Dinesman's multipledwelling problem
You are encouraged to solve this task according to the task description, using any language you may know.
 Task
Solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below.
Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
 The problem
Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors.
 Baker does not live on the top floor.
 Cooper does not live on the bottom floor.
 Fletcher does not live on either the top or the bottom floor.
 Miller lives on a higher floor than does Cooper.
 Smith does not live on a floor adjacent to Fletcher's.
 Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?
Contents
 1 Ada
 2 ALGOL 68
 3 AutoHotkey
 4 AWK
 5 BBC BASIC
 6 Bracmat
 7 C
 8 C#
 9 Clojure
 10 D
 11 EchoLisp
 12 Elixir
 13 Erlang
 14 ERRE
 15 Factor
 16 Forth
 17 Go
 18 Haskell
 19 Icon and Unicon
 20 J
 21 Java
 22 JavaScript
 23 jq
 24 Julia
 25 Kotlin
 26 Lua
 27 Mathematica / Wolfram Language
 28 Perl
 29 Perl 6
 30 Phix
 31 PicoLisp
 32 PowerShell
 33 Prolog
 34 PureBasic
 35 Python
 36 Racket
 37 REXX
 38 Ring
 39 Ruby
 40 Run BASIC
 41 Scala
 42 Sidef
 43 Tcl
 44 uBasic/4tH
 45 UNIX Shell
 46 XPL0
 47 zkl
 48 ZX Spectrum Basic
Ada[edit]
Uses an enum type People to attempt to be naturally reading. Problem is easily changed by altering subtype Floor, type people and the somewhat naturally reading constraints in the Constrained function. If for example you change the floor range to 1..6 and add Superman to people, all possible solutions will be printed.
with Ada.Text_IO; use Ada.Text_IO;
procedure Dinesman is
subtype Floor is Positive range 1 .. 5;
type People is (Baker, Cooper, Fletcher, Miller, Smith);
type Floors is array (People'Range) of Floor;
type PtFloors is access all Floors;
function Constrained (f : PtFloors) return Boolean is begin
if f (Baker) /= Floor'Last and
f (Cooper) /= Floor'First and
Floor'First < f (Fletcher) and f (Fletcher) < Floor'Last and
f (Miller) > f (Cooper) and
abs (f (Smith)  f (Fletcher)) /= 1 and
abs (f (Fletcher)  f (Cooper)) /= 1
then return True; end if;
return False;
end Constrained;
procedure Solve (list : PtFloors; n : Natural) is
procedure Swap (I : People; J : Natural) is
temp : constant Floor := list (People'Val (J));
begin list (People'Val (J)) := list (I); list (I) := temp;
end Swap;
begin
if n = 1 then
if Constrained (list) then
for p in People'Range loop
Put_Line (p'Img & " on floor " & list (p)'Img);
end loop;
end if;
return;
end if;
for i in People'First .. People'Val (n  1) loop
Solve (list, n  1);
if n mod 2 = 1 then Swap (People'First, n  1);
else Swap (i, n  1); end if;
end loop;
end Solve;
thefloors : aliased Floors;
begin
for person in People'Range loop
thefloors (person) := People'Pos (person) + Floor'First;
end loop;
Solve (thefloors'Access, Floors'Length);
end Dinesman;
 Output:
BAKER on floor 3 COOPER on floor 2 FLETCHER on floor 4 MILLER on floor 5 SMITH on floor 1
ALGOL 68[edit]
Algol 68 allows structures containing procedures to have a different procedure for each instance (similar to making each instance a separate derived class in OO languages). This allows for easy specification of the constraints. The constraints for each person could be changed by providing a different PROC(INT)BOOL in the initialisation of the inhabitants. Changing the number of inhabitants would require adding or removing loops from the solution finding code.
# attempt to solve the dinesman Multiple Dwelling problem #
# SETUP #
# special floor values #
INT top floor = 4;
INT bottom floor = 0;
# mode to specify the persons floor constraint #
MODE PERSON = STRUCT( STRING name, REF INT floor, PROC( INT )BOOL ok );
# yields TRUE if the floor of the specified person is OK, FALSE otherwise #
OP OK = ( PERSON p )BOOL: ( ok OF p )( floor OF p );
# yields TRUE if floor is adjacent to other persons floor, FALSE otherwise #
PROC adjacent = ( INT floor, other persons floor )BOOL: floor >= ( other persons floor  1 ) AND floor <= ( other persons floor + 1 );
# displays the floor of an occupant #
PROC print floor = ( PERSON occupant )VOID: print( ( whole( floor OF occupant, 1 ), " ", name OF occupant, newline ) );
# PROBLEM STATEMENT #
# the inhabitants with their floor and constraints #
PERSON baker = ( "Baker", LOC INT := 0, ( INT floor )BOOL: floor /= top floor );
PERSON cooper = ( "Cooper", LOC INT := 0, ( INT floor )BOOL: floor /= bottom floor );
PERSON fletcher = ( "Fletcher", LOC INT := 0, ( INT floor )BOOL: floor /= top floor AND floor /= bottom floor
AND NOT adjacent( floor, floor OF cooper ) );
PERSON miller = ( "Miller", LOC INT := 0, ( INT floor )BOOL: floor > floor OF cooper );
PERSON smith = ( "Smith", LOC INT := 0, ( INT floor )BOOL: NOT adjacent( floor, floor OF fletcher ) );
# SOLUTION #
# "brute force" solution  we run through the possible 5^5 configurations #
# we cold optimise this by e.g. restricting f to bottom floor + 1 TO top floor  1 #
# at the cost of reducing the flexibility of the constraints #
# alternatively, we could add minimum and maximum allowed floors to the PERSON #
# STRUCT and loop through these instead of bottom floor TO top floor #
FOR b FROM bottom floor TO top floor DO
floor OF baker := b;
FOR c FROM bottom floor TO top floor DO
IF b /= c THEN
floor OF cooper := c;
FOR f FROM bottom floor TO top floor DO
IF b /= f AND c /= f THEN
floor OF fletcher := f;
FOR m FROM bottom floor TO top floor DO
IF b /= m AND c /= m AND f /= m THEN
floor OF miller := m;
FOR s FROM bottom floor TO top floor DO
IF b /= s AND c /= s AND f /= s AND m /= s THEN
floor OF smith := s;
IF OK baker AND OK cooper AND OK fletcher AND OK miller AND OK smith
THEN
# found a solution #
print floor( baker );
print floor( cooper );
print floor( fletcher );
print floor( miller );
print floor( smith )
FI
FI
OD
FI
OD
FI
OD
FI
OD
OD
 Output:
2 Baker 1 Cooper 3 Fletcher 4 Miller 0 Smith
AutoHotkey[edit]
See Dinesman's multipledwelling problem/AutoHotkey.
