4-rings or 4-squares puzzle

From Rosetta Code


Task
4-rings or 4-squares puzzle
You are encouraged to solve this task according to the task description, using any language you may know.
Task

Replace       a, b, c, d, e, f,   and   g       with the decimal digits   LOW   ───►   HIGH
such that the sum of the letters inside of each of the four large squares add up to the same sum.

            ╔══════════════╗      ╔══════════════╗
            ║              ║      ║              ║
            ║      a       ║      ║      e       ║
            ║              ║      ║              ║
            ║          ┌───╫──────╫───┐      ┌───╫─────────┐
            ║          │   ║      ║   │      │   ║         │
            ║          │ b ║      ║ d │      │ f ║         │
            ║          │   ║      ║   │      │   ║         │
            ║          │   ║      ║   │      │   ║         │
            ╚══════════╪═══╝      ╚═══╪══════╪═══╝         │
                       │       c      │      │      g      │
                       │              │      │             │
                       │              │      │             │
                       └──────────────┘      └─────────────┘

Show all output here.


  •   Show all solutions for each letter being unique with
        LOW=1     HIGH=7
  •   Show all solutions for each letter being unique with
        LOW=3     HIGH=9
  •   Show only the   number   of solutions when each letter can be non-unique
        LOW=0     HIGH=9


Related task



Ada[edit]

with Ada.Text_IO;
 
procedure Puzzle_Square_4 is
 
procedure Four_Rings (Low, High : in Natural; Unique, Show : in Boolean) is
subtype Test_Range is Natural range Low .. High;
 
type Value_List is array (Positive range <>) of Natural;
function Is_Unique (Values : Value_List) return Boolean is
Count : array (Test_Range) of Natural := (others => 0);
begin
for Value of Values loop
Count (Value) := Count (Value) + 1;
if Count (Value) > 1 then
return False;
end if;
end loop;
return True;
end Is_Unique;
 
function Is_Valid (A,B,C,D,E,F,G : in Natural) return Boolean is
Ring_1 : constant Integer := A + B;
Ring_2 : constant Integer := B + C + D;
Ring_3 : constant Integer := D + E + F;
Ring_4 : constant Integer := F + G;
begin
return
Ring_1 = Ring_2 and
Ring_1 = Ring_3 and
Ring_1 = Ring_4;
end Is_Valid;
 
use Ada.Text_IO;
Count : Natural := 0;
begin
for A in Test_Range loop
for B in Test_Range loop
for C in Test_Range loop
for D in Test_Range loop
for E in Test_Range loop
for F in Test_Range loop
for G in Test_Range loop
if Is_Valid (A,B,C,D,E,F,G) then
if not Unique or (Unique and Is_Unique ((A,B,C,D,E,F,G))) then
Count := Count + 1;
if Show then
Put_Line (A'Image & B'Image & C'Image & D'Image & E'Image & F'Image & G'Image);
end if;
end if;
end if;
end loop;
end loop;
end loop;
end loop;
end loop;
end loop;
end loop;
Put_Line ("There are " & Count'Image &
(if Unique then " unique " else " non-unique ") &
"solutions in " & Low'Image & " .." & High'Image);
New_Line;
end Four_Rings;
 
begin
Four_Rings (Low => 1, High => 7, Unique => True, Show => True);
Four_Rings (Low => 3, High => 9, Unique => True, Show => True);
Four_Rings (Low => 0, High => 9, Unique => False, Show => False);
end Puzzle_Square_4;
Output:
 3 7 2 1 5 4 6
 4 5 3 1 6 2 7
 4 7 1 3 2 6 5
 5 6 2 3 1 7 4
 6 4 1 5 2 3 7
 6 4 5 1 2 7 3
 7 2 6 1 3 5 4
 7 3 2 5 1 4 6
There are  8 unique solutions in  1 .. 7

 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7
There are  4 unique solutions in  3 .. 9

There are  2860 non-unique solutions in  0 .. 9

ALGOL 68[edit]

As with the REXX solution, we use explicit loops to generate the permutations.

BEGIN
# solve the 4 rings or 4 squares puzzle #
# we need to find solutions to the equations: a + b = b + c + d = d + e + f = f + g #
# where a, b, c, d, e, f, g in lo : hi ( not necessarily unique ) #
# depending on show, the solutions will be printed or not #
PROC four rings = ( INT lo, hi, BOOL unique, show )VOID:
BEGIN
INT solutions := 0;
BOOL allow duplicates = NOT unique;
# calculate field width for printinhg solutions #
INT width := -1;
INT max := ABS IF ABS lo > ABS hi THEN lo ELSE hi FI;
WHILE max > 0 DO
width -:= 1;
max OVERAB 10
OD;
# find solutions #
FOR a FROM lo TO hi DO
FOR b FROM lo TO hi DO
IF allow duplicates OR a /= b THEN
INT t = a + b;
FOR c FROM lo TO hi DO
IF allow duplicates OR ( a /= c AND b /= c ) THEN
FOR d FROM lo TO hi DO
IF allow duplicates OR ( a /= d AND b /= d AND c /= d )
THEN
IF b + c + d = t THEN
FOR e FROM lo TO hi DO
IF allow duplicates
OR ( a /= e AND b /= e AND c /= e AND d /= e )
THEN
FOR f FROM lo TO hi DO
IF allow duplicates
OR ( a /= f AND b /= f AND c /= f AND d /= f AND e /= f )
THEN
IF d + e + f = t THEN
FOR g FROM lo TO hi DO
IF allow duplicates
OR ( a /= g AND b /= g AND c /= g AND d /= g AND e /= g AND f /= g )
THEN
IF f + g = t THEN
solutions +:= 1;
IF show THEN
print( ( whole( a, width ), whole( b, width )
, whole( c, width ), whole( d, width )
, whole( e, width ), whole( f, width )
, whole( g, width ), newline
)
)
FI
FI
FI
OD # g #
FI
FI
OD # f #
FI
OD # e #
FI
FI
OD # d #
FI
OD # c #
FI
OD # b #
OD # a # ;
print( ( whole( solutions, 0 )
, IF unique THEN " unique" ELSE " non-unique" FI
, " solutions in "
, whole( lo, 0 )
, " to "
, whole( hi, 0 )
, newline
, newline
)
)
END # four rings # ;
 
# find the solutions as required for the task #
four rings( 1, 7, TRUE, TRUE );
four rings( 3, 9, TRUE, TRUE );
four rings( 0, 9, FALSE, FALSE )
END
Output:
 3 7 2 1 5 4 6
 4 5 3 1 6 2 7
 4 7 1 3 2 6 5
 5 6 2 3 1 7 4
 6 4 1 5 2 3 7
 6 4 5 1 2 7 3
 7 2 6 1 3 5 4
 7 3 2 5 1 4 6
8 unique solutions in 1 to 7

 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7
4 unique solutions in 3 to 9

2860 non-unique solutions in 0 to 9

AppleScript[edit]

Translation of: JavaScript
Translation of: Haskell

(Structured search example)

use framework "Foundation" -- for basic NSArray sort
 
on run
unlines({"rings(true, enumFromTo(1, 7))\n", ¬
map(show, (rings(true, enumFromTo(1, 7)))), ¬
"\nrings(true, enumFromTo(3, 9))\n", ¬
map(show, (rings(true, enumFromTo(3, 9)))), ¬
"\nlength(rings(false, enumFromTo(0, 9)))\n", ¬
show(|length|(rings(false, enumFromTo(0, 9))))})
end run
 
-- RINGS -----------------------------------------------------------------------
 
-- rings :: noRepeatedDigits -> DigitList -> Lists of solutions
-- rings :: Bool -> [Int] -> [[Int]]
on rings(u, digits)
set ds to reverse_(sort(digits))
set h to head(ds)
 
-- QUEEN -------------------------------------------------------------------
script queen
on |λ|(q)
script
on |λ|(x)
x + q ≤ h
end |λ|
end script
set ts to filter(result, ds)
if u then
set bs to delete_(q, ts)
else
set bs to ds
end if
 
-- LEFT BISHOP and its ROOK-----------------------------------------
script leftBishop
on |λ|(lb)
set lRook to lb + q
if lRook > h then
{}
else
if u then
set rbs to difference(ts, {q, lb, lRook})
else
set rbs to ds
end if
 
-- RIGHT BISHOP and its ROOK ---------------------------
script rightBishop
on |λ|(rb)
set rRook to rb + q
if (rRook > h) or (u and (rRook = lb)) then
{}
else
set rookDelta to lRook - rRook
if u then
set ks to difference(ds, ¬
{q, lb, rb, rRook, lRook})
else
set ks to ds
end if
 
-- KNIGHTS LEFT AND RIGHT ------------------
script knights
on |λ|(k)
set k2 to k + rookDelta
 
if elem(k2, ks) and ((not u) or ¬
notElem(k2, ¬
{lRook, k, lb, q, rb, rRook})) then
{{lRook, k, lb, q, rb, k2, rRook}}
else
{}
end if
end |λ|
end script
 
concatMap(knights, ks)
end if
end |λ|
end script
 
concatMap(rightBishop, rbs)
end if
end |λ|
end script
 
concatMap(leftBishop, bs)
end |λ|
end script
 
concatMap(queen, ds)
end rings
 
-- GENERIC FUNCTIONS -----------------------------------------------------------
 
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lst to {}
set lng to length of xs
tell mReturn(f)
repeat with i from 1 to lng
set lst to (lst & |λ|(contents of item i of xs, i, xs))
end repeat
end tell
return lst
end concatMap
 
-- delete :: Eq a => a -> [a] -> [a]
on delete_(x, xs)
set mbIndex to elemIndex(x, xs)
set lng to length of xs
 
if mbIndex is not missing value then
if lng > 1 then
if mbIndex = 1 then
items 2 thru -1 of xs
else if mbIndex = lng then
items 1 thru -2 of xs
else
tell xs to items 1 thru (mbIndex - 1) & ¬
items (mbIndex + 1) thru -1
end if
else
{}
end if
else
xs
end if
end delete_
 
-- difference :: [a] -> [a] -> [a]
on difference(xs, ys)
script mf
on except(a, y)
if a contains y then
my delete_(y, a)
else
a
end if
end except
end script
 
foldl(except of mf, xs, ys)
end difference
 
-- elem :: Eq a => a -> [a] -> Bool
on elem(x, xs)
xs contains x
end elem
 
-- elemIndex :: a -> [a] -> Maybe Int
on elemIndex(x, xs)
set lng to length of xs
repeat with i from 1 to lng
if x = (item i of xs) then return i
end repeat
return missing value
end elemIndex
 
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if n < m then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
 
-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
 
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
 
-- head :: [a] -> a
on head(xs)
if length of xs > 0 then
item 1 of xs
else
missing value
end if
end head
 
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
 
-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
-- notElem :: Eq a => a -> [a] -> Bool
on notElem(x, xs)
xs does not contain x
end notElem
 
-- reverse_ :: [a] -> [a]
on |reverse|:xs
if class of xs is text then
(reverse of characters of xs) as text
else
reverse of xs
end if
end |reverse|:
 
-- show :: a -> String
on show(e)
set c to class of e
if c = list then
script serialized
on |λ|(v)
show(v)
end |λ|
end script
 
"[" & intercalate(", ", map(serialized, e)) & "]"
else if c = record then
script showField
on |λ|(kv)
set {k, ev} to kv
"\"" & k & "\":" & show(ev)
end |λ|
end script
 
"{" & intercalate(", ", ¬
map(showField, zip(allKeys(e), allValues(e)))) & "}"
else if c = date then
"\"" & iso8601Z(e) & "\""
else if c = text then
"\"" & e & "\""
else if (c = integer or c = real) then
e as text
else if c = class then
"null"
else
try
e as text
on error
("«" & c as text) & "»"
end try
end if
end show
 
-- sort :: [a] -> [a]
on sort(xs)
((current application's NSArray's arrayWithArray:xs)'s ¬
sortedArrayUsingSelector:"compare:") as list
end sort
 
-- unlines :: [String] -> String
on unlines(xs)
intercalate(linefeed, xs)
end unlines
Output:
rings(true, enumFromTo(1, 7))

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

rings(true, enumFromTo(3, 9))

[9, 6, 4, 5, 3, 7, 8]
[8, 7, 3, 5, 4, 6, 9]
[9, 6, 5, 4, 3, 8, 7]
[7, 8, 3, 4, 5, 6, 9]

length(rings(false, enumFromTo(0, 9)))

2860

ARM Assembly[edit]

Works with: as version Raspberry Pi
 
 
/* ARM assembly Raspberry PI */
/* program square4.s */
 
/************************************/
/* Constantes */
/************************************/
.equ STDOUT, 1 @ Linux output console
.equ EXIT, 1 @ Linux syscall
.equ WRITE, 4 @ Linux syscall
 
.equ NBBOX, 7
 
/*********************************/
/* 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
 
szCarriageReturn: .asciz "\n************************\n"
 
sMessNbSolution: .ascii "Number of solutions :"
sMessCounter: .fill 11, 1, ' ' @ size => 11
.asciz "\n\n\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
iCounterSol: .skip 4
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r0,#1
mov r1,#7
mov r2,#3 @ 0 = rien 1 = display 2 = count 3 = les deux
bl searchPb
mov r0,#3
mov r1,#9
mov r2,#3 @ 0 = rien 1 = display 2 = count 3 = les deux
bl searchPb
mov r0,#0
mov r1,#9
mov r2,#2 @ 0 = rien 1 = display 2 = count 3 = les deux
bl prepSearchNU
 
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 problèm value not unique */
/******************************************************************/
/* r0 contains start digit */
/* r1 contains end digit */
/* r2 contains action (0 display 1 count) */
prepSearchNU:
push {r3-r12,lr} @ save registers
mov r5,#0 @ counter
mov r12,r0 @ a
1:
mov r11,r0 @ b
2:
mov r10,r0 @ c
3:
mov r9,r0 @ d
4:
add r4,r12,r11 @ a + b reference
add r3,r11,r10
add r3,r9 @ b + c + d
cmp r4,r3
bne 10f
mov r8,r0 @ e
5:
mov r7,r0 @ f
6:
add r3,r9,r8
add r3,r7 @ d + e + f
cmp r3,r4
bne 9f
mov r6,r0 @ g
7:
add r3,r7,r6 @ f + g
cmp r3,r4
bne 8f @ not OK
@ OK
add r5,#1 @ increment counter
 
8:
add r6,#1 @ increment g
cmp r6,r1
ble 7b
9:
add r7,#1 @ increment f
cmp r7,r1
ble 6b
add r8,#1 @ increment e
cmp r8,r1
ble 5b
10:
add r9,#1 @ increment d
cmp r9,r1
ble 4b
add r10,#1 @ increment c
cmp r10,r1
ble 3b
add r11,#1 @ increment b
cmp r11,r1
ble 2b
add r12,#1 @ increment a
cmp r12,r1
ble 1b
 
@ end
tst r2,#0b10 @ print count ?
beq 100f
mov r0,r5 @ counter
ldr r1,iAdrsMessCounter
bl conversion10
ldr r0,iAdrsMessNbSolution
bl affichageMess
 
100:
pop {r3-r12,lr} @ restaur registers
bx lr @return
iAdrsMessCounter: .int sMessCounter
iAdrsMessNbSolution: .int sMessNbSolution
 
/******************************************************************/
/* search problem unique solution */
/******************************************************************/
/* r0 contains start digit */
/* r1 contains end digit */
/* r2 contains action (0 display 1 count) */
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 r5,#0 @ solution counter
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_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 2 firsts squares
ldr r0,iAdriValues_a
ldr r0,[r0,r12,lsl #2]
ldr r1,iAdriValues_b
ldr r1,[r1,r11,lsl #2]
add r4,r0,r1 @ a + b value first square
ldr r0,iAdriValues_c
ldr r0,[r0,r10,lsl #2]
add r7,r1,r0 @ b + c
ldr r1,iAdriValues_d
ldr r1,[r1,r9,lsl #2]
add r7,r1 @ b + c + d
cmp r7,r4 @ equal first square ?
bne 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
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
ldr r0,iAdriValues_d
ldr r0,[r0,r9,lsl #2]
ldr r1,iAdriValues_e
ldr r1,[r1,r8,lsl #2]
add r3,r0,r1 @ d + e
ldr r1,iAdriValues_f
ldr r1,[r1,r7,lsl #2]
add r3,r1 @ de + e + f
cmp r3,r4 @ equal first square ?
bne 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
ldr r0,iAdriValues_f
ldr r0,[r0,r7,lsl #2]
ldr r1,iAdriValues_g
ldr r1,[r1,r6,lsl #2]
add r3,r0,r1 @ f +g
cmp r4,r3 @ equal first square ?
bne 8b
add r5,#1 @ increment counter
ldr r0,[sp,#8] @ load action for two parameter in stack
tst r0,#0b1
beq 9f @ display solution ?
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,iAdrsMessDeb
bl affichageMess
9:
b 8b @ suite
 
90:
ldr r0,[sp,#8] @ load action for two parameter in stack
tst r0,#0b10
beq 100f @ display counter ?
mov r0,r5
ldr r1,iAdrsMessCounter
bl conversion10
ldr r0,iAdrsMessNbSolution
bl affichageMess
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
 
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
iAdrsMessDeb: .int sMessDeb
iAdriCounterSol: .int iCounterSol
/******************************************************************/
/* 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
 
 
Output:
a=3          b=7          c=2          d=1
e=5          f=4          g=6
************************
a=4          b=5          c=3          d=1
e=6          f=2          g=7
************************
a=4          b=7          c=1          d=3
e=2          f=6          g=5
************************
a=5          b=6          c=2          d=3
e=1          f=7          g=4
************************
a=6          b=4          c=1          d=5
e=2          f=3          g=7
************************
a=6          b=4          c=5          d=1
e=2          f=7          g=3
************************
a=7          b=2          c=6          d=1
e=3          f=5          g=4
************************
a=7          b=3          c=2          d=5
e=1          f=4          g=6
************************
Number of solutions :8

a=7          b=8          c=3          d=4
e=5          f=6          g=9
************************
a=8          b=7          c=3          d=5
e=4          f=6          g=9
************************
a=9          b=6          c=4          d=5
e=3          f=7          g=8
************************
a=9          b=6          c=5          d=4
e=3          f=8          g=7
************************
Number of solutions :4

Number of solutions :2860

AWK[edit]

