Solve the no connection puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
You are given a box with eight holes labelled A-to-H, connected by fifteen straight lines in the pattern as shown below:
A B
/│\ /│\
/ │ X │ \
/ │/ \│ \
C───D───E───F
\ │\ /│ /
\ │ X │ /
\│/ \│/
G H
You are also given eight pegs numbered 1-to-8.
- Objective
Place the eight pegs in the holes so that the (absolute) difference between any two numbers connected by any line is greater than one.
- Example
In this attempt:
4 7
/│\ /│\
/ │ X │ \
/ │/ \│ \
8───1───6───2
\ │\ /│ /
\ │ X │ /
\│/ \│/
3 5
Note that 7 and 6 are connected and have a difference of 1, so it is not a solution.
- Task
Produce and show here one solution to the puzzle.
- Related tasks
- See also
No Connection Puzzle (youtube).
11l
<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(‘, ’))</lang>
- Output:
A, B, C, D, E, F, G, H = 3, 4, 7, 1, 8, 2, 5, 6
Ada
This solution is a bit longer than it actually needs to be; however, it uses tasks to find the solution and the used types and solution-generating functions are well-separated, making it more amenable to other solutions or altering it to display all solutions. <lang Ada> With Ada.Text_IO, Connection_Types, Connection_Combinations;
procedure main is
Result : Connection_Types.Partial_Board renames Connection_Combinations;
begin
Ada.Text_IO.Put_Line( Connection_Types.Image(Result) );
end;</lang> <lang Ada>Pragma Ada_2012;
Package Connection_Types with Pure is
-- Name of the nodes. Type Node is (A, B, C, D, E, F, G, H);
-- Type for indicating if a node is connected.
Type Connection_List is array(Node) of Boolean
with Size => 8, Object_Size => 8, Pack;
Function "&"( Left : Connection_List; Right : Node ) return Connection_List;
-- The actual map of the network connections.
Network : Constant Array (Node) of Connection_List:=
(
A => (C|D|E => True, others => False),
B => (D|E|F => True, others => False),
C => (A|D|G => True, others => False),
D => (C|A|B|E|H|G => True, others => False),
E => (D|A|B|F|H|G => True, others => False),
F => (B|E|H => True, others => False),
G => (C|D|E => True, others => False),
H => (D|E|F => True, others => False)
);
-- Values of the nodes. Type Peg is range 1..8;
-- Indicator for which values have been assigned.
Type Used_Peg is array(Peg) of Boolean
with Size => 8, Object_Size => 8, Pack;
Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg;
-- Type describing the layout of the network. Type Partial_Board is array(Node range <>) of Peg; Subtype Board is Partial_Board(Node);
-- Determines if the given board is a solution or partial-solution. Function Is_Solution ( Input : Partial_Board ) return Boolean;
-- Displays the board as text. Function Image ( Input : Partial_Board ) Return String;
End Connection_Types;</lang> <lang Ada> Pragma Ada_2012;
with Connection_Types; use Connection_Types;
Function Connection_Combinations return Partial_Board; </lang> <lang Ada>Pragma Ada_2012;
Package Body Connection_Types is
New_Line : Constant String := ASCII.CR & ASCII.LF;
---------------------
-- Solution Test --
---------------------
Function Is_Solution( Input : Partial_Board ) return Boolean is
(for all Index in Input'Range =>
(for all Connection in Node'Range =>
(if Network(Index)(Connection) and Connection in Input'Range
then abs (Input(Index) - Input(Connection)) > 1
)
)
);
------------------------
-- Concat Operators --
------------------------
Function "&"( Left : Used_Peg; Right : Peg ) return Used_Peg is
begin
return Result : Used_Peg := Left do
Result(Right):= True;
end return;
end "&";
Function "&"(Left : Connection_List; Right : Node) return Connection_List is
begin
Return Result : Connection_List := Left do
Result(Right):= True;
end return;
end "&";
----------------------- -- IMAGE FUNCTIONS -- -----------------------
Function Image(Input : Peg) Return Character is
( Peg'Image(Input)(2) );
Function Image(Input : Peg) Return String is
( 1 => Image(Input) );
Function Image(Input : Partial_Board; Item : Node) Return String is
( 1 => (if Item not in Input'Range then '*' else Image(Input(Item)) ));
Function Image( Input : Partial_Board ) Return String is
A : String renames Image(Input, Connection_Types.A);
B : String renames Image(Input, Connection_Types.B);
C : String renames Image(Input, Connection_Types.C);
D : String renames Image(Input, Connection_Types.D);
E : String renames Image(Input, Connection_Types.E);
F : String renames Image(Input, Connection_Types.F);
G : String renames Image(Input, Connection_Types.G);
H : String renames Image(Input, Connection_Types.H);
begin
return
" "&A&" "&B & New_Line & " /|\ /|\" & New_Line & " / | X | \" & New_Line & " / |/ \| \" & New_Line & " "&C&" - "&D&" - "&E&" - "&F & New_Line & " \ |\ /| /" & New_Line & " \ | X | /" & New_Line & " \|/ \|/" & New_Line & " "&G&" "&H & New_Line;
end Image;
End Connection_Types;</lang> <lang Ada>Function Connection_Combinations return Partial_Board is
begin
Return Result : Board do
declare
-- The Generate task takes two parameters
-- (1) a list of pegs already in use, and
-- (2) a partial-board
-- and, if the state given is a viable yet incomplete solution, it
-- takes a peg and adds it to the state creating a new task with
-- that peg in its used list.
--
-- When a complete solution is found it is copied into result.
task type Generate(
Pegs : not null access Used_Peg:= new Used_Peg'(others => False);
State : not null access Partial_Board:= new Partial_Board'(Node'Last..Node'First => <>)
) is
end Generate;
-- An access to Generate and array thereof, for creating the
-- children tasks.
type Generator is access all Generate;
type Generators is array(Peg range <>) of Generator;
-- Gen handles the actual creation of a new task and state.
Function Gen(P : Peg; G : not null access Generate) return Generator is
begin
return (if G.Pegs(P) then null
else new Generate(
Pegs => new Used_Peg'(G.Pegs.all & P),
State => New Partial_Board'(G.All.State.All & P)
)
);
end;
task body Generate is
begin
if Is_Solution(State.All) then
-- If the state is a partial board, we make children to
-- complete the calculations.
if State'Length <= Node'Pos(Node'Last) then
declare
Subtasks : Constant Generators:=
(
Gen(1, Generate'Access),
Gen(2, Generate'Access),
Gen(3, Generate'Access),
Gen(4, Generate'Access),
Gen(5, Generate'Access),
Gen(6, Generate'Access),
Gen(7, Generate'Access),
Gen(8, Generate'Access)
);
begin
null;
end;
else
Result:= State.All;
end if;
else
-- The current state is not a solution, so we do not continue it.
