Solve the no connection puzzle: Difference between revisions
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{{task|Puzzles}}
You are given a box with eight holes labelled A-<small>to</small>-H, connected by fifteen straight lines in the pattern as shown below:
'''A''' '''B'''
/
/
/
'''C'''
\
\
\
'''G''' '''H'''
You are also given eight pegs numbered 1-<small>to</small>-8.
;Objective:
Place the eight pegs in the holes so that the (absolute) difference between any two numbers connected by any line is <u>greater</u> 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]]
;See also
[https://www.youtube.com/watch?v=AECElyEyZBQ No Connection Puzzle] (youtube).
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight lang="11l">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(‘, ’))</syntaxhighlight>
{{out}}
<pre>
A, B, C, D, E, F, G, H = 3, 4, 7, 1, 8, 2, 5, 6
</pre>
=={{header|Ada}}==
This solution is a bit longer than it actually needs to be; however, it uses tasks to find the solution and the used types and solution-generating functions are well-separated, making it more amenable to other solutions or altering it to display all solutions.
<syntaxhighlight lang="ada">
With
Ada.Text_IO,
Line 55 ⟶ 99:
begin
Ada.Text_IO.Put_Line( Connection_Types.Image(Result) );
end;</
<
Package Connection_Types with Pure is
Line 102 ⟶ 146:
Function Image ( Input : Partial_Board ) Return String;
End Connection_Types;</
<syntaxhighlight lang="ada">
Pragma Ada_2012;
Line 110 ⟶ 154:
Function Connection_Combinations return Partial_Board;
</syntaxhighlight>
<
Package Body Connection_Types is
Line 183 ⟶ 227:
end Image;
End Connection_Types;</
<
begin
Line 257 ⟶ 301:
End return;
End Connection_Combinations;
</syntaxhighlight>
{{out}}
<pre> 4 5
Line 269 ⟶ 313:
3 6
</pre>
=={{header|APL}}==
{{works with|Dyalog APL 17.0 Unicode}}
<syntaxhighlight lang="apl">
perms←{
⍝∇ 20100513/20140818 ra⌈ --()--
1=⍴⍴⍵:⍵[∇ ''⍴⍴⍵]
↑{0∊⍴⍵:⍵ ⋄ (⍺[1]⌷⍵),(1↓⍺)∇ ⍵~⍺[1]⌷⍵}∘(⍳⍵)¨↓⍉1+(⌽⍳⍵)⊤¯1+⍳!⍵
}
solution←{
links← (3 4 5) (4 5 6) (1 4 7) (1 2 3 5 7 8) (1 2 4 6 7 8) (2 5 8) (3 4 5) (4 5 6) ⍝ node i connects with nodes i⊃links
tries←8 perms 8
fails←{1∊{1∊⍵∊¯1 0 1}¨|⍺-¨⍺∘{⍺[⍵]}¨⍵}
⍝ ⍴⍸~tries fails ⍤1⊢links
⍝ 16
solns←⍸~tries fails ⍤1⊢links
tries[''⍴solns;]
}
</syntaxhighlight>
=={{header|ARM Assembly}}==
{{works with|as|Raspberry Pi}}
<syntaxhighlight lang="arm assembly">
/* 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
iAdrszCarriageReturn: .int szCarriageReturn
/******************************************************************/
/* 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]
add r4,#1
add r0,#1
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:
add r10,#1
cmp r10,#NBBOX - 3
bgt 3b
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 4b
ldr r0,iAdriValues_c
ldr r1,iAdriValues_d
mov r2,r10
mov r3,#NBBOX - 2
bl prepValues
mov r9,#-1
5:
add r9,#1
cmp r9,#NBBOX - 4
bgt 4b
@ control d / a b c
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdriValues_a
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 5b
ldr r0,iAdriValues_d
ldr r1,iAdriValues_e
mov r2,r9
mov r3,#NBBOX - 3
bl prepValues
mov r8,#-1
6:
add r8,#1
cmp r8,#NBBOX - 5
bgt 5b
@ control e / a b d
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdriValues_a
ldr r1,[r1,r12,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 6b
ldr r0,iAdriValues_e
ldr r1,iAdriValues_f
mov r2,r8
mov r3,#NBBOX - 4
bl prepValues
mov r7,#-1
7:
add r7,#1
cmp r7,#NBBOX - 6
bgt 6b
@ control f / b e
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 7b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 7b
ldr r0,iAdriValues_f
ldr r1,iAdriValues_g
mov r2,r7
mov r3,#NBBOX - 5
bl prepValues
mov r6,#-1
8:
add r6,#1
cmp r6,#NBBOX - 7
bgt 7b
@ control g / c d e
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdriValues_c
ldr r1,[r1,r10,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 8b
ldr r0,iAdriValues_g
ldr r1,iAdriValues_h
mov r2,r6
mov r3,#NBBOX - 6
bl prepValues
mov r5,#-1
9:
add r5,#1
cmp r5,#NBBOX - 8
bgt 8b
@ control h / d e f
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
ldr r1,iAdriValues_f
ldr r1,[r1,r7,lsl #2]
subs r2,r1,r0
mvnlt r2,r2
addlt r2,#1
cmp r2,#1
beq 9b
@ solution ok display text
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdrsMessValeur_a
bl conversion10
ldr r0,iAdriValues_b
ldr r0,[r0,r11,lsl #2]
ldr r1,iAdrsMessValeur_b
bl conversion10
ldr r0,iAdriValues_c
ldr r0,[r0,r10,lsl #2]
ldr r1,iAdrsMessValeur_c
bl conversion10
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdrsMessValeur_d
bl conversion10
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdrsMessValeur_e
bl conversion10
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdrsMessValeur_f
bl conversion10
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdrsMessValeur_g
bl conversion10
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdrsMessValeur_h
bl conversion10