 
# syntax: GAWK -f 4-RINGS_OR_4-SQUARES_PUZZLE.AWK
# converted from C
BEGIN {
cmd = "SORT /+16"
four_squares(1,7,1,1)
four_squares(3,9,1,1)
four_squares(0,9,0,0)
four_squares(0,6,1,0)
four_squares(2,8,1,0)
exit(0)
}
function four_squares(plo,phi,punique,pshow) {
lo = plo
hi = phi
unique = punique
show = pshow
solutions = 0
print("")
if (show) {
print("A B C D E F G sum A+B B+C+D D+E+F F+G")
print("------------- --- -------------------")
}
acd()
close(cmd)
tmp = (unique) ? "unique" : "non-unique"
printf("%d-%d: %d %s solutions\n",lo,hi,solutions,tmp)
}
function acd() {
for (c=lo; c<=hi; c++) {
for (d=lo; d<=hi; d++) {
if (!unique || c != d) {
a = c + d
if (a >= lo && a <= hi && (!unique || (c != 0 && d != 0))) {
ge()
}
}
}
}
}
function bf() {
for (f=lo; f<=hi; f++) {
if (!unique || (f != a && f != c && f != d && f != g && f != e)) {
b = e + f - c
if (b >= lo && b <= hi && (!unique || (b != a && b != c && b != d && b != g && b != e && b != f))) {
solutions++
if (show) {
printf("%d %d %d %d %d %d %d %4d ",a,b,c,d,e,f,g,a+b) | cmd
printf("%d+%d ",a,b) | cmd
printf("%d+%d+%d ",b,c,d) | cmd
printf("%d+%d+%d ",d,e,f) | cmd
printf("%d+%d\n",f,g) | cmd
}
}
}
}
}
function ge() {
for (e=lo; e<=hi; e++) {
if (!unique || (e != a && e != c && e != d)) {
g = d + e
if (g >= lo && g <= hi && (!unique || (g != a && g != c && g != d && g != e))) {
bf()
}
}
}
}
 
Output:
A B C D E F G  sum  A+B B+C+D D+E+F F+G
-------------  ---  -------------------
4 5 3 1 6 2 7    9  4+5 5+3+1 1+6+2 2+7
7 2 6 1 3 5 4    9  7+2 2+6+1 1+3+5 5+4
3 7 2 1 5 4 6   10  3+7 7+2+1 1+5+4 4+6
6 4 1 5 2 3 7   10  6+4 4+1+5 5+2+3 3+7
6 4 5 1 2 7 3   10  6+4 4+5+1 1+2+7 7+3
7 3 2 5 1 4 6   10  7+3 3+2+5 5+1+4 4+6
4 7 1 3 2 6 5   11  4+7 7+1+3 3+2+6 6+5
5 6 2 3 1 7 4   11  5+6 6+2+3 3+1+7 7+4
1-7: 8 unique solutions

A B C D E F G  sum  A+B B+C+D D+E+F F+G
-------------  ---  -------------------
7 8 3 4 5 6 9   15  7+8 8+3+4 4+5+6 6+9
8 7 3 5 4 6 9   15  8+7 7+3+5 5+4+6 6+9
9 6 4 5 3 7 8   15  9+6 6+4+5 5+3+7 7+8
9 6 5 4 3 8 7   15  9+6 6+5+4 4+3+8 8+7
3-9: 4 unique solutions

0-9: 2860 non-unique solutions

0-6: 4 unique solutions

2-8: 8 unique solutions

Befunge[edit]

This is loosely based on the C algorithm, although many of the conditions have been combined to minimize branching. There is no option to choose whether the results are displayed or not - unique solutions are always displayed, and non-unique solutions just return the solution count.

550" :woL">:#,_&>00p" :hgiH">:#,_&>1+10p" :)n/y( euqinU">:#,_>~>:4v
v!g03!:\*`\g01\!`\g00:p05:+g03:p04:_$30g1+:10g\`v1g<,+$p02%2_|#`*8<
>>+\30g-!+20g*!*00g\#v_$40g1+:10g\`^<<1g00p03<<<_$55+:,\."snoitul"v
v!`\g00::p07:+g04p06:<^<`\g01:+1g06$<_v#!\g00*!*g02++!-g05< v"so"<
>\10g\`*\:::30g-!\40g-!+\50g-!+\60g-! +60g::30g-!\40g-!+\^ >:#,[email protected]
>0g50g.......55+,0vg02+1_80g1+:10g\`!^>>:80p60g+30g-:90p::00g\`!>>v
^9g03g04g06g08g07<_>>0>>^<<*!*g02++!-g07\+!-g06\+!-g05\+!-g04\!-<<\
>>10g\`*\:::::30g-!\40g-!+\50g-!+\60g-!+\70g-!+\80g-!+80g::::30g^^>
Output:
Low: 1
High: 7
Unique (y/n): y

4 7 1 3 2 6 5
6 4 1 5 2 3 7
3 7 2 1 5 4 6
5 6 2 3 1 7 4
7 3 2 5 1 4 6
4 5 3 1 6 2 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4

8 solutions
Low: 3
High: 9
Unique (y/n): y

7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7

4 solutions
Low: 0
High: 9
Unique (y/n): n


2860 solutions

C[edit]

 
#include <stdio.h>
 
#define TRUE 1
#define FALSE 0
 
int a,b,c,d,e,f,g;
int lo,hi,unique,show;
int solutions;
 
void
bf()
{
for (f = lo;f <= hi; f++)
if ((!unique) ||
((f != a) && (f != c) && (f != d) && (f != g) && (f != e)))
{
b = e + f - c;
if ((b >= lo) && (b <= hi) &&
((!unique) || ((b != a) && (b != c) &&
(b != d) && (b != g) && (b != e) && (b != f))))
{
solutions++;
if (show)
printf("%d %d %d %d %d %d %d\n",a,b,c,d,e,f,g);
}
}
}
 
 
void
ge()
{
for (e = lo;e <= hi; e++)
if ((!unique) || ((e != a) && (e != c) && (e != d)))
{
g = d + e;
if ((g >= lo) && (g <= hi) &&
((!unique) || ((g != a) && (g != c) &&
(g != d) && (g != e))))
bf();
}
}
 
void
acd()
{
for (c = lo;c <= hi; c++)
for (d = lo;d <= hi; d++)
if ((!unique) || (c != d))
{
a = c + d;
if ((a >= lo) && (a <= hi) &&
((!unique) || ((c != 0) && (d != 0))))
ge();
}
}
 
 
void
foursquares(int plo,int phi, int punique,int pshow)
{
lo = plo;
hi = phi;
unique = punique;
show = pshow;
solutions = 0;
 
printf("\n");
 
acd();
 
if (unique)
printf("\n%d unique solutions in %d to %d\n",solutions,lo,hi);
else
printf("\n%d non-unique solutions in %d to %d\n",solutions,lo,hi);
}
 
main()
{
foursquares(1,7,TRUE,TRUE);
foursquares(3,9,TRUE,TRUE);
foursquares(0,9,FALSE,FALSE);
}
 

Output

4 7 1 3 2 6 5
6 4 1 5 2 3 7
3 7 2 1 5 4 6
5 6 2 3 1 7 4
7 3 2 5 1 4 6
4 5 3 1 6 2 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4

8 unique solutions in 1 to 7

7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7

4 unique solutions in 3 to 9


2860 non-unique solutions in 0 to 9

C++[edit]

 
//C++14/17
#include <algorithm>//std::for_each
#include <iostream> //std::cout
#include <numeric> //std::iota
#include <vector> //std::vector, save solutions
#include <list> //std::list, for fast erase
 
using std::begin, std::end, std::for_each;
 
//Generates all the valid solutions for the problem in the specified range [from, to)
std::list<std::vector<int>> combinations(int from, int to)
{
if (from > to)
return {}; //Return nothing if limits are invalid
 
auto pool = std::vector<int>(to - from);//Here we'll save our values
std::iota(begin(pool), end(pool), from);//Populates pool
 
auto solutions = std::list<std::vector<int>>{}; //List for the solutions
 
//Brute-force calculation of valid values...
for (auto a : pool)
for (auto b : pool)
for (auto c : pool)
for (auto d : pool)
for (auto e : pool)
for (auto f : pool)
for (auto g : pool)
if ( a == c + d
&& b + c == e + f
&& d + e == g )
solutions.push_back({a, b, c, d, e, f, g});
return solutions;
}
 
//Filter the list generated from "combinations" and return only lists with no repetitions
std::list<std::vector<int>> filter_unique(int from, int to)
{
//Helper lambda to check repetitions:
//If the count is > 1 for an element, there must be a repetition inside the range
auto has_non_unique_values = [](const auto & range, auto target)
{
return std::count( begin(range), end(range), target) > 1;
};
 
//Generates all the solutions...
auto results = combinations(from, to);
 
//For each solution, find duplicates inside
for (auto subrange = cbegin(results); subrange != cend(results); ++subrange)
{
bool repetition = false;
 
//If some element is repeated, repetition becomes true
for (auto x : *subrange)
repetition |= has_non_unique_values(*subrange, x);
 
if (repetition) //If repetition is true, remove the current subrange from the list
{
results.erase(subrange); //Deletes subrange from solutions
--subrange; //Rewind to the last subrange analysed
}
}
 
return results; //Finally return remaining results
}
 
template <class Container> //Template for the sake of simplicity
inline void print_range(const Container & c)
{
for (const auto & subrange : c)
{
std::cout << "[";
for (auto elem : subrange)
std::cout << elem << ' ';
std::cout << "\b]\n";
}
}
 
 
int main()
{
std::cout << "Unique-numbers combinations in range 1-7:\n";
auto solution1 = filter_unique(1, 8);
print_range(solution1);
std::cout << "\nUnique-numbers combinations in range 3-9:\n";
auto solution2 = filter_unique(3,10);
print_range(solution2);
std::cout << "\nNumber of combinations in range 0-9: "
<< combinations(0, 10).size() << "." << std::endl;
 
return 0;
}
 

Output

Unique-numbers combinations in range 1-7:
[3 7 2 1 5 4 6]
[4 5 3 1 6 2 7]
[4 7 1 3 2 6 5]
[5 6 2 3 1 7 4]
[6 4 1 5 2 3 7]
[6 4 5 1 2 7 3]
[7 2 6 1 3 5 4]
[7 3 2 5 1 4 6]

Unique-numbers combinations in range 3-9:
[7 8 3 4 5 6 9]
[8 7 3 5 4 6 9]
[9 6 4 5 3 7 8]
[9 6 5 4 3 8 7]

Number of combinations in range 0-9: 2860.

C#[edit]

Translation of: Java
using System;
using System.Linq;
 
namespace Four_Squares_Puzzle {
class Program {
static void Main(string[] args) {
fourSquare(1, 7, true, true);
fourSquare(3, 9, true, true);
fourSquare(0, 9, false, false);
}
 
private static void fourSquare(int low, int high, bool unique, bool print) {
int count = 0;
 
if (print) {
Console.WriteLine("a b c d e f g");
}
for (int a = low; a <= high; ++a) {
for (int b = low; b <= high; ++b) {
if (notValid(unique, b, a)) continue;
 
int fp = a + b;
for (int c = low; c <= high; ++c) {
if (notValid(unique, c, b, a)) continue;
for (int d = low; d <= high; ++d) {
if (notValid(unique, d, c, b, a)) continue;
if (fp != b + c + d) continue;
 
for (int e = low; e <= high; ++e) {
if (notValid(unique, e, d, c, b, a)) continue;
for (int f = low; f <= high; ++f) {
if (notValid(unique, f, e, d, c, b, a)) continue;
if (fp != d + e + f) continue;
 
for (int g = low; g <= high; ++g) {
if (notValid(unique, g, f, e, d, c, b, a)) continue;
if (fp != f + g) continue;
 
++count;
if (print) {
Console.WriteLine("{0} {1} {2} {3} {4} {5} {6}", a, b, c, d, e, f, g);
}
}
}
}
}
}
}
}
if (unique) {
Console.WriteLine("There are {0} unique solutions in [{1}, {2}]", count, low, high);
}
else {
Console.WriteLine("There are {0} non-unique solutions in [{1}, {2}]", count, low, high);
}
}
 
private static bool notValid(bool unique, int needle, params int[] haystack) {
return unique && haystack.Any(p => p == needle);
}
}
}
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1, 7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3, 9]
There are 2860 non-unique solutions in [0, 9]

Clojure[edit]

(use '[clojure.math.combinatorics]
 
(defn rings [r & {:keys [unique] :or {unique true}}]
(if unique
(apply concat (map permutations (combinations r 7)))
(selections r 7)))
 
(defn four-rings [low high & {:keys [unique] :or {unique true}}]
(for [[a b c d e f g] (rings (range low (inc high)) :unique unique)
 :when (= (+ a b) (+ b c d) (+ d e f) (+ f g))] [a b c d e f g]))
 
Output:
=> (pprint (four-rings 1 7))
([3 7 2 1 5 4 6]
 [4 5 3 1 6 2 7]
 [4 7 1 3 2 6 5]
 [5 6 2 3 1 7 4]
 [6 4 1 5 2 3 7]
 [6 4 5 1 2 7 3]
 [7 2 6 1 3 5 4]
 [7 3 2 5 1 4 6])
nil

=> (pprint (four-rings 3 9))
([7 8 3 4 5 6 9] [8 7 3 5 4 6 9] [9 6 4 5 3 7 8] [9 6 5 4 3 8 7])
nil

=> (count (four-rings 0 9 :unique false))
2860

Common Lisp[edit]

 
(defpackage four-rings
(:use common-lisp)
(:export display-solutions))
(in-package four-rings)
 
(defun correct-answer-p (a b c d e f g)
(let ((v (+ a b)))
(and (equal v (+ b c d))
(equal v (+ d e f))
(equal v (+ f g)))))
 
(defun combinations-if (func len unique min max)
(let ((results nil))
(labels ((inner (cur)
(if (eql (length cur) len)
(when (apply func (reverse cur))
(push cur results))
(dotimes (i (- max min))
(when (or (not unique)
(not (member (+ i min) cur)))
(inner (append (list (+ i min)) cur)))))))
(inner nil))
results))
 
(defun four-rings-solutions (low high unique)
(combinations-if #'correct-answer-p 7 unique low (1+ high)))
 
(defun display-solutions ()
(let ((letters '((a b c d e f g))))
(format t "Low 1, High 7, unique letters: ~%~{~{~3A~}~%~}~%"
(append letters (four-rings-solutions 1 7 t)))
(format t "Low 3, High 9, unique letters: ~%~{~{~3A~}~%~}~%"
(append letters (four-rings-solutions 3 9 t)))
(format t "Number of solutions for Low 0, High 9 non-unique:~%~A~%"
(length (four-rings-solutions 0 9 nil)))))
 

Output:

CL-USER> (four-rings:display-solutions)
Low 1, High 7, unique letters: 
A  B  C  D  E  F  G  
6  4  1  5  2  3  7  
4  5  3  1  6  2  7  
3  7  2  1  5  4  6  
7  3  2  5  1  4  6  
4  7  1  3  2  6  5  
5  6  2  3  1  7  4  
7  2  6  1  3  5  4  
6  4  5  1  2  7  3  

Low 3, High 9, unique letters: 
A  B  C  D  E  F  G  
7  8  3  4  5  6  9  
8  7  3  5  4  6  9  
9  6  4  5  3  7  8  
9  6  5  4  3  8  7  

Number of solutions for Low 0, High 9 non-unique:
2860
NIL

Crystal[edit]

Translation of: Ruby
def check(list)
a, b, c, d, e, f, g = list
first = a + b
{b + c + d, d + e + f, f + g}.all? &.==(first)
end
 
def four_squares(low, high, unique = true, show = unique)
solutions = [] of Array(Int32)
if unique
uniq = "unique"
(low..high).to_a.each_permutation(7, true) { |ary| solutions << ary.clone if check(ary) }
else
uniq = "non-unique"
(low..high).to_a.each_repeated_permutation(7, true) { |ary| solutions << ary.clone if check(ary) }
end
if show
puts " " + ("a".."g").join(" ")
solutions.each { |ary| p ary }
end
puts "#{solutions.size} #{uniq} solutions in #{low} to #{high}"
puts
end
 
{ {1, 7}, {3, 9} }.each do |(low, high)|
four_squares(low, high)
end
four_squares(0, 9, false)

D[edit]

import std.stdio;
 
void main() {
fourSquare(1,7,true,true);
fourSquare(3,9,true,true);
fourSquare(0,9,false,false);
}
 
void fourSquare(int low, int high, bool unique, bool print) {
int count;
 
if (print) {
writeln("a b c d e f g");
}
for (int a=low; a<=high; ++a) {
for (int b=low; b<=high; ++b) {
if (!valid(unique, a, b)) continue;
 
int fp = a+b;
for (int c=low; c<=high; ++c) {
if (!valid(unique, c, a, b)) continue;
for (int d=low; d<=high; ++d) {
if (!valid(unique, d, a, b, c)) continue;
if (fp != b+c+d) continue;
 
for (int e=low; e<=high; ++e) {
if (!valid(unique, e, a, b, c, d)) continue;
for (int f=low; f<=high; ++f) {
if (!valid(unique, f, a, b, c, d, e)) continue;
if (fp != d+e+f) continue;
 
for (int g=low; g<=high; ++g) {
if (!valid(unique, g, a, b, c, d, e, f)) continue;
if (fp != f+g) continue;
 
++count;
if (print) {
writeln(a,' ',b,' ',c,' ',d,' ',e,' ',f,' ',g);
}
}
}
}
}
}
}
}
if (unique) {
writeln("There are ", count, " unique solutions in [",low,",",high,"]");
} else {
writeln("There are ", count, " non-unique solutions in [",low,",",high,"]");
}
}
 
bool valid(bool unique, int needle, int[] haystack ...) {
if (unique) {
foreach (value; haystack) {
if (needle == value) {
return false;
}
}
}
return true;
}
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1,7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3,9]
There are 2860 non-unique solutions in [0,9]

F#[edit]