AWK[edit]
# syntax: GAWK f DINESMANS_MULTIPLEDWELLING_PROBLEM.AWK
BEGIN {
for (Baker=1; Baker<=5; Baker++) {
for (Cooper=1; Cooper<=5; Cooper++) {
for (Fletcher=1; Fletcher<=5; Fletcher++) {
for (Miller=1; Miller<=5; Miller++) {
for (Smith=1; Smith<=5; Smith++) {
if (rules() ~ /^1+$/) {
printf("%d Baker\n",Baker)
printf("%d Cooper\n",Cooper)
printf("%d Fletcher\n",Fletcher)
printf("%d Miller\n",Miller)
printf("%d Smith\n",Smith)
}
}
}
}
}
}
exit(0)
}
function rules( stmt1,stmt2,stmt3,stmt4,stmt5,stmt6,stmt7) {
# The following problem statements may be changed:
#
# Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house
# that contains only five floors numbered 1 (ground) to 5 (top)
stmt1 = Baker!=Cooper && Baker!=Fletcher && Baker!=Miller && Baker!=Smith &&
Cooper!=Fletcher && Cooper!=Miller && Cooper!=Smith &&
Fletcher!=Miller && Fletcher!=Smith &&
Miller!=Smith
stmt2 = Baker != 5 # Baker does not live on the top floor
stmt3 = Cooper != 1 # Cooper does not live on the bottom floor
stmt4 = Fletcher != 5 && Fletcher != 1 # Fletcher does not live on either the top or the bottom floor
stmt5 = Miller > Cooper # Miller lives on a higher floor than does Cooper
stmt6 = abs(SmithFletcher) != 1 # Smith does not live on a floor adjacent to Fletcher's
stmt7 = abs(FletcherCooper) != 1 # Fletcher does not live on a floor adjacent to Cooper's
return(stmt1 stmt2 stmt3 stmt4 stmt5 stmt6 stmt7)
}
function abs(x) { if (x >= 0) { return x } else { return x } }
 Output:
3 Baker 2 Cooper 4 Fletcher 5 Miller 1 Smith
BBC BASIC[edit]
Each of the statements is represented by an equivalent conditional expression (stmt1$, stmt2$ etc.) as indicated in the comments, where the variables Baker, Cooper etc. evaluate to the appropriate floor number. So long as each statement can be expressed in this way, and there is a unique solution, changes to the problem text can be accommodated.
REM Floors are numbered 0 (ground) to 4 (top)
REM "Baker, Cooper, Fletcher, Miller, and Smith live on different floors":
stmt1$ = "Baker<>Cooper AND Baker<>Fletcher AND Baker<>Miller AND " + \
\ "Baker<>Smith AND Cooper<>Fletcher AND Cooper<>Miller AND " + \
\ "Cooper<>Smith AND Fletcher<>Miller AND Fletcher<>Smith AND " + \
\ "Miller<>Smith"
REM "Baker does not live on the top floor":
stmt2$ = "Baker<>4"
REM "Cooper does not live on the bottom floor":
stmt3$ = "Cooper<>0"
REM "Fletcher does not live on either the top or the bottom floor":
stmt4$ = "Fletcher<>0 AND Fletcher<>4"
REM "Miller lives on a higher floor than does Cooper":
stmt5$ = "Miller>Cooper"
REM "Smith does not live on a floor adjacent to Fletcher's":
stmt6$ = "ABS(SmithFletcher)<>1"
REM "Fletcher does not live on a floor adjacent to Cooper's":
stmt7$ = "ABS(FletcherCooper)<>1"
FOR Baker = 0 TO 4
FOR Cooper = 0 TO 4
FOR Fletcher = 0 TO 4
FOR Miller = 0 TO 4
FOR Smith = 0 TO 4
IF EVAL(stmt2$) IF EVAL(stmt3$) IF EVAL(stmt5$) THEN
IF EVAL(stmt4$) IF EVAL(stmt6$) IF EVAL(stmt7$) THEN
IF EVAL(stmt1$) THEN
PRINT "Baker lives on floor " ; Baker
PRINT "Cooper lives on floor " ; Cooper
PRINT "Fletcher lives on floor " ; Fletcher
PRINT "Miller lives on floor " ; Miller
PRINT "Smith lives on floor " ; Smith
ENDIF
ENDIF
ENDIF
NEXT Smith
NEXT Miller
NEXT Fletcher
NEXT Cooper
NEXT Baker
END
 Output:
Baker lives on floor 2 Cooper lives on floor 1 Fletcher lives on floor 3 Miller lives on floor 4 Smith lives on floor 0
Bracmat[edit]
The rules constitute the body of the 'constraints' function. Each statement of the problem is translated into a pattern. Patterns are the rhs of the ':' operator. Constraints can be added or deleted as you like. If the problem is underspecified, for example by deleting one or more patterns, all solutions are output, because the line following the output statement forces Bracmat to backtrack. Patterns are read as follows: the '~' means negation, a '?' is a wildcard that can span zero or more floors, a '' means alternation. If in a pattern there is no wildcard to the left of a person's name, the pattern states that the person must live in the bottom floor. If in a pattern there is no wildcard to the right of a person's name, the pattern states that the person must live in the top floor. If in a pattern name A is left of name B, the pattern states that person A is living in a lower floor than person B. Patterns can be alternated with the '' (OR) operator. The match operator ':', when standing between two patterns, functions as an AND operation, because both sides must match the subject argument 'arg'. The names of the people can be changed to anything, except empty strings. Bracmat supports UTF8 encoded Unicode characters, but falls back to ISO 88591 if a string cannot be parsed as UTF8. If a name contains characters that can be misinterpreted as operators, such as '.' or ' ', the name must be enclosed in double quotes. If there are no reserved characters in a name, double quotes are optional.