ldr r0,iAdrsMessDeb
bl affichageMess
@ display design
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA]
ldr r0,iAdriValues_b
ldr r0,[r0,r11,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA+4]
bl affichageMess
ldr r0,iAdrszMessLine2
bl affichageMess
ldr r0,iAdrszMessLine3
bl affichageMess
ldr r0,iAdrszMessLine4
bl affichageMess
ldr r0,iAdriValues_c
ldr r0,[r0,r10,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA-4]
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA]
ldr r0,iAdriValues_e
ldr r0,[r0,r8,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA+4]
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine5
strb r2,[r0,#POSA+8]
bl affichageMess
ldr r0,iAdrszMessLine6
bl affichageMess
ldr r0,iAdrszMessLine7
bl affichageMess
ldr r0,iAdrszMessLine8
bl affichageMess
ldr r0,iAdriValues_g
ldr r0,[r0,r6,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA]
ldr r0,iAdriValues_h
ldr r0,[r0,r5,lsl #2]
ldr r1,iAdrsConvValue
bl conversion10
ldrb r2,[r1]
ldr r0,iAdrszMessLine1
strb r2,[r0,#POSA+4]
bl affichageMess
//b 9b @ loop for other solution
90:
100:
pop {r0-r12,lr} @ restaur registers
bx lr @return
iAdriValues_a: .int iValues_a
iAdriValues_b: .int iValues_b
iAdriValues_c: .int iValues_c
iAdriValues_d: .int iValues_d
iAdriValues_e: .int iValues_e
iAdriValues_f: .int iValues_f
iAdriValues_g: .int iValues_g
iAdriValues_h: .int iValues_h
iAdrsMessValeur_a: .int sMessValeur_a
iAdrsMessValeur_b: .int sMessValeur_b
iAdrsMessValeur_c: .int sMessValeur_c
iAdrsMessValeur_d: .int sMessValeur_d
iAdrsMessValeur_e: .int sMessValeur_e
iAdrsMessValeur_f: .int sMessValeur_f
iAdrsMessValeur_g: .int sMessValeur_g
iAdrsMessValeur_h: .int sMessValeur_h
iAdrsMessDeb: .int sMessDeb
iAdrsConvValue: .int sConvValue
iAdrszMessLine1: .int szMessLine1
iAdrszMessLine2: .int szMessLine2
iAdrszMessLine3: .int szMessLine3
iAdrszMessLine4: .int szMessLine4
iAdrszMessLine5: .int szMessLine5
iAdrszMessLine6: .int szMessLine6
iAdrszMessLine7: .int szMessLine7
iAdrszMessLine8: .int szMessLine8
/******************************************************************/
/* 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
addne r2,r2,#1 @ else add 1 in the length
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
add r1,#48 @ digit
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]
add r2,#1
add r4,#1
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
add r4,#1 @ next position
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
</syntaxhighlight>
<pre>
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
</pre>
=={{header|AutoHotkey}}==
<syntaxhighlight lang="autohotkey">oGrid := [[ "", "X", "X"] ; setup oGrid
,[ "X", "X", "X", "X"]
,[ "", "X", "X"]]
oNeighbor := [], oCell := [], oRoute := [] , oVisited := [] ; initialize objects
for row, oRow in oGrid
for col, val in oRow
if val ; for each valid cell in oGrid
oNeighbor[row, col] := Neighbors(row, col, oGrid) ; list valid no-connection neighbors
Solve:
for row, oRow in oGrid
for col , val in oRow
if val ; for each valid cell in oGrid
if (oSolution := SolveNoConnect(row, col, 1)).8 ; solve for this cell
break, Solve ; if solution found stop
; show solution
for i , val in oSolution
oCell[StrSplit(val, ":").1 , StrSplit(val, ":").2] := i
A := oCell[1, 2] , B := oCell[1, 3]
C := oCell[2, 1], D := oCell[2, 2] , E := oCell[2, 3], F := oCell[2, 4]
G := oCell[3, 2] , H := oCell[3, 3]
sol =
(
%A% %B%
/|\ /|\
/ | X | \
/ |/ \| \
%C% - %D% - %E% - %F%
\ |\ /| /
\ | X | /
\|/ \|/
%G% %H%
)
MsgBox % sol
return
;-----------------------------------------------------------------------
SolveNoConnect(row, col, val){
global
oRoute.push(row ":" col) ; save route
oVisited[row, col] := true ; mark this cell visited
if oRoute[8] ; if solution found
return true ; end recursion
for each, nn in StrSplit(oNeighbor[row, col], ",") ; for each no-connection neighbor of cell
{
rowX := StrSplit(nn, ":").1 , colX := StrSplit(nn, ":").2 ; get coords of this neighbor
if !oVisited[rowX, colX] ; if not previously visited
{
oVisited[rowX, colX] := true ; mark this cell visited
val++ ; increment
if (SolveNoConnect(rowX, colX, val)) ; recurse
return oRoute ; if solution found return route
}
}
oRoute.pop() ; Solution not found, backtrack oRoute
oVisited[row, col] := false ; Solution not found, remove mark
}
;-----------------------------------------------------------------------
Neighbors(row, col, oGrid){ ; return distant neighbors of oGrid[row,col]
for r , oRow in oGrid
for c, v in oRow
if (v="X") && (abs(row-r) > 1 || abs(col-c) > 1)
list .= r ":"c ","
if (row<>2) && oGrid[row, col]
list .= oGrid[row, col+1] ? row ":" col+1 "," : oGrid[row, col-1] ? row ":" col-1 "," : ""
return Trim(list, ",")
}</syntaxhighlight>
Outputs:<pre>
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6</pre>
=={{header|C}}==
{{trans|Go}}
<syntaxhighlight lang="c">#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;