Null;
end if;
end Generate;
Master : Generate;
begin
null;
end;
End return;
End Connection_Combinations; </lang>
- Output:
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
APL
<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;]
}
</lang>
ARM Assembly
<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
</lang>
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
<lang AutoHotkey>oGrid := [[ "", "X", "X"] ; setup oGrid ,[ "X", "X", "X", "X"] ,[ "", "X", "X"]]
oNeighbor := [], oCell := [], oRoute := [] , oVisited := [] ; initialize objects
for row, oRow in oGrid for col, val in oRow if val ; for each valid cell in oGrid oNeighbor[row, col] := Neighbors(row, col, oGrid) ; list valid no-connection neighbors
Solve: for row, oRow in oGrid for col , val in oRow if val ; for each valid cell in oGrid if (oSolution := SolveNoConnect(row, col, 1)).8 ; solve for this cell break, Solve ; if solution found stop
- show solution
for i , val in oSolution oCell[StrSplit(val, ":").1 , StrSplit(val, ":").2] := i
A := oCell[1, 2] , B := oCell[1, 3] C := oCell[2, 1], D := oCell[2, 2] , E := oCell[2, 3], F := oCell[2, 4] G := oCell[3, 2] , H := oCell[3, 3] sol = (
%A% %B% /|\ /|\ / | X | \ / |/ \| \
%C% - %D% - %E% - %F%
\ |\ /| / \ | X | / \|/ \|/ %G% %H%
) MsgBox % sol return
- -----------------------------------------------------------------------
SolveNoConnect(row, col, val){ global oRoute.push(row ":" col) ; save route oVisited[row, col] := true ; mark this cell visited
if oRoute[8] ; if solution found return true ; end recursion
for each, nn in StrSplit(oNeighbor[row, col], ",") ; for each no-connection neighbor of cell { rowX := StrSplit(nn, ":").1 , colX := StrSplit(nn, ":").2 ; get coords of this neighbor if !oVisited[rowX, colX] ; if not previously visited { oVisited[rowX, colX] := true ; mark this cell visited val++ ; increment if (SolveNoConnect(rowX, colX, val)) ; recurse return oRoute ; if solution found return route } } oRoute.pop() ; Solution not found, backtrack oRoute oVisited[row, col] := false ; Solution not found, remove mark }
- -----------------------------------------------------------------------
Neighbors(row, col, oGrid){ ; return distant neighbors of oGrid[row,col] for r , oRow in oGrid for c, v in oRow if (v="X") && (abs(row-r) > 1 || abs(col-c) > 1) list .= r ":"c "," if (row<>2) && oGrid[row, col] list .= oGrid[row, col+1] ? row ":" col+1 "," : oGrid[row, col-1] ? row ":" col-1 "," : "" return Trim(list, ",") }</lang>
Outputs:
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
C
<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;
}</lang>
- Output:
----- 0 ----- 3 4 7 1 8 2 5 6 ----- 1 ----- 3 5 7 1 8 2 4 6 ----- 2 ----- 3 6 7 1 8 2 4 5 ----- 3 ----- 3 6 7 1 8 2 5 4 ----- 4 ----- 4 3 2 8 1 7 6 5 ----- 5 ----- 4 5 2 8 1 7 6 3 ----- 6 ----- 4 5 7 1 8 2 3 6 ----- 7 ----- 4 6 7 1 8 2 3 5 ----- 8 ----- 5 3 2 8 1 7 6 4 ----- 9 ----- 5 4 2 8 1 7 6 3 ----- 10 ----- 5 4 7 1 8 2 3 6 ----- 11 ----- 5 6 7 1 8 2 3 4 ----- 12 ----- 6 3 2 8 1 7 5 4 ----- 13 ----- 6 3 2 8 1 7 4 5 ----- 14 ----- 6 4 2 8 1 7 5 3 ----- 15 ----- 6 5 2 8 1 7 4 3
C++
<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;
}</lang>
- Output:
----- 0 ----- 3 4 7 1 8 2 5 6 ----- 1 ----- 3 5 7 1 8 2 4 6 ----- 2 ----- 3 6 7 1 8 2 4 5 ----- 3 ----- 3 6 7 1 8 2 5 4 ----- 4 ----- 4 3 2 8 1 7 6 5 ----- 5 ----- 4 5 2 8 1 7 6 3 ----- 6 ----- 4 5 7 1 8 2 3 6 ----- 7 ----- 4 6 7 1 8 2 3 5 ----- 8 ----- 5 3 2 8 1 7 6 4 ----- 9 ----- 5 4 2 8 1 7 6 3 ----- 10 ----- 5 4 7 1 8 2 3 6 ----- 11 ----- 5 6 7 1 8 2 3 4 ----- 12 ----- 6 3 2 8 1 7 5 4 ----- 13 ----- 6 3 2 8 1 7 4 5 ----- 14 ----- 6 4 2 8 1 7 5 3 ----- 15 ----- 6 5 2 8 1 7 4 3
Chapel
<lang chapel>type hole = int; param A : hole = 1; param B : hole = A+1; param C : hole = B+1; param D : hole = C+1; param E : hole = D+1; param F : hole = E+1; param G : hole = F+1; param H : hole = G+1; param starting : int = 0; const holes : domain(hole) = { A,B,C,D,E,F,G,H }; const graph : [holes] domain(hole) = [ A => { C,D,E },
B => { D,E,F },
C => { A,D,G },
D => { A,B,C,E,G,H },
E => { A,B,D,F,G,H },
F => { B,E,H },
G => { C,D,E },
H => { D,E,F }
];
proc check( configuration : [] int, idx : hole ) : bool {
var good = true;
for adj in graph[idx] {
if adj >= idx then continue;
if abs( configuration[idx] - configuration[adj] ) <= 1 {
good = false;
break;
}
}
return good;
}
proc solve( configuration : [] int, pegs : domain(int), idx : hole = A ) : bool {
for value in pegs {
configuration[idx] = value;
if check( configuration, idx ) {
if idx < holes.size {
var prePegs = pegs;
if solve( configuration, prePegs - value, idx + 1 ){
return true;
}
} else {
return true;
}
}
}
configuration[idx] = starting;
return false;
}
proc printBoard( configuration : [] int ){ return "\n " + configuration[A] + " " + configuration[B]+ "\n" + " /|\\ /|\\ \n"+ " / | X | \\ \n"+ " / |/ \\| \\ \n"+ " " + configuration[C] +" - " + configuration[D] + " - " + configuration[E] + " - " + configuration[F] + " \n"+ " \\ |\\ /| / \n"+ " \\ | X | / \n"+ " \\|/ \\|/ \n"+ " " + configuration[G] + " " + configuration[H]+ "\n";
}
proc main(){
var configuration : [holes] int;
for idx in holes do configuration[idx] = starting;
var pegs : domain(int) = {1,2,3,4,5,6,7,8};
solve( configuration, pegs );
writeln( printBoard( configuration ) );
} </lang>
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
D
<lang d>void main() @safe {
import std.stdio, std.math, std.algorithm, std.traits, std.string;
enum Peg { A, B, C, D, E, F, G, H }
immutable Peg[2][15] connections =
[[Peg.A, Peg.C], [Peg.A, Peg.D], [Peg.A, Peg.E],
[Peg.B, Peg.D], [Peg.B, Peg.E], [Peg.B, Peg.F],
[Peg.C, Peg.D], [Peg.D, Peg.E], [Peg.E, Peg.F],
[Peg.G, Peg.C], [Peg.G, Peg.D], [Peg.G, Peg.E],
[Peg.H, Peg.D], [Peg.H, Peg.E], [Peg.H, Peg.F]];
immutable board = r"
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H";
Peg[EnumMembers!Peg.length] perm = [EnumMembers!Peg];
do if (connections[].all!(con => abs(perm[con[0]] - perm[con[1]]) > 1))
return board.tr("ABCDEFGH", "%(%d%)".format(perm)).writeln;
while (perm[].nextPermutation);
}</lang>
- Output:
2 3
/|\ /|\
/ | X | \
/ |/ \| \
6 - 0 - 7 - 1
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Alternative version
Using a simple backtracking.