( Baker Cooper Fletcher Miller Smith:?people
& ( constraints
=
. !arg
: ~(? Baker)
: ~(Cooper ?)
: ~(Fletcher ?? Fletcher)
: ? Cooper ? Miller ?
: ~(? Smith Fletcher ?? Fletcher Smith ?)
: ~(? Cooper Fletcher ?? Fletcher Cooper ?)
)
& ( solution
= floors persons A Z person
. !arg:(?floors.?persons)
& ( !persons:
& constraints$!floors
& out$("Inhabitants, from bottom to top:" !floors)
& ~ { The ~ always fails on evaluation. Here, failure forces Bracmat to backtrack and find all solutions, not just the first one. }
 !persons
: ?A
%?`person
(?Z&solution$(!floors !person.!A !Z))
)
)
& solution$(.!people)
 { After outputting all solutions, the lhs of the  operator fails. The rhs of the  operator, here an empty string, is the final result. }
);
Inhabitants, from bottom to top: Smith Cooper Baker Fletcher Miller
C[edit]
#include <stdio.h>
#include <stdlib.h>
int verbose = 0;
#define COND(a, b) int a(int *s) { return (b); }
typedef int(*condition)(int *);
/* BEGIN problem specific setup */
#define N_FLOORS 5
#define TOP (N_FLOORS  1)
int solution[N_FLOORS] = { 0 };
int occupied[N_FLOORS] = { 0 };
enum tenants {
baker = 0,
cooper,
fletcher,
miller,
smith,
phantom_of_the_opera,
};
const char *names[] = {
"baker",
"cooper",
"fletcher",
"miller",
"smith",
};
COND(c0, s[baker] != TOP);
COND(c1, s[cooper] != 0);
COND(c2, s[fletcher] != 0 && s[fletcher] != TOP);
COND(c3, s[miller] > s[cooper]);
COND(c4, abs(s[smith]  s[fletcher]) != 1);
COND(c5, abs(s[cooper]  s[fletcher]) != 1);
#define N_CONDITIONS 6
condition cond[] = { c0, c1, c2, c3, c4, c5 };
/* END of problem specific setup */
int solve(int person)
{
int i, j;
if (person == phantom_of_the_opera) {
/* check condition */
for (i = 0; i < N_CONDITIONS; i++) {
if (cond[i](solution)) continue;
if (verbose) {
for (j = 0; j < N_FLOORS; j++)
printf("%d %s\n", solution[j], names[j]);
printf("cond %d bad\n\n", i);
}
return 0;
}
printf("Found arrangement:\n");
for (i = 0; i < N_FLOORS; i++)
printf("%d %s\n", solution[i], names[i]);
return 1;
}
for (i = 0; i < N_FLOORS; i++) {
if (occupied[i]) continue;
solution[person] = i;
occupied[i] = 1;
if (solve(person + 1)) return 1;
occupied[i] = 0;
}
return 0;
}
int main()
{
verbose = 0;
if (!solve(0)) printf("Nobody lives anywhere\n");
return 0;
}
 Output:
Found arrangement: 2 baker 1 cooper 3 fletcher 4 miller 0 smith
C, being its compiled self, is not terribly flexible in dynamically changing runtime code content.
Parsing some external problem specification would be one way, but for a small problem, it might as well just recompile with conditions hard coded.
For this program, to change conditions, one needs to edit content between BEGIN and END of problem specific setup.
Those could even be setup in an external file and gets #include
d if need be.
C#[edit]
Constraints as functions solution [edit]
Usage of the DinesmanSolver is very simple. Just feed it a bunch of constraints in the form of functions. (It could also be one function with a bunch of 'and' clauses)
Each tenant is considered an integer from 0 to count.
For each solution, it will output an array of integers that represent the tenants ordered by floor number, from the bottom floor to the top.
public class Program
{
public static void Main()
{
const int count = 5;
const int Baker = 0, Cooper = 1, Fletcher = 2, Miller = 3, Smith = 4;
string[] names = { nameof(Baker), nameof(Cooper), nameof(Fletcher), nameof(Miller), nameof(Smith) };
Func<int[], bool>[] constraints = {
floorOf => floorOf[Baker] != count1,
floorOf => floorOf[Cooper] != 0,
floorOf => floorOf[Fletcher] != count1 && floorOf[Fletcher] != 0,
floorOf => floorOf[Miller] > floorOf[Cooper],
floorOf => Math.Abs(floorOf[Smith]  floorOf[Fletcher]) > 1,
floorOf => Math.Abs(floorOf[Fletcher]  floorOf[Cooper]) > 1,
};
var solver = new DinesmanSolver();
foreach (var tenants in solver.Solve(count, constraints)) {
Console.WriteLine(string.Join(" ", tenants.Select(t => names[t])));
}
}
}
public class DinesmanSolver
{
public IEnumerable<int[]> Solve(int count, params Func<int[], bool>[] constraints) {
foreach (int[] floorOf in Permutations(count)) {
if (constraints.All(c => c(floorOf))) {
yield return Enumerable.Range(0, count).OrderBy(i => floorOf[i]).ToArray();
}
}
}
static IEnumerable<int[]> Permutations(int length) {
if (length == 0) {
yield return new int[0];
yield break;
}
bool forwards = false;
foreach (var permutation in Permutations(length  1)) {
for (int i = 0; i < length; i++) {
yield return permutation.InsertAt(forwards ? i : length  i  1, length  1).ToArray();
}
forwards = !forwards;
}
}
}
static class Extensions
{
public static IEnumerable<T> InsertAt<T>(this IEnumerable<T> source, int position, T newElement) {
if (source == null) throw new ArgumentNullException(nameof(source));
if (position < 0) throw new ArgumentOutOfRangeException(nameof(position));
return InsertAtIterator(source, position, newElement);
}
private static IEnumerable<T> InsertAtIterator<T>(IEnumerable<T> source, int position, T newElement) {
int index = 0;
foreach (T element in source) {
if (index == position) yield return newElement;
yield return element;
index++;
}
if (index < position) throw new ArgumentOutOfRangeException(nameof(position));
if (index == position) yield return newElement;
}
}
 Output:
Smith Cooper Baker Fletcher Miller
Shorter Linq solution [edit]
This challenge is badly stated. It is trivial to state/add any variant as a where clause (and to the enum) in the Linq query. Need more information in order to automatically parse such statements and there is no specification of this in the challenge.