}</syntaxhighlight>
{{out}}
<pre>----- 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</pre>
=={{header|C#}}==
{{trans|Java}}
<syntaxhighlight lang="C#">
using System;
using System.Collections.Generic;
using System.Linq;
public class NoConnection
{
// adopted from Go
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++)
foreach (int peg in links[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]}");
}
}
</syntaxhighlight>
{{out}}
<pre>
6 1
4 3 8 7
2 5
</pre>
=={{header|C++}}==
{{trans|C}}
<syntaxhighlight lang="cpp">#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;
}</syntaxhighlight>
{{out}}
<pre>----- 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</pre>
=={{header|Chapel}}==
<syntaxhighlight lang="chapel">type hole = int;
param A : hole = 1;
param B : hole = A+1;
param C : hole = B+1;
param D : hole = C+1;
param E : hole = D+1;
param F : hole = E+1;
param G : hole = F+1;
param H : hole = G+1;
param starting : int = 0;
const holes : domain(hole) = { A,B,C,D,E,F,G,H };
const graph : [holes] domain(hole) = [ A => { C,D,E },
B => { D,E,F },
C => { A,D,G },
D => { A,B,C,E,G,H },
E => { A,B,D,F,G,H },
F => { B,E,H },
G => { C,D,E },
H => { D,E,F }
];
proc check( configuration : [] int, idx : hole ) : bool {
var good = true;
for adj in graph[idx] {
if adj >= idx then continue;
if abs( configuration[idx] - configuration[adj] ) <= 1 {
good = false;
break;
}
}
return good;
}
proc solve( configuration : [] int, pegs : domain(int), idx : hole = A ) : bool {
for value in pegs {
configuration[idx] = value;
if check( configuration, idx ) {
if idx < holes.size {
var prePegs = pegs;
if solve( configuration, prePegs - value, idx + 1 ){
return true;
}
} else {
return true;
}
}
}
configuration[idx] = starting;
return false;
}
proc printBoard( configuration : [] int ){
return
"\n " + configuration[A] + " " + configuration[B]+ "\n" +
" /|\\ /|\\ \n"+
" / | X | \\ \n"+
" / |/ \\| \\ \n"+
" " + configuration[C] +" - " + configuration[D] + " - " + configuration[E] + " - " + configuration[F] + " \n"+
" \\ |\\ /| / \n"+
" \\ | X | / \n"+
" \\|/ \\|/ \n"+
" " + configuration[G] + " " + configuration[H]+ "\n";
}
proc main(){
var configuration : [holes] int;
for idx in holes do configuration[idx] = starting;
var pegs : domain(int) = {1,2,3,4,5,6,7,8};
solve( configuration, pegs );
writeln( printBoard( configuration ) );
}
</syntaxhighlight>
<pre>
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
</pre>
=={{header|D}}==
<
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],
Line 292 ⟶ 1,437:
G H";
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}}
<pre>
Line 308 ⟶ 1,453:
\|/ \|/
4 5
</pre>
===Alternative version===
Using a simple backtracking.
{{trans|Go}}
<syntaxhighlight lang="d">import std.stdio, std.algorithm, std.conv, std.string, std.typecons;
// Holes A=0, B=1, ..., H=7
// With connections:
const board = r"
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H";
struct Connection { uint a, b; }
immutable Connection[] connections = [
{0, 2}, {0, 3}, {0, 4}, // A to C,D,E
{1, 3}, {1, 4}, {1, 5}, // B to D,E,F
{6, 2}, {6, 3}, {6, 4}, // G to C,D,E
{7, 3}, {7, 4}, {7, 5}, // H to D,E,F
{2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F
];
alias Pegs = uint[8];
int absDiff(in uint a, in uint b) pure nothrow @safe @nogc {
return (a > b) ? (a - b) : (b - a);
}
/** Solution is a simple recursive brute force solver,
it stops at the first found solution.
It returns the solution, the number of positions tested,
and the number of pegs swapped. */
Tuple!(Pegs,"p", uint,"tests", uint,"swaps") solve() pure nothrow @safe @nogc {
uint tests = 0, swaps = 0;
Pegs p = [1, 2, 3, 4, 5, 6, 7, 8];
bool recurse(in uint i) nothrow @safe @nogc {
if (i >= p.length.signed - 1) {
tests++;
return connections.all!(c => absDiff(p[c.a], p[c.b]) > 1);
}
// Try each remain peg from.
foreach (immutable j; i .. p.length) {
swaps++;
swap(p[i], p[j]);
if (recurse(i + 1))
return true;
swap(p[i], p[j]);
}
return false;
}
recurse(0);
return typeof(return)(p, tests, swaps);
}
void main() {
immutable sol = solve();
board.tr("ABCDEFGH", "%(%d%)".format(sol.p)).writeln;
writeln("Tested ", sol.tests, " positions and did ", sol.swaps, " swaps.");
}</syntaxhighlight>
{{out}}
<pre>
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.</pre>
=={{header|Delphi}}==
{{works with|Delphi|6.0}}
{{libheader|SysUtils,StdCtrls}}
<syntaxhighlight lang="Delphi">
{ 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}
Memo.Lines.Add(S);
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;
Memo.Lines.Add(S);
{Show puzzle with values inserted}
ShowPuzzle(Memo);
end;
</syntaxhighlight>
{{out}}
<pre>
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.
</pre>
=={{header|Elixir}}==
{{trans|Ruby}}
This solution uses HLPsolver from [[Solve_a_Hidato_puzzle#Elixir | here]]
<syntaxhighlight lang="elixir"># 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 .
