# 4-rings or 4-squares puzzle

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

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
```

## 11l

Translation of: Python

<lang 11l>F foursquares(lo, hi, unique, show)

```  V solutions = 0
L(c) lo .. hi
L(d) lo .. hi
I !unique | (c != d)
V a = c + d
I a >= lo & a <= hi
I !unique | (c != 0 & d != 0)
L(e) lo .. hi
I !unique | (e !C (a, c, d))
V g = d + e
I g >= lo & g <= hi
I !unique | (g !C (a, c, d, e))
L(f) lo .. hi
I !unique | (f !C (a, c, d, g, e))
V b = e + f - c
I b >= lo & b <= hi
I !unique | (b !C (a, c, d, g, e, f))
solutions++
I show
print(String((a, b, c, d, e, f, g))[1 .< (len)-1])
```
```  V uorn = I unique {‘unique’} E ‘non-unique’
```
```  print(solutions‘ ’uorn‘ solutions in ’lo‘ to ’hi)
print()
```

foursquares(1, 7, 1B, 1B) foursquares(3, 9, 1B, 1B) foursquares(0, 9, 0B, 0B)</lang>

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
```

## AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits

<lang AArch64 Assembly> /* ARM assembly AARCH64 Raspberry PI 3B */ /* program square4_64.s */

/*******************************************/ /* Constantes file */ /*******************************************/ /* for this file see task include a file in language AArch64 assembly*/ .include "../includeConstantesARM64.inc"

.equ NBBOX, 7

/*********************************/ /* Initialized data */ /*********************************/ .data sMessDeb: .asciz "a= @ b= @ c= @ d= @ e= @ f= @ g= @ \n***********************\n"

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

sMessNbSolution: .asciz "Number of solutions : @ \n\n\n"

/*********************************/ /* UnInitialized data */ /*********************************/ .bss .align 8 sZoneConv: .skip 24 qValues_a: .skip 8 * NBBOX qValues_b: .skip 8 * NBBOX - 1 qValues_c: .skip 8 * NBBOX - 2 qValues_d: .skip 8 * NBBOX - 3 qValues_e: .skip 8 * NBBOX - 4 qValues_f: .skip 8 * NBBOX - 5 qValues_g: .skip 8 * NBBOX - 6 qCounterSol: .skip 8

/*********************************/ /* code section */ /*********************************/ .text .global main main: // entry of program

```   mov x0,#1
mov x1,#7
mov x2,#3                    // 0 = rien 1 = display 2 = count 3 = les deux
bl searchPb
mov x0,#3
mov x1,#9
mov x2,#3                    // 0 = rien 1 = display 2 = count 3 = les deux
bl searchPb
mov x0,#0
mov x1,#9
mov x2,#2                    // 0 = rien 1 = display 2 = count 3 = les deux
bl prepSearchNU

```

100: // standard end of the program

```   mov x0, #0                   // return code
mov x8, #EXIT                // request to exit program
svc #0                       // perform the system call

```

/******************************************************************/ /* search problèm value not unique */ /******************************************************************/ /* x0 contains start digit */ /* x1 contains end digit */ /* x2 contains action (0 display 1 count) */ prepSearchNU:

```   stp x12,lr,[sp,-16]!         // save  registres
stp x2,x3,[sp,-16]!          // save  registres
stp x4,x5,[sp,-16]!          // save  registres
stp x6,x7,[sp,-16]!          // save  registres
stp x8,x9,[sp,-16]!          // save  registres
stp x10,fp,[sp,-16]!         // save  registres
mov x5,#0                    // counter
mov x12,x0                   // a
```

1:

```   mov x11,x0                   // b
```

2:

```   mov x10,x0                   // c
```

3:

```   mov x9,x0                    // d
```

4:

```   add x4,x12,x11               // a + b reference
add x3,x3,x9                    // b + c + d
cmp x4,x3
bne 10f
mov x8,x0                    // e
```

5:

```   mov x7,x0                    // f
```

6:

```   add x3,x9,x8
add x3,x3,x7                    // d + e + f
cmp x3,x4
bne 9f
mov x6,x0                    // g
```

7:

```   add x3,x7,x6                 // f + g
cmp x3,x4
bne 8f                       // not OK
// OK

```

8:

```   add x6,x6,1                    // increment g
cmp x6,x1
ble 7b
```

9:

```   add x7,x7,1                   // increment f
cmp x7,x1
ble 6b
cmp x8,x1
ble 5b
```

10:

```   add x9,x9,1                   // increment d
cmp x9,x1
ble 4b
cmp x10,x1
ble 3b
cmp x11,x1
ble 2b
cmp x12,x1
ble 1b

// end
tst x2,#0b10                // print count ?
beq 100f
mov x0,x5                   // counter
bl conversion10
ldr x1,qAdrsZoneConv        // insert conversion in message
bl strInsertAtCharInc
bl affichageMess
```

100:

```   ldp x10,fp,[sp],16          // restaur des  2 registres
ldp x8,x9,[sp],16           // restaur des  2 registres
ldp x6,x7,[sp],16           // restaur des  2 registres
ldp x4,x5,[sp],16           // restaur des  2 registres
ldp x2,x3,[sp],16           // restaur des  2 registres
ldp x12,lr,[sp],16           // restaur des  2 registres
ret
```