 
(* A simple function to generate the sequence
Nigel Galloway: January 31st., 2017 *)

type G = {d:int;x:int;b:int;f:int}
let N n g =
{(max (n-g) n) .. (min (g-n) g)} |> Seq.collect(fun d->{(max (d+n+n) (n+n))..(min (g+g) (d+g+g))} |> Seq.collect(fun x ->
seq{for a in n .. g do for b in n .. g do if (a+b) = x then for c in n .. g do if (b+c+d) = x then yield b} |> Seq.collect(fun b ->
seq{for f in n .. g do for G in n .. g do if (f+G) = x then for e in n .. g do if (f+e+d) = x then yield f} |> Seq.map(fun f -> {d=d;x=x;b=b;f=f}))))
 

Then:

 
printfn "%d" (Seq.length (N 0 9))
 
Output:
2860
 
(* A simple function to generate the sequence with unique values
Nigel Galloway: January 31st., 2017 *)

type G = {d:int;x:int;b:int;f:int}
let N n g =
{(max (n-g) n) .. (min (g-n) g)} |> Seq.filter(fun d -> d <> 0) |> Seq.collect(fun d->{(max (d+n+n) (n+n)) .. (min (g+g) (d+g+g))} |> Seq.collect(fun x ->
seq{for a in n .. g do if a <> d then for b in n .. g do if (a+b) = x && b <> a && b <> d then for c in n .. g do if (b+c+d) = x && c <> d && c <> a && c <> b then yield b} |> Seq.collect(fun b ->
seq{for f in n .. g do if f <> d && f <> b && f <> (x-b) && f <> (x-d-b) then for G in n .. g do if (f+G) = x && G <> d && G <> b && G <> f && G <> (x-b) && G <> (x-d-b) then for e in n .. g do if (f+e+d) = x && e <> d && e <> b && e <> f && e <> G && e <> (x-b) && e <> (x-d-b) then yield f} |> Seq.map(fun f -> {d=d;x=x;b=b;f=f}))))
 

Then:

 
for n in N 1 7 do printfn "%d,%d,%d,%d,%d,%d,%d" (n.x-n.b) n.b (n.x-n.d-n.b) n.d (n.x-n.d-n.f) n.f (n.x-n.f)
 
Output:
4,5,3,1,6,2,7
7,2,6,1,3,5,4
3,7,2,1,5,4,6
6,4,5,1,2,7,3
4,7,1,3,2,6,5
5,6,2,3,1,7,4
6,4,1,5,2,3,7
7,3,2,5,1,4,6

and:

 
for n in N 3 9 do printfn "%d,%d,%d,%d,%d,%d,%d" (n.x-n.b) n.b (n.x-n.d-n.b) n.d (n.x-n.d-n.f) n.f (n.x-n.f)
 
Output:
7,8,3,4,5,6,9
9,6,5,4,3,8,7
8,7,3,5,4,6,9
9,6,4,5,3,7,8

Factor[edit]

This solution uses the backtrack vocabulary — Factor's implementation of John McCarthy's ambiguous operator. In short, we define 7 integers that can take up any value within the range that we give it, such as [3,9], and assign them names a-g. We then test whether the four sums from the puzzle are equal, and if applicable, whether a-g are unique. We send this boolean value to must-be-true and if it's false, then the other possibilities will be explored through the power of continuations.

bag-of is a combinator (higher-order function) that yields every solution in a collection. If we had written 4-rings without using bag-of, it would have returned only the first solution it found.

USING: arrays backtrack formatting grouping kernel locals math
math.ranges prettyprint sequences sequences.generalizations
sets ;
IN: rosetta-code.4-rings
 
:: 4-rings ( lo hi unique? -- seq ) [
7 [ lo hi [a,b] amb-lazy ] replicate
7 firstn :> ( a b c d e f g )
{ a b c d e f g } :> p
a b +
b c d + +
d e f + +
f g +
4array all-equal?
unique? [ p all-unique? and ] when
must-be-true p
] bag-of ;
 
: report ( lo hi unique? -- )
3dup 4-rings over [ dup . ] when length swap "" "non-" ?
"In [%d, %d] there are %d %sunique solutions.\n" printf ;
 
1 7 t report
3 9 t report
0 9 f report
Output:
V{
    { 3 7 2 1 5 4 6 }
    { 4 5 3 1 6 2 7 }
    { 4 7 1 3 2 6 5 }
    { 5 6 2 3 1 7 4 }
    { 6 4 1 5 2 3 7 }
    { 6 4 5 1 2 7 3 }
    { 7 2 6 1 3 5 4 }
    { 7 3 2 5 1 4 6 }
}
In [1, 7] there are 8 unique solutions.
V{
    { 7 8 3 4 5 6 9 }
    { 8 7 3 5 4 6 9 }
    { 9 6 4 5 3 7 8 }
    { 9 6 5 4 3 8 7 }
}
In [3, 9] there are 4 unique solutions.
In [0, 9] there are 2860 non-unique solutions.

Fortran[edit]

This uses the facility standardised in F90 whereby DO-loops can have text labels attached (not in the usual label area) so that the END DO statement can have the corresponding label, and any CYCLE statements can use it also. Similarly, the subroutine's END statement bears the name of the subroutine. This is just syntactic decoration. Rather more useful is extended syntax for dealing with arrays and especially the function ANY for making multiple tests without having to enumerate them in the code. To gain this convenience, the EQUIVALENCE statement makes variables A, B, C, D, E, F, and G occupy the same storage as INTEGER V(7), an array.

One could abandon the use of the named variables in favour of manipulating the array equivalent, and indeed develop code which performs the nested loops via messing with the array, but for simplicity, the individual variables are used. However, tempting though it is to write a systematic sequence of seven nested DO-loops, the variables are not in fact all independent: some are fixed once others are chosen. Just cycling through all the notional possibilities when one only is in fact possible is a bit too much brute-force-and-ignorance, though other problems with other constraints, may encourage such exhaustive stepping. As a result, the code is more tightly bound to the specific features of the problem.

Also standardised in F90 is the $ format code, which specifies that the output line is not to end with the WRITE statement. The problem here is that Fortran does not offer an IF ...FI bracketing construction inside an expression, that would allow something like
WRITE(...) FIRST,LAST,IF (UNIQUE) THEN "Distinct values only" ELSE "Repeated values allowed" FI // "."
so that the correct alternative will be selected. Further, an array (that would hold those two texts) can't be indexed by a LOGICAL variable, and playing with EQUIVALENCE won't help, because the numerical values revealed thereby for .TRUE. and .FALSE. may not be 1 and 0. And anyway, parameters are not allowed to be accessed via EQUIVALENCE to another variable. So, a two-part output, and to reduce the blather, two IF-statements.
      SUBROUTINE FOURSHOW(FIRST,LAST,UNIQUE)	!The "Four Rings" or "Four Squares" puzzle.
Choose values such that A+B = B+C+D = D+E+F = F+G, all being integers in FIRST:LAST...
INTEGER FIRST,LAST !The range of allowed values.
LOGICAL UNIQUE !Solutions need not have unique values.
INTEGER A,B,C,D,E,F,G !Ah, Diophantus of Alexandria.
INTEGER V(7),S,N !Assistants.
EQUIVALENCE (V(1),A),(V(2),B),(V(3),C), !Yes,
1 (V(4),D),(V(5),E),(V(6),F),(V(7),G) !We're all individuals.
WRITE (6,1) FIRST,LAST !Announce: first part.
1 FORMAT (/,"The Four Rings puzzle, over ",I0," to ",I0,".",$) !$: An addendum follows.
IF (UNIQUE) WRITE (6,*) "Distinct values only." !Save on the THEN ... ELSE ... END IF blather.
IF (.NOT.UNIQUE) WRITE (6,*) "Repeated values allowed." !Perhaps the compiler will be smarter.
 
N = 0 !No solutions have been found.
BB:DO B = FIRST,LAST !Start chugging through the possibilities.
CC:DO C = FIRST,LAST !Brute force and ignorance.
IF (UNIQUE .AND. B.EQ.C) CYCLE CC !The first constraint shows up.
DD:DO D = FIRST,LAST !Start by forming B, C, and D.
IF (UNIQUE .AND. ANY(V(2:3).EQ.D)) CYCLE DD !Ignoring A just for now.
S = B + C + D !This is the common sum.
A = S - B !The value of A is not free from BCD.
IF (A < FIRST .OR. A > LAST) CYCLE DD !And it may not be within bounds.
IF (UNIQUE .AND. ANY(V(2:4).EQ.A)) CYCLE DD !Or, if required so, unique.
EE:DO E = FIRST,LAST !Righto, A,B,C,D are valid. Try an E.
IF (UNIQUE .AND. ANY(V(1:4).EQ.E)) CYCLE EE !Precluded already?
F = S - (E + D) !No. So therefore, F is determined.
IF (F < FIRST .OR. F > LAST) CYCLE EE !Acceptable?
IF (UNIQUE .AND. ANY(V(1:5).EQ.F)) CYCLE EE !And, if required, unique?
G = S - F !Yes! So finally, G is determined.
IF (G < FIRST .OR. G > LAST) CYCLE EE !Acceptable?
IF (UNIQUE .AND. ANY(V(1:6).EQ.G)) CYCLE EE !And, if required, unique?
N = N + 1 !Yes! Count a solution set!
IF (UNIQUE) WRITE (6,"(7I3)") V !Show its values.
END DO EE !Consder another E.
END DO DD !Consider another D.
END DO CC !Consider another C.
END DO BB !Consider another B.
WRITE (6,2) N !Announce the count.
2 FORMAT (I9," found.") !Numerous, if no need for distinct values.
END SUBROUTINE FOURSHOW !That was fun!
 
PROGRAM POKE
 
CALL FOURSHOW(1,7,.TRUE.)
CALL FOURSHOW(3,9,.TRUE.)
CALL FOURSHOW(0,9,.FALSE.)
 
END

Output: not in a neat order because the first variable is not determined first.

The Four Rings puzzle, over 1 to 7. Distinct values only.
  7  2  6  1  3  5  4
  7  3  2  5  1  4  6
  6  4  1  5  2  3  7
  6  4  5  1  2  7  3
  4  5  3  1  6  2  7
  5  6  2  3  1  7  4
  4  7  1  3  2  6  5
  3  7  2  1  5  4  6
        8 found.

The Four Rings puzzle, over 3 to 9. Distinct values only.
  9  6  4  5  3  7  8
  9  6  5  4  3  8  7
  8  7  3  5  4  6  9
  7  8  3  4  5  6  9
        4 found.

The Four Rings puzzle, over 0 to 9. Repeated values allowed.
     2860 found.

One might hope that the ANY function will quit as soon as possible and that it will not be invoked if UNIQUE is false, but the modernisers have rejected reliance on short-circuit evaluation and the "help" is quite general on the workings of the ANY function, as also is modern. Here is a sample of the code produced by the Compaq 6.6a Visual Fortran F90/95 compiler, in its normal "debugging" condition and array bound checking of course active...

31:                    IF (UNIQUE .AND. ANY(V(1:6).EQ.G)) CYCLE EE    !And, if required, unique?
00401496   mov         edi,dword ptr [UNIQUE]
00401499   mov         edi,dword ptr [edi]
0040149B   mov         ebx,dword ptr [G (00470380)]
004014A1   mov         eax,0
004014A6   mov         ecx,1
004014AB   mov         dword ptr [ebp-60h],1
004014B2   cmp         dword ptr [ebp-60h],6
004014B6   jg          FOURSHOW+4C4h (004014fc)
004014B8   cmp         ecx,1
004014BB   jl          FOURSHOW+48Ah (004014c2)
004014BD   cmp         ecx,7
004014C0   jle         FOURSHOW+493h (004014cb)
004014C2   xor         esi,esi
004014C4   mov         dword ptr [ebp-6Ch],esi
004014C7   dec         esi
004014C8   bound       esi,qword ptr [ebp-6Ch]
004014CB   imul        esi,ecx,4
004014CE   mov         esi,dword ptr S+4 (00470364)[esi]
004014D4   xor         edx,edx
004014D6   cmp         esi,ebx
004014D8   sete        dl
004014DB   mov         dword ptr [ebp-6Ch],edx
004014DE   mov         edx,eax
004014E0   or          edx,dword ptr [ebp-6Ch]
004014E3   and         edx,1
004014E6   mov         eax,edx
004014E8   neg         eax
004014EA   mov         esi,ecx
004014EC   add         esi,1
004014EF   mov         ecx,esi
004014F1   mov         edx,dword ptr [ebp-60h]
004014F4   add         edx,1
004014F7   mov         dword ptr [ebp-60h],edx
004014FA   jmp         FOURSHOW+47Ah (004014b2)
004014FC   and         edi,eax
004014FE   mov         edx,edi
00401500   and         edx,1
00401503   cmp         edx,0
00401506   jne         FOURSHOW+531h (00401569)
32:                    N = N + 1          !Yes! Count a solution set!
00401508   mov         esi,dword ptr [N (0047035c)]
0040150E   add         esi,1
00401511   mov         dword ptr [N (0047035c)],esi
33:                    IF (UNIQUE) WRITE (6,"(7I3)") V    !Show its values.

I'd rather say nothing at all.

FreeBASIC[edit]

' version 18-03-2017
' compile with: fbc -s console
 
' TRUE/FALSE are built-in constants since FreeBASIC 1.04
' But we have to define them for older versions.
#Ifndef TRUE
#Define FALSE 0
#Define TRUE Not FALSE
#EndIf
 
Sub four_rings(low As Long, high As Long, unique As Long, show As Long)
 
Dim As Long a, b, c, d, e, f, g
Dim As ULong t, total
Dim As ULong l = Len(Str(high))
If l < Len(Str(low)) Then l = Len(Str(low))
 
 
If show = TRUE Then
For a = 97 To 103
Print Space(l); Chr(a);
Next
Print
Print String((l +1) * 7, "=");
Print
End If
 
For a = low To high
For b = low To high
If unique = TRUE Then
If b = a Then Continue For
End If
t = a + b
For c = low To high
If unique = TRUE Then
If c = a OrElse c = b Then Continue For
End If
For d = low To high
If unique = TRUE Then
If d = a OrElse d = b OrElse d = c Then Continue For
End If
If b + c + d = t Then
For e = low To high
If unique = TRUE Then
If e = a OrElse e = b OrElse e = c OrElse e = d Then Continue For
End If
For f = low To high
If unique = TRUE Then
If f = a OrElse f = b OrElse f = c OrElse f = d OrElse f = e Then Continue For
End If
If d + e + f = t Then
For g = low To high
If unique = TRUE Then
If g = a OrElse g = b OrElse g = c OrElse g = d OrElse g = e OrElse g = f Then Continue For
End If
If f + g = t Then
total += 1
If show = TRUE Then
Print Using String(l +1, "#"); a; b; c; d; e; f; g
End If
End If
Next
End If
Next
Next
End If
Next
Next
Next
Next
 
If unique = TRUE Then
Print
Print total; " Unique solutions for "; Str(low); " to "; Str(high)
Else
Print total; " Non unique solutions for "; Str(low); " to "; Str(high)
End If
Print String(40, "-") : Print
End Sub
 
' ------=< MAIN >=------
 
four_rings(1, 7, TRUE, TRUE)
four_rings(3, 9, TRUE, TRUE)
four_rings(0, 9, FALSE, FALSE)
 
' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
 a b c d e f g
==============
 3 7 2 1 5 4 6
 4 5 3 1 6 2 7
 4 7 1 3 2 6 5
 5 6 2 3 1 7 4
 6 4 1 5 2 3 7
 6 4 5 1 2 7 3
 7 2 6 1 3 5 4
 7 3 2 5 1 4 6

8 Unique solutions for 1 to 7
----------------------------------------

 a b c d e f g
==============
 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7

4 Unique solutions for 3 to 9
----------------------------------------

2860 Non unique solutions for 0 to 9
----------------------------------------

Go[edit]

package main
 
import "fmt"
 
func main(){
n, c := getCombs(1,7,true)
fmt.Printf("%d unique solutions in 1 to 7\n",n)
fmt.Println(c)
n, c = getCombs(3,9,true)
fmt.Printf("%d unique solutions in 3 to 9\n",n)
fmt.Println(c)
n, _ = getCombs(0,9,false)
fmt.Printf("%d non-unique solutions in 0 to 9\n",n)
}
 
func getCombs(low,high int,unique bool) (num int,validCombs [][]int){
for a := low; a <= high; a++ {
for b := low; b <= high; b++ {
for c := low; c <= high; c++ {
for d := low; d <= high; d++ {
for e := low; e <= high; e++ {
for f := low; f <= high; f++ {
for g := low; g <= high; g++ {
if validComb(a,b,c,d,e,f,g) {
if !unique || isUnique(a,b,c,d,e,f,g) {
num++
validCombs = append(validCombs,[]int{a,b,c,d,e,f,g})
}
}
}
}
}
}
}
}
}
return
}
func isUnique(a,b,c,d,e,f,g int) (res bool) {
data := make(map[int]int)
data[a]++
data[b]++
data[c]++
data[d]++
data[e]++
data[f]++
data[g]++
return len(data) == 7
}
func validComb(a,b,c,d,e,f,g int) bool{
square1 := a + b
square2 := b + c + d
square3 := d + e + f
square4 := f + g
return square1 == square2 && square2 == square3 && square3 == square4
}
 
Output:
8 unique solutions in 1 to 7
[[3 7 2 1 5 4 6] [4 5 3 1 6 2 7] [4 7 1 3 2 6 5] [5 6 2 3 1 7 4] [6 4 1 5 2 3 7] [6 4 5 1 2 7 3] [7 2 6 1 3 5 4] [7 3 2 5 1 4 6]]
4 unique solutions in 3 to 9
[[7 8 3 4 5 6 9] [8 7 3 5 4 6 9] [9 6 4 5 3 7 8] [9 6 5 4 3 8 7]]
2860 non-unique solutions in 0 to 9

Groovy[edit]

Translation of: Java
class FourRings {
static void main(String[] args) {
fourSquare(1, 7, true, true)
fourSquare(3, 9, true, true)
fourSquare(0, 9, false, false)
}
 
private static void fourSquare(int low, int high, boolean unique, boolean print) {
int count = 0
 
if (print) {
println("a b c d e f g")
}
for (int a = low; a <= high; ++a) {
for (int b = low; b <= high; ++b) {
if (notValid(unique, a, b)) continue
 
int fp = a + b
for (int c = low; c <= high; ++c) {
if (notValid(unique, c, a, b)) continue
for (int d = low; d <= high; ++d) {
if (notValid(unique, d, a, b, c)) continue
if (fp != b + c + d) continue
 
for (int e = low; e <= high; ++e) {
if (notValid(unique, e, a, b, c, d)) continue
for (int f = low; f <= high; ++f) {
if (notValid(unique, f, a, b, c, d, e)) continue
if (fp != d + e + f) continue
 
for (int g = low; g <= high; ++g) {
if (notValid(unique, g, a, b, c, d, e, f)) continue
if (fp != f + g) continue
 