"""
HLPsolver.solve(board, adjacent, false)
|> Enum.sort |> Enum.map(fn {_,cell} -> cell.value end)
|> Enum.zip(~w[A B C D E F G H])
|> Enum.reduce(layout, fn {n,c},acc -> String.replace(acc, c, to_string(n)) end)
|> IO.puts</syntaxhighlight>
{{out}}
<pre>
4 - 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 - 5
</pre>
=={{header|Factor}}==
<syntaxhighlight lang="factor">USING: assocs interpolate io kernel math math.combinatorics
math.ranges math.parser multiline pair-rocket sequences
sequences.generalizations ;
STRING: diagram
${} ${}
/|\ /|\
/ | X | \
/ |/ \| \
${} - ${} - ${} - ${}
\ |\ /| /
\ | X | /
\|/ \|/
${} ${}
;
CONSTANT: adjacency
H{
0 => { 2 3 4 }
1 => { 3 4 5 }
2 => { 0 3 6 }
3 => { 0 1 2 4 6 7 }
4 => { 0 1 3 5 6 7 }
5 => { 1 4 7 }
6 => { 2 3 4 }
7 => { 3 4 5 }
}
: any-consecutive? ( seq n -- ? ) [ - abs 1 = ] curry any? ;
: neighbors ( elt seq i -- seq elt )
adjacency at swap nths swap ;
: solution? ( permutation-seq -- ? )
dup [ neighbors any-consecutive? ] with find-index nip not ;
: find-solution ( -- seq )
8 [1,b] [ solution? ] find-permutation ;
: display-solution ( seq -- )
[ number>string ] map 8 firstn diagram interpolate>string
print ;
: main ( -- ) find-solution display-solution ;
MAIN: main</syntaxhighlight>
{{out}}
<pre>
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</pre>
=={{header|Fortran}}==
{{works with|gfortran|11.2.1}}
<syntaxhighlight lang="fortran">! This is free and unencumbered software released into the public domain,
! via the Unlicense.
! For more information, please refer to <http://unlicense.org/>
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
! already inserted.
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</syntaxhighlight>
The first solution printed:
{{out}}
<pre>
3 4
/│\ /│\
/ │ X │ \
/ │/ \│ \
7───1───8───2
\ │\ /│ /
\ │ X │ /
\│/ \│/
5 6
</pre>
=={{header|Go}}==
A simple recursive brute force solution.
<
import (
Line 401 ⟶ 2,123:
}
return b - a
}</
{{out}}
<pre>
Line 415 ⟶ 2,137:
5 6
Tested 12094 positions and did 20782 swaps.
</pre>
=={{header|Groovy}}==
{{trans|Java}}
<syntaxhighlight lang="groovy">import java.util.stream.Collectors
import java.util.stream.IntStream
class NoConnection {
// adopted from Go
static int[][] links = [
[2, 3, 4], // A to C,D,E
[3, 4, 5], // B to D,E,F
[2, 4], // D to C, E
[5], // E to F
[2, 3, 4], // G to C,D,E
[3, 4, 5], // H to D,E,F
]
static int[] pegs = new int[8]
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]}")
}
}</syntaxhighlight>
{{out}}
<pre> 6 7
2 3 8 1
4 5</pre>
=={{header|Haskell}}==
<syntaxhighlight lang="haskell">import Data.List (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) $
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</syntaxhighlight>
{{Out}}
<pre style="font-size:80%">A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
3 4
7 1 8 2
5 6 </pre>
=={{header|J}}==
Supporting code:
<syntaxhighlight lang="j">holes=:;:'A B C D E F G H'
connections=:".;._2]0 :0
holes e.;:'C D E' NB. A
holes e.;:'D E F' NB. B
holes e.;:'A D G' NB. C
holes e.;:'A B C E G H' NB. D
holes e.;:'A B D F G H' NB. E
holes e.;:'B E H' NB. F
holes e.;:'C D E' NB. G
holes e.;:'D E F' NB. H
)
assert (-:|:) connections NB. catch typos
pegs=: 1+(A.&i.~ !)8
attempt=: [: <./@(-.&0)@,@:| connections * -/~
box=:0 :0
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
)
disp=:verb define
rplc&(,holes;&":&>y) box
)</syntaxhighlight>
Intermezzo:
<syntaxhighlight lang="j"> (#~ 1<attempt"1) pegs
3 4 7 1 8 2 5 6
3 5 7 1 8 2 4 6
3 6 7 1 8 2 4 5
3 6 7 1 8 2 5 4
4 3 2 8 1 7 6 5
4 5 2 8 1 7 6 3
4 5 7 1 8 2 3 6
4 6 7 1 8 2 3 5
5 3 2 8 1 7 6 4
5 4 2 8 1 7 6 3
5 4 7 1 8 2 3 6
5 6 7 1 8 2 3 4
6 3 2 8 1 7 4 5
6 3 2 8 1 7 5 4
6 4 2 8 1 7 5 3
6 5 2 8 1 7 4 3</syntaxhighlight>
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.
<syntaxhighlight lang="j"> disp {. (#~ 1<attempt"1) pegs
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</syntaxhighlight>
'''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):
<syntaxhighlight lang="j"> (#~ 1<attempt"1) pegs
3 5 7 1 8 2 4 6
4 6 7 1 8 2 3 5
5 3 2 8 1 7 6 4
6 4 2 8 1 7 5 3</syntaxhighlight>
The first of these looks like this:
<syntaxhighlight lang="j"> disp {. (#~ 1<attempt"1) pegs
3 - 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 - 6
</syntaxhighlight>
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.
