Department numbers: Difference between revisions
m Applesoft BASIC |
Department numbers in QBasic and XBasic |
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6 4 2
6 5 1</pre>
==={{header|QBasic}}===
<lang qbasic>PRINT "--police-- --sanitation-- --fire--"
FOR police = 2 TO 7 STEP 2
FOR fire = 1 TO 7
IF fire = police THEN GOTO cont
sanitation = 12 - police - fire
IF sanitation = fire OR sanitation = police THEN GOTO cont
IF sanitation >= 1 AND sanitation <= 7 THEN
PRINT USING " # # #"; police; fire; sanitation
END IF
cont:
NEXT fire
NEXT police</lang>
==={{header|XBasic}}===
{{works with|Windows XBasic}}
<lang xbasic>PROGRAM "Depar/num"
DECLARE FUNCTION Entry ()
FUNCTION Entry ()
PRINT "police sanitation fire"
FOR police = 2 TO 7 STEP 2
FOR fire = 1 TO 7
IF fire = police THEN GOTO cont
sanitation = 12 - police - fire
IF sanitation = fire OR sanitation = police THEN GOTO cont
IF sanitation >= 1 AND sanitation <= 7 THEN
PRINT TAB(3); police; TAB(13); fire; TAB(22); sanitation
END IF
cont:
NEXT fire
NEXT police
END FUNCTION
END PROGRAM </lang>
=={{header|C}}==
|
Revision as of 17:21, 13 June 2022
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
There is a highly organized city that has decided to assign a number to each of their departments:
- police department
- sanitation department
- fire department
Each department can have a number between 1 and 7 (inclusive).
The three department numbers are to be unique (different from each other) and must add up to 12.
The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
- Task
Write a computer program which outputs all valid combinations.
Possible output (for the 1st and 14th solutions):
--police-- --sanitation-- --fire-- 2 3 7 6 5 1
11l
<lang 11l>print(‘Police Sanitation Fire’) print(‘----------------------------------’)
L(police) (2..6).step(2)
L(sanitation) 1..7 L(fire) 1..7 I police!=sanitation & sanitation!=fire & fire!=police & police+fire+sanitation==12 print(police"\t\t"sanitation"\t\t"fire)</lang>
- Output:
Police Sanitation Fire ---------------------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
8080 Assembly
<lang 8080asm> org 100h lxi h,obuf ; HL = output buffer mvi b,2 ; B = police pol: mvi c,1 ; C = sanitation san: mvi d,1 ; D = fire fire: mov a,b ; Fire equal to police? cmp d jz next ; If so, invalid combination mov a,c ; Fire equal to sanitation? cmp d jz next ; If so, invalid combination mov a,b ; Total equal to 12? add c add d cpi 12 jnz next ; If not, invalid combination mov a,b ; Combination is valid, add to output call num mov a,c call num mov a,d call num mvi m,13 ; Add a newline to the output inx h mvi m,10 inx h next: mvi a,7 ; Load 7 to compare to inr d ; Next fire number cmp d ; Reached the end? jnc fire ; If not, next fire number inr c ; Otherwise, next sanitation number cmp c ; Reached the end? jnc san ; If not, next sanitation number inr b ; Increment police number twice inr b ; (twice, because it must be even) cmp b ; Reached the end? jnc pol ; If not, next police number mvi m,'$' ; If so, we're done - add CP/M string terminator mvi c,9 ; Print the output string lxi d,ohdr jmp 5 num: adi '0' ; Add number A and space to the output mov m,a inx h mvi m,' ' inx h ret ohdr: db 'P S F',13,10 obuf: equ $ ; Output buffer goes after program</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
8086 Assembly
<lang asm> cpu 8086 bits 16 org 100h section .text mov di,obuf ; Output buffer mov bl,2 ; BL = police pol: mov cl,1 ; CL = sanitation san: mov dl,1 ; DL = fire fire: cmp bl,cl ; Police equal to sanitation? je next ; Invalid combination cmp bl,dl ; Police equal to fire? je next ; Invalid combination cmp cl,dl ; Sanitation equal to fire? je next ; Invalid combination mov al,bl ; Total equal to 12? add al,cl add al,dl cmp al,12 jne next ; If not, invalid combination mov al,bl ; Combination is valid, write the three numbers call num mov al,cl call num mov al,dl call num mov ax,0A0Dh ; And a newline stosw next: mov al,7 ; Load 7 to compare to inc dx ; Increment fire number cmp al,dl ; If 7 or less, jae fire ; next fire number. inc cx ; Otherwise, ncrement sanitation number cmp al,cl ; If 7 or less, jae san ; next sanitation number inc bx ; Increment police number twice inc bx ; (it must be even) cmp al,bl ; If 7 or less, jae pol ; next police number. mov byte [di],'$' ; At the end, terminate the string mov dx,ohdr ; Tell MS-DOS to print it mov ah,9 int 21h ret num: mov ah,' ' ; Space add al,'0' ; Add number to output stosw ; Store number and space ret section .data ohdr: db 'P S F',13,10 ; Header obuf: equ $ ; Place to write output</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Action!
<lang Action!>PROC Main()
BYTE p,s,f
PrintE("P S F") FOR p=2 TO 6 STEP 2 DO FOR s=1 TO 7 DO FOR f=1 TO 7 DO IF p#s AND p#f AND s#f AND p+s+f=12 THEN PrintF("%B %B %B%E",p,s,f) FI OD OD OD
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Ada
<lang Ada>with Ada.Text_IO;
procedure Department_Numbers is
use Ada.Text_IO;
begin
Put_Line (" P S F"); for Police in 2 .. 6 loop for Sanitation in 1 .. 7 loop for Fire in 1 .. 7 loop if Police mod 2 = 0 and Police + Sanitation + Fire = 12 and Sanitation /= Police and Sanitation /= Fire and Police /= Fire then Put_Line (Police'Image & Sanitation'Image & Fire'Image); end if; end loop; end loop; end loop;
end Department_Numbers;</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Aime
<lang aime>integer p, s, f;
p = 0; while ((p += 2) <= 7) {
s = 0; while ((s += 1) <= 7) { f = 0; while ((f += 1) <= 7) { if (p + s + f == 12 && p != s && p != f && s != f) { o_form(" ~ ~ ~\n", p, s, f); } } }
}</lang>
ALGOL 68
As noted in the Fortran sample, once the police and sanitation departments are posited, the fire department value is fixed <lang algol68>BEGIN
# show possible department number allocations for police, sanitation and fire departments # # the police department number must be even, all department numbers in the range 1 .. 7 # # the sum of the department numbers must be 12 # INT max department number = 7; INT department sum = 12; print( ( "police sanitation fire", newline ) ); FOR police FROM 2 BY 2 TO max department number DO FOR sanitation TO max department number DO IF sanitation /= police THEN INT fire = ( department sum - police ) - sanitation; IF fire > 0 AND fire <= max department number AND fire /= sanitation AND fire /= police THEN print( ( whole( police, -6 ) , whole( sanitation, -11 ) , whole( fire, -5 ) , newline ) ) FI FI OD OD
END</lang>
- Output:
police sanitation fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
ALGOL W
<lang algolw>begin
% show possible department number allocations for police, sanitation and fire departments % % the police department number must be even, all department numbers in the range 1 .. 7 % % the sum of the department numbers must be 12 % integer MAX_DEPARTMENT_NUMBER, DEPARTMENT_SUM; MAX_DEPARTMENT_NUMBER := 7; DEPARTMENT_SUM := 12; write( "police sanitation fire" ); for police := 2 step 2 until MAX_DEPARTMENT_NUMBER do begin for sanitation := 1 until MAX_DEPARTMENT_NUMBER do begin IF sanitation not = police then begin integer fire; fire := ( DEPARTMENT_SUM - police ) - sanitation; if fire > 0 and fire <= MAX_DEPARTMENT_NUMBER and fire not = sanitation and fire not = police then begin write( s_w := 0, i_w := 6, police, i_w := 11, sanitation, i_w := 5, fire ) end if_valid_combination end if_sanitation_ne_police end for_sanitation end for_police
end.</lang>
- Output:
police sanitation fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
APL
<lang APL>'PSF'⍪(⊢(⌿⍨)((∪≡⊢)¨↓∧(0=2|1⌷[2]⊢)∧12=+/))↑,⍳3/7</lang>
This prints each triplet of numbers from 1 to 7 for which:
- the elements are unique:
(∪≡⊢)¨↓
- the first element is even:
(0=2|1⌷[2]⊢)
- the sum of the elements is 12:
12=+/
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
AppleScript
Briefly, composing a solution from generic functions:
<lang AppleScript>on run
script on |λ|(x) script on |λ|(y) script on |λ|(z) if y ≠ z and 1 ≤ z and z ≤ 7 then {{x, y, z} as string} else {} end if end |λ| end script concatMap(result, {12 - (x + y)}) --Z end |λ| end script concatMap(result, {1, 2, 3, 4, 5, 6, 7}) --Y end |λ| end script unlines(concatMap(result, {2, 4, 6})) --X
end run
-- 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
-- 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
-- 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
-- unlines :: [String] -> String on unlines(xs)
intercalate(linefeed, xs)
end unlines</lang>
- Output:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
Or more generally:
<lang AppleScript>-- NUMBERING CONSTRAINTS ------------------------------------------------------
-- options :: Int -> Int -> Int -> [(Int, Int, Int)] on options(lo, hi, total)
set ds to enumFromTo(lo, hi) script Xeven on |λ|(x) script Ydistinct on |λ|(y) script ZinRange on |λ|(z) if y ≠ z and lo ≤ z and z ≤ hi then Template:X, y, z else {} end if end |λ| end script concatMap(ZinRange, {total - (x + y)}) -- Z IS IN RANGE end |λ| end script script notX on |λ|(d) d ≠ x end |λ| end script concatMap(Ydistinct, filter(notX, ds)) -- Y IS NOT X end |λ| end script concatMap(Xeven, filter(my even, ds)) -- X IS EVEN
end options
-- TEST -----------------------------------------------------------------------
on run
set xs to options(1, 7, 12) intercalate("\n\n", ¬ {"(Police, Sanitation, Fire)", ¬ unlines(map(show, xs)), ¬ "Number of options: " & |length|(xs)})
end run
-- 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
-- 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
-- even :: Int -> Bool on even(x)
x mod 2 = 0
end even
-- 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
-- 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
-- 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
-- unlines :: [String] -> String on unlines(xs)
intercalate(linefeed, xs)
end unlines</lang>
- Output:
(Police, Sanitation, Fire) [2, 3, 7] [2, 4, 6] [2, 6, 4] [2, 7, 3] [4, 1, 7] [4, 2, 6] [4, 3, 5] [4, 5, 3] [4, 6, 2] [4, 7, 1] [6, 1, 5] [6, 2, 4] [6, 4, 2] [6, 5, 1] Number of options: 14
Arturo
<lang rebol>loop 1..7 'x [
loop 1..7 'y [ loop 1..7 'z [ if all? @[ even? x 12 = sum @[x y z] 3 = size unique @[x y z] ] -> print [x y z] ] ]
]</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
AutoHotkey
<lang AutoHotkey>perm(elements, n, opt:="", Delim:="", str:="", res:="", j:=0, dup:="") { res := IsObject(res) ? res : [], dup := IsObject(dup) ? dup : [] if (n > j) Loop, parse, elements, % Delim res := !(InStr(str, A_LoopField) && !(InStr(opt, "rep"))) ? perm(elements, n, opt, Delim, trim(str Delim A_LoopField, Delim), res, j+1, dup) : res else if !(dup[x := perm_sort(str, Delim)] && (InStr(opt, "comb"))) dup[x] := 1, res.Insert(str) return res, j++ }
perm_sort(str, Delim){ Loop, Parse, str, % Delim res .= A_LoopField "`n" Sort, res, D`n return StrReplace(res, "`n", Delim) }</lang> Example:<lang AutoHotkey>elements := "1234567", n := 3 for k, v in perm(elements, n) if (SubStr(v, 1, 1) + SubStr(v, 2, 1) + SubStr(v, 3, 1) = 12) && (SubStr(v, 1, 1) / 2 = Floor(SubStr(v, 1, 1)/2)) res4 .= v "`n"
MsgBox, 262144, , % res4 return</lang>
Outputs:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
AWK
<lang AWK>
- syntax: GAWK -f DEPARTMENT_NUMBERS.AWK
BEGIN {
print(" # FD PD SD") for (fire=1; fire<=7; fire++) { for (police=1; police<=7; police++) { for (sanitation=1; sanitation<=7; sanitation++) { if (rules() ~ /^1+$/) { printf("%2d %2d %2d %2d\n",++count,fire,police,sanitation) } } } } exit(0)
} function rules( stmt1,stmt2,stmt3) {
stmt1 = fire != police && fire != sanitation && police != sanitation stmt2 = fire + police + sanitation == 12 stmt3 = police % 2 == 0 return(stmt1 stmt2 stmt3)
} </lang>
- Output:
# FD PD SD 1 1 4 7 2 1 6 5 3 2 4 6 4 2 6 4 5 3 2 7 6 3 4 5 7 4 2 6 8 4 6 2 9 5 4 3 10 5 6 1 11 6 2 4 12 6 4 2 13 7 2 3 14 7 4 1
BCPL
<lang bcpl>get "libhdr"
let start() be
for p=2 to 6 by 2 for s=1 to 7 for f=1 to 7 if p~=s & s~=f & p~=f & p+s+f=12 then writef("%N %N %N*N", p, s, f)</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
BASIC
Applesoft BASIC
The Sinclair ZX81 BASIC solution works without any changes.