<lang d>import std.stdio, std.algorithm, std.conv, std.string, std.typecons;
// Holes A=0, B=1, ..., H=7 // With connections: const board = r"
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H";
struct Connection { uint a, b; }
immutable Connection[] connections = [
{0, 2}, {0, 3}, {0, 4}, // A to C,D,E
{1, 3}, {1, 4}, {1, 5}, // B to D,E,F
{6, 2}, {6, 3}, {6, 4}, // G to C,D,E
{7, 3}, {7, 4}, {7, 5}, // H to D,E,F
{2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F
];
alias Pegs = uint[8];
int absDiff(in uint a, in uint b) pure nothrow @safe @nogc {
return (a > b) ? (a - b) : (b - a);
}
/** Solution is a simple recursive brute force solver, it stops at the first found solution. It returns the solution, the number of positions tested, and the number of pegs swapped. */ Tuple!(Pegs,"p", uint,"tests", uint,"swaps") solve() pure nothrow @safe @nogc {
uint tests = 0, swaps = 0; Pegs p = [1, 2, 3, 4, 5, 6, 7, 8];
bool recurse(in uint i) nothrow @safe @nogc {
if (i >= p.length.signed - 1) {
tests++;
return connections.all!(c => absDiff(p[c.a], p[c.b]) > 1);
}
// Try each remain peg from.
foreach (immutable j; i .. p.length) {
swaps++;
swap(p[i], p[j]);
if (recurse(i + 1))
return true;
swap(p[i], p[j]);
}
return false;
}
recurse(0); return typeof(return)(p, tests, swaps);
}
void main() {
immutable sol = solve();
board.tr("ABCDEFGH", "%(%d%)".format(sol.p)).writeln;
writeln("Tested ", sol.tests, " positions and did ", sol.swaps, " swaps.");
}</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
Elixir
This solution uses HLPsolver from here <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</lang>
- Output:
4 - 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 - 5
Factor
<lang factor>USING: assocs interpolate io kernel math math.combinatorics math.ranges math.parser multiline pair-rocket sequences sequences.generalizations ;
STRING: diagram
${} ${}
/|\ /|\
/ | X | \
/ |/ \| \
${} - ${} - ${} - ${}
\ |\ /| /
\ | X | /
\|/ \|/
${} ${}
CONSTANT: adjacency H{
0 => { 2 3 4 }
1 => { 3 4 5 }
2 => { 0 3 6 }
3 => { 0 1 2 4 6 7 }
4 => { 0 1 3 5 6 7 }
5 => { 1 4 7 }
6 => { 2 3 4 }
7 => { 3 4 5 }
}
- any-consecutive? ( seq n -- ? ) [ - abs 1 = ] curry any? ;
- neighbors ( elt seq i -- seq elt )
adjacency at swap nths swap ;
- solution? ( permutation-seq -- ? )
dup [ neighbors any-consecutive? ] with find-index nip not ;
- find-solution ( -- seq )
8 [1,b] [ solution? ] find-permutation ;
- display-solution ( seq -- )
[ number>string ] map 8 firstn diagram interpolate>string print ;
- main ( -- ) find-solution display-solution ;
MAIN: main</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Fortran
<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</lang> The first solution printed:
- Output:
3 4
/│\ /│\
/ │ X │ \
/ │/ \│ \
7───1───8───2
\ │\ /│ /
\ │ X │ /
\│/ \│/
5 6
Go
A simple recursive brute force solution. <lang go>package main
import ( "fmt" "strings" )
func main() { p, tests, swaps := Solution() fmt.Println(p) fmt.Println("Tested", tests, "positions and did", swaps, "swaps.") }
// Holes A=0, B=1, …, H=7 // With connections: const conn = `
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H`
var connections = []struct{ a, b int }{ {0, 2}, {0, 3}, {0, 4}, // A to C,D,E {1, 3}, {1, 4}, {1, 5}, // B to D,E,F {6, 2}, {6, 3}, {6, 4}, // G to C,D,E {7, 3}, {7, 4}, {7, 5}, // H to D,E,F {2, 3}, {3, 4}, {4, 5}, // C-D, D-E, E-F }
type pegs [8]int
// Valid checks if the pegs are a valid solution. // If the absolute difference between any pair of connected pegs is // greater than one it is a valid solution. func (p *pegs) Valid() bool { for _, c := range connections { if absdiff(p[c.a], p[c.b]) <= 1 { return false } } return true }
// Solution is a simple recursive brute force solver, // it stops at the first found solution. // It returns the solution, the number of positions tested, // and the number of pegs swapped. func Solution() (p *pegs, tests, swaps int) { var recurse func(int) bool recurse = func(i int) bool { if i >= len(p)-1 { tests++ return p.Valid() } // Try each remain peg from p[i:] in p[i] for j := i; j < len(p); j++ { swaps++ p[i], p[j] = p[j], p[i] if recurse(i + 1) { return true } p[i], p[j] = p[j], p[i] } return false } p = &pegs{1, 2, 3, 4, 5, 6, 7, 8} recurse(0) return }
func (p *pegs) String() string { return strings.Map(func(r rune) rune { if 'A' <= r && r <= 'H' { return rune(p[r-'A'] + '0') } return r }, conn) }
func absdiff(a, b int) int { if a > b { return a - b } return b - a }</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
Groovy
<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]}")
}
}</lang>
- Output:
6 7 2 3 8 1 4 5
Haskell
<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</lang>
- Output:
A = 3 B = 4 C = 7 D = 1 E = 8 F = 2 G = 5 H = 6 3 4 7 1 8 2 5 6
J
Supporting code:
<lang J>holes=:;:'A B C D E F G H'
connections=:".;._2]0 :0
holes e.;:'C D E' NB. A holes e.;:'D E F' NB. B holes e.;:'A D G' NB. C holes e.;:'A B C E G H' NB. D holes e.;:'A B D F G H' NB. E holes e.;:'B E H' NB. F holes e.;:'C D E' NB. G holes e.;:'D E F' NB. H
) assert (-:|:) connections NB. catch typos
pegs=: 1+(A.&i.~ !)8
attempt=: [: <./@(-.&0)@,@:| connections * -/~
box=:0 :0
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
)
disp=:verb define
rplc&(,holes;&":&>y) box
)</lang>
Intermezzo:
<lang J> (#~ 1<attempt"1) pegs 3 4 7 1 8 2 5 6 3 5 7 1 8 2 4 6 3 6 7 1 8 2 4 5 3 6 7 1 8 2 5 4 4 3 2 8 1 7 6 5 4 5 2 8 1 7 6 3 4 5 7 1 8 2 3 6 4 6 7 1 8 2 3 5 5 3 2 8 1 7 6 4 5 4 2 8 1 7 6 3 5 4 7 1 8 2 3 6 5 6 7 1 8 2 3 4 6 3 2 8 1 7 4 5 6 3 2 8 1 7 5 4 6 4 2 8 1 7 5 3 6 5 2 8 1 7 4 3</lang>
Since there's more than one arrangement where the pegs satisfy the task constraints, and since the task calls for one solution, we will need to pick one of them. We can use the "first" function to satisfy this important constraint.