=={{header|Java}}==
The backtracking is getting tiresome, we'll try a stochastic solution for a change.<br>
{{works with|Java|8}}
<syntaxhighlight lang="java">import static java.lang.Math.abs;
import java.util.*;
import static java.util.stream.Collectors.toList;
import static java.util.stream.IntStream.range;
public class NoConnection {
// adopted from Go
static int[][] links = {
{2, 3, 4}, // A to C,D,E
{3, 4, 5}, // B to D,E,F
{2, 4}, // D to C, E
{5}, // E to F
{2, 3, 4}, // G to C,D,E
{3, 4, 5}, // H to D,E,F
};
static int[] pegs = new int[8];
public static void main(String[] args) {
List<Integer> vals = range(1, 9).mapToObj(i -> i).collect(toList());
do {
Collections.shuffle(vals);
for (int i = 0; i < pegs.length; i++)
pegs[i] = vals.get(i);
} while (!solved());
printResult();
}
static boolean solved() {
for (int i = 0; i < links.length; i++)
for (int peg : links[i])
if (abs(pegs[i] - peg) == 1)
return false;
return true;
}
static void printResult() {
System.out.printf(" %s %s%n", pegs[0], pegs[1]);
System.out.printf("%s %s %s %s%n", pegs[2], pegs[3], pegs[4], pegs[5]);
System.out.printf(" %s %s%n", pegs[6], pegs[7]);
}
}</syntaxhighlight>
(takes about 500 shuffles on average)
<pre> 4 5
2 8 1 7
6 3 </pre>
=={{header|JavaScript}}==
===ES6===
{{Trans|Haskell}}
<syntaxhighlight lang="javascript">(() => {
'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))
.padStart(5, ' ')
))
);
}
// ---------------- 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
const fromEnum = x =>
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();
})();</syntaxhighlight>
{{Out}}
<pre>A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
3 4
7 1 8 2
5 6</pre>
=={{header|jq}}==
{{works with|jq|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'''
<syntaxhighlight lang="jq">
# 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);
</syntaxhighlight>
'''Part 2: The no-connections puzzle for N pegs and holes'''
<syntaxhighlight lang="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 $p
| all( connections[];
(($p[.[0]] - $p[.[1]])|abs) != 1 );
. as $connections | permutations(N) | select(ok($connections));
</syntaxhighlight>
'''Part 3: The 8-peg no-connection puzzle'''
<syntaxhighlight lang="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;</syntaxhighlight>
'''Examples''':
<syntaxhighlight lang="jq"># To print all the solutions:
# solve | pp
# To count the number of solutions:
# count(solve)
# => 16
limit(1; solve) | pp
</syntaxhighlight>
{{output}}
Invocation: jq -n -r -f no_connection.jq
<pre>
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
</pre>
=={{header|Julia}}==
<syntaxhighlight lang="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()
</syntaxhighlight> {{output}} <pre>
Found 16 solutions.
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</pre>
=={{header|Kotlin}}==
{{trans|Go}}
<syntaxhighlight lang="scala">// version 1.2.0
import kotlin.math.abs
// Holes A=0, B=1, …, H=7
// With connections:
const val conn = """
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
"""
val connections = listOf(
0 to 2, 0 to 3, 0 to 4, // A to C, D, E
1 to 3, 1 to 4, 1 to 5, // B to D, E, F
6 to 2, 6 to 3, 6 to 4, // G to C, D, E
7 to 3, 7 to 4, 7 to 5, // H to D, E, F
2 to 3, 3 to 4, 4 to 5 // C-D, D-E, E-F
)
// 'isValid' checks if the pegs are a valid solution.
// If the absolute difference between any pair of connected pegs is
// greater than one it is a valid solution.
fun isValid(pegs: IntArray): Boolean {
for ((a, b) in connections) {
if (abs(pegs[a] - pegs[b]) <= 1) return false
}
return true
}
fun swap(pegs: IntArray, i: Int, j: Int) {
val tmp = pegs[i]
pegs[i] = pegs[j]
pegs[j] = tmp
}
// 'solve' is a simple recursive brute force solver,
// it stops at the first found solution.
// It returns the solution, the number of positions tested,
// and the number of pegs swapped.
fun solve(): Triple<IntArray, Int, Int> {
val pegs = IntArray(8) { it + 1 }
var tests = 0
var swaps = 0
fun recurse(i: Int): Boolean {
if (i >= pegs.size - 1) {
tests++
return isValid(pegs)
}
// Try each remaining peg from pegs[i] onwards
for (j in i until pegs.size) {
swaps++
swap(pegs, i, j)
if (recurse(i + 1)) return true
swap(pegs, i, j)
}
return false
}
recurse(0)
return Triple(pegs, tests, swaps)
}
fun pegsAsString(pegs: IntArray): String {
val ca = conn.toCharArray()
for ((i, c) in ca.withIndex()) {
if (c in 'A'..'H') ca[i] = '0' + pegs[c - 'A']
}
return String(ca)
}
fun main(args: Array<String>) {
val (p, tests, swaps) = solve()
println(pegsAsString(p))
println("Tested $tests positions and did $swaps swaps.")
}</syntaxhighlight>
{{out}}
<pre>
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
</pre>
=={{header|M2000 Interpreter}}==
Final Version, print all solutions (16 from 40320 permutations)
Press space bar to see solutions so far.
<syntaxhighlight lang="m2000 interpreter">
Module no_connection_puzzle {
\\ Holes
Inventory Connections="A":="CDE","B":="DEF","C":="ADG", "D":="ABCEGH"
Append Connections, "E":="ABDFGH","F":="HEB", "G":="CDE","H":="DEF"
Inventory ToDelete, Solutions
\\ eliminate double connnections
con=each(Connections)
While con {
m$=eval$(con, con^)
c$=eval$(con)
If c$="*" Then continue
For i=1 to len(C$) {
d$=mid$(c$,i,1)
r$=Filter$(Connections$(d$), m$)
If r$<>"" Then {
Return connections, d$:=r$
} else {
If m$=connections$(d$) Then {
Return connections, d$:="*" : If not exist(todelete, d$) Then Append todelete, d$
}
}
}
}
con=each(todelete)
While con {
Delete Connections, eval$(con)
}
Inventory Holes
For i=0 to 7 : Append Holes, Chr$(65+i):=i : Next i
CheckValid=lambda Holes, Connections (a$, arr) -> {
val=Array(arr, Holes(a$))
con$=Connections$(a$)
res=True
For i=1 to Len(con$) {
If Abs(Array(Arr, Holes(mid$(con$,i,1)))-val)<2 Then res=False: Exit
}
=res
}
a=(1,2,3,4,5,6,7,8)
h=(,)
solution=(,)
done=False
counter=0
Print "Wait..."