//qAdrsMessCounter: .quad sMessCounter qAdrsMessNbSolution: .quad sMessNbSolution qAdrsZoneConv: .quad sZoneConv /******************************************************************/ /* search problem unique solution */ /******************************************************************/ /* x0 contains start digit */ /* x1 contains end digit */ /* x2 contains action (0 display 1 count) */ searchPb:

```   stp x12,lr,[sp,-16]!         // save  registres
stp x2,x3,[sp,-16]!          // save  registres
stp x4,x5,[sp,-16]!          // save  registres
stp x6,x7,[sp,-16]!          // save  registres
stp x8,x9,[sp,-16]!          // save  registres
stp x10,fp,[sp,-16]!         // save  registres
mov x14,x2                   // save action
// init
ldr x3,qAdrqValues_a         // area value a
mov x4,#0
```

1: // loop init value a

```   str x0,[x3,x4,lsl #3]
cmp x0,x1
ble 1b
mov x5,#0                    // solution counter
mov x12,#-1
```

2:

```   add x12,x12,1                   // increment indice a
cmp x12,#NBBOX-1
bgt 90f
ldr x0,qAdrqValues_a         // area value a
ldr x1,qAdrqValues_b         // area value b
mov x2,x12                   // indice  a
mov x3,#NBBOX                // number of origin values
bl prepValues
mov x11,#-1
```

3:

```   add x11,x11,1                                        // increment indice b
cmp x11,#NBBOX - 2
bgt 2b
ldr x0,qAdrqValues_b                              // area value b
ldr x1,qAdrqValues_c                              // area value c
mov x2,x11                                        // indice b
mov x3,#NBBOX -1                                  // number of origin values
bl prepValues
mov x10,#-1
```

4:

```   add x10,x10,1
cmp x10,#NBBOX - 3
bgt 3b
mov x2,x10
mov x3,#NBBOX - 2
bl prepValues
mov x9,#-1
```

5:

```   add x9,x9,1
cmp x9,#NBBOX - 4
bgt 4b
// control 2 firsts squares
ldr x0,[x0,x12,lsl #3]
ldr x1,[x1,x11,lsl #3]
add x4,x0,x1                               // a + b   value first square
ldr x0,[x0,x10,lsl #3]
add x7,x1,x0                               // b + c
ldr x1,[x1,x9,lsl #3]
add x7,x7,x1                                  // b + c + d
cmp x7,x4                                  // equal first square ?
bne 5b
mov x2,x9
mov x3,#NBBOX - 3
bl prepValues
mov x8,#-1
```

6:

```   add x8,x8,1
cmp x8,#NBBOX - 5
bgt 5b
mov x2,x8
mov x3,#NBBOX - 4
bl prepValues
mov x7,#-1
```

7:

```   add x7,x7,1
cmp x7,#NBBOX - 6
bgt 6b
ldr x0,[x0,x9,lsl #3]
ldr x1,[x1,x8,lsl #3]
add x3,x0,x1                                // d + e
ldr x1,[x1,x7,lsl #3]
add x3,x3,x1                                   // d + e + f
cmp x3,x4                                   // equal first square ?
bne 7b
mov x2,x7
mov x3,#NBBOX - 5
bl prepValues
mov x6,#-1
```

8:

```   add x6,x6,1
cmp x6,#NBBOX - 7
bgt 7b
ldr x0,[x0,x7,lsl #3]
ldr x1,[x1,x6,lsl #3]
cmp x4,x3                                  // equal first square ?
bne 8b

tst x14,#0b1
beq 9f                                     // display solution ?
ldr x0,[x0,x12,lsl #3]
bl conversion10
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x11,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x10,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x9,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x8,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x7,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc
mov x2,x0
ldr x0,[x0,x6,lsl #3]
bl conversion10
mov x0,x2
ldr x1,qAdrsZoneConv            // insert conversion in message
bl strInsertAtCharInc

bl affichageMess
```

9:

```   b 8b    // suite

```

90:

```   tst x14,#0b10
beq 100f                    // display counter ?
mov x0,x5
bl conversion10
ldr x1,qAdrsZoneConv        // insert conversion in message
bl strInsertAtCharInc
bl affichageMess
```

100:

```   ldp x10,fp,[sp],16          // restaur des  2 registres
ldp x8,x9,[sp],16           // restaur des  2 registres
ldp x6,x7,[sp],16           // restaur des  2 registres
ldp x4,x5,[sp],16           // restaur des  2 registres
ldp x2,x3,[sp],16           // restaur des  2 registres
ldp x12,lr,[sp],16           // restaur des  2 registres
ret
```

qAdrsMessDeb: .quad sMessDeb qAdrqCounterSol: .quad qCounterSol /******************************************************************/ /* copy value area and substract value of indice */ /******************************************************************/ /* x0 contains the address of values origin */ /* x1 contains the address of values destination */ /* x2 contains value indice to substract */ /* x3 contains origin values number */ prepValues:

```   stp x1,lr,[sp,-16]!          // save  registres
stp x2,x3,[sp,-16]!          // save  registres
stp x4,x5,[sp,-16]!          // save  registres
stp x6,x7,[sp,-16]!          // save  registres
mov x4,#0                    // indice origin value
mov x5,#0                    // indice destination value
```

1:

```   cmp x4,x2                    // substract indice ?
beq 2f                       // yes -> jump
ldr x6,[x0,x4,lsl #3]        // no -> copy value
str x6,[x1,x5,lsl #3]
add x5,x5,1                  // increment destination indice
```