BASIC256
<lang BASIC256>print "--police-- --sanitation-- --fire--"
for police = 2 to 7 step 2
for fire = 1 to 7 if fire = police then continue for sanitation = 12 - police - fire if sanitation = fire or sanitation = police then continue for if sanitation >= 1 and sanitation <= 7 then print rjust(police, 6); rjust(fire, 13); rjust(sanitation, 12) end if next fire
next police</lang>
- Output:
--police-- --sanitation-- --fire-- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Run BASIC
<lang runbasic>print "police fire sanitation"
for police = 2 to 7 step 2
for fire = 1 to 7 if fire = police then [cont] sanitation = (12-police)-fire if sanitation <= 0 or sanitation > 7 or sanitation = fire or sanitation = police then [cont] print " "; police; chr$(9); fire; chr$(9); sanitation
[cont]
next fire
next police</lang>
PureBasic
<lang PureBasic>OpenConsole() PrintN("--police-- --sanitation-- --fire--")
For police = 2 To 7 Step 2
For fire = 1 To 7 If fire = police: Continue EndIf sanitation = 12 - police - fire If sanitation = fire Or sanitation = police: Continue : EndIf If sanitation >= 1 And sanitation <= 7: PrintN(" " + Str(police) + #TAB$ + #TAB$ + Str(fire) + #TAB$ + #TAB$ + Str(sanitation)) EndIf Next fire
Next police
Input() CloseConsole()</lang>
- Output:
Same as BASIC256 entry.
Yabasic
<lang freebasic>print "--police-- --sanitation-- --fire--"
for police = 2 to 7 step 2
for fire = 1 to 7 if fire = police continue sanitation = 12 - police - fire if sanitation = fire or sanitation = police continue if sanitation >= 1 and sanitation <= 7 print police using "######", fire using "############", sanitation using "###########" next fire
next police</lang>
- Output:
Same as BASIC256 entry.
IS-BASIC
<lang IS-BASIC>100 PRINT "Police","San.","Fire" 110 FOR P=2 TO 7 STEP 2 120 FOR S=1 TO 7 130 IF S<>P THEN 131 LET F=(12-P)-S 140 IF F>0 AND F<=7 AND F<>S AND F<>P THEN PRINT P,S,F 141 END IF 150 NEXT 160 NEXT</lang>
Minimal BASIC
<lang gwbasic> 10 REM Department numbers 20 PRINT "POLICE SANITATION FIRE" 30 FOR P = 2 TO 7 STEP 2 40 FOR S = 1 TO 7 50 IF S = P THEN 90 60 LET F = (12-P)-S 70 IF F <= 0 OR F > 7 OR F = S OR F = P THEN 90 80 PRINT TAB(3); P; TAB(11); S; TAB(19); F 90 NEXT S 100 NEXT P 110 END </lang>
Sinclair ZX81 BASIC
Works with 1k of RAM. This program ought not to need more than minimal changes to be compatible with any old-style BASIC dialect. <lang basic>10 PRINT "POLICE SANITATION FIRE" 20 FOR P=2 TO 7 STEP 2 30 FOR S=1 TO 7 40 IF S=P THEN NEXT S 50 LET F=(12-P)-S 60 IF F>0 AND F<=7 AND F<>S AND F<>P THEN PRINT " ";P;" ";S;" ";F 70 NEXT S 80 NEXT P</lang>
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
BBC BASIC
<lang bbcbasic>REM >deptnums max_dept_num% = 7 dept_sum% = 12 PRINT "police sanitation fire" FOR police% = 2 TO max_dept_num% STEP 2
FOR sanitation% = 1 TO max_dept_num% IF sanitation% <> police% THEN fire% = (dept_sum% - police%) - sanitation% IF fire% > 0 AND fire% <= max_dept_num% AND fire% <> sanitation% AND fire% <> police% THEN PRINT " "; police%; " "; sanitation%; " "; fire% ENDIF NEXT
NEXT END</lang>
- Output:
police sanitation fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
QBasic
<lang qbasic>PRINT "--police-- --sanitation-- --fire--"
FOR police = 2 TO 7 STEP 2
FOR fire = 1 TO 7 IF fire = police THEN GOTO cont sanitation = 12 - police - fire IF sanitation = fire OR sanitation = police THEN GOTO cont IF sanitation >= 1 AND sanitation <= 7 THEN PRINT USING " # # #"; police; fire; sanitation END IF
cont:
NEXT fire
NEXT police</lang>
XBasic
<lang xbasic>PROGRAM "Depar/num"
DECLARE FUNCTION Entry ()
FUNCTION Entry ()
PRINT "police sanitation fire"
FOR police = 2 TO 7 STEP 2 FOR fire = 1 TO 7 IF fire = police THEN GOTO cont sanitation = 12 - police - fire IF sanitation = fire OR sanitation = police THEN GOTO cont IF sanitation >= 1 AND sanitation <= 7 THEN PRINT TAB(3); police; TAB(13); fire; TAB(22); sanitation END IF
cont:
NEXT fire NEXT police
END FUNCTION END PROGRAM </lang>
C
Weird that such a simple task was still not implemented in C, would be great to see some really creative ( obfuscated ) solutions for this one. <lang C>
- include<stdio.h>
int main() { int police,sanitation,fire;
printf("Police Sanitation Fire\n"); printf("----------------------------------");
for(police=2;police<=6;police+=2){ for(sanitation=1;sanitation<=7;sanitation++){ for(fire=1;fire<=7;fire++){ if(police!=sanitation && sanitation!=fire && fire!=police && police+fire+sanitation==12){ printf("\n%d\t\t%d\t\t%d",police,sanitation,fire); } } } }
return 0; } </lang> Output:
Police Sanitation Fire ---------------------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
C#
<lang csharp>using System; public class Program {
public static void Main() { for (int p = 2; p <= 7; p+=2) { for (int s = 1; s <= 7; s++) { int f = 12 - p - s; if (s >= f) break; if (f > 7) continue; if (s == p || f == p) continue; //not even necessary Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}"); Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}"); } } }
}</lang>
- Output:
Police:2, Sanitation:3, Fire:7 Police:2, Sanitation:7, Fire:3 Police:2, Sanitation:4, Fire:6 Police:2, Sanitation:6, Fire:4 Police:4, Sanitation:1, Fire:7 Police:4, Sanitation:7, Fire:1 Police:4, Sanitation:2, Fire:6 Police:4, Sanitation:6, Fire:2 Police:4, Sanitation:3, Fire:5 Police:4, Sanitation:5, Fire:3 Police:6, Sanitation:1, Fire:5 Police:6, Sanitation:5, Fire:1 Police:6, Sanitation:2, Fire:4 Police:6, Sanitation:4, Fire:2
C++
<lang cpp>
- include <iostream>
- include <iomanip>
int main( int argc, char* argv[] ) {
int sol = 1; std::cout << "\t\tFIRE\t\tPOLICE\t\tSANITATION\n"; for( int f = 1; f < 8; f++ ) { for( int p = 1; p < 8; p++ ) { for( int s = 1; s < 8; s++ ) { if( f != p && f != s && p != s && !( p & 1 ) && ( f + s + p == 12 ) ) { std::cout << "SOLUTION #" << std::setw( 2 ) << sol++ << std::setw( 2 ) << ":\t" << std::setw( 2 ) << f << "\t\t " << std::setw( 3 ) << p << "\t\t" << std::setw( 6 ) << s << "\n"; } } } } return 0;
}</lang>
- Output:
FIRE POLICE SANITATION SOLUTION # 1: 1 4 7 SOLUTION # 2: 1 6 5 SOLUTION # 3: 2 4 6 SOLUTION # 4: 2 6 4 SOLUTION # 5: 3 2 7 SOLUTION # 6: 3 4 5 SOLUTION # 7: 4 2 6 SOLUTION # 8: 4 6 2 SOLUTION # 9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
Clojure
<lang clojure>(let [n (range 1 8)]
(for [police n sanitation n fire n :when (distinct? police sanitation fire) :when (even? police) :when (= 12 (+ police sanitation fire))] (println police sanitation fire)))</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
CLU
<lang clu>start_up = proc ()
po: stream := stream$primary_output() stream$putl(po, "P S F\n- - -") for police: int in int$from_to_by(2,7,2) do for sanitation: int in int$from_to(1,7) do for fire: int in int$from_to(1,7) do if police~=sanitation & sanitation~=fire & police~=fire & police+sanitation+fire = 12 then stream$putl(po, int$unparse(police) || " " || int$unparse(sanitation) || " " || int$unparse(fire)) end end end end
end start_up</lang>
- Output:
P S F - - - 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
COBOL
<lang cobol> IDENTIFICATION DIVISION.
PROGRAM-ID. DEPARTMENT-NUMBERS.