++count
if (print) {
printf("%d %d %d %d %d %d %d%n", a, b, c, d, e, f, g)
}
}
}
}
}
}
}
}
if (unique) {
printf("There are %d unique solutions in [%d, %d]%n", count, low, high)
} else {
printf("There are %d non-unique solutions in [%d, %d]%n", count, low, high)
}
}
 
private static boolean notValid(boolean unique, int needle, int ... haystack) {
return unique && Arrays.stream(haystack).anyMatch({ p -> p == needle })
}
}
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1, 7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3, 9]
There are 2860 non-unique solutions in [0, 9]

Haskell[edit]

By exhaustive search[edit]

import Data.List
import Control.Monad
 
perms :: (Eq a) => [a] -> [[a]]
perms [] = [[]]
perms xs = [ x:xr | x <- xs, xr <- perms (xs\\[x]) ]
 
combs :: (Eq a) => Int -> [a] -> [[a]]
combs 0 _ = [[]]
combs n xs = [ x:xr | x <- xs, xr <- combs (n-1) xs ]
 
ringCheck :: [Int] -> Bool
ringCheck [x0, x1, x2, x3, x4, x5, x6] =
v == x1+x2+x3
&& v == x3+x4+x5
&& v == x5+x6
where v = x0 + x1
 
fourRings :: Int -> Int -> Bool -> Bool -> IO ()
fourRings low high allowRepeats verbose = do
let candidates = if allowRepeats
then combs 7 [low..high]
else perms [low..high]
 
solutions = filter ringCheck candidates
 
when verbose $ mapM_ print solutions
 
putStrLn $ show (length solutions)
++ (if allowRepeats then " non" else "")
++ " unique solutions for "
++ show low
++ " to "
++ show high
 
putStrLn ""
 
main = do
fourRings 1 7 False True
fourRings 3 9 False True
fourRings 0 9 True False
Output:
[3,7,2,1,5,4,6]
[4,5,3,1,6,2,7]
[4,7,1,3,2,6,5]
[5,6,2,3,1,7,4]
[6,4,1,5,2,3,7]
[6,4,5,1,2,7,3]
[7,2,6,1,3,5,4]
[7,3,2,5,1,4,6]
8 unique solutions for 1 to 7

[7,8,3,4,5,6,9]
[8,7,3,5,4,6,9]
[9,6,4,5,3,7,8]
[9,6,5,4,3,8,7]
4 unique solutions for 3 to 9

2860 non unique solutions for 0 to 9

By structured search[edit]

For a faster solution (under a third of a second, vs over 25 seconds on this system for the brute force approach above), we can nest a series of smaller and more focused searches from the central digit outwards.

Two things to notice:

  1. If we call the central digit the Queen, then in any solution the Queen plus its left neighbour (left Bishop) must sum to the value of the left Rook (leftmost digit). Symmetrically, the right Rook must be the sum of the Queen and right Bishop.
  2. The difference between the left Rook and the right Rook must be (minus) the difference between the left Knight (between bishop and rook) and the right Knight.


Nesting four bind operators (>>=), we can then build the set of solutions in the order: queens, left bishops and rooks, right bishops and rooks, knights. Probably less readable, but already fast, and could be further optimised.

import Data.List (delete, sortBy, (\\))
import Control.Monad (when)
import Data.Bool (bool)
 
type Rings = [(Int, Int, Int, Int, Int, Int, Int)]
 
rings :: Bool -> [Int] -> Rings
rings u digits =
let ds = sortBy (flip compare) digits
in ds >>= queen u (head ds) ds
 
queen :: Bool -> Int -> [Int] -> Int -> Rings
queen u h ds q =
let ts = filter ((<= h) . (q +)) ds
in bool ds (delete q ts) u >>= leftBishop u q h ts ds
 
leftBishop :: Bool -> Int -> Int -> [Int] -> [Int] -> Int -> Rings
leftBishop u q h ts ds lb =
let lRook = lb + q
in bool
[]
(bool ds (ts \\ [q, lb, lRook]) u >>= rightBishop u q h lb ds lRook)
(lRook <= h)
 
rightBishop :: Bool -> Int -> Int -> Int -> [Int] -> Int -> Int -> Rings
rightBishop u q h lb ds lRook rb =
let rRook = q + rb
in bool
[]
(let ks = bool ds (ds \\ [q, lb, rb, rRook, lRook]) u
in ks >>= knights u (lRook - rRook) lRook lb q rb rRook ks)
((rRook <= h) && (not u || (rRook /= lb)))
 
knights :: Bool -> Int -> Int -> Int -> Int -> Int -> Int -> [Int] -> Int -> Rings
knights u rookDelta lRook lb q rb rRook ks k =
let k2 = k + rookDelta
in [ (lRook, k, lb, q, rb, k2, rRook)
| (k2 `elem` ks) && (not u || notElem k2 [lRook, k, lb, q, rb, rRook]) ]
 
 
-- TEST ---------------------------------------------------
main :: IO ()
main = do
let f (k, xs) = putStrLn k >> nl >> mapM_ print xs >> nl
nl = putStrLn []
mapM_
f
[ ("rings True [1 .. 7]", rings True [1 .. 7])
, ("rings True [3 .. 9]", rings True [3 .. 9])
]
f ("length (rings False [0 .. 9])", [length (rings False [0 .. 9])])
Output:
rings True [1 .. 7]

(7,3,2,5,1,4,6)
(6,4,1,5,2,3,7)
(5,6,2,3,1,7,4)
(4,7,1,3,2,6,5)
(7,2,6,1,3,5,4)
(6,4,5,1,2,7,3)
(4,5,3,1,6,2,7)
(3,7,2,1,5,4,6)

rings True [3 .. 9]

(9,6,4,5,3,7,8)
(8,7,3,5,4,6,9)
(9,6,5,4,3,8,7)
(7,8,3,4,5,6,9)

length (rings False [0 .. 9])

2860

J[edit]

Implementation for the unique version of the puzzle:

fspuz=:dyad define
range=: x+i.1+y-x
lo=. 6+3*x
hi=. _3+2*y
r=.i.0 0
if. lo <: hi do.
for_T.lo ([+[:i.1+-~) hi do.
range2=: (#~ (T-{.range)>:]) range
range3=: (#~ (T-+/2{.range)>:]) range
ab=: (#~ ~:/"1) (,.T-])range2
abc=: ;ab <@([ ,"1 0 -.~)"1/range3
abcd=: (#~ T = +/@}."1) ;abc <@([ ,"1 0 -.~)"1/range3
abcde=: ;abcd <@([ ,"1 0 -.~)"1/range3
abcdef=: (#~ T = +/@(3}.])"1) ;abcde <@([ ,"1 0 -.~)"1/range3
abcdefg=: (#~ T = +/@(5}.])"1) ;abcdef <@([ ,"1 0 -.~)"1/range2
r=.r,(#~ x<:<./"1)(#~ y>:>./"1)abcdefg
end.
end.
)

Implementation for the non-unique version of the puzzle:

fspuz2=:dyad define
range=: x+i.1+y-x
lo=. 3*x
hi=. 2*y
r=.i.0 0
if. lo <: hi do.
for_T.lo ([+[:i.1+-~) hi do.
ab=: (,.T-])range
abc=: ,/ab,"1 0/ range
abcd=: (#~ T = +/@}."1) ,/abc,"1 0/ range
abcde=: ,/abcd,"1 0/ range
abcdef=: (#~ T = +/@(3}.])"1) ,/abcde ,"1 0/ range
abcdefg=: (#~ T = +/@(5}.])"1) ,/abcdef,"1 0/ range
r=.r,(#~ x<:<./"1)(#~ y>:>./"1)abcdefg
end.
end.
)

Task examples:

   1 fspuz 7
4 5 3 1 6 2 7
7 2 6 1 3 5 4
3 7 2 1 5 4 6
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 3 2 5 1 4 6
4 7 1 3 2 6 5
5 6 2 3 1 7 4
3 fspuz 9
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
#0 fspuz2 9
2860

Java[edit]

Uses java 8 features.

import java.util.Arrays;
 
public class FourSquares {
public static void main(String[] args) {
fourSquare(1, 7, true, true);
fourSquare(3, 9, true, true);
fourSquare(0, 9, false, false);
}
 
private static void fourSquare(int low, int high, boolean unique, boolean print) {
int count = 0;
 
if (print) {
System.out.println("a b c d e f g");
}
for (int a = low; a <= high; ++a) {
for (int b = low; b <= high; ++b) {
if (notValid(unique, a, b)) continue;
 
int fp = a + b;
for (int c = low; c <= high; ++c) {
if (notValid(unique, c, a, b)) continue;
for (int d = low; d <= high; ++d) {
if (notValid(unique, d, a, b, c)) continue;
if (fp != b + c + d) continue;
 
for (int e = low; e <= high; ++e) {
if (notValid(unique, e, a, b, c, d)) continue;
for (int f = low; f <= high; ++f) {
if (notValid(unique, f, a, b, c, d, e)) continue;
if (fp != d + e + f) continue;
 
for (int g = low; g <= high; ++g) {
if (notValid(unique, g, a, b, c, d, e, f)) continue;
if (fp != f + g) continue;
 
++count;
if (print) {
System.out.printf("%d %d %d %d %d %d %d%n", a, b, c, d, e, f, g);
}
}
}
}
}
}
}
}
if (unique) {
System.out.printf("There are %d unique solutions in [%d, %d]%n", count, low, high);
} else {
System.out.printf("There are %d non-unique solutions in [%d, %d]%n", count, low, high);
}
}
 
private static boolean notValid(boolean unique, int needle, int... haystack) {
return unique && Arrays.stream(haystack).anyMatch(p -> p == needle);
}
}
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1, 7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3, 9]
There are 2860 non-unique solutions in [0, 9]

JavaScript[edit]

ES6[edit]

Translation of: Haskell
(Structured search version)
(() => {
 
// 4-rings or 4-squares puzzle ------------------------
 
// rings :: noRepeatedDigits -> DigitList -> solutions
// rings :: Bool -> [Int] -> [[Int]]
const rings = (uniq, digits) => {
return 0 < digits.length ? (() => {
const
ns = sortBy(flip(compare), digits),
h = head(ns);
 
// CENTRAL DIGIT :: d
return bindList(
ns,
d => {
const ts = filter(x => (x + d) <= h, ns);
 
// LEFT OF CENTRE :: c and a
return bindList(
uniq ? delete_(d, ts) : ns,
c => {
const a = c + d;
 
// RIGHT OF CENTRE :: e and g
return a > h ? (
[]
) : bindList(uniq ? (
difference(ts, [d, c, a])
) : ns, e => {
const g = d + e;
return ((g > h) || (uniq && (g === c))) ? (
[]
) : (() => {
const
agDelta = a - g,
bfs = uniq ? difference(
ns, [d, c, e, g, a]
) : ns;
 
// MID LEFT, MID RIGHT :: b and f
return bindList(bfs, b => {
const f = b + agDelta;
return elem(f, bfs) && (
!uniq || notElem(f, [
a, b, c, d, e, g
])
) ? ([
[a, b, c, d, e, f, g]
]) : [];
});
})();
});
});
});
})() : []
};
 
 
// TEST -----------------------------------------------
const main = () => {
return unlines([
'rings(true, enumFromTo(1,7))\n',
unlines(map(show, rings(true, enumFromTo(1, 7)))),
 
'\nrings(true, enumFromTo(3, 9))\n',
unlines(map(show, rings(true, enumFromTo(3, 9)))),
 
'\nlength(rings(false, enumFromTo(0, 9)))\n',
length(rings(false, enumFromTo(0, 9)))
.toString(),
''
]);
};
 
// GENERIC FUNCTIONS ----------------------------------
 
// bindList (>>=) :: [a] -> (a -> [b]) -> [b]
const bindList = (xs, mf) => [].concat.apply([], xs.map(mf));
 
// compare :: a -> a -> Ordering
const compare = (a, b) => a < b ? -1 : (a > b ? 1 : 0);
 
// delete_ :: Eq a => a -> [a] -> [a]
const delete_ = (x, xs) =>
xs.length > 0 ? (
(x === xs[0]) ? (
xs.slice(1)
) : [xs[0]].concat(delete_(x, xs.slice(1)))
) : [];
 
// difference :: Eq a => [a] -> [a] -> [a]
const difference = (xs, ys) => {
const s = new Set(ys);
return xs.filter(x => !s.has(x));
};
 
// elem :: Eq a => a -> [a] -> Bool
const elem = (x, xs) => xs.indexOf(x) !== -1;
 
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
 
// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);
 
// flip :: (a -> b -> c) -> b -> a -> c
const flip = f => (a, b) => f.apply(null, [b, a]);
 
// head :: [a] -> a
const head = xs => xs.length ? xs[0] : undefined;
 
// length :: [a] -> Int
const length = xs => xs.length;
 
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
 
// notElem :: Eq a => a -> [a] -> Bool
const notElem = (x, xs) => xs.indexOf(x) === -1;
 
// show :: a -> String
const show = x => JSON.stringify(x); //, null, 2);
 
// sortBy :: (a -> a -> Ordering) -> [a] -> [a]
const sortBy = (f, xs) => xs.sort(f);
 
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
 
 
// MAIN ---
return main();
})();
Output:
rings(true, enumFromTo(1,7))

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

rings(true, enumFromTo(3, 9))

[9,6,4,5,3,7,8]
[8,7,3,5,4,6,9]
[9,6,5,4,3,8,7]
[7,8,3,4,5,6,9]

length(rings(false, enumFromTo(0, 9)))

2860

Julia[edit]

Translation of: Python
 
using Combinatorics
 
function foursquares(low, high, onlyunique=true, showsolutions=true)
integers = collect(low:high)
count = 0
sumsallequal(c) = c[1] + c[2] == c[2] + c[3] + c[4] == c[4] + c[5] + c[6] == c[6] + c[7]
combos = onlyunique ? combinations(integers) :
with_replacement_combinations(integers, 7)
for combo in combos, plist in unique(collect(permutations(combo, 7)))
if sumsallequal(plist)
count += 1
if showsolutions
println("$plist is a solution for the list $integers")
end
end
end
println("""Total $(onlyunique?"unique ":"")solutions for HIGH $high, LOW $low: $count""")
end
 
foursquares(1, 7, true, true)
foursquares(3, 9, true, true)
foursquares(0, 9, false, false)
 
Output:
[3, 7, 2, 1, 5, 4, 6] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[4, 5, 3, 1, 6, 2, 7] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[4, 7, 1, 3, 2, 6, 5] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[5, 6, 2, 3, 1, 7, 4] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[6, 4, 1, 5, 2, 3, 7] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[6, 4, 5, 1, 2, 7, 3] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[7, 2, 6, 1, 3, 5, 4] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
[7, 3, 2, 5, 1, 4, 6] is a solution for the list [1, 2, 3, 4, 5, 6, 7]
Total unique solutions for HIGH 7, LOW 1: 8
[7, 8, 3, 4, 5, 6, 9] is a solution for the list [3, 4, 5, 6, 7, 8, 9]
[8, 7, 3, 5, 4, 6, 9] is a solution for the list [3, 4, 5, 6, 7, 8, 9]
[9, 6, 4, 5, 3, 7, 8] is a solution for the list [3, 4, 5, 6, 7, 8, 9]
[9, 6, 5, 4, 3, 8, 7] is a solution for the list [3, 4, 5, 6, 7, 8, 9]
Total unique solutions for HIGH 9, LOW 3: 4
Total solutions for HIGH 9, LOW 0: 2860

Kotlin[edit]

Translation of: C
// version 1.1.2
 
class FourSquares(
private val lo: Int,
private val hi: Int,
private val unique: Boolean,
private val show: Boolean
) {
private var a = 0
private var b = 0
private var c = 0
private var d = 0
private var e = 0
private var f = 0
private var g = 0
private var s = 0
 
init {
println()
if (show) {
println("a b c d e f g")
println("-------------")
}
acd()
println("\n$s ${if (unique) "unique" else "non-unique"} solutions in $lo to $hi")
}
 
private fun acd() {
c = lo
while (c <= hi) {
d = lo
while (d <= hi) {
if (!unique || c != d) {
a = c + d
if ((a in lo..hi) && (!unique || (c != 0 && d!= 0))) ge()
}
d++
}
c++
}
}
 
private fun bf() {
f = lo
while (f <= hi) {
if (!unique || (f != a && f != c && f != d && f != e && f!= g)) {
b = e + f - c
if ((b in lo..hi) && (!unique || (b != a && b != c && b != d && b != e && b != f && b!= g))) {
s++
if (show) println("$a $b $c $d $e $f $g")
}
}
f++
}
}
 
private fun ge() {
e = lo
while (e <= hi) {
if (!unique || (e != a && e != c && e != d)) {
g = d + e
if ((g in lo..hi) && (!unique || (g != a && g != c && g != d && g != e))) bf()
}
e++
}
}
}
 
fun main(args: Array<String>) {
FourSquares(1, 7, true, true)
FourSquares(3, 9, true, true)
FourSquares(0, 9, false, false)
}
Output:
a b c d e f g
-------------
4 7 1 3 2 6 5
6 4 1 5 2 3 7
3 7 2 1 5 4 6
5 6 2 3 1 7 4
7 3 2 5 1 4 6
4 5 3 1 6 2 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4

8 unique solutions in 1 to 7

a b c d e f g
-------------
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7

4 unique solutions in 3 to 9


2860 non-unique solutions in 0 to 9

Lua[edit]