P(h, a)
sol=Each(Solutions)
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
Sub P(h, a)
If len(a)<=1 Then process(cons(h, a)) : Exit Sub
local b=cons(a)
For i=1 to len(b) {
b=cons(cdr(b),car(b))
P(cons(h,car(b)), cdr(b))
}
End Sub
Sub Process(a)
counter++
Print counter
If keypress(32) Then {
local sol=Each(Solutions)
aa$=Key$
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
}
hole=each(Connections)
done=True
While hole {
If not CheckValid(Eval$(hole, hole^), a) Then done=False : Exit
}
If done Then Append Solutions, Len(Solutions):=a : Print a
End Sub
Sub Disp(a)
Print format$(" {0} {1}", array(a), array(a,1))
Print " /|\ /|\"
Print " / | X | \"
Print " / |/ \| \"
Print Format$("{0} - {1} - {2} - {3}", array(a,2),array(a,3), array(a,4), array(a,5))
Print " \ |\ /| /"
Print " \ | X | /"
Print " \|/ \|/"
Print Format$(" {0} {1}", array(a,6), array(a,7))
End Sub
}
no_connection_puzzle
</syntaxhighlight>
{{out}}
<pre>
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
</pre >
=={{header|m4}}==
{{trans|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.
<syntaxhighlight lang="m4">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</syntaxhighlight>
{{out}}
<pre> 3 4
/|\ /|\
/ | X | \
/ |/ \| \
7---1---8---2
\ |\ /| /
\ | X | /
\|/ \|/
5 6</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
This one simply takes all permutations of the pegs and filters out invalid solutions.
<syntaxhighlight lang="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]];</syntaxhighlight>
{{out}}
<pre> 3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6</pre>
=={{header|Nim}}==
{{trans|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.
<syntaxhighlight lang="nim">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]
discard pegs.findSolution(1, 8)</syntaxhighlight>
{{out}}
<pre>----- 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</pre>
=={{header|Perl}}==
<syntaxhighlight lang="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;
}
}</syntaxhighlight>
{{out}}
<pre>
34
7182
56
</pre>
=={{header|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.
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">txt</span> <span style="color: #0000FF;">=</span> <span style="color: #008000;">"""
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
"""</span>
<span style="color: #000080;font-style:italic;">--constant links = "CA DA DB DC EA EB ED FB FE GC GD GE HD HE HF"</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">links</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #008000;">""</span><span style="color: #0000FF;">,</span><span style="color: #008000;">""</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"A"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"ABC"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"ABD"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"BE"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"CDE"</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"DEF"</span><span style="color: #0000FF;">}</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">integer</span> <span style="color: #000000;">idx</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">sequence</span> <span style="color: #000000;">part</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">object</span> <span style="color: #000000;">res</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">v</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">p</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">v</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">s</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">j</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">links</span><span style="color: #0000FF;">[</span><span style="color: #000000;">idx</span><span style="color: #0000FF;">])</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">p</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">links</span><span style="color: #0000FF;">[</span><span style="color: #000000;">idx</span><span style="color: #0000FF;">][</span><span style="color: #000000;">j</span><span style="color: #0000FF;">]-</span><span style="color: #008000;">'@'</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">abs</span><span style="color: #0000FF;">(</span><span style="color: #000000;">v</span><span style="color: #0000FF;">-</span><span style="color: #000000;">part</span><span style="color: #0000FF;">[</span><span style="color: #000000;">p</span><span style="color: #0000FF;">])<</span><span style="color: #000000;">2</span> <span style="color: #008080;">then</span> <span style="color: #000000;">v</span><span style="color: #0000FF;">=</span><span style="color: #000000;">0</span> <span style="color: #008080;">exit</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">v</span> <span style="color: #008080;">then</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">1</span> <span style="color: #008080;">then</span> <span style="color: #008080;">return</span> <span style="color: #000000;">part</span><span style="color: #0000FF;">&</span><span style="color: #000000;">v</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #000000;">res</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">[</span><span style="color: #000000;">1</span><span style="color: #0000FF;">..</span><span style="color: #000000;">i</span><span style="color: #0000FF;">-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">]&</span><span style="color: #000000;">s</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span><span style="color: #0000FF;">..$],</span><span style="color: #000000;">idx</span><span style="color: #0000FF;">+</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">part</span><span style="color: #0000FF;">&</span><span style="color: #000000;">v</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">if</span> <span style="color: #004080;">sequence</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">then</span> <span style="color: #008080;">return</span> <span style="color: #000000;">res</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">0</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">substitute_all</span><span style="color: #0000FF;">(</span><span style="color: #000000;">txt</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"ABCDEFGH"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">solve</span><span style="color: #0000FF;">(</span><span style="color: #008000;">"12345678"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">""</span><span style="color: #0000FF;">)))</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</pre>
=={{header|Picat}}==
<syntaxhighlight lang="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).</syntaxhighlight>
{{out}}
<pre>
Picat> no_connection_puzzle(_X)
[3,4,7,1,8,2,5,6]
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</pre>
=={{header|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.