DATA DIVISION. WORKING-STORAGE SECTION. 01 BANNER PIC X(24) VALUE "POLICE SANITATION FIRE". 01 COMBINATION. 03 FILLER PIC X(5) VALUE SPACES. 03 POLICE PIC 9. 03 FILLER PIC X(11) VALUE SPACES. 03 SANITATION PIC 9. 03 FILLER PIC X(5) VALUE SPACES. 03 FIRE PIC 9. 01 TOTAL PIC 99.
PROCEDURE DIVISION. BEGIN. DISPLAY BANNER. PERFORM POLICE-LOOP VARYING POLICE FROM 2 BY 2 UNTIL POLICE IS GREATER THAN 6. STOP RUN. POLICE-LOOP. PERFORM SANITATION-LOOP VARYING SANITATION FROM 1 BY 1 UNTIL SANITATION IS GREATER THAN 7. SANITATION-LOOP. PERFORM FIRE-LOOP VARYING FIRE FROM 1 BY 1 UNTIL FIRE IS GREATER THAN 7. FIRE-LOOP. ADD POLICE, SANITATION, FIRE GIVING TOTAL. IF POLICE IS NOT EQUAL TO SANITATION AND POLICE IS NOT EQUAL TO FIRE AND SANITATION IS NOT EQUAL TO FIRE AND TOTAL IS EQUAL TO 12, DISPLAY COMBINATION.</lang>
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Cowgol
<lang cowgol>include "cowgol.coh";
typedef Dpt is int(1, 7);
- print combination if valid
sub print_comb(p: Dpt, s: Dpt, f: Dpt) is
var out: uint8[] := {'*',' ','*',' ','*','\n',0}; out[0] := p + '0'; out[2] := s + '0'; out[4] := f + '0'; if p != s and p != f and f != s and p+s+f == 12 then print(&out[0]); end if;
end sub;
print("P S F\n"); # header
var pol: Dpt := 2; while pol <= 7 loop
var san: Dpt := 1; while san <= 7 loop var fire: Dpt := 1; while fire <= 7 loop print_comb(pol, san, fire); fire := fire + 1; end loop; san := san + 1; end loop; pol := pol + 2;
end loop;</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
D
<lang D> import std.stdio, std.range;
void main() {
int sol = 1; writeln("\t\tFIRE\t\tPOLICE\t\tSANITATION"); foreach( f; iota(1,8) ) { foreach( p; iota(1,8) ) { foreach( s; iota(1,8) ) { if( f != p && f != s && p != s && !( p & 1 ) && ( f + s + p == 12 ) ) { writefln("SOLUTION #%2d:\t%2d\t\t%3d\t\t%6d", sol++, f, p, s); } } } }
} </lang> Output:
FIRE POLICE SANITATION SOLUTION # 1: 1 4 7 SOLUTION # 2: 1 6 5 SOLUTION # 3: 2 4 6 SOLUTION # 4: 2 6 4 SOLUTION # 5: 3 2 7 SOLUTION # 6: 3 4 5 SOLUTION # 7: 4 2 6 SOLUTION # 8: 4 6 2 SOLUTION # 9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
Delphi
<lang Delphi> program Department_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
var
i, j, k, count: Integer;
begin
writeln('Police Sanitation Fire'); writeln('------ ---------- ----'); count := 0; i := 2; while i < 7 do begin for j := 1 to 7 do begin if j = i then Continue; for k := 1 to 7 do begin if (k = i) or (k = j) then Continue; if i + j + k <> 12 then Continue; writeln(format(' %d %d %d', [i, j, k])); inc(count); end; end; inc(i, 2); end; writeln(#10, count, ' valid combinations'); readln;
end.</lang>
Draco
<lang draco>proc main() void:
byte police, sanitation, fire;
writeln("Police Sanitation Fire"); for police from 2 by 2 upto 7 do for sanitation from 1 upto 7 do for fire from 1 upto 7 do if police /= sanitation and police /= fire and sanitation /= fire and police + sanitation + fire = 12 then writeln(police:6, " ", sanitation:10, " ", fire:4) fi od od od
corp</lang>
- Output:
Police Sanitation Fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Elixir
<lang elixir> IO.puts("P - F - S") for p <- [2,4,6],
f <- 1..7, s <- 1..7, p != f and p != s and f != s and p + f + s == 12 do "#{p} - #{f} - #{s}" end |> Enum.each(&IO.puts/1)
</lang> <lang elixir> P - F - S 2 - 3 - 7 2 - 4 - 6 2 - 6 - 4 2 - 7 - 3 4 - 1 - 7 4 - 2 - 6 4 - 3 - 5 4 - 5 - 3 4 - 6 - 2 4 - 7 - 1 6 - 1 - 5 6 - 2 - 4 6 - 4 - 2 6 - 5 - 1 </lang>
Excel
LAMBDA
Binding the name departmentNumbers to the following expression in the Name Manager of the Excel WorkBook:
(See LAMBDA: The ultimate Excel worksheet function)
<lang lisp>departmentNumbers =validRows(
LAMBDA(ab, LET( x, INDEX(ab, 0, 1), y, INDEX(ab, 0, 2), z, 12 - (x + y), IF(y <> z, IF(1 <= z, IF(7 >= z, CHOOSE({1, 2, 3}, x, y, z), NA() ), NA() ), NA() ) ) )( cartesianProduct({2;4;6})( SEQUENCE(7, 1, 1, 1) ) )
)</lang>
and also assuming the following generic bindings in the Name Manager for the WorkBook:
<lang lisp>cartesianProduct =LAMBDA(xs,
LAMBDA(ys, LET( ny, ROWS(ys), ixs, SEQUENCE(ROWS(xs) * ny, 2, 1, 1),
IF(0 <> MOD(ixs, 2), INDEX(xs, 1 + QUOTIENT(ixs, ny * 2) ), LET( r, MOD(QUOTIENT(ixs, 2), ny),
INDEX(ys, IF(0 = r, ny, r)) ) ) ) )
)
validRows
=LAMBDA(xs,
LET( ixs, SEQUENCE(ROWS(xs), 1, 1, 1), valids, FILTER( ixs, LAMBDA(i, NOT(ISNA(INDEX(xs, i))) )(ixs) ),
INDEX( xs, valids, SEQUENCE( 1, COLUMNS(xs) ) ) )
)</lang>
- Output:
The formula in cell B2 below defines a two-dimensional array which populates the range B2:D15:
fx | =departmentNumbers | ||||
---|---|---|---|---|---|
A | B | C | D | ||
1 | Valid department numbers | ||||
2 | 2 | 3 | 7 | ||
3 | 2 | 4 | 6 | ||
4 | 2 | 6 | 4 | ||
5 | 2 | 7 | 3 | ||
6 | 4 | 1 | 7 | ||
7 | 4 | 2 | 6 | ||
8 | 4 | 3 | 5 | ||
9 | 4 | 5 | 3 | ||
10 | 4 | 6 | 2 | ||
11 | 4 | 7 | 1 | ||
12 | 6 | 1 | 5 | ||
13 | 6 | 2 | 4 | ||
14 | 6 | 4 | 2 | ||
15 | 6 | 5 | 1 |
F#
<lang fsharp> // A function to generate department numbers. Nigel Galloway: May 2nd., 2018 type dNum = {Police:int; Fire:int; Sanitation:int} let fN n=n.Police%2=0&&n.Police+n.Fire+n.Sanitation=12&&n.Police<>n.Fire&&n.Police<>n.Sanitation&&n.Fire<>n.Sanitation List.init (7*7*7) (fun n->{Police=n%7+1;Fire=(n/7)%7+1;Sanitation=(n/49)+1})|>List.filter fN|>List.iter(printfn "%A") </lang>
- Output:
{Police = 6; Fire = 5; Sanitation = 1;} {Police = 4; Fire = 7; Sanitation = 1;} {Police = 6; Fire = 4; Sanitation = 2;} {Police = 4; Fire = 6; Sanitation = 2;} {Police = 4; Fire = 5; Sanitation = 3;} {Police = 2; Fire = 7; Sanitation = 3;} {Police = 6; Fire = 2; Sanitation = 4;} {Police = 2; Fire = 6; Sanitation = 4;} {Police = 6; Fire = 1; Sanitation = 5;} {Police = 4; Fire = 3; Sanitation = 5;} {Police = 4; Fire = 2; Sanitation = 6;} {Police = 2; Fire = 4; Sanitation = 6;} {Police = 4; Fire = 1; Sanitation = 7;} {Police = 2; Fire = 3; Sanitation = 7;}
Factor
<lang factor>USING: formatting io kernel math math.combinatorics math.ranges sequences sets ; IN: rosetta-code.department-numbers
7 [1,b] 3 <k-permutations> [ [ first even? ] [ sum 12 = ] bi and ] filter
"{ Police, Sanitation, Fire }" print nl [ "%[%d, %]\n" printf ] each</lang>
- Output:
{ Police, Sanitation, Fire } { 2, 3, 7 } { 2, 4, 6 } { 2, 6, 4 } { 2, 7, 3 } { 4, 1, 7 } { 4, 2, 6 } { 4, 3, 5 } { 4, 5, 3 } { 4, 6, 2 } { 4, 7, 1 } { 6, 1, 5 } { 6, 2, 4 } { 6, 4, 2 } { 6, 5, 1 }
Fermat
<lang fermat>!!'Police Sanitation Fire'; !!'------|----------|----'; for p = 2 to 6 by 2 do
for s = 1 to 7 do for f = 1 to 7 do if p+f+s=12 and f<>p and f<>s and s<>p then !!(' ',p,' ',s,' ',f); fi od od od;</lang>
- Output:
Police Sanitation Fire ------|----------|----
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
FOCAL
<lang focal>01.10 F P=2,2,6;F S=1,7;F G=1,7;D 2 01.20 Q
02.10 I (P-S)2.2,2.6,2.2 02.20 I (P-G)2.3,2.6,2.3 02.30 I (S-G)2.4,2.6,2.4 02.40 I (P+S+G-12)2.6,2.5,2.6 02.50 T %1,P,S,G,! 02.60 R</lang>
- Output:
= 2= 3= 7 = 2= 4= 6 = 2= 6= 4 = 2= 7= 3 = 4= 1= 7 = 4= 2= 6 = 4= 3= 5 = 4= 5= 3 = 4= 6= 2 = 4= 7= 1 = 6= 1= 5 = 6= 2= 4 = 6= 4= 2 = 6= 5= 1
Forth
<lang Forth>\ if department numbers are valid, print them on a single line
- fire ( pol san fir -- )
2dup = if 2drop drop exit then 2 pick over = if 2drop drop exit then rot . swap . . cr ;
\ tries to assign numbers with given policeno and sanitationno \ and fire = 12 - policeno - sanitationno
- sanitation ( pol san -- )
2dup = if 2drop exit then \ no repeated numbers 12 over - 2 pick - \ calculate fireno dup 1 < if 2drop drop exit then \ cannot be less than 1 dup 7 > if 2drop drop exit then \ cannot be more than 7 fire ;
\ tries to assign numbers with given policeno \ and sanitation = 1, 2, 3, ..., or 7
- police ( pol -- )
8 1 do dup i sanitation loop drop ;
\ tries to assign numbers with police = 2, 4, or 6
- departments cr \ leave input line
8 2 do i police 2 +loop ;</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 ok
Fortran
This uses the ability standardised in F90 of labelling a DO-loop so that its start and end are linked by usage of the same name, with this checked by the compiler. Further, in avoiding the use of the dreaded GO TO statement, the CYCLE statement can be employed instead with the same effect, and it too can bear the same name so that it is clear which loop is involved. These names prefix the DO-loop, and so, force some additional indentation. They are not statement labels and must be unique themselves. Notably, they cannot be the same text as the name of the index variable for their DO-loop, unlike the lead given by BASIC with its FOR I ... NEXT I
arrangement.
