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[[Category:Puzzles]] {{task}}
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 the number '''12'''.
The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
;Task: Write a program which outputs all valid combinations.
Possible output:
1 2 9
5 3 4
11l
{{trans|C}}
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)
{{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
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);
}
}
}
}
ALGOL 68
As noted in the Fortran sample, once the police and sanitation departments are posited, the fire department value is fixed
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
{{out}}
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
{{Trans|ALGOL 68}}
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.
{{out}}
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
AppleScript
Briefly, composing a solution from generic functions:
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
{{Out}}
237
246
264
273
417
426
435
453
462
471
615
624
642
651
Or more generally: {{Trans|JavaScript}} {{Trans|Haskell}}
-- 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
{{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
{{Out}}
(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
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)
}
Example:
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
Outputs:
237
246
264
273
417
426
435
453
462
471
615
624
642
651
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)
}
{{out}}
# 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
BASIC
==={{header|IS-BASIC}}===
=
## 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.
```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
{{out}}
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
= {{trans|ALGOL 68}}
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
{{out}}
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
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.
#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;
}
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#
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}");
}
}
}
}
{{out}}
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++
#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;
}
{{out}}
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
(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)))
{{Out}}
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
{{Trans|C++}}
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);
}
}
}
}
}
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
=={{header|F_Sharp|F#}}==
// 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")
{{out}}
{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
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
{{out}}
{ 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 }
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.
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.
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
' 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
{{out}}
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
'''[https://gambas-playground.proko.eu/?gist=cdfeadc26aaeb0e23f4523626b6fe7c9 Click this link to run this code]'''
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
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
{{trans|Kotlin}}
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)
}
{{out}}
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
Haskell
Bare minimum:
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)]
_ -> []
or, resugaring this into list comprehension format:
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 ]
Do notation:
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 []
Unadorned brute force – more than enough at this small scale:
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] ]
{{Out}}
(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:
options :: Int -> Int -> Int -> [(Int, Int, Int)]
options lo hi total =
(\ds ->
filter even ds >>=
\x ->
filter (/= x) ds >>=
\y ->
[total - (x + y)] >>=
\z ->
case y /= z && lo <= z && z <= hi of
True -> [(x, y, z)]
_ -> [])
[lo .. hi]
-- TEST -----------------------------------------------------------------------
main :: IO ()
main = do
let xs = options 1 7 12
putStrLn "(Police, Sanitation, Fire)\n"
mapM_ print xs
mapM_ putStrLn ["\nNumber of options: ", show (length xs)]
Reaching again for a little more syntactic sugar, the options function above could also be re-written either as a list comprehension,
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 ]
or in Do notation:
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)
{{Out}}
(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:'''
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
'''Example usage:'''
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
Alternate approach
(#~ 12=+/"1) 1+3 comb 7 [ load'stats'
1 4 7
1 5 6
2 3 7
2 4 6
3 4 5
Note that we are only showing the distinct [[combinations]] here, not all [[permutations]] of those combinations.
Java
{{trans|Kotlin}}
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);
}
}
{{out}}
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:
(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');
})();
{{Out}}
(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: {{Trans|Haskell}}
(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);
})();
{{Out}}
(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:
(() => {
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
return '(Police, Sanitation, Fire)\n' +
concatMap(x =>
concatMap(y =>
concatMap(z =>
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');
})();
{{Out}}
(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: {{Trans|Haskell}}
(() => {
'use strict';
// NUMBERING CONSTRAINTS --------------------------------------------------
// options :: Int -> Int -> Int -> [(Int, Int, Int)]
const options = (lo, hi, total) => {
const
bind = flip(concatMap),
ds = enumFromTo(lo, hi);
return bind(filter(even, ds),
x => bind(filter(d => d !== x, ds),
y => bind([total - (x + y)],
z => (z !== y && lo <= z && z <= hi) ? [
[x, y, z]
] : []
)
)
)
};
// GENERIC FUNCTIONS ------------------------------------------------------
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
// even :: Integral a => a -> Bool
const even = n => n % 2 === 0;
// 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]);
// length :: [a] -> Int
const length = xs => xs.length;
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
// show :: a -> String
const show = x => JSON.stringify(x) //, null, 2);
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
// TEST -------------------------------------------------------------------
const xs = options(1, 7, 12);
return '(Police, Sanitation, Fire)\n\n' +
unlines(map(show, xs)) +
'\n\nNumber of options: ' + length(xs);
})();
{{Out}}
(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 first can easily be made more efficient by adding some pruning.
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'''
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: $fire, police: $police, sanitation: $sanitation}
'''In Brief'''
{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 )
'''Pruning'''
range(1;8) as $fire
| (range(1;8) | select(. != $fire)) as $police
| (range(1;8) | select(. != $fire and . != $police)) as $sanitation
| {fire: $fire, police: $police, sanitation: $sanitation}
| select( ([.[]] | add) == 12)
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())
{{out}}
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
// 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")
}
{{out}}
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
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 ) )
{{out}}
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
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();
{{out|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
Select[Permutations[Range[7], {3}], Total[#] == 12 && EvenQ[First[#]] &]
{{out}}
{{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
{{Trans|C++}}
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;
};
};
};
};
}
}
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
=={{header|Modula-2}}==
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.