<syntaxhighlight lang="prolog">:- use_module(library(clpfd)).
edge(a, c).
edge(a, d).
edge(a, e).
edge(b, d).
edge(b, e).
edge(b, f).
edge(c, d).
edge(c, g).
edge(d, e).
edge(d, g).
edge(d, h).
edge(e, f).
edge(e, g).
edge(e, h).
edge(f, h).
connected(A, B) :-
( edge(A,B); edge(B, A)).
no_connection_puzzle(Vs) :-
% construct the arranged list of the nodes
bagof(A, B^(edge(A,B); edge(B, A)), Lst),
sort(Lst, L),
length(L, Len),
% construct the list of the values
length(Vs, Len),
Vs ins 1..Len,
all_distinct(Vs),
% two connected nodes must have values different for more than 1
set_constraints(L, Vs),
label(Vs).
set_constraints([], []).
set_constraints([H | T], [VH | VT]) :-
set_constraint(H, T, VH, VT),
set_constraints(T, VT).
set_constraint(_, [], _, []).
set_constraint(H, [H1 | T1], V, [VH | VT]) :-
connected(H, H1),
( V - VH #> 1; VH - V #> 1),
set_constraint(H, T1, V, VT).
set_constraint(H, [H1 | T1], V, [_VH | VT]) :-
\+connected(H, H1),
set_constraint(H, T1, V, VT).
</syntaxhighlight>
Output :
<pre> ?- 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.
</pre>
=={{header|Python}}==
A brute force search solution.
<
from itertools import permutations
from enum import Enum
Line 445 ⟶ 3,479:
if __name__ == '__main__':
solutions = solve()
print("A, B, C, D, E, F, G, H =", ', '.join(str(i) for i in solutions[0]))</
{{out}}
Line 454 ⟶ 3,488:
Add the following code after that above:
<
"""Prettyprint a solution"""
boardformat = r"""
Line 474 ⟶ 3,508:
for i, s in enumerate(solutions, 1):
print("\nSolution", i, end='')
pp(s)</
;Extra output:
Line 652 ⟶ 3,686:
\|/ \|/
4 3</pre>
=={{header|Racket}}==
<syntaxhighlight lang="racket">#lang racket
;; Solve the no connection puzzle. Tim Brown Oct. 2014
;; absolute difference of a and b if they are both true
(define (and- a b) (and a b (abs (- a b))))
;; Finds the differences of all established connections in the network
(define (network-diffs (A #f) (B #f) (C #f) (D #f) (E #f) (F #f) (G #f) (H #f))
(list (and- A C) (and- A D) (and- A E)
(and- B D) (and- B E) (and- B F)
(and- C D) (and- C G)
(and- D E) (and- D G) (and- D H)
(and- E F) (and- E G) (and- E H)
(and- F G)))
;; Make sure there is “no connection” in the network N; return N if good
(define (good-network? N)
(and (for/and ((d (filter values (apply network-diffs N)))) (> d 1)) N))
;; possible optimisation is to reverse the arguments to network-diffs, reverse the return value from
;; this function and make this a cons but we're pretty quick here as it is.
(define (find-good-network pegs (n/w null))
(if (null? pegs) n/w
(for*/or ((p pegs))
(define n/w+ (append n/w (list p)))
(and (good-network? n/w+)
(find-good-network (remove p pegs =) n/w+)))))
(define (render-puzzle pzl)
(apply printf (regexp-replace* "O" #<<EOS
O O
/|\ /|\
/ | X | \
/ |/ \| \
O - O - O - O
\ |\ /| /
\ | X | /
\|/ \|/
O O~%
EOS
"~a") pzl))
(render-puzzle (find-good-network '(1 2 3 4 5 6 7 8)))</syntaxhighlight>
{{out}}
<pre> 3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6</pre>
=={{header|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:
* [[Solve a Hidato puzzle#Raku|Solve a Hidato puzzle]]
* [[Solve a Hopido puzzle#Raku|Solve a Hopido puzzle]]
* [[Solve a Holy Knight's tour#Raku|Solve a Holy Knight's tour]]
* [[Solve a Numbrix puzzle#Raku|Solve a Numbrix puzzle]]
* [[Solve the no connection puzzle#Raku|Solve the no connection puzzle]]
The idiosyncratic adjacency diagram is dealt with by the simple expedient of bending the two vertical lines <tt>||</tt> into two bows <tt>)(</tt>, such that adjacency can be calculated simply as a distance of 2 or less.
<syntaxhighlight lang="raku" line>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";
}</syntaxhighlight>
{{out}}
<pre> 4 3
2 8 1 7
6 5
18 tries</pre>
=={{header|Red}}==
===Basic version===
<syntaxhighlight lang="red">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]
]
]
]
; start with and empty list
check []
</syntaxhighlight>
{{out}}
<pre>Here is a solution: 3 4 7 1 8 2 5 6
(halted)
</pre>
===With graphics===
<syntaxhighlight lang="red">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"]
]
; start with and empty list
check []
</syntaxhighlight>
{{out}}
[https://raw.githubusercontent.com/Palaing/redlib/master/games/images/noconnexion.png graphical output image]
=={{header|REXX}}==
===unannotated solutions===
<
parse arg limit .
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
@.7 = 'G C D E' /* ║ \│/ \│/ ║ */
@.8 = 'H D E F' /* ║ G H ║ */
cnt= 0 /* ╚═══════════════════════════╝ */
pegs=
do b=1 for pegs;
do c=1 for pegs;
do d=1 for pegs;
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 _
cnt= cnt + 1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
say /*display a blank line to
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. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
?:
nH= nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return
nL= nn-1
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK.*/</syntaxhighlight>
{{out|output|text= when using the default input:}}
<pre>
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6
Line 716 ⟶ 4,021:
found 1 solution.