The method is just to generate all the possibilities, discarding those that fail the specified tests. However, the requirement that the codes add up to twelve means that after the first two are chosen the third is determined, and blandly looping through all the possibilities is too much brute force and ignorance, though other collections of rules could make that bearable.
Since the modernisers of Fortran made a point of specifying that it does not specify the manner of evaluation of compound boolean expressions, specifically, that there is to be no reliance on Short-circuit_evaluation, both parts of the compound expression of the line labelled 5 "may" be evaluated even though the first may have determined the result. Prior to the introduction of LOGICAL variables with F66, one employed integer arithmetic as is demonstrated in the arithmetic-IF test of the line labelled 6. On the B6700, this usage ran faster than the corresponding boolean expression - possibly because there was no test for short-circuiting the expression when the first part of a multiply was zero...
Note that the syntax enables two classes of labels: the old-style numerical label in columns one to five, and the special label-like prefix of a DO-loop that is not in columns one to five. And yes, a line can have both. <lang Fortran> INTEGER P,S,F !Department codes for Police, Sanitation, and Fire. Values 1 to 7 only.
1 PP:DO P = 2,7,2 !The police demand an even number. They're special and use violence. 2 SS:DO S = 1,7 !The sanitation department accepts any value. 3 IF (P.EQ.S) CYCLE SS !But it must differ from the others. 4 F = 12 - (P + S) !The fire department accepts any number, but the sum must be twelve. 5 IF (F.LE.0 .OR. F.GT.7) CYCLE SS !Ensure that the only option is within range. 6 IF ((F - S)*(F - P)) 7,8,7 !And F is to differ from S and from P 7 WRITE (6,"(3I2)") P,S,F !If we get here, we have a possible set. 8 END DO SS !Next S 9 END DO PP !Next P. END !Well, that was straightforward.</lang>
Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
FreeBASIC
<lang freebasic>' version 15-08-2017 ' compile with: fbc -s console
Dim As Integer fire, police, sanitation
Print "police fire sanitation" Print "----------------------"
For police = 2 To 7 Step 2
For fire = 1 To 7 If fire = police Then Continue For sanitation = 12 - police - fire If sanitation = fire Or sanitation = police Then Continue For If sanitation >= 1 And sanitation <= 7 Then Print Using " # # # "; police; fire; sanitation End If Next
Next
' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End</lang>
- Output:
police fire sanitation ---------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Gambas
Click this link to run this code <lang gambas>Public Sub Main() Dim siC0, siC1, siC2 As Short Dim sOut As New String[] Dim sTemp As String
For siC0 = 2 To 6 Step 2
For siC1 = 1 To 7 For siC2 = 1 To 7 If sic0 + siC1 + siC2 = 12 Then If siC0 <> siC1 And siC1 <> siC2 And siC0 <> siC2 Then sOut.Add(Str(siC0) & Str(siC1) & Str(siC2)) End If Next Next
Next
Print "\tPolice\tFire\tSanitation" siC0 = 0
For Each sTemp In sOut
Inc sic0 Print "[" & Format(Str(siC0), "00") & "]\t" & Left(sTemp, 1) & "\t" & Mid(sTemp, 2, 1) & "\t" & Right(sTemp, 1)
Next
End</lang> Output:
Police Fire Sanitation [01] 2 3 7 [02] 2 4 6 [03] 2 6 4 [04] 2 7 3 [05] 4 1 7 [06] 4 2 6 [07] 4 3 5 [08] 4 5 3 [09] 4 6 2 [10] 4 7 1 [11] 6 1 5 [12] 6 2 4 [13] 6 4 2 [14] 6 5 1
Go
<lang go>package main
import "fmt"
func main() {
fmt.Println("Police Sanitation Fire") fmt.Println("------ ---------- ----") count := 0 for i := 2; i < 7; i += 2 { for j := 1; j < 8; j++ { if j == i { continue } for k := 1; k < 8; k++ { if k == i || k == j { continue } if i + j + k != 12 { continue } fmt.Printf(" %d %d %d\n", i, j, k) count++ } } } fmt.Printf("\n%d valid combinations\n", count)
}</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
Groovy
<lang groovy>class DepartmentNumbers {
static void main(String[] args) { println("Police Sanitation Fire") println("------ ---------- ----") int count = 0 for (int i = 2; i <= 6; i += 2) { for (int j = 1; j <= 7; ++j) { if (j == i) continue for (int k = 1; k <= 7; ++k) { if (k == i || k == j) continue if (i + j + k != 12) continue println(" $i $j $k") count++ } } } println() println("$count valid combinations") }
}</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
GW-BASIC
<lang gwbasic>10 PRINT "Police Sanitation Fire" 20 PRINT "------|----------|----" 30 FOR P = 2 TO 7 STEP 2 40 FOR S = 1 TO 7 50 IF S = P THEN GOTO 100 60 FOR F = 1 TO 7 70 IF S = F OR F = P THEN GOTO 90 80 IF P+S+F = 12 THEN PRINT USING " # # #";P;F;S 90 NEXT F 100 NEXT S 110 NEXT P</lang>
Haskell
Bare minimum: <lang haskell>main :: IO () main =
mapM_ print $ [2, 4, 6] >>= \x -> [1 .. 7] >>= \y -> [12 - (x + y)] >>= \z -> case y /= z && 1 <= z && z <= 7 of True -> [(x, y, z)] _ -> []</lang>
or, resugaring this into list comprehension format: <lang Haskell>main :: IO () main =
mapM_ print [ (x, y, z) | x <- [2, 4, 6] , y <- [1 .. 7] , z <- [12 - (x + y)] , y /= z && 1 <= z && z <= 7 ]</lang>
Do notation: <lang Haskell>main :: IO () main =
mapM_ print $ do x <- [2, 4, 6] y <- [1 .. 7] z <- [12 - (x + y)] if y /= z && 1 <= z && z <= 7 then [(x, y, z)] else []</lang>
Unadorned brute force – more than enough at this small scale: <lang Haskell>import Data.List (nub)
main :: IO () main =
let xs = [1 .. 7] in mapM_ print $ xs >>= \x -> xs >>= \y -> xs >>= \z -> [ (x, y, z) | even x && 3 == length (nub [x, y, z]) && 12 == sum [x, y, z] ]</lang>
- Output:
(2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1)
Or, more generally: <lang Haskell>-------------------- DEPARTMENT NUMBERS ------------------
options :: Int -> Int -> Int -> [(Int, Int, Int)] options lo hi total =
( \ds -> filter even ds >>= \x -> filter (/= x) ds >>= \y -> [total - (x + y)] >>= \z -> [ (x, y, z) | y /= z && lo <= z && z <= hi ] ) [lo .. hi]
TEST -------------------------
main :: IO () main =
let xs = options 1 7 12 in putStrLn "(Police, Sanitation, Fire)\n" >> mapM_ print xs >> mapM_ putStrLn [ "\nNumber of options: ", show (length xs) ]</lang>
Reaching again for a little more syntactic sugar, the options function above could also be re-written either as a list comprehension, <lang Haskell>options :: Int -> Int -> Int -> [(Int, Int, Int)] options lo hi total =
let ds = [lo .. hi] in [ (x, y, z) | x <- filter even ds , y <- filter (/= x) ds , let z = total - (x + y) , y /= z && lo <= z && z <= hi ]</lang>
or in Do notation: <lang haskell>import Control.Monad (guard)
options :: Int -> Int -> Int -> [(Int, Int, Int)] options lo hi total =
let ds = [lo .. hi] in do x <- filter even ds y <- filter (/= x) ds let z = total - (x + y) guard $ y /= z && lo <= z && z <= hi return (x, y, z)</lang>
- Output:
(Police, Sanitation, Fire) (2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1) Number of options: 14
J
Solution: <lang j>require 'stats' permfrom=: ,/@(perm@[ {"_ 1 comb) NB. get permutations of length x from y possible items
alluniq=: # = #@~. NB. check items are unique addto12=: 12 = +/ NB. check items add to 12 iseven=: -.@(2&|) NB. check items are even policeeven=: {.@iseven NB. check first item is even conditions=: policeeven *. addto12 *. alluniq
Validnums=: >: i.7 NB. valid Department numbers
getDeptNums=: [: (#~ conditions"1) Validnums {~ permfrom</lang> Example usage: <lang j> 3 getDeptNums 7 4 1 7 4 7 1 6 1 5 6 5 1 2 3 7 2 7 3 2 4 6 2 6 4 4 2 6 4 6 2 6 2 4 6 4 2 4 3 5 4 5 3</lang>
Alternate approach
<lang J> (/:"#. 2|]) (#~ 12=+/"1) 1+3 comb 7 [ load'stats' 4 1 7 6 1 5 2 3 7 2 4 6 4 3 5</lang>
Note that we are only showing the distinct valid combinations here, not all valid permutations of those combinations.