<lang J> disp {. (#~ 1<attempt"1) pegs
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
</lang>
Video
If we follow the video and also connect A and B as well as G and H, we get only four solutions (which we can see are reflections / rotations of each other):
<lang J> (#~ 1<attempt"1) pegs 3 5 7 1 8 2 4 6 4 6 7 1 8 2 3 5 5 3 2 8 1 7 6 4 6 4 2 8 1 7 5 3</lang>
The first of these looks like this:
<lang J> disp {. (#~ 1<attempt"1) pegs
3 - 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 - 6
</lang>
For this puzzle, we can also see that the solution can be described as: put the starting and ending numbers in the middle - everything else follows from there. It's perhaps interesting that we get this solution even if we do not explicitly put that logic into our code - it's built into the puzzle itself and is still the only solution no matter how we arrive there.
Java
The backtracking is getting tiresome, we'll try a stochastic solution for a change.
<lang java>import static java.lang.Math.abs; import java.util.*; import static java.util.stream.Collectors.toList; import static java.util.stream.IntStream.range;
public class NoConnection {
// adopted from Go
static int[][] links = {
{2, 3, 4}, // A to C,D,E
{3, 4, 5}, // B to D,E,F
{2, 4}, // D to C, E
{5}, // E to F
{2, 3, 4}, // G to C,D,E
{3, 4, 5}, // H to D,E,F
};
static int[] pegs = new int[8];
public static void main(String[] args) {
List<Integer> vals = range(1, 9).mapToObj(i -> i).collect(toList());
do {
Collections.shuffle(vals);
for (int i = 0; i < pegs.length; i++)
pegs[i] = vals.get(i);
} while (!solved());
printResult(); }
static boolean solved() {
for (int i = 0; i < links.length; i++)
for (int peg : links[i])
if (abs(pegs[i] - peg) == 1)
return false;
return true;
}
static void printResult() {
System.out.printf(" %s %s%n", pegs[0], pegs[1]);
System.out.printf("%s %s %s %s%n", pegs[2], pegs[3], pegs[4], pegs[5]);
System.out.printf(" %s %s%n", pegs[6], pegs[7]);
}
}</lang> (takes about 500 shuffles on average)
4 5
2 8 1 7
6 3
JavaScript
ES6
<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();
})();</lang>
- Output:
A = 3
B = 4
C = 7
D = 1
E = 8
F = 2
G = 5
H = 6
3 4
7 1 8 2
5 6
jq
We present a generate-and-test solver for a slightly more general version of the problem, in which there are N pegs and holes, and in which the connectedness of holes is defined by an array such that holes i and j are connected if and only if [i,j] is a member of the array.
The jq index origin is 0, and so in the following, the pegs and holes are internally numbered from 0 to (N-1) inclusive. That is, we interpret a permutation, p, of 0 .. (N-1) as meaning that the i-th peg is numbered p[i], for i in 0 .. (N-1).
However the pretty-print function shows solutions using the 1-to-8 numbering scheme for pegs, and the A-to-H lettering scheme for holes.
Part 1: Generic functions <lang jq># Short-circuit determination of whether (a|condition)
- 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);</lang>
Part 2: The no-connections puzzle for N pegs and holes <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 $p
| forall( connections;
(($p[.[0]] - $p[.[1]])|abs) != 1 );
. as $connections | permutations(N) | select(ok($connections);</lang>
Part 3: The 8-peg no-connection puzzle <lang jq># The connectedness matrix:
- 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;</lang>
Examples: <lang 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 $s else . end);
one(solve) | pp </lang>
- Output:
<lang sh>$ jq -n -r -f no_connection.jq
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4</lang>
Julia
<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()
</lang>
- Output:
Found 16 solutions.
3 4 /|\ /|\ / | X | \ / |/ \| \ 7 - 1 - 8 - 2 \ |\ /| / \ | X | / \|/ \|/ 5 6
Kotlin
<lang scala>// version 1.2.0
import kotlin.math.abs
// Holes A=0, B=1, …, H=7 // With connections: const val conn = """
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
"""
val connections = listOf(
0 to 2, 0 to 3, 0 to 4, // A to C, D, E 1 to 3, 1 to 4, 1 to 5, // B to D, E, F 6 to 2, 6 to 3, 6 to 4, // G to C, D, E 7 to 3, 7 to 4, 7 to 5, // H to D, E, F 2 to 3, 3 to 4, 4 to 5 // C-D, D-E, E-F
)
// 'isValid' checks if the pegs are a valid solution. // If the absolute difference between any pair of connected pegs is // greater than one it is a valid solution. fun isValid(pegs: IntArray): Boolean {
for ((a, b) in connections) {
if (abs(pegs[a] - pegs[b]) <= 1) return false
}
return true
}
fun swap(pegs: IntArray, i: Int, j: Int) {
val tmp = pegs[i] pegs[i] = pegs[j] pegs[j] = tmp
}
// 'solve' is a simple recursive brute force solver, // it stops at the first found solution. // It returns the solution, the number of positions tested, // and the number of pegs swapped.
fun solve(): Triple<IntArray, Int, Int> {
val pegs = IntArray(8) { it + 1 }
var tests = 0
var swaps = 0
fun recurse(i: Int): Boolean {
if (i >= pegs.size - 1) {
tests++
return isValid(pegs)
}
// Try each remaining peg from pegs[i] onwards
for (j in i until pegs.size) {
swaps++
swap(pegs, i, j)
if (recurse(i + 1)) return true
swap(pegs, i, j)
}
return false
}
recurse(0) return Triple(pegs, tests, swaps)
}
fun pegsAsString(pegs: IntArray): String {
val ca = conn.toCharArray()
for ((i, c) in ca.withIndex()) {
if (c in 'A'..'H') ca[i] = '0' + pegs[c - 'A']
}
return String(ca)
}
fun main(args: Array<String>) {
val (p, tests, swaps) = solve()
println(pegsAsString(p))
println("Tested $tests positions and did $swaps swaps.")