</pre>
{{out|output|text= when using the default input of: <tt> 999 </tt>}}
<pre>
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6
Line 740 ⟶ 4,044:
===annotated solutions===
Usage note: if the '''limit''' (the 1<sup>st</sup> argument) is negative, a diagram (node graph) is shown.
<syntaxhighlight lang="rexx">/*REXX program solves the "no─connection" puzzle (the puzzle has eight pegs). */
@abc= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */
@.
@.
@.
@.
@.
@.
@.
@.
cnt= 0 /* ╚═══════════════════════════╝ */
do
do
__=
subs=
!._.0= subs
pegs= pegs - 1
do
do
do
do
do
do
end /*
end /*
end /*
say
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
end /*cn*/ /* [↑] see if there're any duplicates.*/
nL= nn-1
$= !.node.ch; fn= value($) /*the node name and its current peg #.*/
return 0 /*the subroutine arg value passed is OK*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
showNodes:
show= 0
do box=1 for sourceline() while oLimit<0 /*Negative? Then display the diagram. */
xw= sourceline(box) /*get a source line of this program. */
if \show
xb=
xt= xb /*get a working copy of the box. */
xt= translate(xt, value(@), @) /*substitute the peg name with a value.*/
end /*jx*/ /* [↑] graph is limited to 26 nodes.*/
say _ xb _ _ xt /*display
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</syntaxhighlight>
{{out|output|text= when using the default inputs of: <tt> -1 </tt>}}
<pre>
╔═══════════════════════════╗ ╔═══════════════════════════╗
Line 830 ⟶ 4,137:
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ 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
</pre>
=={{header|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]:
<
# Solve No Connection Puzzle
#
Line 862 ⟶ 4,195:
g.board[H[0]][H[1]].adj = [A,B,C,G]
g.solve
</syntaxhighlight>
{{out}}
<pre>
Line 874 ⟶ 4,207:
2 8 1 7
6 4
</pre>
=={{header|Scala}}==
{{libheader|Scala sub-repositories}}
{{Out}}Best seen in running your browser either by [https://scalafiddle.io/sf/Ub2LEup/0 ScalaFiddle (ES aka JavaScript, non JVM)] or [https://scastie.scala-lang.org/ZXGSJLFEQe21Frh4Vwp0WA Scastie (remote JVM)].
<syntaxhighlight lang="scala">object NoConnection extends App {
private def links = Seq(
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 =
links.indices.forall(
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")
}
printResult(genRandom.dropWhile(!notSolved(links, _)).head)
}</syntaxhighlight>
=={{header|Tcl}}==
{{tcllib|struct::list}}
<syntaxhighlight lang="tcl">package require Tcl 8.6
package require struct::list
proc haveAdjacent {a b c d e f g h} {
expr {
[edge $a $c] ||
[edge $a $d] ||
[edge $a $e] ||
[edge $b $d] ||
[edge $b $e] ||
[edge $b $f] ||
[edge $c $d] ||
[edge $c $g] ||
[edge $d $e] ||
[edge $d $g] ||
[edge $d $h] ||
[edge $e $f] ||
[edge $e $g] ||
[edge $e $h] ||
[edge $f $h]
}
}
proc edge {x y} {
expr {abs($x-$y) == 1}
}
set layout [string trim {
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
} \n]
struct::list foreachperm p {1 2 3 4 5 6 7 8} {
if {![haveAdjacent {*}$p]} {
puts [string map [join [
lmap name {A B C D E F G H} val $p {list $name $val}
]] $layout]
break
}
}</syntaxhighlight>
{{out}}
<pre> 3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6</pre>
=={{header|Wren}}==
{{trans|Kotlin}}
{{libheader|Wren-dynamic}}
<syntaxhighlight lang="wren">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.")</syntaxhighlight>
{{out}}
<pre>
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
</pre>
=={{header|XPL0}}==
<syntaxhighlight lang="xpl0">include c:\cxpl\codes;
int Hole, Max, I;
char Box(8), Str;
def A, B, C, D, E, F, G, H;
[for Hole:= 0 to 7 do Box(Hole):= Hole+1;
Max:= 7;
while abs(Box(D)-Box(A)) < 2 or abs(Box(D)-Box(C)) < 2 or
abs(Box(D)-Box(G)) < 2 or abs(Box(D)-Box(E)) < 2 or
abs(Box(A)-Box(C)) < 2 or abs(Box(C)-Box(G)) < 2 or
abs(Box(G)-Box(E)) < 2 or abs(Box(E)-Box(A)) < 2 or
abs(Box(E)-Box(B)) < 2 or abs(Box(E)-Box(H)) < 2 or
abs(Box(E)-Box(F)) < 2 or abs(Box(B)-Box(D)) < 2 or
abs(Box(D)-Box(H)) < 2 or abs(Box(H)-Box(F)) < 2 or
abs(Box(F)-Box(B)) < 2 do
loop [I:= Box(0); \next permutation
for Hole:= 0 to Max-1 do Box(Hole):= Box(Hole+1);
Box(Max):= I;
if I # Max+1 then [Max:= 7; quit]
else Max:= Max-1];
Str:= "
# #
/|\ /|\
/ | X | \
/ |/ \| \
# - # - # - #
\ |\ /| /
\ | 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);
]</syntaxhighlight>
{{out}}
<pre>
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
</pre>
=={{header|zkl}}==
{{trans|D}}
<
connections:=T(
T(PegA, PegC), T(PegA, PegD), T(PegA, PegE),
Line 884 ⟶ 4,460:
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
#<<< // 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.
break; // comment out to see all 16 solutions
}</syntaxhighlight>
The filter1 method stops on the first True, so it acts like a conditional or.
{{out}}
<pre>
/|\ /|\
/ | X | \
/ |/ \| \
\ |\ /| /
\ | X | /
\|/ \|/
</pre>
| |||