2:

```  add x4,x4,1                   // increment origin indice
cmp x4,x3                     // end ?
blt 1b
```

100:

```   ldp x6,x7,[sp],16           // restaur des  2 registres
ldp x4,x5,[sp],16           // restaur des  2 registres
ldp x2,x3,[sp],16           // restaur des  2 registres
ldp x1,lr,[sp],16          // restaur des  2 registres
ret
```

/********************************************************/ /* File Include fonctions */ /********************************************************/ /* for this file see task include a file in language AArch64 assembly */ .include "../includeARM64.inc" </lang>

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
```

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;</lang>

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

As with the REXX solution, we use explicit loops to generate the permutations. <lang algol68>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</lang>

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

Translation of: JavaScript

(Structured search example) <lang applescript>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))

-- 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
Template: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

```   if length of xs > 0 then
item 1 of xs
else
missing value
end if
```

-- 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</lang>

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```

## Applesoft BASIC

Translation of: C

<lang gwbasic> 100 TRUE = NOT FALSE

```110 PLO = 1:PHI = 7:PUNIQUE = TRUE:PSHOW = TRUE: GOSUB 150"FOURSQUARES"
120 PLO = 3:PHI = 9:PUNIQUE = TRUE:PSHOW = TRUE: GOSUB 150"FOURSQUARES"
130 PLO = 0:PHI = 9:PUNIQUE = FALSE:PSHOW = FALSE: GOSUB 150"FOURSQUARES"
140  END
150 LO = PLO
160 HI = PHI
170 UNIQUE = PUNIQUE
180 SHOW = PSHOW
190 S = 0: REM SOLUTIONS
200  PRINT
210  GOSUB 270"ACD"
220  PRINT
230  PRINT S" ";
240  IF  NOT UNIQUE THEN  PRINT "NON-";
250  PRINT "UNIQUE SOLUTIONS IN "LO" TO "HI
260  RETURN
270  FOR C = LO TO HI
280      FOR D = LO TO HI
290          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  GOSUB 320"GE"
300  NEXT D,C
310  RETURN
320  FOR E = LO TO HI
330      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  GOSUB 360"BF"
340  NEXT E
350  RETURN
360  FOR F = LO TO HI
370      IF (( NOT UNIQUE) OR ((F <  > A) AND (F <  > C) AND (F <  > D) AND (F <  > G) AND (F <  > E))) THEN  GOSUB 400
380  NEXT F
390  RETURN
400 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 S = S + 1: IF (SHOW) THEN  PRINT A" "B" "C" "D" "E" "F" "G
410  RETURN</lang>
```

## ARM Assembly

Works with: as version Raspberry Pi

<lang ARM Assembly>

/* 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

```

/******************************************************************/ /* 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,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
```

8:

```   add r6,#1                                     @ increment g
cmp r6,r1
ble 7b
```

9:

```   add r7,#1                                     @ increment f
cmp r7,r1
ble 6b
cmp r8,r1
ble 5b
```

10:

```   add r9,#1                                     @ increment d
cmp r9,r1
ble 4b
cmp r10,r1
ble 3b
cmp r11,r1
ble 2b
cmp r12,r1
ble 1b
```
```   @ end
tst r2,#0b10                                    @ print count ?
beq 100f
mov r0,r5                                       @ counter
bl conversion10
bl affichageMess
```

100:

```   pop {r3-r12,lr}                                 @ restaur registers
bx lr                                           @return
```

/******************************************************************/ /* 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]
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
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,[r0,r12,lsl #2]
ldr r1,[r1,r11,lsl #2]
add r4,r0,r1                               @ a + b   value first square
ldr r0,[r0,r10,lsl #2]
add r7,r1,r0                               @ b + c
ldr r1,[r1,r9,lsl #2]
add r7,r1                                  @ b + c + d
cmp r7,r4                                  @ equal first square ?
bne 5b
mov r2,r9
mov r3,#NBBOX - 3
bl prepValues
mov r8,#-1
```

6:

```   add r8,#1
cmp r8,#NBBOX - 5
bgt 5b
mov r2,r8
mov r3,#NBBOX - 4
bl prepValues
mov r7,#-1
```

7:

```   add r7,#1
cmp r7,#NBBOX - 6
bgt 6b
ldr r0,[r0,r9,lsl #2]
ldr r1,[r1,r8,lsl #2]
add r3,r0,r1                                @ d + e
ldr r1,[r1,r7,lsl #2]
add r3,r1                                   @ de + e + f
cmp r3,r4                                   @ equal first square ?
bne 7b
mov r2,r7
mov r3,#NBBOX - 5
bl prepValues
mov r6,#-1
```

8:

```   add r6,#1
cmp r6,#NBBOX - 7
bgt 7b
ldr r0,[r0,r7,lsl #2]
ldr r1,[r1,r6,lsl #2]
cmp r4,r3                                  @ equal first square ?
bne 8b
ldr r0,[sp,#8]                             @ load action for two parameter in stack
tst r0,#0b1
beq 9f                                     @ display solution ?
ldr r0,[r0,r12,lsl #2]
bl conversion10
ldr r0,[r0,r11,lsl #2]
bl conversion10
ldr r0,[r0,r10,lsl #2]
bl conversion10
ldr r0,[r0,r9,lsl #2]
bl conversion10
ldr r0,[r0,r8,lsl #2]
bl conversion10
ldr r0,[r0,r7,lsl #2]
bl conversion10
ldr r0,[r0,r6,lsl #2]
bl conversion10
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
bl conversion10
bl affichageMess
```