using System;
using System.Collections.Generic;
using static System.Linq.Enumerable;
static class Program
{
enum Tenants { Baker = 0, Cooper = 1, Fletcher = 2, Miller = 3, Smith = 4 };
static void Main()
{
var count = Enum.GetNames(typeof(Tenants)).Length;
var top = count  1;
var solve =
from f in Range(0, count).Permutations()
let floors = f.ToArray()
where floors[(int)Tenants.Baker] != top //r1
where floors[(int)Tenants.Cooper] != 0 //r2
where floors[(int)Tenants.Fletcher] != top && floors[(int)Tenants.Fletcher] != 0 //r3
where floors[(int)Tenants.Miller] > floors[(int)Tenants.Cooper] //r4
where Math.Abs(floors[(int)Tenants.Smith]  floors[(int)Tenants.Fletcher]) !=1 //r5
where Math.Abs(floors[(int)Tenants.Fletcher]  floors[(int)Tenants.Cooper]) !=1 //r6
select floors;
var solved = solve.First();
var output = Range(0,count).OrderBy(i=>solved[i]).Select(f => ((Tenants)f).ToString());
Console.WriteLine(String.Join(" ", output));
Console.Read();
}
public static IEnumerable<IEnumerable<T>> Permutations<T>(this IEnumerable<T> values)
{
if (values.Count() == 1)
return values.ToSingleton();
return values.SelectMany(v => Permutations(values.Except(v.ToSingleton())), (v, p) => p.Prepend(v));
}
public static IEnumerable<T> ToSingleton<T>(this T item) { yield return item; }
}
Output:
Smith Cooper Baker Fletcher Miller
Clojure[edit]
This solution uses the contributed package clojure.core.logic, a miniKanrenbased logic solver (and contributed clojure.tools.macro as well). The "setup" part of this code defines relational functions (or constraints) for testing "immediately above", "higher", and "on nonadjacent floors". These are used (along with the package's "permuteo" constraint) to define a constraint dinesmano which searches for all the resident orders that satisfy the criteria. The criteria are listed in onetoone correspondence with the problem statement. The problem statement could be changed to any mixture of these constraint types, and additional constraint functions could be defined as necessary. The final part of the code searches for all solutions and prints them out.
(ns rosettacode.dinesman
(:use [clojure.core.logic]
[clojure.tools.macro :as macro]))
; whether x is immediately above (left of) y in list s; uses pattern matching on s
(defne aboveo [x y s]
([_ _ (x y . ?rest)])
([_ _ [_ . ?rest]] (aboveo x y ?rest)))
; whether x is on a higher floor than y
(defne highero [x y s]
([_ _ (x . ?rest)] (membero y ?rest))
([_ _ (_ . ?rest)] (highero x y ?rest)))
; whether x and y are on nonadjacent floors
(defn nonadjacento [x y s]
(conda
((aboveo x y s) fail)
((aboveo y x s) fail)
(succeed)))
(defn dinesmano [rs]
(macro/symbolmacrolet [_ (lvar)]
(all
(permuteo ['Baker 'Cooper 'Fletcher 'Miller 'Smith] rs)
(aboveo _ 'Baker rs) ;someone lives above Baker
(aboveo 'Cooper _ rs) ;Cooper lives above someone
(aboveo 'Fletcher _ rs)
(aboveo _ 'Fletcher rs)
(highero 'Miller 'Cooper rs)
(nonadjacento 'Smith 'Fletcher rs)
(nonadjacento 'Fletcher 'Cooper rs))))
(let [solns (run* [q] (dinesmano q))]
(println "solution count:" (count solns))
(println "solution(s) highest to lowest floor:")
(doseq [soln solns] (println " " soln)))
 Output:
solution count: 1 solution(s) highest to lowest floor: (Miller Fletcher Baker Cooper Smith)
D[edit]
This code uses the second lazy permutations function of Permutations#Lazy_version.
As for flexibility: the solve code works with an arbitrary number of people and predicates.
import std.stdio, std.math, std.algorithm, std.traits, permutations2;
void main() {
enum Names { Baker, Cooper, Fletcher, Miller, Smith }
immutable(bool function(in Names[]) pure nothrow)[] predicates = [
s => s[Names.Baker] != s.length  1,
s => s[Names.Cooper] != 0,
s => s[Names.Fletcher] != 0 && s[Names.Fletcher] != s.length1,
s => s[Names.Miller] > s[Names.Cooper],
s => abs(s[Names.Smith]  s[Names.Fletcher]) != 1,
s => abs(s[Names.Cooper]  s[Names.Fletcher]) != 1];
permutations([EnumMembers!Names])
.filter!(solution => predicates.all!(pred => pred(solution)))
.writeln;
}
 Output:
[[Fletcher, Cooper, Miller, Smith, Baker]]
Simpler Version[edit]
void main() {
import std.stdio, std.math, std.algorithm, permutations2;
["Baker", "Cooper", "Fletcher", "Miller", "Smith"]
.permutations
.filter!(s =>
s.countUntil("Baker") != 4 && s.countUntil("Cooper") &&
s.countUntil("Fletcher") && s.countUntil("Fletcher") != 4 &&
s.countUntil("Miller") > s.countUntil("Cooper") &&
abs(s.countUntil("Smith")  s.countUntil("Fletcher")) != 1 &&
abs(s.countUntil("Cooper")  s.countUntil("Fletcher")) != 1)
.writeln;
}
The output is the same.