Java
<lang Java>public class DepartmentNumbers {
public static void main(String[] args) { System.out.println("Police Sanitation Fire"); System.out.println("------ ---------- ----"); int count = 0; for (int i = 2; i <= 6; i += 2) { for (int j = 1; j <= 7; ++j) { if (j == i) continue; for (int k = 1; k <= 7; ++k) { if (k == i || k == j) continue; if (i + j + k != 12) continue; System.out.printf(" %d %d %d\n", i, j, k); count++; } } } System.out.printf("\n%d valid combinations", count); }
}</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
JavaScript
ES5
Briefly: <lang JavaScript>(function () {
'use strict';
// concatMap :: (a -> [b]) -> [a] -> [b] function concatMap(f, xs) { return [].concat.apply([], xs.map(f)); };
return '(Police, Sanitation, Fire)\n' + concatMap(function (x) { return concatMap(function (y) { return concatMap(function (z) { return z !== y && 1 <= z && z <= 7 ? [ [x, y, z] ] : []; }, [12 - (x + y)]); }, [1, 2, 3, 4, 5, 6, 7]); }, [2, 4, 6]) .map(JSON.stringify) .join('\n');
})();</lang>
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1]
Or, more generally:
<lang JavaScript>(function () {
'use strict';
// NUMBERING CONSTRAINTS --------------------------------------------------
// options :: Int -> Int -> Int -> [(Int, Int, Int)] function options(lo, hi, total) { var bind = flip(concatMap), ds = enumFromTo(lo, hi);
return bind(filter(even, ds), function (x) { // X is even, return bind(filter(function (d) { return d !== x; }, ds), function (y) { // Y is distinct from X, return bind([total - (x + y)], function (z) { // Z sums with x and y to total, and is in ds. return z !== y && lo <= z && z <= hi ? [ [x, y, z] ] : []; })})})};
// GENERIC FUNCTIONS ------------------------------------------------------
// concatMap :: (a -> [b]) -> [a] -> [b] function concatMap(f, xs) { return [].concat.apply([], xs.map(f)); };
// enumFromTo :: Int -> Int -> [Int] function enumFromTo(m, n) { return Array.from({ length: Math.floor(n - m) + 1 }, function (_, i) { return m + i; }); };
// even :: Integral a => a -> Bool function even(n) { return n % 2 === 0; };
// filter :: (a -> Bool) -> [a] -> [a] function filter(f, xs) { return xs.filter(f); };
// flip :: (a -> b -> c) -> b -> a -> c function flip(f) { return function (a, b) { return f.apply(null, [b, a]); }; };
// length :: [a] -> Int function length(xs) { return xs.length; };
// map :: (a -> b) -> [a] -> [b] function map(f, xs) { return xs.map(f); };
// show :: a -> String function show(x) { return JSON.stringify(x); }; //, null, 2);
// unlines :: [String] -> String function unlines(xs) { return xs.join('\n'); };
// TEST ------------------------------------------------------------------- var xs = options(1, 7, 12); return '(Police, Sanitation, Fire)\n\n' + unlines(map(show, xs)) + '\n\nNumber of options: ' + length(xs);
})();</lang>
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1] Number of options: 14
ES6
Briefly: <lang JavaScript>(() => {
"use strict";
const label = "(Police, Sanitation, Fire)", solutions = [2, 4, 6] .flatMap( x => [1, 2, 3, 4, 5, 6, 7] .flatMap( y => [12 - (x + y)] .flatMap( z => z !== y && 1 <= z && z <= 7 ? [ [x, y, z] ] : [] ) ) ) .map(JSON.stringify) .join("\n");
return `${label}\n${solutions}`;
})();</lang>
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1]
Or, more generally, by composition of generic functions:
<lang JavaScript>(() => {
"use strict";
// -------------- NUMBERING CONSTRAINTS --------------
// options :: Int -> Int -> Int -> [(Int, Int, Int)] const options = lo => hi => total => { const bind = xs => f => xs.flatMap(f), ds = enumFromTo(lo)(hi);
return bind(ds.filter(even))( x => bind(ds.filter(d => d !== x))( y => bind([total - (x + y)])( z => (z !== y && lo <= z && z <= hi) ? [ [x, y, z] ] : [] ) ) ); };
// ---------------------- TEST ----------------------- const main = () => { const label = "(Police, Sanitation, Fire)", solutions = options(1)(7)(12), n = solutions.length, list = solutions .map(JSON.stringify) .join("\n");
return ( `${label}\n\n${list}\n\nNumber of options: ${n}` ); };
// ---------------- GENERIC FUNCTIONS ----------------
// enumFromTo :: Int -> Int -> [Int] const enumFromTo = m => n => Array.from({ length: 1 + n - m }, (_, i) => m + i);
// even :: Integral a => a -> Bool const even = n => n % 2 === 0;
// MAIN --- return main();
})();</lang>
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1] Number of options: 14
jq
In this section, we present three solutions.
The first illustrates how a straightforward generate-and-test algorithm using familiar for-loops can be translated into jq.
The second illustrates how essentially the same algorithm can be written in a more economical way, without sacrificing comprehensibility.
The third illustrates how the built-in function `combinations/1' can be used to achieve greater efficiency.
The solutions in all cases are presented as a stream of JSON objects such as:
{"fire":1,"police":4,"sanitation":7}
as these are self-explanatory, though it would be trivial to present them in another format. For brevity, the solutions are omitted here.
Nested for-loop <lang jq>def check(fire; police; sanitation):
(fire != police) and (fire != sanitation) and (police != sanitation) and (fire + police + sanitation == 12) and (police % 2 == 0);
range(1;8) as $fire | range(1;8) as $police | range(1;8) as $sanitation | select( check($fire; $police; $sanitation) ) | {$fire, $police, $sanitation}</lang>
In Brief <lang jq>{fire: range(1;8), police: range(1;8), sanitation: range(1;8)} | select( .fire != .police and .fire != .sanitation and .police != .sanitation
and .fire + .police + .sanitation == 12 and .police % 2 == 0 )</lang>
combinations <lang jq>
[range(1;8)] | combinations(3) | select( add == 12 and .[1] % 2 == 0) | {fire: .[0], police: .[1], sanitation: .[2]}</lang>
Julia
<lang julia>using Printf
function findsolution(rng=1:7)
rst = Matrix{Int}(0, 3) for p in rng, f in rng, s in rng if p != s != f != p && p + s + f == 12 && iseven(p) rst = [rst; p s f] end end return rst
end
function printsolutions(sol::Matrix{Int})
println(" Pol. Fire San.") println(" ---- ---- ----") for row in 1:size(sol, 1) @printf("%2i | %4i%7i%7i\n", row, sol[row, :]...) end
end
printsolutions(findsolution()) </lang>
- Output:
Pol. Fire San. ---- ---- ---- 1 | 2 7 3 2 | 2 6 4 3 | 2 4 6 4 | 2 3 7 5 | 4 7 1 6 | 4 6 2 7 | 4 5 3 8 | 4 3 5 9 | 4 2 6 10 | 4 1 7 11 | 6 5 1 12 | 6 4 2 13 | 6 2 4 14 | 6 1 5
Kotlin
<lang scala>// version 1.1.2
fun main(args: Array<String>) {
println("Police Sanitation Fire") println("------ ---------- ----") var count = 0 for (i in 2..6 step 2) { for (j in 1..7) { if (j == i) continue for (k in 1..7) { if (k == i || k == j) continue if (i + j + k != 12) continue println(" $i $j $k") count++ } } } println("\n$count valid combinations")
}</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
Lua
<lang lua> print( "Fire", "Police", "Sanitation" ) sol = 0 for f = 1, 7 do
for p = 1, 7 do for s = 1, 7 do if s + p + f == 12 and p % 2 == 0 and f ~= p and f ~= s and p ~= s then print( f, p, s ); sol = sol + 1 end end end
end print( string.format( "\n%d solutions found", sol ) ) </lang>
- Output:
Fire Police Sanitation 1 4 7 1 6 5 2 4 6 2 6 4 3 2 7 3 4 5 4 2 6 4 6 2 5 4 3 5 6 1 6 2 4 6 4 2 7 2 3 7 4 1 14 solutions found
MAD
<lang MAD> NORMAL MODE IS INTEGER
PRINT COMMENT $ POLICE SANITATION FIRE$ THROUGH LOOP, FOR P=2, 2, P.G.7 THROUGH LOOP, FOR S=1, 1, S.G.7 THROUGH LOOP, FOR F=1, 1, F.G.7 WHENEVER P.E.S .OR. P.E.F .OR. S.E.F, TRANSFER TO LOOP WHENEVER P+S+F .E. 12, PRINT FORMAT OCC, P, S, F
LOOP CONTINUE
VECTOR VALUES OCC = $I6,S2,I10,S2,I4*$ END OF PROGRAM</lang>
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Maple
<lang Maple>#determines if i, j, k are exclusive numbers exclusive_numbers := proc(i, j, k) if (i = j) or (i = k) or (j = k) then return false; end if; return true; end proc;
- outputs all possible combinations of numbers that statisfy given conditions
department_numbers := proc() local i, j, k; printf("Police Sanitation Fire\n"); for i to 7 do for j to 7 do k := 12 - i - j; if (k <= 7) and (k >= 1) and (i mod 2 = 0) and exclusive_numbers(i,j,k) then printf("%d %d %d\n", i, j, k); end if; end do; end do; end proc;
department_numbers(); </lang>
- Output:
Police Sanitation Fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Mathematica /Wolfram Language
<lang Mathematica>Select[Permutations[Range[7], {3}], Total[#] == 12 && EvenQ[First[#]] &]</lang>
- Output:
{{2, 3, 7}, {2, 4, 6}, {2, 6, 4}, {2, 7, 3}, {4, 1, 7}, {4, 2, 6}, {4, 3, 5}, {4, 5, 3}, {4, 6, 2}, {4, 7, 1}, {6, 1, 5}, {6, 2, 4}, {6, 4, 2}, {6, 5, 1}}
Modula-2
<lang modula2>MODULE DepartmentNumbers; FROM Conversions IMPORT IntToStr; FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteInt(num : INTEGER); VAR str : ARRAY[0..16] OF CHAR; BEGIN
IntToStr(num,str); WriteString(str);
END WriteInt;
VAR i,j,k,count : INTEGER; BEGIN
count:=0;
WriteString("Police Sanitation Fire"); WriteLn; WriteString("------ ---------- ----"); WriteLn;
FOR i:=2 TO 6 BY 2 DO FOR j:=1 TO 7 DO IF j=i THEN CONTINUE; END; FOR k:=1 TO 7 DO IF (k=i) OR (k=j) THEN CONTINUE; END; IF i+j+k # 12 THEN CONTINUE; END; WriteString(" "); WriteInt(i); WriteString(" "); WriteInt(j); WriteString(" "); WriteInt(k); WriteLn; INC(count); END; END; END;
WriteLn; WriteInt(count); WriteString(" valid combinations"); WriteLn;
ReadChar;
END DepartmentNumbers.</lang>
Mercury
<lang mercury>:- module department_numbers.
- - interface.
- - import_module io.
- - pred main(io::di, io::uo) is cc_multi.
- - implementation.
- - import_module int, list, solutions, string.
main(!IO) :-
io.print_line("P S F", !IO), unsorted_aggregate(department_number, print_solution, !IO).
- - pred print_solution({int, int, int}::in, io::di, io::uo) is det.
print_solution({P, S, F}, !IO) :-
io.format("%d %d %d\n", [i(P), i(S), i(F)], !IO).