Translation of: D
function valid(unique,needle,haystack)
if unique then
for _,value in pairs(haystack) do
if needle == value then
return false
end
end
end
return true
end
 
function fourSquare(low,high,unique,prnt)
count = 0
if prnt then
print("a", "b", "c", "d", "e", "f", "g")
end
for a=low,high do
for b=low,high do
if valid(unique, a, {b}) then
fp = a + b
for c=low,high do
if valid(unique, c, {a, b}) then
for d=low,high do
if valid(unique, d, {a, b, c}) and fp == b + c + d then
for e=low,high do
if valid(unique, e, {a, b, c, d}) then
for f=low,high do
if valid(unique, f, {a, b, c, d, e}) and fp == d + e + f then
for g=low,high do
if valid(unique, g, {a, b, c, d, e, f}) and fp == f + g then
count = count + 1
if prnt then
print(a, b, c, d, e, f, g)
end
end
end
end
end
end
end
end
end
end
end
end
end
end
if unique then
print(string.format("There are %d unique solutions in [%d, %d]", count, low, high))
else
print(string.format("There are %d non-unique solutions in [%d, %d]", count, low, high))
end
end
 
fourSquare(1,7,true,true)
fourSquare(3,9,true,true)
fourSquare(0,9,false,false)
Output:
a       b       c       d       e       f       g
3       7       2       1       5       4       6
4       5       3       1       6       2       7
4       7       1       3       2       6       5
5       6       2       3       1       7       4
6       4       1       5       2       3       7
6       4       5       1       2       7       3
7       2       6       1       3       5       4
7       3       2       5       1       4       6
There are 8 unique solutions in [1, 7]
a       b       c       d       e       f       g
7       8       3       4       5       6       9
8       7       3       5       4       6       9
9       6       4       5       3       7       8
9       6       5       4       3       8       7
There are 4 unique solutions in [3, 9]
There are 2860 non-unique solutions in [0, 9]

Modula-2[edit]

MODULE FourSquare;
FROM Conversions IMPORT IntToStr;
FROM Terminal IMPORT *;
 
PROCEDURE WriteInt(num : INTEGER);
VAR str : ARRAY[0..16] OF CHAR;
BEGIN
IntToStr(num,str);
WriteString(str);
END WriteInt;
 
PROCEDURE four_square(low, high : INTEGER; unique, print : BOOLEAN);
VAR count : INTEGER;
VAR a, b, c, d, e, f, g : INTEGER;
VAR fp : INTEGER;
BEGIN
count:=0;
 
IF print THEN
WriteString('a b c d e f g');
WriteLn;
END;
FOR a:=low TO high DO
FOR b:=low TO high DO
IF unique AND (b=a) THEN CONTINUE; END;
 
fp:=a+b;
FOR c:=low TO high DO
IF unique AND ((c=a) OR (c=b)) THEN CONTINUE; END;
FOR d:=low TO high DO
IF unique AND ((d=a) OR (d=b) OR (d=c)) THEN CONTINUE; END;
IF fp # b+c+d THEN CONTINUE; END;
 
FOR e:=low TO high DO
IF unique AND ((e=a) OR (e=b) OR (e=c) OR (e=d)) THEN CONTINUE; END;
FOR f:=low TO high DO
IF unique AND ((f=a) OR (f=b) OR (f=c) OR (f=d) OR (f=e)) THEN CONTINUE; END;
IF fp # d+e+f THEN CONTINUE; END;
 
FOR g:=low TO high DO
IF unique AND ((g=a) OR (g=b) OR (g=c) OR (g=d) OR (g=e) OR (g=f)) THEN CONTINUE; END;
IF fp # f+g THEN CONTINUE; END;
 
INC(count);
IF print THEN
WriteInt(a);
WriteString(' ');
WriteInt(b);
WriteString(' ');
WriteInt(c);
WriteString(' ');
WriteInt(d);
WriteString(' ');
WriteInt(e);
WriteString(' ');
WriteInt(f);
WriteString(' ');
WriteInt(g);
WriteLn;
END;
END;
END;
END;
END;
END;
END;
END;
IF unique THEN
WriteString('There are ');
WriteInt(count);
WriteString(' unique solutions in [');
WriteInt(low);
WriteString(', ');
WriteInt(high);
WriteString(']');
WriteLn;
ELSE
WriteString('There are ');
WriteInt(count);
WriteString(' non-unique solutions in [');
WriteInt(low);
WriteString(', ');
WriteInt(high);
WriteString(']');
WriteLn;
END;
END four_square;
 
BEGIN
four_square(1,7,TRUE,TRUE);
four_square(3,9,TRUE,TRUE);
four_square(0,9,FALSE,FALSE);
ReadChar; (* Wait so results can be viewed. *)
END FourSquare.

Nim[edit]

Adapted from Rust version.

func isUnique(a, b, c, d, e, f, g: uint8): bool =
a != b and a != c and a != d and a != e and a != f and a != g and
b != c and b != d and b != e and b != f and b != g and
c != d and c != e and c != f and c != g and
d != e and d != f and d != f and
e != f and e != g and
f != g
 
func isSolution(a, b, c, d, e, f, g: uint8): bool =
let sum = a + b
sum == b + c + d and sum == d + e + f and sum == f + g
 
func fourSquares(l, h: uint8, unique: bool): seq[array[7, uint8]] =
for a in l..h:
for b in l..h:
for c in l..h:
for d in l..h:
for e in l..h:
for f in l..h:
for g in l..h:
if (not unique or isUnique(a, b, c, d, e, f, g)) and
isSolution(a, b, c, d, e, f, g):
result &= [a, b, c, d, e, f, g]
 
proc printFourSquares(l, h: uint8, unique = true) =
let solutions = fourSquares(l, h, unique)
 
if unique:
for s in solutions:
echo s
 
echo solutions.len, (if unique: " " else: " non-"), "unique solutions in ",
l, " to ", h, " range\n"
 
when isMainModule:
printFourSquares(1, 7)
printFourSquares(3, 9)
printFourSquares(0, 9, unique = false)
Output:
[3, 7, 2, 1, 5, 4, 6]
[4, 5, 3, 1, 6, 2, 7]
[4, 7, 1, 3, 2, 6, 5]
[5, 6, 2, 3, 1, 7, 4]
[6, 4, 1, 5, 2, 3, 7]
[6, 4, 5, 1, 2, 7, 3]
[7, 2, 6, 1, 3, 5, 4]
[7, 3, 2, 5, 1, 4, 6]
8 unique solutions in 1 to 7 range

[7, 8, 3, 4, 5, 6, 9]
[8, 7, 3, 5, 4, 6, 9]
[9, 6, 4, 5, 3, 7, 8]
[9, 6, 5, 4, 3, 8, 7]
4 unique solutions in 3 to 9 range

2860 non-unique solutions in 0 to 9 range

Pascal[edit]

Works with: Free Pascal

There are so few solutions of 7 consecutive numbers, so I used a modified version, to get all the expected solutions at once.

program square4;
{$MODE DELPHI}
{$R+,O+}
const
LoDgt = 0;
HiDgt = 9;
type
tchkset = set of LoDgt..HiDgt;
tSol = record
solMin : integer;
solDat : array[1..7] of integer;
end;
 
var
sum,a,b,c,d,e,f,g,cnt,uniqueCount : NativeInt;
sol : array of tSol;
 
procedure SolOut;
var
i,j,mn: NativeInt;
Begin
mn := 0;
repeat
writeln(mn:3,' ...',mn+6:3);
For i := Low(sol) to High(sol) do
with sol[i] do
IF solMin = mn then
Begin
For j := 1 to 7 do
write(solDat[j]:3);
writeln;
end;
writeln;
inc(mn);
until mn > HiDgt-6;
end;
 
function CheckUnique:Boolean;
var
i,sum,mn: NativeInt;
chkset : tchkset;
 
Begin
chkset:= [];
include(chkset,a);include(chkset,b);include(chkset,c);
include(chkset,d);include(chkset,e);include(chkset,f);
include(chkset,g);
sum := 0;
For i := LoDgt to HiDgt do
IF i in chkset then
inc(sum);
 
result := sum = 7;
IF result then
begin
inc(uniqueCount);
//find the lowest entry
mn:= LoDgt;
For i := LoDgt to HiDgt do
IF i in chkset then
Begin
mn := i;
BREAK;
end;
// are they consecutive
For i := mn+1 to mn+6 do
IF NOT(i in chkset) then
EXIT;
 
setlength(sol,Length(sol)+1);
with sol[high(sol)] do
Begin
solMin:= mn;
solDat[1]:= a;solDat[2]:= b;solDat[3]:= c;
solDat[4]:= d;solDat[5]:= e;solDat[6]:= f;
solDat[7]:= g;
end;
end;
end;
 
Begin
cnt := 0;
uniqueCount := 0;
For a:= LoDgt to HiDgt do
Begin
For b := LoDgt to HiDgt do
Begin
sum := a+b;
//a+b = b+c+d => a = c+d => d := a-c
For c := a-LoDgt downto LoDgt do
begin
d := a-c;
e := sum-d;
IF e>HiDgt then
e:= HiDgt;
For e := e downto LoDgt do
begin
f := sum-e-d;
IF f in [loDGt..Hidgt]then
Begin
g := sum-f;
IF g in [loDGt..Hidgt]then
Begin
inc(cnt);
CheckUnique;
end;
end;
end;
end;
end;
end;
SolOut;
writeln(' solution count for ',loDgt,' to ',HiDgt,' = ',cnt);
writeln('unique solution count for ',loDgt,' to ',HiDgt,' = ',uniqueCount);
end.
Output:
  0 ...  6
  4  2  3  1  5  0  6
  5  1  3  2  4  0  6
  6  0  5  1  3  2  4
  6  0  4  2  3  1  5

  1 ...  7
  3  7  2  1  5  4  6
  4  5  3  1  6  2  7
  4  7  1  3  2  6  5
  5  6  2  3  1  7  4
  6  4  5  1  2  7  3
  6  4  1  5  2  3  7
  7  2  6  1  3  5  4
  7  3  2  5  1  4  6

  2 ...  8
  5  7  3  2  6  4  8
  5  8  3  2  4  7  6
  5  8  2  3  4  6  7
  6  7  4  2  3  8  5
  7  4  5  2  6  3  8
  7  6  4  3  2  8  5
  8  3  6  2  5  4  7
  8  4  6  2  3  7  5

  3 ...  9
  7  8  3  4  5  6  9
  8  7  3  5  4  6  9
  9  6  5  4  3  8  7
  9  6  4  5  3  7  8

       solution count for 0 to 9 = 2860
unique solution count for 0 to 9 = 192

Perl[edit]

Relying on the modules ntheory and Set::CrossProduct to generate the tuples needed. Both are supply results via iterators, particularly important in the latter case, to avoid gobbling too much memory.

Library: ntheory
use ntheory qw/forperm/;
use Set::CrossProduct;
 
sub four_sq_permute {
my($list) = @_;
my @solutions;
forperm {
@c = @$list[@_];
push @solutions, [@c] if check(@c);
} @$list;
print +@solutions . " unique solutions found using: " . join(', ', @$list) . "\n";
return @solutions;
}
 
sub four_sq_cartesian {
my(@list) = @_;
my @solutions;
my $iterator = Set::CrossProduct->new( [(@list) x 7] );
while( my $c = $iterator->get ) {
push @solutions, [@$c] if check(@$c);
}
print +@solutions . " non-unique solutions found using: " . join(', ', @{@list[0]}) . "\n";
return @solutions;
}
 
sub check {
my(@c) = @_;
$a = $c[0] + $c[1];
$b = $c[1] + $c[2] + $c[3];
$c = $c[3] + $c[4] + $c[5];
$d = $c[5] + $c[6];
$a == $b and $a == $c and $a == $d;
}
 
sub display {
my(@solutions) = @_;
my $fmt = "%2s " x 7 . "\n";
printf $fmt, ('a'..'g');
printf $fmt, @$_ for @solutions;
print "\n";
}
 
display four_sq_permute( [1..7] );
display four_sq_permute( [3..9] );
display four_sq_permute( [8, 9, 11, 12, 17, 18, 20, 21] );
four_sq_cartesian( [0..9] );
Output:
8 unique solutions found using: 1, 2, 3, 4, 5, 6, 7
 a  b  c  d  e  f  g
 3  7  2  1  5  4  6
 4  5  3  1  6  2  7
 4  7  1  3  2  6  5
 5  6  2  3  1  7  4
 6  4  1  5  2  3  7
 6  4  5  1  2  7  3
 7  2  6  1  3  5  4
 7  3  2  5  1  4  6

4 unique solutions found using: 3, 4, 5, 6, 7, 8, 9
 a  b  c  d  e  f  g
 7  8  3  4  5  6  9
 8  7  3  5  4  6  9
 9  6  4  5  3  7  8
 9  6  5  4  3  8  7

8 unique solutions found using: 8, 9, 11, 12, 17, 18, 20, 21
 a  b  c  d  e  f  g
17 21  8  9 11 18 20
17 21  9  8 12 18 20
20 18  8 12  9 17 21
20 18 11  9  8 21 17
20 18 11  9 12 17 21
20 18 12  8  9 21 17
21 17  9 12  8 18 20
21 17 12  9 11 18 20

2860 non-unique solutions found using: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Perl 6[edit]

Works with: Rakudo version 2016.12
sub four-squares ( @list, :$unique=1, :$show=1 ) {
 
my @solutions;
 
for $unique.&combos -> @c {
@solutions.push: @c if [==]
@c[0] + @c[1],
@c[1] + @c[2] + @c[3],
@c[3] + @c[4] + @c[5],
@c[5] + @c[6];
}
 
say +@solutions, ($unique ?? ' ' !! ' non-'), "unique solutions found using {join(', ', @list)}.\n";
 
my $f = "%{@list.max.chars}s";
 
say join "\n", (('a'..'g').fmt: $f), @solutions».fmt($f), "\n" if $show;
 
multi combos ( $ where so * ) { @list.combinations(7).map: |*.permutations }
 
multi combos ( $ where not * ) { [X] @list xx 7 }
}
 
# TASK
four-squares( [1..7] );
four-squares( [3..9] );
four-squares( [8, 9, 11, 12, 17, 18, 20, 21] );
four-squares( [0..9], :unique(0), :show(0) );
Output:
8 unique solutions found using 1, 2, 3, 4, 5, 6, 7.

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


4 unique solutions found using 3, 4, 5, 6, 7, 8, 9.

a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7


8 unique solutions found using 8, 9, 11, 12, 17, 18, 20, 21.

 a  b  c  d  e  f  g
17 21  8  9 11 18 20
20 18 11  9  8 21 17
17 21  9  8 12 18 20
20 18  8 12  9 17 21
20 18 12  8  9 21 17
21 17  9 12  8 18 20
20 18 11  9 12 17 21
21 17 12  9 11 18 20


2860 non-unique solutions found using 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Picat[edit]

import cp.
 
main =>
puzzle_all(1, 7, true, Sol1),
foreach(Sol in Sol1) println(Sol) end,
nl,
 
puzzle_all(3, 9, true, Sol2),
foreach(Sol in Sol2) println(Sol) end,
nl,
 
puzzle_all(0, 9, false, Sol3),
println(len=Sol3.len),
nl.
 
puzzle_all(Min, Max, Distinct, LL) =>
L = [A,B,C,D,E,F,G],
L :: Min..Max,
if Distinct then
all_different(L)
else
true
end,
T #= A+B,
T #= B+C+D,
T #= D+E+F,
T #= F+G,
 % Another approach:
 % Sums = $[A+B,B+C+D,D+E+F,F+G],
 % foreach(I in 2..Sums.len) Sums[I] #= Sums[I-1] end,
LL = solve_all(L).
 

Test:

Picat> main
[3,7,2,1,5,4,6]
[4,5,3,1,6,2,7]
[4,7,1,3,2,6,5]
[5,6,2,3,1,7,4]
[6,4,1,5,2,3,7]
[6,4,5,1,2,7,3]
[7,2,6,1,3,5,4]
[7,3,2,5,1,4,6]

[7,8,3,4,5,6,9]
[8,7,3,5,4,6,9]
[9,6,4,5,3,7,8]
[9,6,5,4,3,8,7]

len = 2860


Phix[edit]

integer solutions
 
procedure check(sequence set, bool show)
integer {a,b,c,d,e,f,g} = set, ab = a+b
if ab=b+d+c and ab=d+e+f and ab=f+g then
solutions += 1
if show then
 ?set
end if
end if
end procedure
 
procedure foursquares(integer lo, integer hi, bool uniq, bool show)
sequence set = repeat(lo,7)
solutions = 0
if uniq then
for i=1 to 7 do
set[i] = lo+i-1
end for
for i=1 to factorial(7) do
check(permute(i,set),show)
end for
else
integer done = 0
while not done do
check(set,show)
for i=1 to 7 do
set[i] += 1
if set[i]<=hi then exit end if
if i=7 then
done = 1
exit
end if
set[i] = lo
end for
end while
end if
printf(1,"%d solutions\n",solutions)
end procedure
foursquares(1,7,uniq:=True,show:=True)
foursquares(3,9,True,True)
foursquares(0,9,False,False)
Output:
{6,4,5,1,2,7,3}
{3,7,2,1,5,4,6}
{6,4,1,5,2,3,7}
{4,7,1,3,2,6,5}
{7,3,2,5,1,4,6}
{5,6,2,3,1,7,4}
{4,5,3,1,6,2,7}
{7,2,6,1,3,5,4}
8 solutions
{7,8,3,4,5,6,9}
{8,7,3,5,4,6,9}
{9,6,4,5,3,7,8}
{9,6,5,4,3,8,7}
4 solutions
2860 solutions

PL/SQL[edit]

Works with: Oracle
 
CREATE TABLE allints (v NUMBER);
CREATE TABLE results
(
a NUMBER,
b NUMBER,
c NUMBER,
d NUMBER,
e NUMBER,
f NUMBER,
g NUMBER
);
 
CREATE OR REPLACE PROCEDURE foursquares(lo NUMBER,hi NUMBER,uniq BOOLEAN,show BOOLEAN)
AS
a NUMBER;
b NUMBER;
c NUMBER;
d NUMBER;
e NUMBER;
f NUMBER;
g NUMBER;
out_line VARCHAR2(2000);
 
CURSOR results_cur IS
SELECT
a,
b,
c,
d,
e,
f,
g
FROM
results
ORDER BY
a,b,c,d,e,f,g;
 
results_rec results_cur%ROWTYPE;
 
solutions NUMBER;
uorn VARCHAR2(2000);
BEGIN
solutions := 0;
DELETE FROM allints;
DELETE FROM results;
FOR i IN lo..hi LOOP
INSERT INTO allints VALUES (i);
END LOOP;
COMMIT;
 