100:

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

```   push {r1-r6,lr}                                @ save  registres
mov r4,#0                                      @ indice origin value
mov r5,#0                                      @ indice destination value
```

1:

```   cmp r4,r2                                      @ substract indice ?
beq 2f                                         @ yes -> jump
ldr r6,[r0,r4,lsl #2]                          @ no -> copy value
str r6,[r1,r5,lsl #2]
add r5,#1                                      @ increment destination indice
```

2:

```  add r4,#1                                       @ increment origin indice
cmp r4,r3                                       @ end ?
blt 1b
```

100:

```   pop {r1-r6,lr}                                 @ restaur registres
bx lr                                          @return
```

/******************************************************************/ /* display text with size calculation */ /******************************************************************/ /* r0 contains the address of the message */ affichageMess:

```   push {r0,r1,r2,r7,lr}                          @ save  registres
mov r2,#0                                      @ counter length
```

1: @ loop length calculation

```   ldrb r1,[r0,r2]                                @ read octet start position + index
cmp r1,#0                                      @ if 0 its over
bne 1b                                         @ and loop
@ so here r2 contains the length of the message
mov r1,r0                                      @ address message in r1
mov r0,#STDOUT                                 @ code to write to the standard output Linux
mov r7, #WRITE                                 @ code call system "write"
svc #0                                         @ call systeme
pop {r0,r1,r2,r7,lr}                           @ restaur des  2 registres */
bx lr                                          @ return
```

/******************************************************************/ /* Converting a register to a decimal unsigned */ /******************************************************************/ /* r0 contains value and r1 address area */ /* r0 return size of result (no zero final in area) */ /* area size => 11 bytes */ .equ LGZONECAL, 10 conversion10:

```   push {r1-r4,lr}                                 @ save registers
mov r3,r1
mov r2,#LGZONECAL
```

1: @ start loop

```   bl divisionpar10U                               @ unsigned  r0 <- dividende. quotient ->r0 reste -> r1
strb r1,[r3,r2]                                 @ store digit on area
cmp r0,#0                                       @ stop if quotient = 0
subne r2,#1                                     @ else previous position
bne 1b                                          @ and loop
@ and move digit from left of area
mov r4,#0
```

2:

```   ldrb r1,[r3,r2]
strb r1,[r3,r4]
cmp r2,#LGZONECAL
ble 2b
@ and move spaces in end on area
mov r0,r4                                         @ result length
mov r1,#' '                                       @ space
```

3:

```   strb r1,[r3,r4]                                   @ store space in area
cmp r4,#LGZONECAL
ble 3b                                            @ loop if r4 <= area size

```

100:

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

```

/***************************************************/ /* division par 10 unsigned */ /***************************************************/ /* r0 dividende */ /* r0 quotient */ /* r1 remainder */ divisionpar10U:

```   push {r2,r3,r4, lr}
mov r4,r0                                          @ save value
ldr r3,iMagicNumber                                @ r3 <- magic_number    raspberry 1 2
umull r1, r2, r3, r0                               @ r1<- Lower32Bits(r1*r0) r2<- Upper32Bits(r1*r0)
mov r0, r2, LSR #3                                 @ r2 <- r2 >> shift 3
add r2,r0,r0, lsl #2                               @ r2 <- r0 * 5
sub r1,r4,r2, lsl #1                               @ r1 <- r4 - (r2 * 2)  = r4 - (r0 * 10)
pop {r2,r3,r4,lr}
bx lr                                              @ leave function
```

iMagicNumber: .int 0xCCCCCCCD

</lang>

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

<lang AWK>

1. syntax: GAWK -f 4-RINGS_OR_4-SQUARES_PUZZLE.AWK
2. 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()
}
}
}
```

} </lang>

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

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.

<lang befunge>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-!+\^ >:#,_@ >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^^></lang>

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

<lang C>

1. include <stdio.h>
1. define TRUE 1
2. 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);
```

} </lang> 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#

Translation of: Java

<lang csharp>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);
}
}
```

}</lang>

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]```

## C++

<lang cpp> //C++14/17

1. include <algorithm>//std::for_each
2. include <iostream> //std::cout
3. include <numeric> //std::iota
4. include <vector> //std::vector, save solutions
5. 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;
```

} </lang> 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.

```

## Clojure

<lang clojure>(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]))
```

</lang>

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

<lang lisp> (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)))))
```

</lang> 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

Translation of: Ruby

<lang 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)</lang>

## D

<lang D>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;
```

}</lang>

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]```

See #Pascal

## F#

<lang fsharp> (* 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}))))
```

</lang> Then: <lang fsharp> printfn "%d" (Seq.length (N 0 9)) </lang>

Output:
```2860
```

<lang fsharp> (* 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}))))
```

</lang> Then: <lang fsharp> 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) </lang>

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: <lang fsharp> 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) </lang>

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

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. <lang factor>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</lang>

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

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 <lang Fortran>WRITE(...) FIRST,LAST,IF (UNIQUE) THEN "Distinct values only" ELSE "Repeated values allowed" FI // "."</lang> 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. <lang Fortran> 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  </lang>
```

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
004014EF   mov         ecx,esi
004014F1   mov         edx,dword ptr [ebp-60h]
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)]
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

<lang freebasic>' 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.