- - pred department_number({int, int, int}::out) is nondet.
department_number({Police, Sanitation, Fire}) :-
list.member(Police, [2, 4, 6]), list.member(Sanitation, 1 .. 7), list.member(Fire, 1 .. 7), Police \= Sanitation, Police \= Fire, Sanitation \= Fire, Police + Sanitation + Fire = 12.</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Nim
<lang Nim>type Solution = tuple[p, s, f: int]
iterator solutions(max, total: Positive): Solution =
for p in countup(2, max, 2): for s in 1..max: if s == p: continue let f = total - p - s if f notin [p, s] and f in 1..max: yield (p, s, f)
echo "P S F" for sol in solutions(7, 12):
echo sol.p, " ", sol.s, " ", sol.f</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Objeck
<lang objeck>class Program {
function : Main(args : String[]) ~ Nil { sol := 1; "\t\tFIRE\tPOLICE\tSANITATION"->PrintLine(); for( f := 1; f < 8; f+=1; ) { for( p := 1; p < 8; p+=1; ) { for( s:= 1; s < 8; s+=1; ) { if( f <> p & f <> s & p <> s & ( p and 1 ) = 0 & ( f + s + p = 12 ) ) { "SOLUTION #{$sol}: \t{$f}\t{$p}\t{$s}"->PrintLine(); sol += 1; }; }; }; }; }
}</lang>
Output:
FIRE POLICE SANITATION SOLUTION #1: 1 4 7 SOLUTION #2: 1 6 5 SOLUTION #3: 2 4 6 SOLUTION #4: 2 6 4 SOLUTION #5: 3 2 7 SOLUTION #6: 3 4 5 SOLUTION #7: 4 2 6 SOLUTION #8: 4 6 2 SOLUTION #9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
OCaml
<lang OCaml>(*
* Caution: This is my first Ocaml program and anyone with Ocaml experience probably thinks it's horrible * So please don't use this as an example for "good ocaml code" see it more as * "this is what my first lines of ocaml might look like" * * The only reason im publishing this is that nobody has yet submitted an example in ocaml *)
(* sfp is just a convenience to put a combination if sanitation (s) fire (f) and police (p) department in one record*)
type sfp = {s : int; f : int; p : int}
(* Convenience Function to print a single sfp Record *) let print_sfp e =
Printf.printf "%d %d %d\n" e.s e.f e.p
(* Convenience Function to print a list of sfp Records*) let print_sfp_list l =
l |> List.iter print_sfp
(* Computes sum of list l *) let sum l = List.fold_left (+) 0 l
(* checks if element e is in list l *) let element_in_list e l =
l |> List.find_map (fun x -> if x == e then Some(e) else None) <> None
(* returns a list with only the unique elements of list l *) let uniq l =
let rec uniq_helper acc l = match l with | [] -> acc | h::t -> if element_in_list h t then uniq_helper acc t else uniq_helper (h::acc) t in uniq_helper [] l |> List.rev
(* checks wheter or not list l only contains unique elements *) let is_uniq l = uniq l = l
(* computes all combinations for a given list of sanitation, fire & police departments
im not very proud of this function...maybe someone with some experience can clean it up? ;)
- )
let department_numbers sl fl pl =
sl |> List.fold_left (fun aa s -> fl |> List.fold_left (fun fa f -> pl |> List.fold_left (fun pa p -> if sum [s;f;p] == 12 && is_uniq [s;f;p] then {s = s; f = f; p = p} :: pa else pa) [] |> List.append fa) [] |> List.append aa) []
(* "main" function *)
let _ =
let s = [1;2;3;4;5;6;7] in let f = [1;2;3;4;5;6;7] in let p = [2;4;6] in let result = department_numbers s f p in print_endline "S F P"; print_sfp_list result;
</lang>
- Output:
S F P 1 5 6 1 7 4 2 4 6 2 6 4 3 5 4 3 7 2 4 2 6 4 6 2 5 1 6 5 3 4 6 2 4 6 4 2 7 1 4 7 3 2
PARI/GP
<lang parigp>forstep(p=2,6,2, for(f=1,7, s=12-p-f; if(p!=f && p!=s && f!=s && s>0 && s<8, print(p" "f" "s))))</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Perl
<lang Perl>
- !/usr/bin/perl
my @even_numbers;
for (1..7) {
if ( $_ % 2 == 0) { push @even_numbers, $_; }
}
print "Police\tFire\tSanitation\n";
foreach my $police_number (@even_numbers) {
for my $fire_number (1..7) { for my $sanitation_number (1..7) { if ( $police_number + $fire_number + $sanitation_number == 12 && $police_number != $fire_number && $fire_number != $sanitation_number && $sanitation_number != $police_number) { print "$police_number\t$fire_number\t$sanitation_number\n"; } } }
} </lang>
Above Code cleaned up and shortened
<lang Perl>
- !/usr/bin/perl
use strict; # Not necessary but considered good perl style use warnings; # this one too
print "Police\t-\tFire\t-\tSanitation\n"; for my $p ( 1..7 ) # Police Department {
for my $f ( 1..7) # Fire Department { for my $s ( 1..7 ) # Sanitation Department { if ( $p % 2 == 0 && $p + $f + $s == 12 && $p != $f && $f != $s && $s != $p && $f != $s) # Check if the combination of numbers is valid { print "$p\t-\t$f\t-\t$s\n"; } } }
} </lang>
Output: <lang Perl> Police - Fire - Sanitation 2 - 3 - 7 2 - 4 - 6 2 - 6 - 4 2 - 7 - 3 4 - 1 - 7 4 - 2 - 6 4 - 3 - 5 4 - 5 - 3 4 - 6 - 2 4 - 7 - 1 6 - 1 - 5 6 - 2 - 4 6 - 4 - 2 6 - 5 - 1 </lang>
Alternate with Regex
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Department_numbers use warnings;
print "P S F\n\n";
'246 1234567 1234567' =~
/(.).* \s .*?(?!\1)(.).* \s .*(?!\1)(?!\2)(.) (??{$1+$2+$3!=12}) (?{ print "@{^CAPTURE}\n" })(*FAIL)/x;</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Alternate with Glob
<lang perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Department_numbers use warnings;
print "P S F\n\n";
print tr/+/ /r, "\n" for
grep !/(\d).*\1/ && 12 == eval, glob '{2,4,6}' . '+{1,2,3,4,5,6,7}' x 2;</lang>
Output same as with Regex
Phix
printf(1,"Police Sanitation Fire\n") printf(1,"------ ---------- ----\n") integer solutions = 0 for police=2 to 7 by 2 do for sanitation=1 to 7 do if sanitation!=police then integer fire = 12-(police+sanitation) if fire>=1 and fire<=7 and fire!=police and fire!=sanitation then printf(1," %d %d %d\n", {police,sanitation,fire}) solutions += 1 end if end if end for end for printf(1,"\n%d solutions found\n", solutions)
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 solutions found
PHP
<lang PHP><?php
$valid = 0; for ($police = 2 ; $police <= 6 ; $police += 2) {
for ($sanitation = 1 ; $sanitation <= 7 ; $sanitation++) { $fire = 12 - $police - $sanitation; if ((1 <= $fire) and ($fire <= 7) and ($police != $sanitation) and ($sanitation != $fire)) { echo 'Police: ', $police, ', Sanitation: ', $sanitation, ', Fire: ', $fire, PHP_EOL; $valid++; } }
} echo $valid, ' valid combinations found.', PHP_EOL;</lang>
- Output:
Police: 2, Sanitation: 3, Fire: 7 Police: 2, Sanitation: 4, Fire: 6 Police: 2, Sanitation: 6, Fire: 4 Police: 2, Sanitation: 7, Fire: 3 Police: 4, Sanitation: 1, Fire: 7 Police: 4, Sanitation: 2, Fire: 6 Police: 4, Sanitation: 3, Fire: 5 Police: 4, Sanitation: 5, Fire: 3 Police: 4, Sanitation: 6, Fire: 2 Police: 4, Sanitation: 7, Fire: 1 Police: 6, Sanitation: 1, Fire: 5 Police: 6, Sanitation: 2, Fire: 4 Police: 6, Sanitation: 4, Fire: 2 Police: 6, Sanitation: 5, Fire: 1 14 valid combinations found.
Picat
Constraint model
<lang Picat>import cp.
go ?=>
N = 7, Sols = findall([P,S,F], department_numbers(N, P,S,F)), println(" P S F"), foreach([P,S,F] in Sols) printf("%2d %2d %2d\n",P,S,F) end, nl, printf("Number of solutions: %d\n", Sols.len), nl.
go => true.
department_numbers(N, Police,Sanitation,Fire) =>
Police :: 1..N, Sanitation :: 1..N, Fire :: 1..N, all_different([Police,Sanitation,Fire]), Police + Sanitation + Fire #= 12, Police mod 2 #= 0, solve([Police,Sanitation,Fire]).</lang>
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 Number of solutions: 14
Loop
<lang Picat>go2 => department_numbers2(N) =>
println(" P S F"), foreach(P in 1..N, P mod 2 == 0) foreach(S in 1..N, P != S) foreach(F in 1..N, F != P, F != S, P + S + F == 12) printf("%2d %2d %2d\n",P,S,F) end end end.</lang>
List comprehension
<lang Picat>import util.
department_numbers3(N) =>
println("P S F"), L = [[P.to_string,S.to_string,F.to_string] : P in 1..N, P mod 2 == 0, S in 1..N, P != S, F in 1..N, F != P, F != S, P + S + F == 12], println(map(L,join).join("\n")).</lang>
Prolog style
<lang Picat>go :-
println("P F S"), assign(Police, Fire, Sanitation), printf("%w %w %w\n", Police, Fire, Sanitation), fail, nl.
dept(X) :- between(1, 7, X).
police(X) :- member(X, [2, 4, 6]). fire(X) :- dept(X). san(X) :- dept(X).
assign(A, B, C) :-
police(A), fire(B), san(C), A != B, A != C, B != C, 12 is A + B + C.</lang>
PicoLisp
<lang PicoLisp>(de numbers NIL
(co 'numbers (let N 7 (for P N (for S N (for F N (yield (list P S F)) ) ) ) ) ) )
(de departments NIL
(use (L) (while (setq L (numbers)) (or (bit? 1 (car L)) (= (car L) (cadr L)) (= (car L) (caddr L)) (= (cadr L) (caddr L)) (<> 12 (apply + L)) (println L) ) ) ) )
(departments)</lang>
- Output:
(2 3 7) (2 4 6) (2 6 4) (2 7 3) (4 1 7) (4 2 6) (4 3 5) (4 5 3) (4 6 2) (4 7 1) (6 1 5) (6 2 4) (6 4 2) (6 5 1)
Pilog
<lang PicoLisp>(be departments (@Pol @Fire @San)
(member @Pol (2 4 6)) (for @Fire 1 7) (for @San 1 7) (different @Pol @Fire) (different @Pol @San) (different @Fire @San) (^ @ (= 12 (+ (-> @Pol) (-> @Fire) (-> @San)) ) ) )</lang>
- Output:
: (? (departments @Police @Fire @Sanitation)) @Police=2 @Fire=3 @Sanitation=7 @Police=2 @Fire=4 @Sanitation=6 @Police=2 @Fire=6 @Sanitation=4 @Police=2 @Fire=7 @Sanitation=3 @Police=4 @Fire=1 @Sanitation=7 @Police=4 @Fire=2 @Sanitation=6 @Police=4 @Fire=3 @Sanitation=5 @Police=4 @Fire=5 @Sanitation=3 @Police=4 @Fire=6 @Sanitation=2 @Police=4 @Fire=7 @Sanitation=1 @Police=6 @Fire=1 @Sanitation=5 @Police=6 @Fire=2 @Sanitation=4 @Police=6 @Fire=4 @Sanitation=2 @Police=6 @Fire=5 @Sanitation=1 -> NIL
Prolog
<lang prolog> dept(X) :- between(1, 7, X).
police(X) :- member(X, [2, 4, 6]). fire(X) :- dept(X). san(X) :- dept(X).
assign(A, B, C) :-
police(A), fire(B), san(C), A =\= B, A =\= C, B =\= C, 12 is A + B + C.
main :-
write("P F S"), nl, forall(assign(Police, Fire, Sanitation), format("~w ~w ~w~n", [Police, Fire, Sanitation])), halt.