}</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
M2000 Interpreter
Final Version, print all solutions (16 from 40320 permutations)
Press space bar to see solutions so far. <lang M2000 Interpreter> Module no_connection_puzzle {
\\ Holes
Inventory Connections="A":="CDE","B":="DEF","C":="ADG", "D":="ABCEGH"
Append Connections, "E":="ABDFGH","F":="HEB", "G":="CDE","H":="DEF"
Inventory ToDelete, Solutions
\\ eliminate double connnections
con=each(Connections)
While con {
m$=eval$(con, con^)
c$=eval$(con)
If c$="*" Then continue
For i=1 to len(C$) {
d$=mid$(c$,i,1)
r$=Filter$(Connections$(d$), m$)
If r$<>"" Then {
Return connections, d$:=r$
} else {
If m$=connections$(d$) Then {
Return connections, d$:="*" : If not exist(todelete, d$) Then Append todelete, d$
}
}
}
}
con=each(todelete)
While con {
Delete Connections, eval$(con)
}
Inventory Holes
For i=0 to 7 : Append Holes, Chr$(65+i):=i : Next i
CheckValid=lambda Holes, Connections (a$, arr) -> {
val=Array(arr, Holes(a$))
con$=Connections$(a$)
res=True
For i=1 to Len(con$) {
If Abs(Array(Arr, Holes(mid$(con$,i,1)))-val)<2 Then res=False: Exit
}
=res
}
a=(1,2,3,4,5,6,7,8)
h=(,)
solution=(,)
done=False
counter=0
Print "Wait..."
P(h, a)
sol=Each(Solutions)
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
Sub P(h, a)
If len(a)<=1 Then process(cons(h, a)) : Exit Sub
local b=cons(a)
For i=1 to len(b) {
b=cons(cdr(b),car(b))
P(cons(h,car(b)), cdr(b))
}
End Sub
Sub Process(a)
counter++
Print counter
If keypress(32) Then {
local sol=Each(Solutions)
aa$=Key$
While sol {
Print "Solution:";sol^+1
Disp(Eval(Solutions))
aa$=Key$
}
}
hole=each(Connections)
done=True
While hole {
If not CheckValid(Eval$(hole, hole^), a) Then done=False : Exit
}
If done Then Append Solutions, Len(Solutions):=a : Print a
End Sub
Sub Disp(a)
Print format$(" {0} {1}", array(a), array(a,1))
Print " /|\ /|\"
Print " / | X | \"
Print " / |/ \| \"
Print Format$("{0} - {1} - {2} - {3}", array(a,2),array(a,3), array(a,4), array(a,5))
Print " \ |\ /| /"
Print " \ | X | /"
Print " \|/ \|/"
Print Format$(" {0} {1}", array(a,6), array(a,7))
End Sub
} no_connection_puzzle
</lang>
- Output:
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
m4
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.
<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</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7---1---8---2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Mathematica/Wolfram Language
This one simply takes all permutations of the pegs and filters out invalid solutions. <lang Mathematica>sol = Fold[
Select[#,
Function[perm, Abs[perm[[#21]] - perm[[#22]]] > 1]] &,
Permutations[
Range[8]], {{1, 3}, {1, 4}, {1, 5}, {2, 4}, {2, 5}, {2, 6}, {3,
4}, {3, 7}, {4, 5}, {4, 7}, {4, 8}, {5, 6}, {5, 7}, {5, 8}, {6,
8}}]1;
Print[StringForm[
" `` ``\n /|\\ /|\\\n / | X | \\\n / |/ \\| \\\n`` - `` \
- `` - ``\n \\ |\\ /| /\n \\ | X | /\n \\|/ \\|/\n `` ``",
Sequence @@ sol]];</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Nim
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.
<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)</lang>
- Output:
----- 1 ----- 3 4 7 1 8 2 5 6 ----- 2 ----- 3 5 7 1 8 2 4 6 ----- 3 ----- 3 6 7 1 8 2 4 5 ----- 4 ----- 3 6 7 1 8 2 5 4 ----- 5 ----- 4 3 2 8 1 7 6 5 ----- 6 ----- 4 5 2 8 1 7 6 3 ----- 7 ----- 4 5 7 1 8 2 3 6 ----- 8 ----- 4 6 7 1 8 2 3 5 ----- 9 ----- 5 3 2 8 1 7 6 4 ----- 10 ----- 5 4 2 8 1 7 6 3 ----- 11 ----- 5 4 7 1 8 2 3 6 ----- 12 ----- 5 6 7 1 8 2 3 4 ----- 13 ----- 6 3 2 8 1 7 5 4 ----- 14 ----- 6 3 2 8 1 7 4 5 ----- 15 ----- 6 4 2 8 1 7 5 3 ----- 16 ----- 6 5 2 8 1 7 4 3
Perl
<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 $p =~ tr/-/ /r;
exit;
}
}</lang>
- Output:
34 7182 56
Phix
Brute force solution. I ordered the links highest letter first, then grouped by start letter to eliminate things asap. Nothing to eliminate when placing A and B, when placing C, check that CA>1, when placing D, check that DA,DB,DC are all >1, etc. <lang Phix> constant txt = """
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H"""
--constant links = "CA DA DB DC EA EB ED FB FE GC GD GE HD HE HF" constant links = {"","","A","ABC","ABD","BE","CDE","DEF"}
function solve(sequence s, integer idx, sequence part) object res integer v, p
for i=1 to length(s) do
v = s[i]
for j=1 to length(links[idx]) do
p = links[idx][j]-'@'
if abs(v-part[p])<2 then v=0 exit end if
end for
if v then
if length(s)=1 then return part&v end if
res = solve(s[1..i-1]&s[i+1..$],idx+1,part&v)
if sequence(res) then return res end if
end if
end for
return 0
end function
printf(1,substitute_all(txt,"ABCDEFGH",solve("12345678",1,"")))</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Picat
<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).
</lang> Test:
Picat> no_connection_puzzle(_X)
[3,4,7,1,8,2,5,6]
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Prolog
Works with SWi-Prolog with module clpfd written by Markus Triska
We first compute a list of nodes, with sort this list, and we attribute a value at the nodes. <lang Prolog>:- use_module(library(clpfd)).
edge(a, c). edge(a, d). edge(a, e). edge(b, d). edge(b, e). edge(b, f). edge(c, d). edge(c, g). edge(d, e). edge(d, g). edge(d, h). edge(e, f). edge(e, g). edge(e, h). edge(f, h).
connected(A, B) :- ( edge(A,B); edge(B, A)).
no_connection_puzzle(Vs) :- % construct the arranged list of the nodes bagof(A, B^(edge(A,B); edge(B, A)), Lst), sort(Lst, L), length(L, Len),
% construct the list of the values length(Vs, Len), Vs ins 1..Len, all_distinct(Vs),
% two connected nodes must have values different for more than 1 set_constraints(L, Vs), label(Vs).
set_constraints([], []).
set_constraints([H | T], [VH | VT]) :- set_constraint(H, T, VH, VT), set_constraints(T, VT).
set_constraint(_, [], _, []). set_constraint(H, [H1 | T1], V, [VH | VT]) :- connected(H, H1), ( V - VH #> 1; VH - V #> 1), set_constraint(H, T1, V, VT).
set_constraint(H, [H1 | T1], V, [_VH | VT]) :- \+connected(H, H1), set_constraint(H, T1, V, VT).