PARI/GP
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))))
{{out}}
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
#!/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";
}
}
}
}
Above Code cleaned up and shortened
#!/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";
}
}
}
}
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
Perl 6
for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) {
say <police fire sanitation> Z=> .list;
}
}
{{out}}
(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)
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)
{{out}}
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
Python
Procedural
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()
{{out}}
\ 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'': {{Works with|Python|3}}
'''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()
{{Out}}
('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: {{Works with|Python|3.7}}
'''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()
{{Out}}
('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
Racket
We filter the Cartesian product of the lists of candidate department numbers.
#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))))
{{out}}
'((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))
REXX
bare bones
/*REXX program finds/displays all possible variants of (3) department numbering puzzle.*/
say 'police fire sanitation' /*display a crude title for the output.*/
do p=2 to 7 by 2 /*try numbers for the police department*/
do f=1 for 7 /* " " " " fire " */
do s=1 for 7; $=p+f+s /* " " " " sanitation " */
if f\==p & s\==p & s\==f & $==12 then say center(p,6) center(f,5) center(s,10)
end /*s*/
end /*f*/
end /*p*/ /*stick a fork in it, we're all done. */
{{out|output|text= when using the default inputs:}}
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
options and optimizing
A little extra code was added to allow the specification for the high department number as well as the sum.
Two optimizing statements were added (for speed), but for this simple puzzle they aren't needed.
Also, extra code was added to nicely format a title (header) for the output, as well as displaying the number of solutions found.
/*REXX program finds/displays all possible variants of (3) department numbering puzzle.*/
parse arg high sum . /*obtain optional arguments from the CL*/
if high=='' | high=="," then high= 7 /*Not specified? Then use the default.*/
if sum=='' | sum=="," then sum=12 /* " " " " " " */
@pd= ' police '; @fd= " fire " ; @sd= ' sanitation ' /*define names of departments.*/
@dept= ' department '; L=length(@dept) /*literal; and also its length*/
#=0 /*initialize the number of solutions. */
do PD=2 by 2 to high /*try numbers for the police department*/
do FD=1 for high /* " " " " fire " */
if FD==PD then iterate /*Same FD# & PD#? They must be unique.*/
if FD+PD>sum-1 then iterate PD /*Is sum too large? Try another PD#. */ /* ◄■■■■■■ optimizing code*/
do SD=1 for high /*try numbers for the sanitation dept. */
if SD==PD | SD==FD then iterate /*Is SD# ¬unique? They must be unique,*/
$=PD+FD+SD /*compute sum of department numbers. */
if $> sum then iterate FD /*Is the sum too high? Try another FD#*/ /* ◄■■■■■■ optimizing code*/
if $\==sum then iterate /*Is the sum ¬correct? " " SD#*/
#=# + 1 /*bump the number of solutions (so far)*/
if #==1 then do /*Is this the 1st solution? Show hdr.*/
say center(@pd, L) center(@fd, L) center(@sd, L)
say copies(center( @dept, L)' ', 3)
say copies(center('number', L)' ', 3)
say center('', L, "═") center('', L, "═") center('', L, "═")
end
say center(PD, L) center(FD, L) center(SD, L) /*display a solution.*/
end /*SD*/
end /*FD*/
end /*PD*/
say /*display a blank line before the #sols*/
if #==0 then #= 'no' /*use a better word for bupkis. */
say # "solutions found." /*stick a fork in it, we're all done. */
{{out|output|text= when using the default inputs:}}
police fire sanitation
department department department
number number number
════════════ ════════════ ════════════
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
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
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
(1..7).to_a.permutation(3){|p| puts p.join if p.first.even? && p.sum == 12 }
{{Output}}
237
246
264
273
417
426
435
453
462
471
615
624
642
651
Rust
{{trans|C}}
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);
}
}
}
}
}
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") )
}
{{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
{{trans|Perl 6}}
@(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(" "))
}
})
{{out}}
["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:
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({ $0 })
for result in res {
print(result)
}
Iterative approach:
var res = [(Int, Int, Int)]()
for x in [2, 4, 6] {
for y in 1...7 where x != y {
let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else {
continue
}
res.append((x, y, z))
}
}
for result in res {
print(result)
}
{{out}}
(2, 3, 7)
(2, 4, 6)
(2, 6, 4)
(2, 7, 3)
(4, 1, 7)
(4, 2, 6)
(4, 3, 5)
(4, 5, 3)
(4, 6, 2)
(4, 7, 1)
(6, 1, 5)
(6, 2, 4)
(6, 4, 2)
(6, 5, 1)
Tcl
Since Tool Command Language is a multi-paradigm language, very different solutions are possible.
VERSION A - using procedures and list operations
# Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$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."
VERSION B - using simple for loops with number literals
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."
VERSION C - using simple for loops with number variables
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."
VERSION D - using list filter with lambda expressions
# 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 $numbersOk] valid combinations found."
{{output}} All four versions (A, B, C and D) produce the same result:
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
14 valid combinations found.
Visual Basic .NET
{{trans|C#}}
Module Module1
Sub Main()
For p = 2 To 7 Step 2
For s = 1 To 7
Dim f = 12 - p - s
If s >= f Then
Exit For
End If
If f > 7 Then
Continue For
End If
If s = p OrElse f = p Then
Continue For 'not even necessary
End If
Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}")
Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}")
Next
Next
End Sub
End Module
{{out}}
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
zkl
Utils.Helpers.pickNFrom(3,[1..7].walk()) // 35 combos
.filter(fcn(numbers){ numbers.sum(0)==12 }) // which all sum to 12 (==5)
.println();
{{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:
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) }
{{out}}
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