?- main. </lang>
- Output:
P F S 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Python
Procedural
<lang python>from itertools import permutations
def solve():
c, p, f, s = "\\,Police,Fire,Sanitation".split(',') print(f"{c:>3} {p:^6} {f:^4} {s:^10}") c = 1 for p, f, s in permutations(range(1, 8), r=3): if p + s + f == 12 and p % 2 == 0: print(f"{c:>3}: {p:^6} {f:^4} {s:^10}") c += 1
if __name__ == '__main__':
solve()</lang>
- Output:
\ Police Fire Sanitation 1: 2 3 7 2: 2 4 6 3: 2 6 4 4: 2 7 3 5: 4 1 7 6: 4 2 6 7: 4 3 5 8: 4 5 3 9: 4 6 2 10: 4 7 1 11: 6 1 5 12: 6 2 4 13: 6 4 2 14: 6 5 1
Composition of pure functions
Expressing the options directly and declaratively in terms of a bind operator, without importing permutations:
<lang python>Department numbers
from itertools import (chain) from operator import (ne)
- options :: Int -> Int -> Int -> [(Int, Int, Int)]
def options(lo, hi, total):
Eligible integer triples. ds = enumFromTo(lo)(hi) return bind(filter(even, ds))( lambda x: bind(filter(curry(ne)(x), ds))( lambda y: bind([total - (x + y)])( lambda z: [(x, y, z)] if ( z != y and lo <= z <= hi ) else [] ) ) )
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Test
xs = options(1, 7, 12) print(('Police', 'Sanitation', 'Fire')) for tpl in xs: print(tpl) print('\nNo. of options: ' + str(len(xs)))
- GENERIC ABSTRACTIONS ------------------------------------
- bind (>>=) :: [a] -> (a -> [b]) -> [b]
def bind(xs):
List monad injection operator. Two computations sequentially composed, with any value produced by the first passed as an argument to the second. return lambda f: list( chain.from_iterable( map(f, xs) ) )
- curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
A curried function derived from an uncurried function. return lambda a: lambda b: f(a, b)
- enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
Integer enumeration from m to n. return lambda n: list(range(m, 1 + n))
- even :: Int -> Bool
def even(x):
True if x is an integer multiple of two. return 0 == x % 2
if __name__ == '__main__':
main()</lang>
- Output:
('Police', 'Sanitation', 'Fire') (2, 3, 7) (2, 4, 6) (2, 6, 4) (2, 7, 3) (4, 1, 7) (4, 2, 6) (4, 3, 5) (4, 5, 3) (4, 6, 2) (4, 7, 1) (6, 1, 5) (6, 2, 4) (6, 4, 2) (6, 5, 1) No. of options: 14
List comprehension
Nested bind (or concatMap) expressions (like those above) can also be translated into list comprehension notation:
<lang python>Department numbers
from operator import ne
- options :: Int -> Int -> Int -> [(Int, Int, Int)]
def options(lo, hi, total):
Eligible triples. ds = enumFromTo(lo)(hi) return [ (x, y, z) for x in filter(even, ds) for y in filter(curry(ne)(x), ds) for z in [total - (x + y)] if y != z and lo <= z <= hi ]
- Or with less tightly-constrained generation,
- and more winnowing work downstream:
- options2 :: Int -> Int -> Int -> [(Int, Int, Int)]
def options2(lo, hi, total):
Eligible triples. ds = enumFromTo(lo)(hi) return [ (x, y, z) for x in ds for y in ds for z in [total - (x + y)] if even(x) and y not in [x, z] and lo <= z <= hi ]
- GENERIC -------------------------------------------------
- curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
A curried function derived from an uncurried function. return lambda a: lambda b: f(a, b)
- enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
Integer enumeration from m to n. return lambda n: list(range(m, 1 + n))
- even :: Int -> Bool
def even(x):
True if x is an integer multiple of two. return 0 == x % 2
- unlines :: [String] -> String
def unlines(xs):
A single string derived by the intercalation of a list of strings with the newline character. return '\n'.join(xs)
- TEST ----------------------------------------------------
- main :: IO ()
def main():
Test
xs = options(1, 7, 12) print(('Police', 'Sanitation', 'Fire')) print(unlines(map(str, xs))) print('\nNo. of options: ' + str(len(xs)))
if __name__ == '__main__':
main()</lang>
- Output:
('Police', 'Sanitation', 'Fire') (2, 3, 7) (2, 4, 6) (2, 6, 4) (2, 7, 3) (4, 1, 7) (4, 2, 6) (4, 3, 5) (4, 5, 3) (4, 6, 2) (4, 7, 1) (6, 1, 5) (6, 2, 4) (6, 4, 2) (6, 5, 1) No. of options: 14
Adapted from C# Example
<lang python>
- We start with the Police Department.
- Range is the start, stop, and step. This returns only even numbers.
for p in range(2, 7, 2):
#Next, the Sanitation Department. A simple range. for s in range(1, 7): # And now the Fire Department. After determining the Police and Fire # numbers we just have to subtract those from 12 to get the FD number. f = 12 - p -s if s >= f: break elif f > 7: continue print("Police: ", p, " Sanitation:", s, " Fire: ", f) print("Police: ", p, " Sanitation:", f, " Fire: ", s)
</lang>
- Output:
Police: 2 Sanitation: 3 Fire: 7 Police: 2 Sanitation: 7 Fire: 3 Police: 2 Sanitation: 4 Fire: 6 Police: 2 Sanitation: 6 Fire: 4 Police: 4 Sanitation: 1 Fire: 7 Police: 4 Sanitation: 7 Fire: 1 Police: 4 Sanitation: 2 Fire: 6 Police: 4 Sanitation: 6 Fire: 2 Police: 4 Sanitation: 3 Fire: 5 Police: 4 Sanitation: 5 Fire: 3 Police: 6 Sanitation: 1 Fire: 5 Police: 6 Sanitation: 5 Fire: 1 Police: 6 Sanitation: 2 Fire: 4 Police: 6 Sanitation: 4 Fire: 2
Quackery
<lang Quackery> [ 2dup = iff
[ 2drop drop ] done dip over swap over = iff [ 2drop drop ] done rot echo sp swap echo sp echo cr ] is fire ( pol san fir --> )
[ 2dup = iff 2drop done 12 over - dip over swap - dup 1 < iff [ 2drop drop ] done dup 7 > iff [ 2drop drop ] done fire ] is sanitation ( pol san --> )
[ 7 times [ dup i^ 1+ sanitation ] drop ] is police ( pol --> )
[ cr ' [ 2 4 6 ] witheach police ] is departments ( --> )
departments</lang>
- Output:
2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
R
We solve this task in two lines. The rest of the code is to make the result look nice. <lang rsplus>allPermutations <- setNames(expand.grid(seq(2, 7, by = 2), 1:7, 1:7), c("Police", "Sanitation", "Fire")) solution <- allPermutations[which(rowSums(allPermutations)==12 & apply(allPermutations, 1, function(x) !any(duplicated(x)))),] solution <- solution[order(solution$Police, solution$Sanitation),] row.names(solution) <- paste0("Solution #", seq_len(nrow(solution)), ":") print(solution)</lang>
- Output:
Police Sanitation Fire Solution #1: 2 3 7 Solution #2: 2 4 6 Solution #3: 2 6 4 Solution #4: 2 7 3 Solution #5: 4 1 7 Solution #6: 4 2 6 Solution #7: 4 3 5 Solution #8: 4 5 3 Solution #9: 4 6 2 Solution #10: 4 7 1 Solution #11: 6 1 5 Solution #12: 6 2 4 Solution #13: 6 4 2 Solution #14: 6 5 1
Racket
We filter the Cartesian product of the lists of candidate department numbers.
<lang racket>#lang racket (cons '(police fire sanitation)
(filter (λ (pfs) (and (not (check-duplicates pfs)) (= 12 (apply + pfs)) pfs)) (cartesian-product (range 2 8 2) (range 1 8) (range 1 8))))
</lang>
- Output:
'((police fire sanitation) (2 3 7) (2 4 6) (2 6 4) (2 7 3) (4 1 7) (4 2 6) (4 3 5) (4 5 3) (4 6 2) (4 7 1) (6 1 5) (6 2 4) (6 4 2) (6 5 1))
Raku
(formerly Perl 6) <lang perl6>for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) { say <police fire sanitation> Z=> .list; }
} </lang>
- Output:
(police => 4 fire => 1 sanitation => 7) (police => 4 fire => 7 sanitation => 1) (police => 6 fire => 1 sanitation => 5) (police => 6 fire => 5 sanitation => 1) (police => 2 fire => 3 sanitation => 7) (police => 2 fire => 7 sanitation => 3) (police => 2 fire => 4 sanitation => 6) (police => 2 fire => 6 sanitation => 4) (police => 4 fire => 2 sanitation => 6) (police => 4 fire => 6 sanitation => 2) (police => 6 fire => 2 sanitation => 4) (police => 6 fire => 4 sanitation => 2) (police => 4 fire => 3 sanitation => 5) (police => 4 fire => 5 sanitation => 3)
REXX
This REXX example essentially uses brute force approach for so simple a puzzle. <lang rexx>/*REXX program finds/displays all possible variants of (3) department numbering puzzle.*/ say 'police sanitation fire' /*display simple title for the output*/ say '══════ ══════════ ════' /* " head separator " " " */
- =0 /*number of solutions found (so far). */
do p=1 for 7; if p//2 then iterate /*try numbers for the police department*/ do s=1 for 7; if s==p then iterate /* " " " " fire " */ do f=1 for 7; if f==s then iterate /* " " " " sanitation " */ if p + s + f \== 12 then iterate /*check if sum of department nums ¬= 12*/ #= # + 1 /*bump count of the number of solutions*/ say center(p,6) center(s,10) center(f,4) /*display one possible solution. */ end /*s*/ end /*f*/ end /*p*/
say '══════ ══════════ ════' /* " head separator " " " */ say /*stick a fork in it, we're all done. */ say # ' solutions found.' /*also, show the # of solutions found. */</lang>
- output when using the default inputs:
police sanitation fire ══════ ══════════ ════ 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 ══════ ══════════ ════ 14 solutions found.