</lang> Output :
?- no_connection_puzzle(Vs). Vs = [4, 3, 2, 8, 1, 7, 6, 5] . 27 ?- setof(Vs, no_connection_puzzle(Vs), R), length(R, Len). R = [[3, 4, 7, 1, 8, 2, 5, 6], [3, 5, 7, 1, 8, 2, 4|...], [3, 6, 7, 1, 8, 2|...], [3, 6, 7, 1, 8|...], [4, 3, 2, 8|...], [4, 5, 2|...], [4, 5|...], [4|...], [...|...]|...], Len = 16.
Python
A brute force search solution. <lang python>from __future__ import print_function from itertools import permutations from enum import Enum
A, B, C, D, E, F, G, H = Enum('Peg', 'A, B, C, D, E, F, G, H')
connections = ((A, C), (A, D), (A, E),
(B, D), (B, E), (B, F),
(G, C), (G, D), (G, E),
(H, D), (H, E), (H, F),
(C, D), (D, E), (E, F))
def ok(conn, perm):
"""Connected numbers ok?""" this, that = (c.value - 1 for c in conn) return abs(perm[this] - perm[that]) != 1
def solve():
return [perm for perm in permutations(range(1, 9))
if all(ok(conn, perm) for conn in connections)]
if __name__ == '__main__':
solutions = solve()
print("A, B, C, D, E, F, G, H =", ', '.join(str(i) for i in solutions[0]))</lang>
- Output:
A, B, C, D, E, F, G, H = 3, 4, 7, 1, 8, 2, 5, 6
- All solutions pretty printed
Add the following code after that above: <lang python>def pp(solution):
"""Prettyprint a solution"""
boardformat = r"""
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H"""
for letter, number in zip("ABCDEFGH", solution):
boardformat = boardformat.replace(letter, str(number))
print(boardformat)
if __name__ == '__main__':
for i, s in enumerate(solutions, 1):
print("\nSolution", i, end=)
pp(s)</lang>
- Extra output
Solution 1
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Solution 2
3 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 6
Solution 3
3 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Solution 4
3 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 4
Solution 5
4 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 5
Solution 6
4 5
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 3
Solution 7
4 5
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
Solution 8
4 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 5
Solution 9
5 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 4
Solution 10
5 4
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
6 3
Solution 11
5 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 6
Solution 12
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
Solution 13
6 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
4 5
Solution 14
6 3
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
5 4
Solution 15
6 4
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
5 3
Solution 16
6 5
/|\ /|\
/ | X | \
/ |/ \| \
2 - 8 - 1 - 7
\ |\ /| /
\ | X | /
\|/ \|/
4 3
Racket
<lang racket>#lang racket
- Solve the no connection puzzle. Tim Brown Oct. 2014
- absolute difference of a and b if they are both true
(define (and- a b) (and a b (abs (- a b))))
- Finds the differences of all established connections in the network
(define (network-diffs (A #f) (B #f) (C #f) (D #f) (E #f) (F #f) (G #f) (H #f))
(list (and- A C) (and- A D) (and- A E)
(and- B D) (and- B E) (and- B F)
(and- C D) (and- C G)
(and- D E) (and- D G) (and- D H)
(and- E F) (and- E G) (and- E H)
(and- F G)))
- Make sure there is “no connection” in the network N; return N if good
(define (good-network? N)
(and (for/and ((d (filter values (apply network-diffs N)))) (> d 1)) N))
- possible optimisation is to reverse the arguments to network-diffs, reverse the return value from
- this function and make this a cons but we're pretty quick here as it is.
(define (find-good-network pegs (n/w null))
(if (null? pegs) n/w
(for*/or ((p pegs))
(define n/w+ (append n/w (list p)))
(and (good-network? n/w+)
(find-good-network (remove p pegs =) n/w+)))))
(define (render-puzzle pzl)
(apply printf (regexp-replace* "O" #<<EOS O O /|\ /|\ / | X | \ / |/ \| \
O - O - O - O
\ |\ /| / \ | X | / \|/ \|/ O O~%
EOS
"~a") pzl))
(render-puzzle (find-good-network '(1 2 3 4 5 6 7 8)))</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
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
- Solve a Hopido puzzle
- Solve a Holy Knight's tour
- Solve a Numbrix puzzle
- Solve the no connection puzzle
The idiosyncratic adjacency diagram is dealt with by the simple expedient of bending the two vertical lines || into two bows )(, such that adjacency can be calculated simply as a distance of 2 or less. <lang perl6>my @adjacent = gather -> $y, $x {
take [$y,$x] if abs($x|$y) > 2;
} for flat -5 .. 5 X -5 .. 5;
solveboard q:to/END/;
. _ . . _ . . . . . . . _ . _ 1 . _ . . . . . . . _ . . _ . END
sub solveboard($board) {
my $max = +$board.comb(/\w+/); my $width = $max.chars;
my @grid;
my @known;
my @neigh;
my @degree;
@grid = $board.lines.map: -> $line {
[ $line.words.map: { /^_/ ?? 0 !! /^\./ ?? Rat !! $_ } ]
}
sub neighbors($y,$x --> List) {
eager gather for @adjacent {
my $y1 = $y + .[0];
my $x1 = $x + .[1];
take [$y1,$x1] if defined @grid[$y1][$x1];
}
}
for ^@grid -> $y {
for ^@grid[$y] -> $x {
if @grid[$y][$x] -> $v {
@known[$v] = [$y,$x];
}
if @grid[$y][$x].defined {
@neigh[$y][$x] = neighbors($y,$x);
@degree[$y][$x] = +@neigh[$y][$x];
}
}
}
print "\e[0H\e[0J";
my $tries = 0;
try_fill 1, @known[1];
sub try_fill($v, $coord [$y,$x] --> Bool) {
return True if $v > $max;
$tries++;
my $old = @grid[$y][$x];
return False if +$old and $old != $v;
return False if @known[$v] and @known[$v] !eqv $coord;
@grid[$y][$x] = $v; # conjecture grid value
print "\e[0H"; # show conjectured board
for @grid -> $r {
say do for @$r {
when Rat { ' ' x $width }
when 0 { '_' x $width }
default { .fmt("%{$width}d") }
}
}
my @neighbors = @neigh[$y][$x][];
my @degrees;
for @neighbors -> \n [$yy,$xx] {
my $d = --@degree[$yy][$xx]; # conjecture new degrees
push @degrees[$d], n; # and categorize by degree
}
for @degrees.grep(*.defined) -> @ties {
for @ties.reverse { # reverse works better for this hidato anyway
return True if try_fill $v + 1, $_;
}
}
for @neighbors -> [$yy,$xx] {
++@degree[$yy][$xx]; # undo degree conjectures
}
@grid[$y][$x] = $old; # undo grid value conjecture
return False;
}
say "$tries tries";
}</lang>
- Output:
4 3
2 8 1 7
6 5
18 tries
Red
Basic version
<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 [] </lang>
- Output:
Here is a solution: 3 4 7 1 8 2 5 6 (halted)
With graphics
<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 [] </lang>
- Output:
REXX
unannotated solutions
<lang rexx>/*REXX program solves the "no─connection" puzzle (the puzzle has eight pegs). */ parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */ if limit== | limit=="," then limit= 1 /* ║ A B ║ */
