = links[idx][j]-'@' if abs(v-part[p])<2 then v=0 exit end if end for if v then if length(s)=1 then return part&v end if res = solve(s[1..i-1]&s[i+1..\$],idx+1,part&v) if sequence(res) then return res end if end if end for return 0 end function printf(1,substitute_all(txt,"ABCDEFGH",solve("12345678",1,""))) ``` {{out}} ```txt 3 4 /|\ /|\ / | X | \ / |/ \| \ 7 - 1 - 8 - 2 \ |\ /| / \ | X | / \|/ \|/ 5 6 ``` ## Picat ```Picat import cp. no_connection_puzzle(X) => N = 8, X = new_list(N), X :: 1..N, Graph = {{1,2}, {1,3}, {1,4}, {2,1}, {2,3}, {2,5}, {2,6}, {3,2}, {3,4}, {3,6}, {3,7}, {4,1}, {4,3}, {4,6}, {4,7}, {5,2}, {5,3}, {5,6}, {5,8}, {6,2}, {6,3}, {6,4}, {6,5}, {6,7}, {6,8}, {7,3}, {7,4}, {7,6}, {7,8}, {8,5}, {8,6}, {8,7}}, 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). ``` Test: ```txt Picat> no_connection_puzzle(_X) [2,5,8,6,3,1,4,7] 2 5 /|\ /|\ / | X | \ / |/ \| \ 8 - 6 - 3 - 1 \ |\ /| / \ | X | / \|/ \|/ 4 7 ``` ## 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. ```Prolog :- use_module(library(clpfd)). edge(a, c). edge(a, d). edge(a, e). edge(b, d). edge(b, e). edge(b, f). edge(c, d). edge(c, g). edge(d, e). edge(d, g). edge(d, h). edge(e, f). edge(e, g). edge(e, h). edge(f, h). connected(A, B) :- ( edge(A,B); edge(B, A)). no_connection_puzzle(Vs) :- % construct the arranged list of the nodes bagof(A, B^(edge(A,B); edge(B, A)), Lst), sort(Lst, L), length(L, Len), % construct the list of the values length(Vs, Len), Vs ins 1..Len, all_distinct(Vs), % two connected nodes must have values different for more than 1 set_constraints(L, Vs), label(Vs). set_constraints([], []). set_constraints([H | T], [VH | VT]) :- set_constraint(H, T, VH, VT), set_constraints(T, VT). set_constraint(_, [], _, []). set_constraint(H, [H1 | T1], V, [VH | VT]) :- connected(H, H1), ( V - VH #> 1; VH - V #> 1), set_constraint(H, T1, V, VT). set_constraint(H, [H1 | T1], V, [_VH | VT]) :- \+connected(H, H1), set_constraint(H, T1, V, VT). ``` Output : ```txt ?- 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. ```python from __future__ import print_function from itertools import permutations from enum import Enum A, B, C, D, E, F, G, H = Enum('Peg', 'A, B, C, D, E, F, G, H') connections = ((A, C), (A, D), (A, E), (B, D), (B, E), (B, F), (G, C), (G, D), (G, E), (H, D), (H, E), (H, F), (C, D), (D, E), (E, F)) def ok(conn, perm): """Connected numbers ok?""" this, that = (c.value - 1 for c in conn) return abs(perm[this] - perm[that]) != 1 def solve(): return [perm for perm in permutations(range(1, 9)) if all(ok(conn, perm) for conn in connections)] if __name__ == '__main__': solutions = solve() print("A, B, C, D, E, F, G, H =", ', '.join(str(i) for i in solutions[0])) ``` {{out}} ```txt 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: ```python def pp(solution): """Prettyprint a solution""" boardformat = r""" A B /|\ /|\ / | X | \ / |/ \| \ C - D - E - F \ |\ /| / \ | X | / \|/ \|/ G H""" for letter, number in zip("ABCDEFGH", solution): boardformat = boardformat.replace(letter, str(number)) print(boardformat) if __name__ == '__main__': for i, s in enumerate(solutions, 1): print("\nSolution", i, end='') pp(s) ``` ;Extra output:
```Solution 1
3   4
/|\ /|\
/ | X | \
/  |/ \|  \
7 - 1 - 8 - 2
\  |\ /|  /
\ | X | /
\|/ \|/
5   6

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

## Racket

```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" #<_  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:

```txt

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

found  1  solution.

```

'''output'''   when using the input of:    999

```txt

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
/*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 input of:    -3

```txt

╔═══════════════════════════╗              ╔═══════════════════════════╗
║          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 [http://rosettacode.org/wiki/Solve_a_Hidato_puzzle#With_Warnsdorff HLPSolver]:

```ruby

#  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 = < !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

{{tcllib|struct::list}}

```tcl
package require Tcl 8.6
package require struct::list

proc haveAdjacent {a b c d e f g h} {
expr {
[edge \$a \$c] ||
[edge \$a \$d] ||
[edge \$a \$e] ||
[edge \$b \$d] ||
[edge \$b \$e] ||
[edge \$b \$f] ||
[edge \$c \$d] ||
[edge \$c \$g] ||
[edge \$d \$e] ||
[edge \$d \$g] ||
[edge \$d \$h] ||
[edge \$e \$f] ||
[edge \$e \$g] ||
[edge \$e \$h] ||
[edge \$f \$h]
}
}
proc edge {x y} {
expr {abs(\$x-\$y) == 1}
}

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

{{out}}

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

## XPL0

```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);
]
```

{{out}}

```txt

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

```

## zkl

{{trans|D}}

```zkl
const PegA=0, PegB=1, PegC=2, PegD=3, PegE=4, PegF=5, PegG=6, PegH=7;
connections:=T(
T(PegA, PegC), T(PegA, PegD), T(PegA, PegE),
T(PegB, PegD), T(PegB, PegE), T(PegB, PegF),
T(PegC, PegD), T(PegD, PegE), T(PegE, PegF),
T(PegG, PegC), T(PegG, PegD), T(PegG, PegE),
T(PegH, PegD), T(PegH, PegE), T(PegH, PegF) );
CZ:=connections.len();
#<<<   // 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.
{{out}}

```txt

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

```

```