Ring
<lang ring> sanitation= 0 see "police fire sanitation" + nl
for police = 2 to 7 step 2
for fire = 1 to 7 if fire = police loop ok sanitation = 12 - police - fire if sanitation = fire or sanitation = police loop ok if sanitation >= 1 and sanitation <= 7 see " " + police + " " + fire + " " + sanitation + nl ok next
next </lang> Output:
police fire sanitation 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Ruby
<lang ruby> (1..7).to_a.permutation(3){|p| puts p.join if p.first.even? && p.sum == 12 } </lang>
- Output:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
Rust
<lang rust>extern crate num_iter; fn main() {
println!("Police Sanitation Fire"); println!("----------------------");
for police in num_iter::range_step(2, 7, 2) { for sanitation in 1..8 { for fire in 1..8 { if police != sanitation && sanitation != fire && fire != police && police + fire + sanitation == 12 { println!("{:6}{:11}{:4}", police, sanitation, fire); } } } }
}</lang>
Scala
<lang scala>val depts = {
(1 to 7).permutations.map{ n => (n(0),n(1),n(2)) }.toList.distinct // All permutations of possible department numbers .filter{ n => n._1 % 2 == 0 } // Keep only even numbers favored by Police Chief .filter{ n => n._1 + n._2 + n._3 == 12 } // Keep only numbers that add to 12
}
{ println( "(Police, Sanitation, Fire)") println( depts.mkString("\n") ) } </lang>
- Output:
(Police, Sanitation, Fire) (2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1)
Sidef
<lang ruby>@(1..7)->combinations(3, {|*a|
a.sum == 12 || next a.permutations {|*b| b[0].is_even || next say (%w(police fire sanitation) ~Z b -> join(" ")) }
})</lang>
- Output:
["police", 4] ["fire", 1] ["sanitation", 7] ["police", 4] ["fire", 7] ["sanitation", 1] ["police", 6] ["fire", 1] ["sanitation", 5] ["police", 6] ["fire", 5] ["sanitation", 1] ["police", 2] ["fire", 3] ["sanitation", 7] ["police", 2] ["fire", 7] ["sanitation", 3] ["police", 2] ["fire", 4] ["sanitation", 6] ["police", 2] ["fire", 6] ["sanitation", 4] ["police", 4] ["fire", 2] ["sanitation", 6] ["police", 4] ["fire", 6] ["sanitation", 2] ["police", 6] ["fire", 2] ["sanitation", 4] ["police", 6] ["fire", 4] ["sanitation", 2] ["police", 4] ["fire", 3] ["sanitation", 5] ["police", 4] ["fire", 5] ["sanitation", 3]
Swift
Functional approach:
<lang swift>let res = [2, 4, 6].map({x in
return (1...7) .filter({ $0 != x }) .map({y -> (Int, Int, Int)? in let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else { return nil }
return (x, y, z) }).compactMap({ $0 })
}).flatMap({ $0 })
for result in res {
print(result)
}</lang>
Iterative approach:
<lang swift>var res = [(Int, Int, Int)]()
for x in [2, 4, 6] {
for y in 1...7 where x != y { let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else { continue }
res.append((x, y, z)) }
}
for result in res {
print(result)
}</lang>
- Output:
(2, 3, 7) (2, 4, 6) (2, 6, 4) (2, 7, 3) (4, 1, 7) (4, 2, 6) (4, 3, 5) (4, 5, 3) (4, 6, 2) (4, 7, 1) (6, 1, 5) (6, 2, 4) (6, 4, 2) (6, 5, 1)
Tcl
Since Tool Command Language is a multi-paradigm language, very different solutions are possible.
VERSION A - using procedures and list operations <lang tcl>
- Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$i <= $max} {incr i} { lappend l $i } return $l
}
- Procedure named "anyEqual" returns true if any elements are equal,
- false otherwise.
proc anyEqual l {
if {[llength [lsort -unique $l]] != [llength $l]} { return 1 } return 0
}
- Procedure named "odd" tells whether a value is odd or not.
proc odd n {
expr $n %2 != 0
}
- Procedure named "sum" sums its parameters.
proc sum args {
expr [join $args +]
}
- Create lists of candidate numbers using proc ".."
set sanitation [.. 7] set fire $sanitation
- Filter even numbers for police stations (remove odd ones).
set police [lmap e $sanitation {
if [odd $e] continue set e
}]
- Try all combinations and display acceptable ones.
set valid 0 foreach p $police {
foreach s $sanitation { foreach f $fire { # Check for equal elements in list. if [anyEqual [list $p $s $f]] continue # Check for sum of list elements. if {[sum $p $s $f] != 12} continue puts "$p $s $f" incr valid } }
} puts "$valid valid combinations found." </lang>
VERSION B - using simple for loops with number literals <lang tcl> set valid 0 for {set police 2} {$police <= 6} {incr police 2} {
for {set sanitation 1} {$sanitation <= 7} {incr sanitation} { if {$police == $sanitation} continue for {set fire 1} {$fire <= 7} {incr fire} { if {$police == $fire || $sanitation == $fire} continue if {[expr $police + $sanitation + $fire] != 12} continue puts "$police $sanitation $fire" incr valid } }
} puts "$valid valid combinations found." </lang>
VERSION C - using simple for loops with number variables <lang tcl> set min 1 set max 7 set valid 0 for {set police $min} {$police <= $max} {incr police} {
if {[expr $police % 2] == 1} continue ;# filter even numbers for police for {set sanitation $min} {$sanitation <= $max} {incr sanitation} { if {$police == $sanitation} continue for {set fire $min} {$fire <= $max} {incr fire} { if {$police == $fire || $sanitation == $fire} continue if {[expr $police + $sanitation + $fire] != 12} continue puts "$police $sanitation $fire" incr valid } }
}
puts "$valid valid combinations found." </lang>
VERSION D - using list filter with lambda expressions <lang tcl>
- Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$i <= $max} {incr i} { lappend l $i } return $l
}
- Procedure named "..." returns list of n lists of integers from 1 to max.
proc ... {max n} {
foreach i [.. $n] { lappend result [.. $max] } return $result
}
- Procedure named "crossProduct" returns cross product of lists
proc crossProduct {listOfLists} {
set result [list [list]] foreach factor $listOfLists { set newResult {} foreach combination $result { foreach elt $factor { lappend newResult [linsert $combination end $elt] } } set result $newResult } return $result
}
- Procedure named "filter" filters list elements by using a
- condition λ (lambda) expression
proc filter {l condition} {
return [lmap el $l { if {![apply $condition $el]} continue set el }]
}
- Here the fun using lambda expressions begins. The following is the main program.
- Set λ expressions
set λPoliceEven {_ {expr [lindex $_ 0] % 2 == 0}} set λNoEquals {_ {expr [llength [lsort -unique $_]] == [llength $_]}} set λSumIs12 {_ {expr [join $_ +] == 12}}
- Create all combinations and filter acceptable ones
set numbersOk [filter [filter [filter [crossProduct [... 7 3]] ${λPoliceEven}] ${λSumIs12}] ${λNoEquals}] puts [join $numbersOk \n] puts "[llength $numbersOk] valid combinations found." </lang>
- Output:
All four versions (A, B, C and D) produce the same result:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations found.
Tiny BASIC
<lang tinybasic> PRINT "Police Sanitation Fire"
PRINT "------|----------|----"
10 LET P = P + 2
LET S = 0
20 LET S = S + 1
LET F = 0 IF S = P THEN GOTO 20
30 LET F = F + 1
IF S = F THEN GOTO 30 IF F = P THEN GOTO 30 IF P + S + F = 12 THEN PRINT " ",P," ", S," ", F IF F < 7 THEN GOTO 30 IF S < 7 THEN GOTO 20 IF P < 6 THEN GOTO 10</lang>
Transd
<lang scheme>#lang transd
MainModule : {
_start: (lambda (lout "Police | Sanit. | Fire") (for i in Range(1 8) where (not (mod i 2)) do (for j in Range(1 8) where (neq i j) do (for k in Range(1 8) where (and (neq i k) (neq j k)) do (if (eq (+ i j k) 12) (lout i " " j " " k))))) )
}</lang>
- Output:
Police | Sanit. | Fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
TypeScript
<lang javascript> // Department numbers console.log(`POLICE SANITATION FIRE`); let f: number; for (var p = 2; p <= 7; p += 2) {
for (var s = 1; s <= 7; s++) { if (s != p) { f = (12 - p) - s; if ((f > 0) && (f <= 7) && (f != s) && (f != p)) console.log(` ${p} ${s} ${f}`); } }
} </lang>
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Visual Basic .NET
<lang vbnet>Module Module1
Sub Main() For p = 2 To 7 Step 2 For s = 1 To 7 Dim f = 12 - p - s If s >= f Then Exit For End If If f > 7 Then Continue For End If If s = p OrElse f = p Then Continue For 'not even necessary End If Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}") Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}") Next Next End Sub
End Module</lang>
- Output:
Police:2, Sanitation:3, Fire:7 Police:2, Sanitation:7, Fire:3 Police:2, Sanitation:4, Fire:6 Police:2, Sanitation:6, Fire:4 Police:4, Sanitation:1, Fire:7 Police:4, Sanitation:7, Fire:1 Police:4, Sanitation:2, Fire:6 Police:4, Sanitation:6, Fire:2 Police:4, Sanitation:3, Fire:5 Police:4, Sanitation:5, Fire:3 Police:6, Sanitation:1, Fire:5 Police:6, Sanitation:5, Fire:1 Police:6, Sanitation:2, Fire:4 Police:6, Sanitation:4, Fire:2
Vlang
<lang go>fn main() {
println("Police Sanitation Fire") println("------ ---------- ----") mut count := 0 for i := 2; i < 7; i += 2 { for j in 1..8 { if j == i { continue } for k in 1..8 { if k == i || k == j { continue } if i + j + k != 12 { continue } println(" $i $j $k") count++ } } } println("\n$count valid combinations")
}</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
VTL-2
<lang VTL2>10 P=2 20 S=1 30 F=1 40 #=0<((P=S)+(P=F)+(S=F)+(P+S+F=12=0))*110 50 ?=P 60 $=32 70 ?=S 80 $=32 90 ?=F 100 ?="" 110 F=F+1 120 #=F<8*40 130 S=S+1 140 #=S<8*30 150 P=P+2 160 #=P<8*20</lang>
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Wren
<lang ecmascript>System.print("Police Sanitation Fire") System.print("------ ---------- ----") var count = 0 for (h in 1..3) {
var i = h * 2 for (j in 1..7) { if (j != i) { for (k in 1..7) { if ((k != i && k != j) && (i + j + k == 12) ) { System.print(" %(i) %(j) %(k)") count = count + 1 } } } }
} System.print("\n%(count) valid combinations")</lang>
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
XPL0
<lang xpl0> \Department numbers code CrLf=9, IntIn=10, IntOut=11, Text=12; integer P, S, F;
begin Text(0, "POLICE SANITATION FIRE"); CrLf(0); P:= 2; while P <= 7 do
begin for S:= 1, 7 do if S # P then begin F:= (12 - P) - S; if (F > 0) & (F <= 7) & (F # S) & (F # P) then begin Text(0, " "); IntOut(0, P); Text(0, " "); IntOut(0, S); Text(0, " "); IntOut(0, F); CrLf(0) end end; P:= P + 2 end;
end </lang>
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
zkl
<lang zkl>Utils.Helpers.pickNFrom(3,[1..7].walk()) // 35 combos .filter(fcn(numbers){ numbers.sum(0)==12 }) // which all sum to 12 (==5) .println();</lang>
- Output:
L(L(1,4,7),L(1,5,6),L(2,3,7),L(2,4,6),L(3,4,5))
Note: The sum of three odd numbers is odd, so a+b+c=12 means at least one even nmber (1 even, two odd or 3 even). Futher, 2a+b=12, a,b in (2,4,6) has one solution: a=2,b=4
For a table with repeated solutions using nested loops: <lang zkl>println("Police Fire Sanitation"); foreach p,f,s in ([2..7,2], [1..7], [1..7])
{ if((p!=s!=f) and p+f+s==12) println(p,"\t",f,"\t",s) }</lang>
- Output:
Police Fire Sanitation 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
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