1. Ifndef TRUE
``` #Define FALSE 0
#Define TRUE Not FALSE
```
1. 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</lang>

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
----------------------------------------```

## FutureBasic

This simple example uses old-style, length-limited Pascal strings for formatting to make it easier to compare with similar code posted here for this task. However, FB more commonly uses Apple's modern and superior Core Foundation strings. <lang futurebasic> local fn FourRings( low as long, high as long, unique as BOOL, show as BOOL )

``` long          a, b, c, d, e, f, g
unsigned long t, total = 0
unsigned long l = len\$( str\$(high) )

if l < len\$( str\$(low) ) then l = len\$( str\$( low) )

if ( show == YES )
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 == YES )
if b == a then continue
end if
t = a + b
for c = low to high
if unique == YES
if c == a or c == b then continue
end if
for d = low to high
if unique == YES
if d == a or d == b or d == c then continue
end if
if b + c + d == t
for e = low to high
if unique == YES
if e == a or e == b or e == c or e == d then continue
end if
for f = low to high
if unique == YES
if f == a or f == b or f == c or f == d or f == e then continue
end if
if ( d + e + f == t )
for g = low to high
if unique == YES
if g == a or g == b or g == c or g == d or g == e or g == f then continue
end if
if ( f + g == t )
total += 1
if( show == YES )
printf @"%3d%3d%3d%3d%3d%3d%3d", a, b, c, d, e, f, g
end if
end if
next
end if
next
next
end if
next
next
next
next

if ( unique == YES )
print
print total; " unique solutions for"; str\$(low); " to"; str\$(high)
print string\$(30, "-") : print
else
print total; " non-unique solutions for"; str\$(low); " to"; str\$(high)
print string\$(36, "-") : print
end if
```

end fn

window 1, @"4 Rings", ( 0, 0, 350, 400 )

fn FourRings( 1, 7, YES, YES ) fn FourRings( 3, 9, YES, YES ) fn FourRings( 0, 9, NO, NO )

HandleEvents </lang>

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

<lang go>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 } </lang>

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

Translation of: Java

<lang groovy>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 })
}
```

}</lang>

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]```

#### By exhaustive search

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</lang>
```
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

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. <lang haskell>import Data.List (delete, sortBy, (\\))

4 RINGS OR 4 SQUARES PUZZLE --------------

type Rings = [(Int, Int, Int, Int, Int, Int, Int)]

rings :: Bool -> [Int] -> Rings rings u digits =

``` ((>>=) <*> (queen u =<< head))
(sortBy (flip compare) digits)
```

queen :: Bool -> Int -> [Int] -> Int -> Rings queen u h ds q = xs >>= leftBishop u q h ts ds

``` where
ts = filter ((<= h) . (q +)) ds
xs
| u = delete q ts
| otherwise = ds
```

leftBishop ::

``` Bool ->
Int ->
Int ->
[Int] ->
[Int] ->
Int ->
Rings
```

leftBishop u q h ts ds lb

``` | lRook <= h = xs >>= rightBishop u q h lb ds lRook
| otherwise = []
where
lRook = lb + q
xs
| u = ts \\ [q, lb, lRook]
| otherwise = ds
```

rightBishop ::

``` Bool ->
Int ->
Int ->
Int ->
[Int] ->
Int ->
Int ->
Rings
```

rightBishop u q h lb ds lRook rb

``` | (rRook <= h) && (not u || (rRook /= lb)) =
let ks
| u = (ds \\ [q, lb, rb, rRook, lRook])
| otherwise = ds
in ks
>>= knights
u
(lRook - rRook)
lRook
lb
q
rb
rRook
ks
| otherwise = []
where
rRook = q + rb
```

knights ::

``` Bool ->
Int ->
Int ->
Int ->
Int ->
Int ->
Int ->
[Int] ->
Int ->
Rings
```

knights u rookDelta lRook lb q rb rRook ks k =

``` [ (lRook, k, lb, q, rb, k2, rRook)
| (k2 `elem` ks)
&& ( not u
|| notElem
k2
[lRook, k, lb, q, rb, rRook]
)
]
where
k2 = k + rookDelta
```

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])]
)</lang>
```
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

Implementation for the unique version of the puzzle:

``` 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.
```

)</lang>

Implementation for the non-unique version of the puzzle:

``` 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.
```

)</lang>

<lang J> 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</lang>

## Java

Uses java 8 features. <lang Java>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);
}
```

}</lang>

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

### ES6

(Structured search version)

<lang javascript>(() => {

```   // 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),
```
```           // 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();
```

})();</lang>

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```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

Since jq is built on back-tracking and optimizes the tail-recursion involved here, this entry will focus on generic solutiond for problems of this sort. Specifically, the number of boxes is unrestricted.

#### N boxes with arbitrary overlaps

In this section, an arbitrary pattern of overlaps can be specified as follows.

