Department numbers
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 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
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
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
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
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>
or 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>
- 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>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)]</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
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
Gambas
<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
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>
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>(() => {
// 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');
})();</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 = 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);
})();</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
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
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>
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
bare bones
<lang rexx>/*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. */</lang>
- output 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. <lang rexx>/*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. */</lang>
- output 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.
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)
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