IF uniq = TRUE THEN
INSERT INTO results
SELECT
a.v a,
b.v b,
c.v c,
d.v d,
e.v e,
f.v f,
g.v g
FROM
allints a, allints b, allints c,allints d,
allints e, allints f, allints g
WHERE
a.v NOT IN (b.v,c.v,d.v,e.v,f.v,g.v) AND
b.v NOT IN (c.v,d.v,e.v,f.v,g.v) AND
c.v NOT IN (d.v,e.v,f.v,g.v) AND
d.v NOT IN (e.v,f.v,g.v) AND
e.v NOT IN (f.v,g.v) AND
f.v NOT IN (g.v) AND
a.v = c.v + d.v AND
g.v = d.v + e.v AND
b.v = e.v + f.v - c.v
ORDER BY
a,b,c,d,e,f,g;
uorn := ' unique solutions in ';
ELSE
INSERT INTO results
SELECT
a.v a,
b.v b,
c.v c,
d.v d,
e.v e,
f.v f,
g.v g
FROM
allints a, allints b, allints c,allints d,
allints e, allints f, allints g
WHERE
a.v = c.v + d.v AND
g.v = d.v + e.v AND
b.v = e.v + f.v - c.v
ORDER BY
a,b,c,d,e,f,g;
uorn := ' non-unique solutions in ';
END IF;
COMMIT;
 
OPEN results_cur;
LOOP
FETCH results_cur INTO results_rec;
EXIT WHEN results_cur%notfound;
a := results_rec.a;
b := results_rec.b;
c := results_rec.c;
d := results_rec.d;
e := results_rec.e;
f := results_rec.f;
g := results_rec.g;
 
solutions := solutions + 1;
IF show = TRUE THEN
out_line := TO_CHAR(a) || ' ';
out_line := out_line || ' ' || TO_CHAR(b) || ' ';
out_line := out_line || ' ' || TO_CHAR(c) || ' ';
out_line := out_line || ' ' || TO_CHAR(d) || ' ';
out_line := out_line || ' ' || TO_CHAR(e) || ' ';
out_line := out_line || ' ' || TO_CHAR(f) ||' ';
out_line := out_line || ' ' || TO_CHAR(g);
END IF;
 
DBMS_OUTPUT.put_line(out_line);
END LOOP;
CLOSE results_cur;
out_line := TO_CHAR(solutions) || uorn;
out_line := out_line || TO_CHAR(lo) || ' to ' || TO_CHAR(hi);
DBMS_OUTPUT.put_line(out_line);
 
END;
/
 

Output

SQL> execute foursquares(1,7,TRUE,TRUE);
3  7  2  1  5  4  6                                                             
4  5  3  1  6  2  7                                                             
4  7  1  3  2  6  5                                                             
5  6  2  3  1  7  4                                                             
6  4  1  5  2  3  7                                                             
6  4  5  1  2  7  3                                                             
7  2  6  1  3  5  4                                                             
7  3  2  5  1  4  6                                                             
8 unique solutions in 1 to 7                                                    

PL/SQL procedure successfully completed.

SQL> execute foursquares(3,9,TRUE,TRUE);
7  8  3  4  5  6  9                                                             
8  7  3  5  4  6  9                                                             
9  6  4  5  3  7  8                                                             
9  6  5  4  3  8  7                                                             
4 unique solutions in 3 to 9                                                    

PL/SQL procedure successfully completed.

SQL> execute foursquares(0,9,FALSE,FALSE);
2860 non-unique solutions in 0 to 9                                             

PL/SQL procedure successfully completed.

Prolog[edit]

Works with SWI-Prolog 7.5.8

 
:- use_module(library(clpfd)).
 
% main predicate
my_sum(Min, Max, Top, LL):-
L = [A,B,C,D,E,F,G],
L ins Min..Max,
( Top == 0
-> all_distinct(L)
; true),
R #= A+B,
R #= B+C+D,
R #= D+E+F,
R #= F+G,
setof(L, labeling([ff], L), LL).
 
 
my_sum_1(Min, Max) :-
my_sum(Min, Max, 0, LL),
maplist(writeln, LL).
 
my_sum_2(Min, Max, Len) :-
my_sum(Min, Max, 1, LL),
length(LL, Len).
 

Output

 ?- my_sum_1(1,7).
[3,7,2,1,5,4,6]
[4,5,3,1,6,2,7]
[4,7,1,3,2,6,5]
[5,6,2,3,1,7,4]
[6,4,1,5,2,3,7]
[6,4,5,1,2,7,3]
[7,2,6,1,3,5,4]
[7,3,2,5,1,4,6]
true.

 ?- my_sum_1(3,9).
[7,8,3,4,5,6,9]
[8,7,3,5,4,6,9]
[9,6,4,5,3,7,8]
[9,6,5,4,3,8,7]
true.

 ?- my_sum_2(0,9,N).
N = 2860.

Python[edit]

Procedural[edit]

Itertools[edit]

import itertools
 
def all_equal(a,b,c,d,e,f,g):
return a+b == b+c+d == d+e+f == f+g
 
def foursquares(lo,hi,unique,show):
solutions = 0
if unique:
uorn = "unique"
citer = itertools.combinations(range(lo,hi+1),7)
else:
uorn = "non-unique"
citer = itertools.combinations_with_replacement(range(lo,hi+1),7)
 
for c in citer:
for p in set(itertools.permutations(c)):
if all_equal(*p):
solutions += 1
if show:
print str(p)[1:-1]
 
print str(solutions)+" "+uorn+" solutions in "+str(lo)+" to "+str(hi)
print

Output

foursquares(1,7,True,True)
4, 5, 3, 1, 6, 2, 7
3, 7, 2, 1, 5, 4, 6
5, 6, 2, 3, 1, 7, 4
4, 7, 1, 3, 2, 6, 5
6, 4, 5, 1, 2, 7, 3
7, 3, 2, 5, 1, 4, 6
7, 2, 6, 1, 3, 5, 4
6, 4, 1, 5, 2, 3, 7
8 unique solutions in 1 to 7


foursquares(3,9,True,True)
7, 8, 3, 4, 5, 6, 9
9, 6, 4, 5, 3, 7, 8
8, 7, 3, 5, 4, 6, 9
9, 6, 5, 4, 3, 8, 7
4 unique solutions in 3 to 9


foursquares(0,9,False,False)
2860 non-unique solutions in 0 to 9

Generators[edit]

Faster solution without itertools

 
def foursquares(lo,hi,unique,show):
 
def acd_iter():
"""
Iterates through all the possible valid values of
a, c, and d.
 
a = c + d
"""

for c in range(lo,hi+1):
for d in range(lo,hi+1):
if (not unique) or (c <> d):
a = c + d
if a >= lo and a <= hi:
if (not unique) or (c <> 0 and d <> 0):
yield (a,c,d)
 
def ge_iter():
"""
Iterates through all the possible valid values of
g and e.
 
g = d + e
"""

for e in range(lo,hi+1):
if (not unique) or (e not in (a,c,d)):
g = d + e
if g >= lo and g <= hi:
if (not unique) or (g not in (a,c,d,e)):
yield (g,e)
 
def bf_iter():
"""
Iterates through all the possible valid values of
b and f.
 
b = e + f - c
"""

for f in range(lo,hi+1):
if (not unique) or (f not in (a,c,d,g,e)):
b = e + f - c
if b >= lo and b <= hi:
if (not unique) or (b not in (a,c,d,g,e,f)):
yield (b,f)
 
solutions = 0
acd_itr = acd_iter()
for acd in acd_itr:
a,c,d = acd
ge_itr = ge_iter()
for ge in ge_itr:
g,e = ge
bf_itr = bf_iter()
for bf in bf_itr:
b,f = bf
solutions += 1
if show:
print str((a,b,c,d,e,f,g))[1:-1]
if unique:
uorn = "unique"
else:
uorn = "non-unique"
 
print str(solutions)+" "+uorn+" solutions in "+str(lo)+" to "+str(hi)
print
Output
foursquares(1,7,True,True)
4, 7, 1, 3, 2, 6, 5
6, 4, 1, 5, 2, 3, 7
3, 7, 2, 1, 5, 4, 6
5, 6, 2, 3, 1, 7, 4
7, 3, 2, 5, 1, 4, 6
4, 5, 3, 1, 6, 2, 7
6, 4, 5, 1, 2, 7, 3
7, 2, 6, 1, 3, 5, 4
8 unique solutions in 1 to 7


foursquares(3,9,True,True)
7, 8, 3, 4, 5, 6, 9
8, 7, 3, 5, 4, 6, 9
9, 6, 4, 5, 3, 7, 8
9, 6, 5, 4, 3, 8, 7
4 unique solutions in 3 to 9


foursquares(0,9,False,False)
2860 non-unique solutions in 0 to 9

Functional[edit]

Translation of: Haskell
Translation of: JavaScript
Works with: Python version 3.7
'''4-rings or 4-squares puzzle'''
 
from itertools import chain
 
 
# rings :: noRepeatedDigits -> DigitList -> Lists of solutions
# rings :: Bool -> [Int] -> [[Int]]
def rings(uniq):
'''Sets of unique or non-unique integer values
(drawn from the `digits` argument)
for each of the seven names [a..g] such that:
(a + b) == (b + c + d) == (d + e + f) == (f + g)
'''

def go(digits):
ns = sorted(digits, reverse=True)
h = ns[0]
 
# CENTRAL DIGIT :: d
def central(d):
xs = list(filter(lambda x: h >= (d + x), ns))
 
# LEFT NEIGHBOUR AND LEFTMOST :: c and a
def left(c):
a = c + d
if a > h:
return []
else:
# RIGHT NEIGHBOUR AND RIGHTMOST :: e and g
def right(e):
g = d + e
if ((g > h) or (uniq and (g == c))):
return []
else:
agDelta = a - g
bfs = difference(ns)(
[d, c, e, g, a]
) if uniq else ns
 
# MID LEFT AND RIGHT :: b and f
def midLeftRight(b):
f = b + agDelta
return [[a, b, c, d, e, f, g]] if (
(f in bfs) and (
(not uniq) or (
f not in [a, b, c, d, e, g]
)
)
) else []
 
# CANDIDATE DIGITS BOUND TO POSITIONS [a .. g] --------
 
return concatMap(midLeftRight)(bfs)
 
return concatMap(right)(
difference(xs)([d, c, a]) if uniq else ns
)
 
return concatMap(left)(
delete(d)(xs) if uniq else ns
)
 
return concatMap(central)(ns)
 
return lambda digits: go(digits) if digits else []
 
 
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Testing unique digits [1..7], [3..9] and unrestricted digits'''
 
print(main.__doc__ + ':\n')
print(unlines(map(
lambda tpl: '\nrings' + repr(tpl) + ':\n\n' + unlines(
map(repr, uncurry(rings)(*tpl))
), [
(True, enumFromTo(1)(7)),
(True, enumFromTo(3)(9))
]
)))
tpl = (False, enumFromTo(0)(9))
print(
'\n\nlen(rings' + repr(tpl) + '):\n\n' +
str(len(uncurry(rings)(*tpl)))
)
 
 
# GENERIC -------------------------------------------------
 
# concatMap :: (a -> [b]) -> [a] -> [b]
def concatMap(f):
'''A concatenated list over which a function has been mapped.
The list monad can be derived by using a function f which
wraps its output in a list,
(using an empty list to represent computational failure).
'''

return lambda xs: list(
chain.from_iterable(map(f, xs))
)
 
 
# delete :: Eq a => a -> [a] -> [a]
def delete(x):
'''xs with the first of any instances of x removed.'''
def go(xs):
xs.remove(x)
return xs
return lambda xs: go(list(xs)) if (
x in xs
) else list(xs)
 
 
# difference :: Eq a => [a] -> [a] -> [a]
def difference(xs):
'''All elements of ys except any also found in xs'''
def go(ys):
s = set(ys)
return [x for x in xs if x not in s]
return lambda ys: go(ys)
 
 
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
 
 
# uncurry :: (a -> b -> c) -> ((a, b) -> c)
def uncurry(f):
'''A function over a pair of arguments,
derived from a vanilla or curried function.
'''

return lambda x, y: f(x)(y)
 
 
# unlines :: [String] -> String
def unlines(xs):
'''A single string formed by the intercalation
of a list of strings with the newline character.
'''

return '\n'.join(xs)
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
Testing unique digits [1..7], [3..9] and unrestricted digits:

rings(True, [1, 2, 3, 4, 5, 6, 7]):

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

rings(True, [3, 4, 5, 6, 7, 8, 9]):

[9, 6, 4, 5, 3, 7, 8]
[8, 7, 3, 5, 4, 6, 9]
[9, 6, 5, 4, 3, 8, 7]
[7, 8, 3, 4, 5, 6, 9]


len(rings(False, [0, 1, 2, 3, 4, 5, 6, 7, 8, 9])):

2860

R[edit]

Function "perms" is a modified version of the "permutations" function from the "gtools" R package.

# 4 rings or 4 squares puzzle
 
perms <- function (n, r, v = 1:n, repeats.allowed = FALSE) {
if (repeats.allowed)
sub <- function(n, r, v) {
if (r == 1)
matrix(v, n, 1)
else if (n == 1)
matrix(v, 1, r)
else {
inner <- Recall(n, r - 1, v)
cbind(rep(v, rep(nrow(inner), n)), matrix(t(inner),
ncol = ncol(inner), nrow = nrow(inner) * n,
byrow = TRUE))
}
}
else sub <- function(n, r, v) {
if (r == 1)
matrix(v, n, 1)
else if (n == 1)
matrix(v, 1, r)
else {
X <- NULL
for (i in 1:n) X <- rbind(X, cbind(v[i], Recall(n - 1, r - 1, v[-i])))
X
}
}
X <- sub(n, r, v[1:n])
 
result <- vector(mode = "numeric")
 
for(i in 1:nrow(X)){
y <- X[i, ]
x1 <- y[1] + y[2]
x2 <- y[2] + y[3] + y[4]
x3 <- y[4] + y[5] + y[6]
x4 <- y[6] + y[7]
if(x1 == x2 & x2 == x3 & x3 == x4) result <- rbind(result, y)
}
return(result)
}
 
print_perms <- function(n, r, v = 1:n, repeats.allowed = FALSE, table.out = FALSE) {
a <- perms(n, r, v, repeats.allowed)
colnames(a) <- rep("", ncol(a))
rownames(a) <- rep("", nrow(a))
if(!repeats.allowed){
print(a)
cat(paste('\n', nrow(a), 'unique solutions from', min(v), 'to', max(v)))
} else {
cat(paste('\n', nrow(a), 'non-unique solutions from', min(v), 'to', max(v)))
}
}
 
registerS3method("print_perms", "data.frame", print_perms)
 
print_perms(7, 7, repeats.allowed = FALSE, table.out = TRUE)
print_perms(7, 7, v = 3:9, repeats.allowed = FALSE, table.out = TRUE)
print_perms(10, 7, v = 0:9, repeats.allowed = TRUE, table.out = FALSE)
 
 
Output:
              
 3 7 2 1 5 4 6
 4 5 3 1 6 2 7
 4 7 1 3 2 6 5
 5 6 2 3 1 7 4
 6 4 1 5 2 3 7
 6 4 5 1 2 7 3
 7 2 6 1 3 5 4
 7 3 2 5 1 4 6

 8 unique solutions from 1 to 7
             
 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7

 4 unique solutions from 3 to 9

 2860 non-unique solutions from 0 to 9

REXX[edit]

fast version[edit]

This REXX version is faster than the more idiomatic version, but is longer (statement-wise) and
a bit easier to read (visualize).

/*REXX pgm solves the 4-rings puzzle,  where letters represent unique (or not) digits). */
arg LO HI unique show . /*the ARG statement capitalizes args.*/
if LO=='' | LO=="," then LO=1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI=7 /* " " " " " " */
if unique=='' | unique==',' | unique=='UNIQUE' then unique=1 /*unique letter solutions*/
else unique=0 /*non-unique " */
if show=='' | show==',' | show=='SHOW' then show=1 /*noshow letter solutions*/
else show=0 /* show " " */
w=max(3, length(LO), length(HI) ) /*maximum width of any number found. */
bar=copies('═', w) /*define a horizontal bar (for title). */
times=HI - LO + 1 /*calculate number of times to loop. */
#=0 /*number of solutions found (so far). */
do a=LO for times
do b=LO for times
if unique then if b==a then iterate
do c=LO for times
if unique then do; if c==a then iterate
if c==b then iterate
end
do d=LO for times
if unique then do; if d==a then iterate
if d==b then iterate
if d==c then iterate
end
do e=LO for times
if unique then do; if e==a then iterate
if e==b then iterate
if e==c then iterate
if e==d then iterate
end
do f=LO for times
if unique then do; if f==a then iterate
if f==b then iterate
if f==c then iterate
if f==d then iterate
if f==e then iterate
end
do g=LO for times
if unique then do; if g==a then iterate
if g==b then iterate
if g==c then iterate
if g==d then iterate
if g==e then iterate
if g==f then iterate
end
sum=a+b
if f+g\==sum then iterate
if b+c+d\==sum then iterate
if d+e+f\==sum then iterate
#=# + 1 /*bump the count of solutions.*/
if #==1 then call align 'a', 'b', 'c', 'd', 'e', 'f', 'g'
if #==1 then call align bar, bar, bar, bar, bar, bar, bar
call align a, b, c, d, e, f, g
end /*g*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say
_= ' non-unique'
if unique then _= ' unique '
say # _ 'solutions found.'
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
align: parse arg a1,a2,a3,a4,a5,a6,a7
if show then say left('',9) center(a1,w) center(a2,w) center(a3,w) center(a4,w),
center(a5,w) center(a6,w) center(a7,w)
return
output   when using the default inputs:     1   7
           a   b   c   d   e   f   g
          ═══ ═══ ═══ ═══ ═══ ═══ ═══
           3   7   2   1   5   4   6
           4   5   3   1   6   2   7
           4   7   1   3   2   6   5
           5   6   2   3   1   7   4
           6   4   1   5   2   3   7
           6   4   5   1   2   7   3
           7   2   6   1   3   5   4
           7   3   2   5   1   4   6

8  unique  solutions found.
output   when using the input of:     3   9
           a   b   c   d   e   f   g
          ═══ ═══ ═══ ═══ ═══ ═══ ═══
           7   8   3   4   5   6   9
           8   7   3   5   4   6   9
           9   6   4   5   3   7   8
           9   6   5   4   3   8   7

4  unique  solutions found.
output   when using the input of:     0   9   non-unique   noshow
2860  non-unique solutions found.

idiomatic version[edit]

This REXX version is slower than the faster version   (because of the multiple   if   clauses.