EchoLisp[edit]
The problem is solved using the amb library. The solution separates the constrainst procedure from the solver procedure. The solver does not depend on names, number of floors. This flexibility allows to easily add floors, names, constraints. See Antoinette example below, Antoinette is very close ❤️ to Cooper, and wants a prime numbered floor.
Setup  Solver[edit]
(require 'hash)
(require' amb)
;;
;; Solver
;;
(define (dwellingpuzzle context names floors H)
;; each amb calls gives a floor to a name
(for ((name names))
(hashset H name (amb context floors)))
;; They live on different floors.
(ambrequire (distinct? (ambchoices context)))
(constraints floors H) ;; may fail and backtrack
;; result returned to ambrun
(for/list ((name names))
(cons name (hashref H name)))
;; (ambfail) is possible here to see all solutions
)
(define (task names)
(ambrun dwellingpuzzle
(ambmakecontext)
names
(iota (length names)) ;; list of floors : 0,1, ....
(makehash)) ;; hash table : "name" > floor
)
Problem data  constraints[edit]
(define names '("baker" "cooper" "fletcher" "miller" "smith" ))
(definesyntaxrule (floor name) (hashref H name))
(definesyntaxrule (touch a b) (= (abs ( (hashref H a) (hashref H b))) 1))
(define (constraints floors H)
(define top (1 (length floors)))
;; Baker does not live on the top floor.
(ambrequire (!= (floor "baker") top))
;; Cooper does not live on the bottom floor.
(ambrequire (!= (floor "cooper") 0))
;; Fletcher does not live on either the top or the bottom floor.
(ambrequire (!= (floor "fletcher") top))
(ambrequire (!= (floor "fletcher") 0))
;; Miller lives on a higher floor than does Cooper.
(ambrequire (> (floor "miller") (floor "cooper")))
;; Smith does not live on a floor adjacent to Fletcher's.
(ambrequire (not (touch "smith" "fletcher")))
;; Fletcher does not live on a floor adjacent to Cooper's.
(ambrequire (not (touch "fletcher" "cooper")))
)
 Output:
(task names)
→ ((baker . 2) (cooper . 1) (fletcher . 3) (miller . 4) (smith . 0))
Changing data  constraints[edit]
;; add a name/floor
(define names '("baker" "cooper" "fletcher" "miller" "smith" "antoinette"))
(define (constraints floors H)
;; ... same as above, add the following
;; Antoinette does not like 💔 Smith
(ambrequire (not (touch "smith" "antoinette")))
;; Antoinette is very close ❤️ to Cooper
(ambrequire (touch "cooper" "antoinette"))
;; Antoinette wants a prime numbered floor
(ambrequire (prime? (floor "antoinette")))
)
 Output:
(task names)
→ ((baker . 0) (cooper . 1) (fletcher . 3) (miller . 4) (smith . 5) (antoinette . 2))
Elixir[edit]
Simple solution:
defmodule Dinesman do
def problem do
names = ~w( Baker Cooper Fletcher Miller Smith )a
predicates = [fn(c)> :Baker != List.last(c) end,
fn(c)> :Cooper != List.first(c) end,
fn(c)> :Fletcher != List.first(c) && :Fletcher != List.last(c) end,
fn(c)> floor(c, :Miller) > floor(c, :Cooper) end,
fn(c)> abs(floor(c, :Smith)  floor(c, :Fletcher)) != 1 end,
fn(c)> abs(floor(c, :Cooper)  floor(c, :Fletcher)) != 1 end]
permutation(names)
> Enum.filter(fn candidate >
Enum.all?(predicates, fn predicate > predicate.(candidate) end)
end)
> Enum.each(fn name_list >
Enum.with_index(name_list)
> Enum.each(fn {name,i} > IO.puts "#{name} lives on #{i+1}" end)
end)
end
defp floor(c, name), do: Enum.find_index(c, fn x > x == name end)
defp permutation([]), do: [[]]
defp permutation(list), do: (for x < list, y < permutation(list  [x]), do: [xy])
end
Dinesman.problem
 Output:
Smith lives on 1 Cooper lives on 2 Baker lives on 3 Fletcher lives on 4 Miller lives on 5
Erlang[edit]
The people is an argument list. The rules is an argument list of options. Only rules that have a function in the program can be in the options. The design of the rules can be argued. Perhaps {cooper, does_not_live_on, 0}, etc, would be better for people unfamiliar with lisp.
module( dinesman_multiple_dwelling ).
export( [solve/2, task/0] ).
solve( All_persons, Rules ) >
[house(Bottom_floor, B, C, D, Top_floor)  Bottom_floor < All_persons, B < All_persons, C < All_persons, D < All_persons, Top_floor < All_persons,
lists:all( fun (Fun) > Fun( house(Bottom_floor, B, C, D, Top_floor) ) end, rules( Rules ))].
task() >
All_persons = [baker, cooper, fletcher, miller, smith],
Rules = [all_on_different_floors, {not_lives_on_floor, 4, baker}, {not_lives_on_floor, 0, cooper}, {not_lives_on_floor, 4, fletcher}, {not_lives_on_floor, 0, fletcher},
{on_higher_floor, miller, cooper}, {not_adjacent, smith, fletcher}, {not_adjacent, fletcher, cooper}],
[House] = solve( All_persons, Rules ),
[io:fwrite("~p lives on floor ~p~n", [lists:nth(X, House), X  1])  X < lists:seq(1,5)].
house( A, B, C, D, E ) > [A, B, C, D, E].
is_all_on_different_floors( [A, B, C, D, E] ) >
A =/= B andalso A =/= C andalso A =/= D andalso A =/= E
andalso B =/= C andalso B =/= D andalso B =/= E
andalso C =/= D andalso C =/= E
andalso D =/= E.