/* ║ /│\ /│\ ║ */
@. = /* ║ / │ \/ │ \ ║ */ @.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */ @.2 = 'B D E F' /* ║ / │/ \│ \ ║ */ @.3 = 'C A D G' /* ║ C────D────E────F ║ */ @.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */ @.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */ @.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */ @.7 = 'G C D E' /* ║ \│/ \│/ ║ */ @.8 = 'H D E F' /* ║ G H ║ */ cnt= 0 /* ╚═══════════════════════════╝ */
do pegs=1 while @.pegs\==; _= word(@.pegs, 1)
subs= 0
do #=1 for words(@.pegs) -1 /*create list of node paths.*/
__= word(@.pegs, # + 1); if __>_ then iterate
subs= subs + 1; !._.subs= __
end /*#*/
!._.0= subs /*assign the number of the node paths. */
end /*pegs*/
pegs= pegs - 1 /*the number of pegs to be seated. */
_= ' ' /*_ is used for indenting the output.*/
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
say _ 'a='a _ "b="||b _ 'c='c _ "d="d _ 'e='e _ "f="f _ 'g='g _ "h="h
cnt= cnt + 1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/ s= left('s', cnt\==1) /*handle the case of plurals (or not).*/ say 'found ' cnt " solution"s'.' /*display the number of solutions found*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ ?: parse arg node; nn= value(node)
nH= nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there any are duplicates.*/
nL= nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$= !.node.ch; fn= value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK.*/</lang>
- output when using the default input:
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6 found 1 solution.
- output when using the default input of: 999
a=3 b=4 c=7 d=1 e=8 f=2 g=5 h=6
a=3 b=5 c=7 d=1 e=8 f=2 g=4 h=6
a=3 b=6 c=7 d=1 e=8 f=2 g=4 h=5
a=3 b=6 c=7 d=1 e=8 f=2 g=5 h=4
a=4 b=3 c=2 d=8 e=1 f=7 g=6 h=5
a=4 b=5 c=2 d=8 e=1 f=7 g=6 h=3
a=4 b=5 c=7 d=1 e=8 f=2 g=3 h=6
a=4 b=6 c=7 d=1 e=8 f=2 g=3 h=5
a=5 b=3 c=2 d=8 e=1 f=7 g=6 h=4
a=5 b=4 c=2 d=8 e=1 f=7 g=6 h=3
a=5 b=4 c=7 d=1 e=8 f=2 g=3 h=6
a=5 b=6 c=7 d=1 e=8 f=2 g=3 h=4
a=6 b=3 c=2 d=8 e=1 f=7 g=4 h=5
a=6 b=3 c=2 d=8 e=1 f=7 g=5 h=4
a=6 b=4 c=2 d=8 e=1 f=7 g=5 h=3
a=6 b=5 c=2 d=8 e=1 f=7 g=4 h=3
found 16 solutions.
annotated solutions
Usage note: if the limit (the 1st argument) is negative, a diagram (node graph) is shown. <lang rexx>/*REXX program solves the "no─connection" puzzle (the puzzle has eight pegs). */ @abc= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' parse arg limit . /*number of solutions wanted.*/ /* ╔═══════════════════════════╗ */ if limit== | limit=="," then limit= 1 /* ║ A B ║ */ oLimit= limit; limit= abs(limit) /* ║ /│\ /│\ ║ */ @. = /* ║ / │ \/ │ \ ║ */ @.1 = 'A C D E' /* ║ / │ /\ │ \ ║ */ @.2 = 'B D E F' /* ║ / │/ \│ \ ║ */ @.3 = 'C A D G' /* ║ C────D────E────F ║ */ @.4 = 'D A B C E G' /* ║ \ │\ /│ / ║ */ @.5 = 'E A B D F H' /* ║ \ │ \/ │ / ║ */ @.6 = 'F B E H' /* ║ \ │ /\ │ / ║ */ @.7 = 'G C D E' /* ║ \│/ \│/ ║ */ @.8 = 'H D E F' /* ║ G H ║ */ cnt= 0 /* ╚═══════════════════════════╝ */
do pegs=1 while @.pegs\==; _= word(@.pegs, 1)
subs= 0
do #=1 for words(@.pegs) -1 /*create list of node paths.*/
__= word(@.pegs, #+1); if __>_ then iterate
subs= subs + 1; !._.subs= __
end /*#*/
!._.0= subs /*assign the number of the node paths. */
end /*pegs*/
pegs= pegs - 1 /*the number of pegs to be seated. */
_= ' ' /*_ is used for indenting the output. */
do a=1 for pegs; if ?('A') then iterate
do b=1 for pegs; if ?('B') then iterate
do c=1 for pegs; if ?('C') then iterate
do d=1 for pegs; if ?('D') then iterate
do e=1 for pegs; if ?('E') then iterate
do f=1 for pegs; if ?('F') then iterate
do g=1 for pegs; if ?('G') then iterate
do h=1 for pegs; if ?('H') then iterate
call showNodes
cnt= cnt + 1; if cnt==limit then leave a
end /*h*/
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say /*display a blank line to the terminal.*/ s= left('s', cnt\==1) /*handle the case of plurals (or not).*/ say 'found ' cnt " solution"s'.' /*display the number of solutions found*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ ?: parse arg node; nn= value(node)
nH= nn+1
do cn=c2d('A') to c2d(node)-1; if value( d2c(cn) )==nn then return 1
end /*cn*/ /* [↑] see if there're any duplicates.*/
nL= nn-1
do ch=1 for !.node.0 /* [↓] see if there any ¬= ±1 values.*/
$= !.node.ch; fn= value($) /*the node name and its current peg #.*/
if nL==fn | nH==fn then return 1 /*if ≡ ±1, then the node can't be used.*/
end /*ch*/ /* [↑] looking for suitable number. */
return 0 /*the subroutine arg value passed is OK*/
/*──────────────────────────────────────────────────────────────────────────────────────*/ showNodes: _= left(, 5) /*_ is used for padding the output. */ show= 0 /*indicates no graph has been found yet*/
do box=1 for sourceline() while oLimit<0 /*Negative? Then display the diagram. */
xw= sourceline(box) /*get a source line of this program. */
p2= lastpos('*', xw) /*the position of last asterisk.*/
p1= lastpos('*', xw, max(1, p2-1) ) /* " " " penultimate " */
if pos('╔', xw)\==0 then show= 1 /*Have found the top-left box corner ? */
if \show then iterate /*Not found? Then skip this line. */
xb= substr(xw, p1+1, p2-p1-2) /*extract the "box" part of line. */
xt= xb /*get a working copy of the box. */
do jx=1 for pegs /*do a substitution for all the pegs. */
@= substr(@abc, jx, 1) /*get the name of the peg (A ──► Z). */
xt= translate(xt, value(@), @) /*substitute the peg name with a value.*/
end /*jx*/ /* [↑] graph is limited to 26 nodes.*/
say _ xb _ _ xt /*display one line of the graph. */
if pos('╝', xw)\==0 then return /*Is this last line of graph? Then stop*/
end /*box*/
say _ 'a='a _ "b="||b _ 'c='c _ "d="d _ ' e='e _ "f="f _ 'g='g _ "h="h return</lang>
- output when using the default inputs of: -1
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 4 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 5 6 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 5 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 4 6 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
╔═══════════════════════════╗ ╔═══════════════════════════╗
║ A B ║ ║ 3 6 ║
║ /│\ /│\ ║ ║ /│\ /│\ ║
║ / │ \/ │ \ ║ ║ / │ \/ │ \ ║
║ / │ /\ │ \ ║ ║ / │ /\ │ \ ║
║ / │/ \│ \ ║ ║ / │/ \│ \ ║
║ C────D────E────F ║ ║ 7────1────8────2 ║
║ \ │\ /│ / ║ ║ \ │\ /│ / ║
║ \ │ \/ │ / ║ ║ \ │ \/ │ / ║
║ \ │ /\ │ / ║ ║ \ │ /\ │ / ║
║ \│/ \│/ ║ ║ \│/ \│/ ║
║ G H ║ ║ 4 5 ║
╚═══════════════════════════╝ ╚═══════════════════════════╝
found 3 solutions.