Note that the REXX language doesn't have short-circuits   (when executing multiple clauses in   if   (and other)   statements.

/*REXX pgm solves the 4-rings puzzle,  where letters represent unique (or not) digits). */
arg LO HI unique show . /*the ARG statement capitalizes args.*/
if LO=='' | LO=="," then LO=1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI=7 /* " " " " " " */
if unique=='' | unique==',' | unique=='UNIQUE' then u=1 /*unique letter solutions*/
else u=0 /*non-unique " */
if show=='' | show==',' | show=='SHOW' then show=1 /*noshow letter solutions*/
else show=0 /* show " " */
w=max(3, length(LO), length(HI) ) /*maximum width of any number found. */
bar=copies('═', w) /*define a horizontal bar (for title). */
times=HI - LO + 1 /*calculate number of times to loop. */
#=0 /*number of solutions found (so far). */
do a=LO for times
do b=LO for times; if u then if b==a then iterate
do c=LO for times; if u then if c==a|c==b then iterate
do d=LO for times; if u then if d==a|d==b|d==c then iterate
do e=LO for times; if u then if e==a|e==b|e==c|e==d then iterate
do f=LO for times; if u then if f==a|f==b|f==c|f==d|f==e then iterate
do g=LO for times; if u then if g==a|g==b|g==c|g==d|g==e|g==f then iterate
sum=a+b
if f+g==sum & b+c+d==sum & d+e+f==sum then #=#+1 /*bump # of solutions.*/
else iterate /*sum not equal, no─go*/
if #==1 then call align 'a', 'b', 'c', 'd', 'e', 'f', 'g'
if #==1 then call align bar, bar, bar, bar, bar, bar, bar
call align a, b, c, d, e, f, g
end /*g*/ /*for 1st time, show title*/
end /*f*/
end /*e*/
end /*d*/
end /*c*/
end /*b*/
end /*a*/
say
_= ' non-unique'
if u then _= ' unique '
say # _ 'solutions found.'
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
align: parse arg a1,a2,a3,a4,a5,a6,a7
if show then say left('',9) center(a1,w) center(a2,w) center(a3,w) center(a4,w),
center(a5,w) center(a6,w) center(a7,w)
return
output   is identical to the faster REXX version.


Ruby[edit]

def four_squares(low, high, unique=true, show=unique)
f = -> (a,b,c,d,e,f,g) {[a+b, b+c+d, d+e+f, f+g].uniq.size == 1}
if unique
uniq = "unique"
solutions = [*low..high].permutation(7).select{|ary| f.call(*ary)}
else
uniq = "non-unique"
solutions = [*low..high].repeated_permutation(7).select{|ary| f.call(*ary)}
end
if show
puts " " + [*"a".."g"].join(" ")
solutions.each{|ary| p ary}
end
puts "#{solutions.size} #{uniq} solutions in #{low} to #{high}"
puts
end
 
[[1,7], [3,9]].each do |low, high|
four_squares(low, high)
end
four_squares(0, 9, false)
Output:
 a  b  c  d  e  f  g
[3, 7, 2, 1, 5, 4, 6]
[4, 5, 3, 1, 6, 2, 7]
[4, 7, 1, 3, 2, 6, 5]
[5, 6, 2, 3, 1, 7, 4]
[6, 4, 1, 5, 2, 3, 7]
[6, 4, 5, 1, 2, 7, 3]
[7, 2, 6, 1, 3, 5, 4]
[7, 3, 2, 5, 1, 4, 6]
8 unique solutions in 1 to 7

 a  b  c  d  e  f  g
[7, 8, 3, 4, 5, 6, 9]
[8, 7, 3, 5, 4, 6, 9]
[9, 6, 4, 5, 3, 7, 8]
[9, 6, 5, 4, 3, 8, 7]
4 unique solutions in 3 to 9

2860 non-unique solutions in 0 to 9

Rust[edit]

 
#![feature(inclusive_range_syntax)]
 
fn is_unique(a: u8, b: u8, c: u8, d: u8, e: u8, f: u8, g: u8) -> bool {
a != b && a != c && a != d && a != e && a != f && a != g &&
b != c && b != d && b != e && b != f && b != g &&
c != d && c != e && c != f && c != g &&
d != e && d != f && d != g &&
e != f && e != g &&
f != g
}
 
fn is_solution(a: u8, b: u8, c: u8, d: u8, e: u8, f: u8, g: u8) -> bool {
a + b == b + c + d &&
b + c + d == d + e + f &&
d + e + f == f + g
}
 
fn four_squares(low: u8, high: u8, unique: bool) -> Vec<Vec<u8>> {
let mut results: Vec<Vec<u8>> = Vec::new();
 
for a in low..=high {
for b in low..=high {
for c in low..=high {
for d in low..=high {
for e in low..=high {
for f in low..=high {
for g in low..=high {
if (!unique || is_unique(a, b, c, d, e, f, g)) &&
is_solution(a, b, c, d, e, f, g) {
results.push(vec![a, b, c, d, e, f, g]);
}
}
}
}
}
}
}
}
results
}
 
fn print_results(solutions: &Vec<Vec<u8>>) {
for solution in solutions {
println!("{:?}", solution)
}
}
 
fn print_results_summary(solutions: usize, low: u8, high: u8, unique: bool) {
let uniqueness = if unique {
"unique"
} else {
"non-unique"
};
println!("{} {} solutions in {} to {} range", solutions, uniqueness, low, high)
}
 
fn uniques(low: u8, high: u8) {
let solutions = four_squares(low, high, true);
print_results(&solutions);
print_results_summary(solutions.len(), low, high, true);
}
 
fn nonuniques(low: u8, high: u8) {
let solutions = four_squares(low, high, false);
print_results_summary(solutions.len(), low, high, false);
}
 
fn main() {
uniques(1, 7);
println!();
uniques(3, 9);
println!();
nonuniques(0, 9);
}
 
Output:
[3, 7, 2, 1, 5, 4, 6]
[4, 5, 3, 1, 6, 2, 7]
[4, 7, 1, 3, 2, 6, 5]
[5, 6, 2, 3, 1, 7, 4]
[6, 4, 1, 5, 2, 3, 7]
[6, 4, 5, 1, 2, 7, 3]
[7, 2, 6, 1, 3, 5, 4]
[7, 3, 2, 5, 1, 4, 6]
8 unique solutions in 1 to 7 range

[7, 8, 3, 4, 5, 6, 9]
[8, 7, 3, 5, 4, 6, 9]
[9, 6, 4, 5, 3, 7, 8]
[9, 6, 5, 4, 3, 8, 7]
4 unique solutions in 3 to 9 range

2860 non-unique solutions in 0 to 9 range

Scala[edit]

Translation of: Java
object FourRings {
def isValid(unique: Boolean, needle: Integer, haystack: Integer*): Boolean
= !unique || !haystack.contains(needle)
 
def fourSquare(low: Int, high: Int, unique: Boolean, print: Boolean): Unit = {
var count = 0
if (print) {
println("a b c d e f g")
}
(low to high).foreach(a => (low to high).foreach(b => if (isValid(unique, a, b)) {
val fp = a + b
(low to high).foreach(c => if (isValid(unique, c, a, b)) {
(low to high).foreach(d => if (isValid(unique, d, a, b, c) && fp == b + c + d) {
(low to high).foreach(e => if (isValid(unique, e, a, b, c, d)) {
(low to high).foreach(f => if (isValid(unique, f, a, b, c, d, e) && fp == d + e + f) {
(low to high).foreach(g => if (isValid(unique, g, a, b, c, d, e, f) && fp == f + g) {
count = count + 1
if (print) {
printf("%d %d %d %d %d %d %d%n", a, b, c, d, e, f, g)
}
})
})
})
})
})
}))
if (unique) {
printf("There are %d unique solutions in [%d, %d]%n", count, low, high)
} else {
printf("There are %d non-unique solutions in [%d, %d]%n", count, low, high)
}
}
 
def main(args: Array[String]): Unit = {
fourSquare(1, 7, unique = true, print = true)
fourSquare(3, 9, unique = true, print = true)
fourSquare(0, 9, unique = false, print = false)
}
}
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1, 7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3, 9]
There are 2860 non-unique solutions in [0, 9]

Scheme[edit]

 
(import (scheme base)
(scheme write)
(srfi 1))
 
;; return all combinations of size elements from given set
(define (combinations size set unique?)
(if (zero? size)
(list '())
(let loop ((base-combns (combinations (- size 1) set unique?))
(results '())
(items set))
(cond ((null? base-combns) ; end, as no base-combinations to process
results)
((null? items) ; check next base-combination
(loop (cdr base-combns)
results
set))
((and unique? ; ignore if wanting list unique
(member (car items) (car base-combns) =))
(loop base-combns
results
(cdr items)))
(else ; keep the new combination
(loop base-combns
(cons (cons (car items) (car base-combns))
results)
(cdr items)))))))
 
;; checks if all 4 sums are the same
(define (solution? a b c d e f g)
(= (+ a b)
(+ b c d)
(+ d e f)
(+ f g)))
 
;; Tasks
(display "Solutions: LOW=1 HIGH=7\n")
(display (filter (lambda (combination) (apply solution? combination))
(combinations 7 (iota 7 1) #t))) (newline)
 
(display "Solutions: LOW=3 HIGH=9\n")
(display (filter (lambda (combination) (apply solution? combination))
(combinations 7 (iota 7 3) #t))) (newline)
 
(display "Solution count: LOW=0 HIGH=9 non-unique\n")
(display (count (lambda (combination) (apply solution? combination))
(combinations 7 (iota 10 0) #f))) (newline)
 
Output:
Solutions: LOW=1 HIGH=7
((4 5 3 1 6 2 7) (6 4 1 5 2 3 7) (3 7 2 1 5 4 6) (7 3 2 5 1 4 6) (4 7 1 3 2 6 5) (7 2 6 1 3 5 4) (5 6 2 3 1 7 4) (6 4 5 1 2 7 3))
Solutions: LOW=3 HIGH=9
((7 8 3 4 5 6 9) (8 7 3 5 4 6 9) (9 6 4 5 3 7 8) (9 6 5 4 3 8 7))
Solution count: LOW=0 HIGH=9 non-unique
2860

Sidef[edit]

Translation of: Perl 6
func four_squares (list, unique=true, show=true) {
 
var solutions = []
 
func check(c) {
solutions << c if ([
c[0] + c[1],
c[1] + c[2] + c[3],
c[3] + c[4] + c[5],
c[5] + c[6],
].uniq.len == 1)
}
 
if (unique) {
list.combinations(7, {|*a|
a.permutations { |*c|
check(c)
}
})
} else {
7.of { list }.cartesian {|*c|
check(c)
}
}
 
say (solutions.len,
(unique ? ' ' : ' non-'),
"unique solutions found using #{list.join(', ')}.\n")
 
if (show) {
var f = "%#{list.max.len+1}s"
say ("\n".join(
('a'..'g').map{f % _}.join,
solutions.map{ .map{f % _}.join }...
), "\n")
}
}
 
# TASK
four_squares(@(1..7))
four_squares(@(3..9))
four_squares([8, 9, 11, 12, 17, 18, 20, 21])
four_squares(@(0..9), unique: false, show: false)
Output:
8 unique solutions found using 1, 2, 3, 4, 5, 6, 7.

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

4 unique solutions found using 3, 4, 5, 6, 7, 8, 9.

 a b c d e f g
 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7

8 unique solutions found using 8, 9, 11, 12, 17, 18, 20, 21.

  a  b  c  d  e  f  g
 17 21  8  9 11 18 20
 20 18 11  9  8 21 17
 17 21  9  8 12 18 20
 20 18  8 12  9 17 21
 20 18 12  8  9 21 17
 21 17  9 12  8 18 20
 20 18 11  9 12 17 21
 21 17 12  9 11 18 20

2860 non-unique solutions found using 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Simula[edit]

BEGIN
 
INTEGER PROCEDURE GETCOMBS(LOW, HIGH, UNIQUE, COMBS);
INTEGER LOW, HIGH;
INTEGER ARRAY COMBS;
BOOLEAN UNIQUE;
BEGIN
INTEGER A, B, C, D, E, F, G;
INTEGER NUM;
 
BOOLEAN PROCEDURE ISUNIQUE(A, B, C, D, E, F, G);
INTEGER A, B, C, D, E, F, G;
BEGIN
INTEGER ARRAY DATA(LOW:HIGH);
INTEGER I;
 
FOR I := LOW STEP 1 UNTIL HIGH DO
DATA(I) := -1;
 
FOR I := A, B, C, D, E, F, G DO
IF DATA(I) = -1
THEN DATA(I) := 1
ELSE GOTO L;
 
ISUNIQUE := TRUE;
L:
END;
 
PROCEDURE ADDCOMB;
BEGIN
NUM := NUM + 1;
COMBS(NUM, LOW + 0) := A;
COMBS(NUM, LOW + 1) := B;
COMBS(NUM, LOW + 2) := C;
COMBS(NUM, LOW + 3) := D;
COMBS(NUM, LOW + 4) := E;
COMBS(NUM, LOW + 5) := F;
COMBS(NUM, LOW + 6) := G;
END;
 
FOR A := LOW STEP 1 UNTIL HIGH DO
FOR B := LOW STEP 1 UNTIL HIGH DO
FOR C := LOW STEP 1 UNTIL HIGH DO
FOR D := LOW STEP 1 UNTIL HIGH DO
FOR E := LOW STEP 1 UNTIL HIGH DO
FOR F := LOW STEP 1 UNTIL HIGH DO
FOR G := LOW STEP 1 UNTIL HIGH DO
BEGIN
IF VALIDCOMB(A, B, C, D, E, F, G) THEN
BEGIN
IF UNIQUE THEN
BEGIN IF ISUNIQUE(A, B, C, D, E, F, G) THEN ADDCOMB END
ELSE ADDCOMB;
END;
END;
GETCOMBS := NUM;
END;
 
 
BOOLEAN PROCEDURE VALIDCOMB(A, B, C, D, E, F, G);
INTEGER A, B, C, D, E, F, G;
BEGIN
INTEGER SQUARE1, SQUARE2, SQUARE3, SQUARE4;
 
SQUARE1 := A + B;
SQUARE2 := B + C + D;
SQUARE3 := D + E + F;
SQUARE4 := F + G;
VALIDCOMB := SQUARE1 = SQUARE2 AND SQUARE2 = SQUARE3 AND SQUARE3 = SQUARE4
END;
 
COMMENT ----- MAIN PROGRAM ----- ;
 
INTEGER ARRAY LO(1:3);
INTEGER ARRAY HI(1:3);
BOOLEAN ARRAY UQ(1:3);
INTEGER I;
 
LO(1) := 1; HI(1) := 7; UQ(1) := TRUE;
LO(2) := 3; HI(2) := 9; UQ(2) := TRUE;
LO(3) := 0; HI(3) := 9; UQ(3) := FALSE;
 
FOR I := 1 STEP 1 UNTIL 3 DO
BEGIN
INTEGER LOW, HIGH;
BOOLEAN UNIQ;
 
LOW := LO(I); HIGH := HI(I); UNIQ := UQ(I);
BEGIN
INTEGER ARRAY VALIDCOMBS(1:8000, LOW:HIGH);
INTEGER N;
 
N := GETCOMBS(LOW, HIGH, UNIQ, VALIDCOMBS);
OUTINT(N, 0);
IF UNIQ THEN OUTTEXT(" UNIQUE");
OUTTEXT(" SOLUTIONS IN ");
OUTINT(LOW, 0); OUTTEXT(" TO ");
OUTINT(HIGH, 0);
OUTIMAGE;
IF I < 3 THEN
BEGIN INTEGER I, J;
FOR I := 1 STEP 1 UNTIL N DO
BEGIN
OUTTEXT("[");
FOR J := LOW STEP 1 UNTIL HIGH DO
OUTINT(VALIDCOMBS(I, J), 2);
OUTTEXT(" ]");
OUTIMAGE;
END;
END;
END;
END;
 
END.
 
Output:
8 UNIQUE SOLUTIONS IN 1 TO 7
[ 3 7 2 1 5 4 6 ]
[ 4 5 3 1 6 2 7 ]
[ 4 7 1 3 2 6 5 ]
[ 5 6 2 3 1 7 4 ]
[ 6 4 1 5 2 3 7 ]
[ 6 4 5 1 2 7 3 ]
[ 7 2 6 1 3 5 4 ]
[ 7 3 2 5 1 4 6 ]
4 UNIQUE SOLUTIONS IN 3 TO 9
[ 7 8 3 4 5 6 9 ]
[ 8 7 3 5 4 6 9 ]
[ 9 6 4 5 3 7 8 ]
[ 9 6 5 4 3 8 7 ]
2860 SOLUTIONS IN 0 TO 9

SQL PL[edit]

Works with: Db2 LUW
version 9.7 or higher.