We will associate the letters "a", "b", ... with the integers 0, 1,... so that each box can be represented as an array of integers; the puzzle configuration can then be characterized by an array of such arrays. For the particular puzzle under consideration, the characteristic array is:

[[0,1], [1,2,3], [3,4,5], [5,6]]

The solution in this subsection is quite efficient for the family of problems based on permutations, but as is shown, can also be used without the permutation constraint. <lang jq># Generate a stream of all the permutations of the input array def permutations:

``` if length == 0 then []
else
range(0;length) as \$i
| [.[\$i]] + (del(.[\$i])|permutations)
end ;
```
1. Permutations of a ... n inclusive

def permutations(a;n):

``` [range(a;n+1)] | permutations;
```
1. value of a box
2. Input: the table of values

def valueOfBox(\$box):

``` [ .[ \$box[] ]] | add;
```

def allEqual(\$boxes):

``` . as \$values
| valueOfBox(\$boxes[0]) as \$sum
| all(\$boxes[1:][]; . as \$box | \$values | valueOfBox(\$box) == \$sum);
```

def combinations(\$m; \$n; \$size):

``` [range(0; \$size) | [range(\$m; \$n)]] | combinations;
```

def count(s): reduce s as \$x (null; .+1);

1. a=0, b=1, etc

def boxes: [[0,1], [1,2,3], [3,4,5], [5,6]];

``` "1 to 7:",
(permutations(1;7) | select(allEqual(boxes))),
"\n3 to 9:",
(permutations(3;9) | select(allEqual(boxes))),
"\n0 to 9:\n\(count(permutations(0;9) | select(allEqual(boxes))))",
"\nThere are \(count(combinations(0;10;7) | select(allEqual(boxes)))) solutions for 0 to 9 with replacement."
```

Output:
```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]

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]

There are 1152 distinct solutions for 0 to 9.

There are 2860 solutions for 0 to 9 with replacement.
```

#### N boxes with one overlap between adjacent boxes and no uniqueness constraint

In this subsection, an efficient solution for the N-boxes puzzle in the case of non-uniqueness (i.e. unrestricted choice of values within the specified range) is given. It is assumed, however, that each box (except for the last) has exactly one overlap with its successor.

For consistency with the prior section, the pattern can be specified in the same way, i.e. as a characteristic array, which for the specific problem at hand could be: [[0,1], [1,2,3], [3,4,5], [5,6]].

<lang jq># rings/3 assumes that each box (except for the last) has exactly one overlap with its successor.

1. Input: ignored.
2. Output: a stream of solutions, i.e. a stream of arrays.
3. \$boxes is an array of boxes, each box being a flat array.
4. \$min and \$max define the range of permissible values of items in the boxes (inclusive)

def rings(\$boxes; \$min; \$max):

``` def inrange: \$min <= . and . <= \$max;

# The following helper function deals with the case when the global per-box sum (\$sum) is known.
# Input: an array representing the solution so far, or null.
# Output: the input plus the solution corresponding to the first argument.
# \$this is the sum of the previous items in the first box, or 0.
def solve(\$boxes; \$this; \$sum):

# The following is a helper function for handling the case when:
# *  \$sum is known
# *  \$boxes[0] | length == 1, and
# *  \$boxes|length>1
def lastInBox(\$boxes; \$this):
. as \$in
| (\$boxes[1:] | (.[0] |= .[1:])) as \$bx
# the first entry in the next box must be the same:
| (\$sum - \$this) as \$next
| select(\$next | inrange)
| (. + [\$next]) | solve( \$bx; \$next; \$sum) ;
```
```   . as \$in
| if \$boxes|length == 0 then \$in
else \$boxes[0] as \$box
| if \$box|length == 0
```

then solve( \$boxes[1:]; 0; \$sum )

```       elif \$box|length == 1
# is this the last box?
then if \$boxes|length == 1
then (\$sum - \$this) as \$next
| select(\$next | inrange)
| \$in + [\$next]
else lastInBox(\$boxes; \$this)
end
else # \$box|length > 1
range(\$min; \$max + 1) as \$first
| select( (\$this + \$first) <= \$sum)
| (\$in + [\$first]) | solve( [\$box[1:]] + \$boxes[1:]; \$this + \$first; \$sum)
end
end ;

. as \$in
| \$boxes[0] as \$box
| (\$boxes[1:] | .[0] |= .[1:]) as \$bx
| [range(0; \$box|length) | [range(\$min; \$max + 1)]]
| combinations
```

def count(s): reduce s as \$x (null; .+1);</lang> The specific task <lang jq># a=0, b=1, etc def boxes: [[0,1], [1,2,3], [3,4,5], [5,6]];

count(rings(boxes; 0; 9))</lang>

Output:
```2860
```

## Julia

Translation of: Python

<lang julia> 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) </lang>

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

Translation of: C

<lang scala>// 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)
```

}</lang>

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

Translation of: D

<lang lua>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)</lang>

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]```

## Mathematica/Wolfram Language

<lang Mathematica>{low, high} = {1, 7}; SolveValues[{a + b == b + c + d == d + e + f == f + g, low <= a <= high,

``` low <= b <= high, low <= c <= high, low <= d <= high,
low <= e <= high, low <= f <= high, low <= g <= high,
a != b != c != d != e != f != g}, {a, b, c, d, e, f, g}, Integers]
```

{low, high} = {3, 9}; SolveValues[{a + b == b + c + d == d + e + f == f + g, low <= a <= high,

``` low <= b <= high, low <= c <= high, low <= d <= high,
low <= e <= high, low <= f <= high, low <= g <= high,
a != b != c != d != e != f != g}, {a, b, c, d, e, f, g}, Integers]
```

{low, high} = {0, 9}; SolveValues[{a + b == b + c + d == d + e + f == f + g, low <= a <= high,