is_not_adjacent( Person1, Person2, House ) >
is_not_below( Person1, Person2, House ) andalso is_not_below( Person2, Person1, House ).
is_not_below( _Person1, _Person2, [_Person] ) > true;
is_not_below( Person1, Person2, [Person1, Person2  _T] ) > false;
is_not_below( Person1, Person2, [_Person  T] ) > is_not_below( Person1, Person2, T ).
is_on_higher_floor( Person1, _Person2, [Person1  _T] ) > false;
is_on_higher_floor( _Person1, Person2, [Person2  _T] ) > true;
is_on_higher_floor( Person1, Person2, [_Person  T] ) > is_on_higher_floor( Person1, Person2, T ).
rules( Rules ) > lists:map( fun rules_fun/1, Rules ).
rules_fun( all_on_different_floors ) > fun is_all_on_different_floors/1;
rules_fun( {not_lives_on_floor, N, Person} ) > fun (House) > Person =/= lists:nth(N + 1, House) end;
rules_fun( {on_higher_floor, Person1, Person2} ) > fun (House) > is_on_higher_floor( Person1, Person2, House ) end;
rules_fun( {not_below, Person1, Person2} ) > fun (House) > is_not_below( Person1, Person2, House ) end;
rules_fun( {not_adjacent, Person1, Person2} ) > fun (House) > is_not_adjacent( Person1, Person2, House ) end.
 Output:
8> dinesman_multiple_dwelling:task(). smith lives on floor 0 cooper lives on floor 1 baker lives on floor 2 fletcher lives on floor 3 miller lives on floor 4
ERRE[edit]
PROGRAM DINESMAN
BEGIN
! Floors are numbered 0 (ground) to 4 (top)
! "Baker, Cooper, Fletcher, Miller, and Smith live on different floors":
stmt1$="Baker<>Cooper AND Baker<>Fletcher AND Baker<>Miller AND "+"Baker<>Smith AND Cooper<>Fletcher AND Cooper<>Miller AND "+"Cooper<>Smith AND Fletcher<>Miller AND Fletcher<>Smith AND "+"Miller<>Smith"
! "Baker does not live on the top floor":
stmt2$="Baker<>4"
! "Cooper does not live on the bottom floor":
stmt3$="Cooper<>0"
! "Fletcher does not live on either the top or the bottom floor":
stmt4$="Fletcher<>0 AND Fletcher<>4"
! "Miller lives on a higher floor than does Cooper":
stmt5$="Miller>Cooper"
! "Smith does not live on a floor adjacent to Fletcher's":
stmt6$="ABS(SmithFletcher)<>1"
! "Fletcher does not live on a floor adjacent to Cooper's":
stmt7$="ABS(FletcherCooper)<>1"
FOR Baker=0 TO 4 DO
FOR Cooper=0 TO 4 DO
FOR Fletcher=0 TO 4 DO
FOR Miller=0 TO 4 DO
FOR Smith=0 TO 4 DO
IF Baker<>4 AND Cooper<>0 AND Miller>Cooper THEN
IF Fletcher<>0 AND Fletcher<>4 AND ABS(SmithFletcher)<>1 AND ABS(FletcherCooper)<>1 THEN
IF Baker<>Cooper AND Baker<>Fletcher AND Baker<>Miller AND Baker<>Smith AND Cooper<>Fletcher AND Cooper<>Miller AND Cooper<>Smith AND Fletcher<>Miller AND Fletcher<>Smith AND Miller<>Smith THEN
PRINT("Baker lives on floor ";Baker)
PRINT("Cooper lives on floor ";Cooper)
PRINT("Fletcher lives on floor ";Fletcher)
PRINT("Miller lives on floor ";Miller)
PRINT("Smith lives on floor ";Smith)
END IF
END IF
END IF
END FOR ! Smith
END FOR ! Miller
END FOR ! Fletcher
END FOR ! Cooper
END FOR ! Baker
END PROGRAM
 Output:
Baker lives on floor 2 Cooper lives on floor 1 Fletcher lives on floor 3 Miller lives on floor 4 Smith lives on floor 0
Factor[edit]
All rules are encoded in the ``meetsconstraints?`` word. Any variations to the rules requires modifying ``meetsconstraints?``
USING: kernel
combinators.shortcircuit
math math.combinatorics math.ranges
sequences
qw prettyprint ;
IN: rosetta.dinesman
: /= ( x y  ? ) = not ;
: fifth ( seq  elt ) 4 swap nth ;
: meetsconstraints? ( seq  ? )
{
[ first 5 /= ] ! Baker does not live on the top floor.
[ second 1 /= ] ! Cooper does not live on the bottom floor.
[ third { 1 5 } member? not ] ! Fletcher does not live on either the top or bottom floor.
[ [ fourth ] [ second ] bi > ] ! Miller lives on a higher floor than does Cooper.
[ [ fifth ] [ third ] bi  abs 1 /= ] ! Smith does not live on a floor adjacent to Fletcher's.
[ [ third ] [ second ] bi  abs 1 /= ] ! Fletcher does not live on a floor adjacent to Cooper's.
} 1&& ;
: solutions (  seq )
5 [1,b] allpermutations [ meetsconstraints? ] filter ;
: >names ( seq  seq )
[ 1  qw{ baker cooper fletcher miller smith } nth ] map ;
: dinesman (  )
solutions [ >names . ] each ;
 Output:
{ "fletcher" "cooper" "miller" "smith" "baker" }
Forth[edit]
This solution takes advantage of several of Forth's strengths. Forth is able to picture a number in any base from 2 to 36.
This program simply iterates through all numbers between 01234 and 43210 (base 5). To see whether this is a permutation worth testing, a binary mask is generated. If all 5 bits are set (31 decimal), this is a possible candidate. Then all ASCII digits of the generated number are converted back to numbers by subtracting the value of ASCII "0". Finally each of the conditions is tested.
All conditions are confined to a single word. The algorithm "as is" will work up to 10 floors. After that, we have to take into consideration that characters AZ are used as digits. That will work up to 36 floors.