Ruby
Be it Golden Frogs jumping on trancendental lilly pads, or a Knight on a board, or square pegs into round holes this is essentially a Hidato Like Problem, so I use HLPSolver: <lang ruby>
- Solve No Connection Puzzle
- Nigel_Galloway
- October 6th., 2014
require 'HLPSolver' ADJACENT = 0,0 A,B,C,D,E,F,G,H = [0,1],[0,2],[1,0],[1,1],[1,2],[1,3],[2,1],[2,2]
board1 = <<EOS
. 0 0 . 0 0 1 0 . 0 0 .
EOS g = HLPsolver.new(board1) g.board[A[0]][A[1]].adj = [B,G,H,F] g.board[B[0]][B[1]].adj = [A,C,G,H] g.board[C[0]][C[1]].adj = [B,E,F,H] g.board[D[0]][D[1]].adj = [F] g.board[E[0]][E[1]].adj = [C] g.board[F[0]][F[1]].adj = [A,C,D,G] g.board[G[0]][G[1]].adj = [A,B,F,H] g.board[H[0]][H[1]].adj = [A,B,C,G] g.solve </lang>
- Output:
Problem:
0 0
0 0 1 0
0 0
Solution:
5 3
2 8 1 7
6 4
Scala
- Output:
Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
<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)
}</lang>
Tcl
<lang tcl>package require Tcl 8.6 package require struct::list
proc haveAdjacent {a b c d e f g h} {
expr {
[edge $a $c] || [edge $a $d] || [edge $a $e] || [edge $b $d] || [edge $b $e] || [edge $b $f] || [edge $c $d] || [edge $c $g] || [edge $d $e] || [edge $d $g] || [edge $d $h] || [edge $e $f] || [edge $e $g] || [edge $e $h] || [edge $f $h]
}
} proc edge {x y} {
expr {abs($x-$y) == 1}
}
set layout [string trim {
A B
/|\ /|\
/ | X | \
/ |/ \| \
C - D - E - F
\ |\ /| /
\ | X | /
\|/ \|/
G H
} \n] struct::list foreachperm p {1 2 3 4 5 6 7 8} {
if {![haveAdjacent {*}$p]} {
puts [string map [join [ lmap name {A B C D E F G H} val $p {list $name $val} ]] $layout] break
}
}</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Wren
<lang ecmascript>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.")</lang>
- Output:
3 4
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
5 6
Tested 12094 positions and did 20782 swaps.
XPL0
<lang XPL0>include c:\cxpl\codes;
int Hole, Max, I; char Box(8), Str; def A, B, C, D, E, F, G, H; [for Hole:= 0 to 7 do Box(Hole):= Hole+1; Max:= 7; while abs(Box(D)-Box(A)) < 2 or abs(Box(D)-Box(C)) < 2 or
abs(Box(D)-Box(G)) < 2 or abs(Box(D)-Box(E)) < 2 or
abs(Box(A)-Box(C)) < 2 or abs(Box(C)-Box(G)) < 2 or
abs(Box(G)-Box(E)) < 2 or abs(Box(E)-Box(A)) < 2 or
abs(Box(E)-Box(B)) < 2 or abs(Box(E)-Box(H)) < 2 or
abs(Box(E)-Box(F)) < 2 or abs(Box(B)-Box(D)) < 2 or
abs(Box(D)-Box(H)) < 2 or abs(Box(H)-Box(F)) < 2 or
abs(Box(F)-Box(B)) < 2 do
loop [I:= Box(0); \next permutation
for Hole:= 0 to Max-1 do Box(Hole):= Box(Hole+1);
Box(Max):= I;
if I # Max+1 then [Max:= 7; quit]
else Max:= Max-1];
Str:= "
# # /|\ /|\ / | X | \ / |/ \| \
- - # - # - #
\ |\ /| / \ | X | / \|/ \|/ # #
"; Hole:= 0; I:= 0; repeat if Str(I)=^# then [Str(I):= Box(Hole)+^0; Hole:= Hole+1];
I:= I+1;
until Hole = 8; Text(0, Str); ]</lang>
- Output:
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4
zkl
<lang zkl>const PegA=0, PegB=1, PegC=2, PegD=3, PegE=4, PegF=5, PegG=6, PegH=7; connections:=T(
T(PegA, PegC), T(PegA, PegD), T(PegA, PegE), T(PegB, PegD), T(PegB, PegE), T(PegB, PegF), T(PegC, PegD), T(PegD, PegE), T(PegE, PegF), T(PegG, PegC), T(PegG, PegD), T(PegG, PegE), T(PegH, PegD), T(PegH, PegE), T(PegH, PegF) );
CZ:=connections.len();
- <<< // 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
}</lang> The filter1 method stops on the first True, so it acts like a conditional or.
- Output:
5 6
/|\ /|\
/ | X | \
/ |/ \| \
7 - 1 - 8 - 2
\ |\ /| /
\ | X | /
\|/ \|/
3 4