```  low <= b <= high, low <= c <= high, low <= d <= high,
low <= e <= high, low <= f <= high, low <= g <= high}, {a, b, c, d,
e, f, g}, Integers] // Length</lang>
```
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}}

{{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```

## Modula-2

<lang modula2>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.</lang>

## Nim

Adapted from Rust version. <lang nim>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)</lang>
```
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

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. <lang pascal>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.</lang>

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

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

<lang perl>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] );</lang>

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```

### With Recursion

<lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/4-rings_or_4-squares_puzzle use warnings;

for ( [1 .. 7], [3 .. 9] )

``` {
print "for @\$_\n\n";
findunique( \$_ );
print "\n";
}
```

my \$count = 0; findcount(); print "count of non-unique 0-9: \$count\n";

sub findunique

``` {
my @allowed = @{ shift @_ };
if( @_ == 4 ) { \$_[0] == \$_[2] + \$_[3] or return }
elsif( @_ == 6 ) { \$_[1] + \$_[2] == \$_[4] + \$_[5] or return }
elsif( @_ == 7 ) { \$_[3] + \$_[4] == \$_[6] and print "@_\n"; return }
for my \$n ( @allowed )
{
findunique( [ grep \$n != \$_, @allowed ], @_, \$n );
}
}
```

sub findcount

``` {
if( @_ == 4 ) { \$_[0] == \$_[2] + \$_[3] or return }
elsif( @_ == 6 ) { \$_[1] + \$_[2] == \$_[4] + \$_[5] or return }
elsif( @_ == 7 ) { \$_[3] + \$_[4] == \$_[6] and \$count++; return }
findcount( @_, \$_ ) for 0 .. 9;
}</lang>
```
Output:
```for 1 2 3 4 5 6 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

for 3 4 5 6 7 8 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

count of non-unique 0-9: 2860
```

## Phix

```-- demo/rosetta/4_rings_or_4_squares_puzzle.exw
with javascript_semantics
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, hi, bool uniq, 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
```

## Picat

<lang Picat>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).</lang>
```
Output:
```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```

## PL/SQL

Works with: Oracle

<lang plsql> 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; / </lang> 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

Works with SWI-Prolog 7.5.8 <lang Prolog>

- 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).
```

</lang> 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

### Procedural

#### Itertools

<lang Python>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</lang>
```

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

Faster solution without itertools <lang Python> 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</lang>
```

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

Translation of: JavaScript
Works with: Python version 3.7

<lang python>4-rings or 4-squares puzzle

from itertools import chain

1. rings :: noRepeatedDigits -> DigitList -> Lists of solutions
2. 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 []
```

1. TEST ----------------------------------------------------
2. 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)))
)
```

1. GENERIC -------------------------------------------------
1. 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))
)
```

1. 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)
```

1. 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)
```

1. enumFromTo :: (Int, Int) -> [Int]

def enumFromTo(m):

```   Integer enumeration from m to n.
return lambda n: list(range(m, 1 + n))
```

1. 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)
```

1. 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)
```

1. MAIN ---

if __name__ == '__main__':

```   main()</lang>
```
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

Function "perms" is a modified version of the "permutations" function from the "gtools" R package. <lang R># 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)

</lang>

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
```

## Racket

Using a folder, so we can count as well as produce lists of results

<lang racket>#lang racket

(define solution? (match-lambda [(list a b c d e f g) (= (+ a b) (+ b c d) (+ d e f) (+ f g))]))

(define (fold-4-rings-or-4-squares-puzzle lo hi kons k0)

``` (for*/fold ((k k0))
((combination (in-combinations (range lo (add1 hi)) 7))
(permutation (in-permutations combination))
#:when (solution? permutation))
(kons permutation k)))
```

(fold-4-rings-or-4-squares-puzzle 1 7 cons null) (fold-4-rings-or-4-squares-puzzle 3 9 cons null) (fold-4-rings-or-4-squares-puzzle 0 9 (λ (ignored-solution count) (add1 count)) 0)</lang>

Output:
```'((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))
'((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))
192```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2016.12

<lang perl6>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 }
```

}

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) );</lang>

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.
```

## REXX

### fast version

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

1. =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</lang>
```
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

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. <lang rexx>/*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. */

1. =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</lang>
```
output   is identical to the faster REXX version.

## Ruby

<lang ruby>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)</lang>

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

<lang rust>

1. ![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);
```

} </lang>

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

Translation of: Java

<lang scala>object FourRings {

``` def fourSquare(low: Int, high: Int, unique: Boolean, print: Boolean): Unit = {
def isValid(needle: Integer, haystack: Integer*) = !unique || !haystack.contains(needle)
```
```   if (print) println("a b c d e f g")
```
```   var count = 0
for {
a <- low to high
b <- low to high if isValid(a, b)
fp = a + b
c <- low to high if isValid(c, a, b)
d <- low to high if isValid(d, a, b, c) && fp == b + c + d
e <- low to high if isValid(e, a, b, c, d)
f <- low to high if isValid(f, a, b, c, d, e) && fp == d + e + f
g <- low to high if isValid(g, a, b, c, d, e, f) && fp == f + g
} {
count += 1
if (print) println(s"\$a \$b \$c \$d \$e \$f \$g")
}

println(s"There are \$count \${if(unique) "unique" else "non-unique"} solutions in [\$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)
}
```

}</lang>

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

<lang scheme> (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)))
```

(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)
```

</lang>

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

Translation of: Raku

<lang ruby>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")
}
```

}

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)</lang>

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

<lang simula>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
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. </lang>

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

Works with: Db2 LUW

version 9.7 or higher.