Although this is not ANS Forth, one should have little trouble converting it.
0 enum baker \ enumeration of all tenants
enum cooper
enum fletcher
enum miller
constant smith
create names \ names of all the tenants
," Baker"
," Cooper"
," Fletcher"
," Miller"
," Smith" \ get name, type it
does> swap cells + @c count type ." lives in " ;
5 constant #floor \ number of floors
#floor 1 constant top \ top floor
0 constant bottom \ we're counting the floors from 0
: [email protected] [email protected] [char] 0  ; ( a  n)
: floor chars over + [email protected] ; ( a n1  a n2)
\ is it a valid permutation?
: perm? ( n  a f)
#floor base ! 0 swap s>d <# #floor 0 ?do # loop #>
over >r bounds do 1 i [email protected] lshift + loop
31 = r> swap decimal \ create binary mask and check
;
\ test a solution
: solution? ( a  a f)
baker floor top <> \ baker on top floor?
if cooper floor bottom <> \ cooper on the bottom floor?
if fletcher floor dup bottom <> swap top <> and
if cooper floor swap miller floor rot >
if smith floor swap fletcher floor rot  abs 1 <>
if cooper floor swap fletcher floor rot  abs 1 <>
if true exit then \ we found a solution!
then
then
then
then
then false \ nice try, no cigar..
;
( a )
: .solution #floor 0 do i names i chars over + [email protected] 1+ emit cr loop drop ;
\ main routine
: dinesman ( )
2932 194 do
i perm? if solution? if .solution leave else drop then else drop then
loop
; \ show the solution
dinesman
 Output:
Baker lives in 3 Cooper lives in 2 Fletcher lives in 4 Miller lives in 5 Smith lives in 1
Go[edit]
package main
import "fmt"
// The program here is restricted to finding assignments of tenants (or more
// generally variables with distinct names) to floors (or more generally
// integer values.) It finds a solution assigning all tenants and assigning
// them to different floors.
// Change number and names of tenants here. Adding or removing names is
// allowed but the names should be distinct; the code is not written to handle
// duplicate names.
var tenants = []string{"Baker", "Cooper", "Fletcher", "Miller", "Smith"}
// Change the range of floors here. The bottom floor does not have to be 1.
// These should remain nonnegative integers though.
const bottom = 1
const top = 5
// A type definition for readability. Do not change.
type assignments map[string]int
// Change rules defining the problem here. Change, add, or remove rules as
// desired. Each rule should first be commented as human readable text, then
// coded as a function. The function takes a tentative partial list of
// assignments of tenants to floors and is free to compute anything it wants
// with this information. Other information available to the function are
// package level defintions, such as top and bottom. A function returns false
// to say the assignments are invalid.
var rules = []func(assignments) bool{
// Baker does not live on the top floor
func(a assignments) bool {
floor, assigned := a["Baker"]
return !assigned  floor != top
},
// Cooper does not live on the bottom floor
func(a assignments) bool {
floor, assigned := a["Cooper"]
return !assigned  floor != bottom
},
// Fletcher does not live on either the top or the bottom floor
func(a assignments) bool {
floor, assigned := a["Fletcher"]
return !assigned  (floor != top && floor != bottom)
},
// Miller lives on a higher floor than does Cooper
func(a assignments) bool {
if m, assigned := a["Miller"]; assigned {
c, assigned := a["Cooper"]
return !assigned  m > c
}
return true
},
// Smith does not live on a floor adjacent to Fletcher's
func(a assignments) bool {
if s, assigned := a["Smith"]; assigned {
if f, assigned := a["Fletcher"]; assigned {
d := s  f
return d*d > 1
}
}
return true
},
// Fletcher does not live on a floor adjacent to Cooper's
func(a assignments) bool {
if f, assigned := a["Fletcher"]; assigned {
if c, assigned := a["Cooper"]; assigned {
d := f  c
return d*d > 1
}
}
return true
},
}
// Assignment program, do not change. The algorithm is a depth first search,
// tentatively assigning each tenant in order, and for each tenant trying each
// unassigned floor in order. For each tentative assignment, it evaluates all
// rules in the rules list and backtracks as soon as any one of them fails.
//
// This algorithm ensures that the tenative assignments have only names in the
// tenants list, only floor numbers from bottom to top, and that tentants are
// assigned to different floors. These rules are hard coded here and do not
// need to be coded in the the rules list above.
func main() {
a := assignments{}
var occ [top + 1]bool
var df func([]string) bool
df = func(u []string) bool {
if len(u) == 0 {
return true
}
tn := u[0]
u = u[1:]
f:
for f := bottom; f <= top; f++ {
if !occ[f] {
a[tn] = f
for _, r := range rules {
if !r(a) {
delete(a, tn)
continue f
}
}
occ[f] = true
if df(u) {
return true
}
occ[f] = false
delete(a, tn)
}
}
return false
}
if !df(tenants) {
fmt.Println("no solution")
return
}
for t, f := range a {
fmt.Println(t, f)
}
}
 Output:
Baker 3 Cooper 2 Fletcher 4 Miller 5 Smith 1
Haskell[edit]
The List monad is perfect for this kind of problem. One can express the problem statements in a very natural and concise way:
import Data.List (permutations)
import Control.Monad (guard)
dinesman :: [(Int,Int,Int,Int,Int)]
dinesman = do
 baker, cooper, fletcher, miller, smith are integers representing
 the floor that each person lives on, from 1 to 5
 Baker, Cooper, Fletcher, Miller, and Smith live on different floors
 of an apartment house that contains only five floors.