With SQL PL:

 
--#SET TERMINATOR @
 
SET SERVEROUTPUT ON @
 
CREATE TABLE ALL_INTS (
V INTEGER
)@
 
CREATE TABLE RESULTS (
A INTEGER,
B INTEGER,
C INTEGER,
D INTEGER,
E INTEGER,
F INTEGER,
G INTEGER
)@
 
CREATE OR REPLACE PROCEDURE FOUR_SQUARES(
IN LO INTEGER,
IN HI INTEGER,
IN UNIQ SMALLINT,
--IN UNIQ BOOLEAN,
IN SHOW SMALLINT)
--IN SHOW BOOLEAN)
BEGIN
DECLARE A INTEGER;
DECLARE B INTEGER;
DECLARE C INTEGER;
DECLARE D INTEGER;
DECLARE E INTEGER;
DECLARE F INTEGER;
DECLARE G INTEGER;
DECLARE OUT_LINE VARCHAR(2000);
DECLARE I SMALLINT;
 
DECLARE SOLUTIONS INTEGER;
DECLARE UORN VARCHAR(2000);
 
SET SOLUTIONS = 0;
DELETE FROM ALL_INTS;
DELETE FROM RESULTS;
SET I = LO;
WHILE (I <= HI) DO
INSERT INTO ALL_INTS VALUES (I);
SET I = I + 1;
END WHILE;
COMMIT;
 
-- Computes unique solutions.
IF (UNIQ = 0) THEN
--IF (UNIQ = TRUE) THEN
INSERT INTO RESULTS
SELECT
A.V A, B.V B, C.V C, D.V D, E.V E, F.V F, G.V G
FROM
ALL_INTS A, ALL_INTS B, ALL_INTS C, ALL_INTS D, ALL_INTS E, ALL_INTS F,
ALL_INTS G
WHERE
A.V NOT IN (B.V, C.V, D.V, E.V, F.V, G.V)
AND B.V NOT IN (C.V, D.V, E.V, F.V, G.V)
AND C.V NOT IN (D.V, E.V, F.V, G.V)
AND D.V NOT IN (E.V, F.V, G.V)
AND E.V NOT IN (F.V, G.V)
AND F.V NOT IN (G.V)
AND A.V = C.V + D.V
AND G.V = D.V + E.V
AND B.V = E.V + F.V - C.V
ORDER BY
A, B, C, D, E, F, G;
SET UORN = ' unique solutions in ';
ELSE
-- Compute non-unique solutions.
INSERT INTO RESULTS
SELECT
A.V A, B.V B, C.V C, D.V D, E.V E, F.V F, G.V G
FROM
ALL_INTS A, ALL_INTS B, ALL_INTS C, ALL_INTS D, ALL_INTS E, ALL_INTS F,
ALL_INTS G
WHERE
A.V = C.V + D.V
AND G.V = D.V + E.V
AND B.V = E.V + F.V - C.V
ORDER BY
A, B, C, D, E, F, G;
SET UORN = ' non-unique solutions in ';
END IF;
COMMIT;
 
-- Counts the possible solutions.
FOR v AS c CURSOR FOR
SELECT
A, B, C, D, E, F, G
FROM RESULTS
ORDER BY
A, B, C, D, E, F, G
DO
SET SOLUTIONS = SOLUTIONS + 1;
-- Shows the results.
IF (SHOW = 0) THEN
--IF (SHOW = TRUE) THEN
SET OUT_LINE = A || ' ' || B || ' ' || C || ' ' || D || ' ' || E || ' '
|| F ||' ' || G;
CALL DBMS_OUTPUT.PUT_LINE(OUT_LINE);
END IF;
END FOR;
 
SET OUT_LINE = SOLUTIONS || UORN || LO || ' to ' || HI;
CALL DBMS_OUTPUT.PUT_LINE(OUT_LINE);
END
@
 
CALL FOUR_SQUARES(1, 7, 0, 0)@
CALL FOUR_SQUARES(3, 9, 0, 0)@
CALL FOUR_SQUARES(0, 9, 1, 1)@
 

Output:

db2 [email protected]
db2 => CREATE TABLE ALL_INTS ( V INTEGER )
DB20000I  The SQL command completed successfully.

db2 => CREATE TABLE RESULTS ( A INTEGER, B INTEGER, C INTEGER, D INTEGER, E INTEGER, F INTEGER, G INTEGER )
DB20000I  The SQL command completed successfully.

db2 => CREATE OR REPLACE PROCEDURE FOUR_SQUARES(
...
db2 (cont.) => END @
DB20000I  The SQL command completed successfully.

db2 => CALL FOUR_SQUARES(1, 7, 0, 0)

  Return Status = 0

3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
8 unique solutions in 1 TO 7

db2 => CALL FOUR_SQUARES(3, 9, 0, 0)

  Return Status = 0

7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
4 unique solutions in 3 TO 9

CALL FOUR_SQUARES(0, 9, 1, 1)

  Return Status = 0

2860 non-unique solutions in 0 TO 9

Stata[edit]

Use the program perm in the Permutations task for the first two questions, as it's fast enough. Use joinby for the third.

perm 7
rename * (a b c d e f g)
list if a==c+d & b+c==e+f & d+e==g, noobs sep(50)
 
+---------------------------+
| a b c d e f g |
|---------------------------|
| 3 7 2 1 5 4 6 |
| 4 5 3 1 6 2 7 |
| 4 7 1 3 2 6 5 |
| 5 6 2 3 1 7 4 |
| 6 4 1 5 2 3 7 |
| 6 4 5 1 2 7 3 |
| 7 2 6 1 3 5 4 |
| 7 3 2 5 1 4 6 |
+---------------------------+
 
foreach var of varlist _all {
replace `var'=`var'+2
}
list if a==c+d & b+c==e+f & d+e==g, noobs sep(50)
 
+---------------------------+
| a b c d e f g |
|---------------------------|
| 7 8 3 4 5 6 9 |
| 8 7 3 5 4 6 9 |
| 9 6 4 5 3 7 8 |
| 9 6 5 4 3 8 7 |
+---------------------------+
 
clear
set obs 10
gen b=_n-1
gen q=1
save temp, replace
rename b c
joinby q using temp
rename b d
joinby q using temp
rename b e
gen a=c+d
gen g=d+e
drop if a>9 | g>9
joinby q using temp
gen f=b+c-e
drop if f<0 | f>9
drop q
order a b c d e f g
erase temp.dta
count
2,860

Tcl[edit]

This task is a good opportunity to practice metaprogramming in Tcl. The procedure compile_4rings builds a lambda expression which takes values for {a b c d e f g} as parameters and returns true if those values satisfy the specified expressions ($exprs). This approach lets the bytecode compiler optimise our code.

For the final challenge, we vary the code generation a bit in compile_4rings_hard: instead of a lambda taking parameters, this generates a nested loop that searches exhaustively through the possible values for each variable.

The puzzle can be varied freely by changing the values of $vars and $exprs specified at the top of the script.

set vars {a b c d e f g}
set exprs {
{$a+$b}
{$b+$c+$d}
{$d+$e+$f}
{$f+$g}
}
 
proc permute {xs} {
if {[llength $xs] < 2} {
return $xs
}
set i -1
foreach x $xs {
incr i
set rest [lreplace $xs $i $i]
foreach rest [permute $rest] {
lappend res [list $x {*}$rest]
}
}
return $res
}
 
proc range {a b} {
set a [uplevel 1 [list expr $a]]
set b [uplevel 1 [list expr $b]]
set res {}
while {$a <= $b} {
lappend res $a
incr a
}
return $res
}
 
proc compile_4rings {vars exprs} {
set script "set _ \[[list expr [lindex $exprs 0]]\]\n"
foreach expr [lrange $exprs 1 end] {
append script "if {\$_ != $expr} {return false}\n"
}
append script "return true\n"
list $vars $script
}
 
proc solve_4rings {vars exprs range} {
set lambda [compile_4rings $vars $exprs]
foreach values [permute $range] {
if {[apply $lambda {*}$values]} {
puts " $values"
}
}
}
 
proc compile_4rings_hard {vars exprs values} {
append script "set _ \[[list expr [lindex $exprs 0]]\]\n"
foreach expr [lrange $exprs 1 end] {
append script "if {\$_ != $expr} {continue}\n"
}
append script "incr res\n"
foreach var $vars {
set script [list foreach $var $values $script]
}
set script "set res 0\n$script\nreturn \$res"
list {} $script
}
 
proc solve_4rings_hard {vars exprs range} {
apply [compile_4rings_hard $vars $exprs $range]
}
 
puts "# Combinations of 1..7:"
solve_4rings $vars $exprs [range 1 7]
puts "# Combinations of 3..9:"
solve_4rings $vars $exprs [range 3 9]
puts "# Number of solutions, free over 0..9:"
puts [solve_4rings_hard $vars $exprs [range 0 9]]
Output:
# Combinations of 1..7:
 3 7 2 1 5 4 6
 4 5 3 1 6 2 7
 4 7 1 3 2 6 5
 5 6 2 3 1 7 4
 6 4 1 5 2 3 7
 6 4 5 1 2 7 3
 7 2 6 1 3 5 4
 7 3 2 5 1 4 6
# Combinations of 3..9:
 7 8 3 4 5 6 9
 8 7 3 5 4 6 9
 9 6 4 5 3 7 8
 9 6 5 4 3 8 7
# Number of solutions, free over 0..9:
2860

X86 Assembly[edit]

See 4-rings_or_4-squares_puzzle/X86 Assembly

VBA[edit]

Translation of: C
Dim a As Integer, b As Integer, c As Integer, d As Integer
Dim e As Integer, f As Integer, g As Integer
Dim lo As Integer, hi As Integer, unique As Boolean, show As Boolean
Dim solutions As Integer
Private Sub bf()
For f = lo To hi
If ((Not unique) Or _
((f <> a And f <> c And f <> d And f <> g And f <> e))) Then
b = e + f - c
If ((b >= lo) And (b <= hi) And _
((Not unique) Or ((b <> a) And (b <> c) And _
(b <> d) And (b <> g) And (b <> e) And (b <> f)))) Then
solutions = solutions + 1
If show Then Debug.Print a; b; c; d; e; f; g
End If
End If
Next
End Sub
Private Sub ge()
For e = lo To hi
If ((Not unique) Or ((e <> a) And (e <> c) And (e <> d))) Then
g = d + e
If ((g >= lo) And (g <= hi) And _
((Not unique) Or ((g <> a) And (g <> c) And _
(g <> d) And (g <> e)))) Then
bf
End If
End If
Next
End Sub
Private Sub acd()
For c = lo To hi
For d = lo To hi
If ((Not unique) Or (c <> d)) Then
a = c + d
If ((a >= lo) And (a <= hi) And _
((Not unique) Or ((c <> 0) And (d <> 0)))) Then
ge
End If
End If
Next d
Next c
End Sub
Private Sub foursquares(plo As Integer, phi As Integer, punique As Boolean, pshow As Boolean)
lo = plo
hi = phi
unique = punique
show = pshow
solutions = 0
acd
Debug.Print
If unique Then
Debug.Print solutions; " unique solutions in"; lo; "to"; hi
Else
Debug.Print solutions; " non-unique solutions in"; lo; "to"; hi
End If
End Sub
Public Sub program()
Call foursquares(1, 7, True, True)
Debug.Print
Call foursquares(3, 9, True, True)
Call foursquares(0, 9, False, False)
End Sub
 
Output:
4  7  1  3  2  6  5 
6  4  1  5  2  3  7 
3  7  2  1  5  4  6 
5  6  2  3  1  7  4 
7  3  2  5  1  4  6 
4  5  3  1  6  2  7 
6  4  5  1  2  7  3 
7  2  6  1  3  5  4 

8  unique solutions in 1 to 7 

7  8  3  4  5  6  9 
8  7  3  5  4  6  9 
9  6  4  5  3  7  8 
9  6  5  4  3  8  7 

4  unique solutions in 3 to 9 

2860  non-unique solutions in 0 to 9 

Visual Basic .NET[edit]

Similar to the other brute-force algorithims, but with a couple of enhancements. A "used" list is maintained to simplify checking of the nested variables overlap. Also the d, f and g For Each loops are constrained by the other variables instead of blindly going through all combinations.

Module Module1
 
Dim CA As Char() = "0123456789ABC".ToCharArray()
 
Sub FourSquare(lo As Integer, hi As Integer, uni As Boolean, sy As Char())
If sy IsNot Nothing Then Console.WriteLine("a b c d e f g" & vbLf & "-------------")
Dim r = Enumerable.Range(lo, hi - lo + 1).ToList(), u As New List(Of Integer),
t As Integer, cn As Integer = 0
For Each a In r
u.Add(a)
For Each b In r
If uni AndAlso u.Contains(b) Then Continue For
u.Add(b)
t = a + b
For Each c In r : If uni AndAlso u.Contains(c) Then Continue For
u.Add(c)
For d = a - c To a - c
If d < lo OrElse d > hi OrElse uni AndAlso u.Contains(d) OrElse
t <> b + c + d Then Continue For
u.Add(d)
For Each e In r
If uni AndAlso u.Contains(e) Then Continue For
u.Add(e)
For f = b + c - e To b + c - e
If f < lo OrElse f > hi OrElse uni AndAlso u.Contains(f) OrElse
t <> d + e + f Then Continue For
u.Add(f)
For g = t - f To t - f : If g < lo OrElse g > hi OrElse
uni AndAlso u.Contains(g) Then Continue For
cn += 1 : If sy IsNot Nothing Then _
Console.WriteLine("{0} {1} {2} {3} {4} {5} {6}",
sy(a), sy(b), sy(c), sy(d), sy(e), sy(f), sy(g))
Next : u.Remove(f) : Next : u.Remove(e) : Next : u.Remove(d)
Next : u.Remove(c) : Next : u.Remove(b) : Next : u.Remove(a)
Next : Console.WriteLine("{0} {1}unique solutions for [{2},{3}]{4}",
cn, If(uni, "", "non-"), lo, hi, vbLf)
End Sub
 
Sub main()
fourSquare(1, 7, True, CA)
fourSquare(3, 9, True, CA)
fourSquare(0, 9, False, Nothing)
fourSquare(5, 12, True, CA)
End Sub
 
End Module
Output:
Added the zkl example for [5,12]
a b c d e f g
-------------
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
8 unique solutions for [1,7]

a b c d e f g
-------------
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
4 unique solutions for [3,9]

2860 non-unique solutions for [0,9]

a b c d e f g
-------------
B 9 6 5 7 8 C
B A 6 5 7 9 C
C 8 7 5 6 9 B
C 9 7 5 6 A B
4 unique solutions for [5,12]

Yabasic[edit]

Translation of: D
fourSquare(1,7,true,true)
fourSquare(3,9,true,true)
fourSquare(0,9,false,false)
 
 
sub fourSquare(low, high, unique, prin)
local count, a, b, c, d, e, f, g, fp
 
if (prin) print "a b c d e f g"
 
for a = low to high
for b = low to high
if (not valid(unique, a, b)) continue
 
fp = a+b
for c = low to high
if (not valid(unique, c, a, b)) continue
for d = low to high
if (not valid(unique, d, a, b, c)) continue
if (fp <> b+c+d) continue
 
for e = low to high
if (not valid(unique, e, a, b, c, d)) continue
for f = low to high
if (not valid(unique, f, a, b, c, d, e)) continue
if (fp <> d+e+f) continue
 
for g = low to high
if (not valid(unique, g, a, b, c, d, e, f)) continue
if (fp <> f+g) continue
 
count = count + 1
if (prin) print a," ",b," ",c," ",d," ",e," ",f," ",g
next
next
next
next
next
next
next
if (unique) then
print "There are ", count, " unique solutions in [",low,",",high,"]"
else
print "There are ", count, " non-unique solutions in [",low,",",high,"]"
end if
end sub
 
sub valid(unique, needle, n1, n2, n3, n4, n5, n6)
local i
 
if (unique) then
for i = 1 to numparams - 2
switch i
case 1: if needle = n1 return false : break
case 2: if needle = n2 return false : break
case 3: if needle = n3 return false : break
case 4: if needle = n4 return false : break
case 5: if needle = n5 return false : break
case 6: if needle = n6 return false : break
end switch
next
end if
return true
end sub
Output:
a b c d e f g
3 7 2 1 5 4 6
4 5 3 1 6 2 7
4 7 1 3 2 6 5
5 6 2 3 1 7 4
6 4 1 5 2 3 7
6 4 5 1 2 7 3
7 2 6 1 3 5 4
7 3 2 5 1 4 6
There are 8 unique solutions in [1,7]
a b c d e f g
7 8 3 4 5 6 9
8 7 3 5 4 6 9
9 6 4 5 3 7 8
9 6 5 4 3 8 7
There are 4 unique solutions in [3,9]
There are 2860 non-unique solutions in [0,9]

zkl[edit]

    // unique: No repeated numbers in solution
fcn fourSquaresPuzzle(lo=1,hi=7,unique=True){ //-->list of solutions
_assert_(0<=lo and hi<36);
notUnic:=fcn(a,b,c,etc){ abc:=vm.arglist; // use base 36, any repeated character?
abc.apply("toString",36).concat().unique().len()!=abc.len()
};
s:=List(); // solutions
foreach a,b,c in ([lo..hi],[lo..hi],[lo..hi]){ // chunk to reduce unique
if(unique and notUnic(a,b,c)) continue; // solution space. Slow VM
foreach d,e in ([lo..hi],[lo..hi]){ // -->for d { for e {} }
if(unique and notUnic(a,b,c,d,e)) continue;
foreach f,g in ([lo..hi],[lo..hi]){
if(unique and notUnic(a,b,c,d,e,f,g)) continue;
sqr1,sqr2,sqr3,sqr4 := a+b,b+c+d,d+e+f,f+g;
if((sqr1==sqr2==sqr3) and sqr1==sqr4) s.append(T(a,b,c,d,e,f,g));
}
}
}
s
}
fcn show(solutions,msg){
if(not solutions){ println("No solutions for",msg); return(); }
 
println(solutions.len(),msg," solutions found:");
w:=(1).max(solutions.pump(List,(0).max,"numDigits")); // max width of any number found
fmt:=" " + "%%%ds ".fmt(w)*7; // eg " %1s %1s %1s %1s %1s %1s %1s"
println(fmt.fmt(["a".."g"].walk().xplode()));
println("-"*((w+1)*7 + 1)); // calculate the width of horizontal bar
foreach s in (solutions){ println(fmt.fmt(s.xplode())) }
}
fourSquaresPuzzle() : show(_," unique (1-7)"); println();
fourSquaresPuzzle(3,9) : show(_," unique (3-9)"); println();
fourSquaresPuzzle(5,12) : show(_," unique (5-12)"); println();
println(fourSquaresPuzzle(0,9,False).len(), // 10^7 possibilities
" non-unique (0-9) solutions found.");
Output:
8 unique (1-7) solutions found:
 a b c d e f g 
---------------
 3 7 2 1 5 4 6 
 4 5 3 1 6 2 7 
 4 7 1 3 2 6 5 
 5 6 2 3 1 7 4 
 6 4 1 5 2 3 7 
 6 4 5 1 2 7 3 
 7 2 6 1 3 5 4 
 7 3 2 5 1 4 6 

4 unique (3-9) solutions found:
 a b c d e f g 
---------------
 7 8 3 4 5 6 9 
 8 7 3 5 4 6 9 
 9 6 4 5 3 7 8 
 9 6 5 4 3 8 7 

4 unique (5-12) solutions found:
  a  b  c  d  e  f  g 
----------------------
 11  9  6  5  7  8 12 
 11 10  6  5  7  9 12 
 12  8  7  5  6  9 11 
 12  9  7  5  6 10 11 

2860 non-unique (0-9) solutions found.