With SQL PL: <lang 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)@ </lang> Output:

```db2 -td@
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

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

<lang stata>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</lang>
```

## Tcl

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.

<lang Tcl>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]]</lang>

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```

## VBA

Translation of: C

<lang vb>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 </lang>

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

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. <lang vbnet>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
For Each b In r
If uni AndAlso u.Contains(b) Then Continue For
t = a + b
For Each c In r : If uni AndAlso u.Contains(c) Then Continue For
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
For Each e In r
If uni AndAlso u.Contains(e) Then Continue For
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
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</lang>

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]```

## Vlang

Translation of: Go

<lang vlang>fn main(){ mut n, mut c := get_combs(1,7,true) println("\$n unique solutions in 1 to 7") println(c) n, c = get_combs(3,9,true) println("\$n unique solutions in 3 to 9") println(c) n, _ = get_combs(0,9,false) println("\$n non-unique solutions in 0 to 9") }

fn get_combs(low int,high int,unique bool) (int, [][]int) {

```   mut num := 0
mut valid_combs := [][]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 valid_comb(a,b,c,d,e,f,g) { if !unique || is_unique(a,b,c,d,e,f,g) { num++ valid_combs << [a,b,c,d,e,f,g] } } } } } } } } } return num, valid_combs } fn is_unique(a int,b int,c int,d int,e int,f int,g int) bool { mut data := map[int]int{} data[a]++ data[b]++ data[c]++ data[d]++ data[e]++ data[f]++ data[g]++ return data.len == 7 } fn valid_comb(a int,b int,c int,d int,e int,f int,g int) bool { square1 := a + b square2 := b + c + d square3 := d + e + f square4 := f + g return square1 == square2 && square2 == square3 && square3 == square4 }</lang>

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
```

## Wren

Translation of: C
Library: Wren-fmt

<lang ecmascript>import "/fmt" for Fmt

var a = 0 var b = 0 var c = 0 var d = 0 var e = 0 var f = 0 var g = 0

var lo var hi var unique var show var solutions

var bf = Fn.new {

```   f = lo
while (f <= hi) {
if (!unique || (f != a && f != c && f != d && f != e && f != g)) {
b = e + f - c
if (b >= lo && b <= hi &&
(!unique || (b != a && b != c && b != d && b != e && b != f && b != g))) {
solutions = solutions + 1
if (show) Fmt.lprint("\$d \$d \$d \$d \$d \$d \$d", [a, b, c, d, e, f, g])
}
}
f = f + 1
}
```

}

var ge = Fn.new {

```   e = lo
while (e <= hi) {
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.call()
}
e = e + 1
}
```

}

var acd = Fn.new {

```   c = lo
while (c <= hi) {
d = lo
while (d <= hi) {
if (!unique || c != d) {
a = c + d
if (a >= lo && a <= hi && (!unique || (c != 0 && d != 0))) ge.call()
}
d = d + 1
}
c = c + 1
}
```

}

var foursquares = Fn.new { |plo, phi, punique, pshow|

```   lo = plo
hi = phi
unique = punique
show = pshow
solutions = 0
if (show) {
System.print("\na b c d e f g")
System.print("-------------")
}
acd.call()
if (unique) {
Fmt.print("\n\$d unique solutions in \$d to \$d", solutions, lo, hi)
} else {
Fmt.print("\n\$d non-unique solutions in \$d to \$d\n", solutions, lo, hi)
}
```

}

foursquares.call(1, 7, true, true) foursquares.call(3, 9, true, true) foursquares.call(0, 9, false, false)</lang>

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
```

## XPL0

<lang XPL0>int Show, Low, High, Digit(7\a..g\), Count; proc Rings(Level); int Level; \of recursion int D, Temp, I, Set; [for D:= Low to High do

```   [Digit(Level):= D;
if Level < 7-1 then Rings(Level+1)
else [  Temp:= Digit(0) + Digit(1); \solution?
if Temp = Digit(1) + Digit(2) + Digit(3) and
Temp = Digit(3) + Digit(4) + Digit(5) and
Temp = Digit(5) + Digit(6) then
[Count:= Count+1;
if Show then
[Set:= 0;           \digits must be unique
for I:= 0 to 7-1 do
Set:= Set ! 1<<Digit(I);
if Set = %111_1111 << Low then
[for I:= 0 to 7-1 do
[IntOut(0, Digit(I));  ChOut(0, ^ )];
CrLf(0);
];
];
];
];
];
```

];

[Show:= true; Low:= 1; High:= 7; Rings(0); CrLf(0); Low:= 3; High:= 9; Rings(0); CrLf(0); Show:= false; Low:= 0; High:= 9; Count:= 0; Rings(0); IntOut(0, Count); CrLf(0); ]</lang>

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

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
```

## Yabasic

Translation of: D

<lang Yabasic>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</lang>

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

<lang zkl> // 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
```

}</lang> <lang zkl>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.");</lang>
```
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.
```