[baker, cooper, fletcher, miller, smith] < permutations [1..5]
 Baker does not live on the top floor.
guard $ baker /= 5
 Cooper does not live on the bottom floor.
guard $ cooper /= 1
 Fletcher does not live on either the top or the bottom floor.
guard $ fletcher /= 5 && fletcher /= 1
 Miller lives on a higher floor than does Cooper.
guard $ miller > cooper
 Smith does not live on a floor adjacent to Fletcher's.
guard $ abs (smith  fletcher) /= 1
 Fletcher does not live on a floor adjacent to Cooper's.
guard $ abs (fletcher  cooper) /= 1
 Where does everyone live?
return (baker, cooper, fletcher, miller, smith)
main :: IO ()
main = do
print $ head dinesman  print first solution: (3,2,4,5,1)
print dinesman  print all solutions (only one): [(3,2,4,5,1)]
Or as a list comprehension:
import Data.List (permutations)
print [ ("Baker lives on " ++ show b
, "Cooper lives on " ++ show c
, "Fletcher lives on " ++ show f
, "Miller lives on " ++ show m
, "Smith lives on " ++ show s)  [b,c,f,m,s] < permutations [1..5], b/=5,c/=1,f/=1,f/=5,m>c,abs(sf)>1,abs(cf)>1]
 Output:
[("Baker lives on 3","Cooper lives on 2","Fletcher lives on 4","Miller lives on 5","Smith lives on 1")]
Icon and Unicon[edit]
This solution uses string invocation to call operators and the fact the Icon/Unicon procedures are first class values. The procedure names could also be given as strings and it would be fairly simple to read the names and all the rules directly from a file. Each name and rule recurses and relies on the inherent backtracking in the language to achieve the goal.
The rules explicitly call stop() after showing the solution. Removing the stop would cause the solver to try all possible cases and report all possible solutions (if there were multiple ones).
invocable all
global nameL, nameT, rules
procedure main() # Dinesman
nameT := table()
nameL := ["Baker", "Cooper", "Fletcher", "Miller", "Smith"]
rules := [ [ distinct ],
[ "~=", "Baker", top() ],
[ "~=", "Cooper", bottom() ],
[ "~=", "Fletcher", top() ],
[ "~=", "Fletcher", bottom() ],
[ ">", "Miller", "Cooper" ],
[ notadjacent, "Smith", "Fletcher" ],
[ notadjacent, "Fletcher", "Cooper" ],
[ showsolution ],
[ stop ] ]
if not solve(1) then
write("No solution found.")
end
procedure dontstop() # use if you want to search for all solutions
end
procedure showsolution() # show the soluton
write("The solution is:")
every write(" ",n := !nameL, " lives in ", nameT[n])
return
end
procedure eval(n) # evaluate a rule
r := copy(rules[ntop()])
every r[i := 2 to *r] := rv(r[i])
if get(r)!r then suspend
end
procedure rv(x) # return referenced value if it exists
return \nameT[x]  x
end
procedure solve(n) # recursive solver
if n > top() then { # apply rules
if n <= top() + *rules then
( eval(n) & solve(n+1) )  fail
}
else # setup locations
(( nameT[nameL[n]] := bottom() to top() ) & solve(n + 1))  fail
return
end
procedure distinct(a,b) # ensure each name is distinct
if nameT[n := !nameL] = nameT[n ~== key(nameT)] then fail
suspend
end
procedure notadjacent(n1,n2) # ensure n1,2 are not adjacent
if not adjacent(n1,n2) then suspend
end
procedure adjacent(n1,n2) # ensure n1,2 are adjacent
if abs(n1  n2) = 1 then suspend
end
procedure bottom() # return bottom
return if *nameL > 0 then 1 else 0
end
procedure top() # return top
return *nameL
end
 Output:
The solution is: Baker lives in 3 Cooper lives in 2 Fletcher lives in 4 Miller lives in 5 Smith lives in 1
J[edit]
This problem asks us to pick from one of several possibilities. We can represent these possibilities as permutations of the residents' initials, arranged in order from lowest floor to top floor:
possible=: ((i.!5) A. i.5) { 'BCFMS'
Additionally, we are given a variety of constraints which eliminate some possibilities:
possible=: (#~ 'B' ~: {:"1) possible NB. Baker not on top floor
possible=: (#~ 'C' ~: {."1) possible NB. Cooper not on bottom floor
possible=: (#~ 'F' ~: {:"1) possible NB. Fletcher not on top floor
possible=: (#~ 'F' ~: {."1) possible NB. Fletcher not on bottom floor
possible=: (#~ </@i."1&'CM') possible NB. Miller on higher floor than Cooper
possible=: (#~ 0 = +/@E."1~&'SF') possible NB. Smith not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FS') possible NB. Fletcher not immediately below Smith
possible=: (#~ 0 = +/@E."1~&'CF') possible NB. Cooper not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FC') possible NB. Fletcher not immediately below Cooper
The answer is thus:
possible
SCBFM
(bottom floor) Smith, Cooper, Baker, Fletcher, Miller (top floor)
Java[edit]
Code:
import java.util.*;
class DinesmanMultipleDwelling {
private static void generatePermutations(String[] apartmentDwellers, Set<String> set, String curPermutation) {
for (String s : apartmentDwellers) {
if (!curPermutation.contains(s)) {
String nextPermutation = curPermutation + s;
if (nextPermutation.length() == apartmentDwellers.length) {
set.add(nextPermutation);
} else {
generatePermutations(apartmentDwellers, set, nextPermutation);
}
}
}
}
private static boolean topFloor(String permutation, String person) { //Checks to see if the person is on the top floor
return permutation.endsWith(person);
}
private static boolean bottomFloor(String permutation, String person) {//Checks to see if the person is on the bottom floor
return permutation.startsWith(person);
}
public static boolean livesAbove(String permutation, String upperPerson, String lowerPerson) {//Checks to see if the person lives above the other person
return permutation.indexOf(upperPerson) > permutation.indexOf(lowerPerson);
}
public static boolean adjacent(String permutation, String person1, String person2) { //checks to see if person1 is adjacent to person2
return (Math.abs(permutation.indexOf(person1)  permutation.indexOf(person2)) == 1);
}
private static boolean isPossible(String s) {
/*
What this does should be obvious...proper explaination can be given if needed
Conditions here Switching any of these to ! or reverse will change what is given as a result
example
if(topFloor(s, "B"){
}
to
if(!topFloor(s, "B"){
}
or the opposite
if(!topFloor(s, "B"){
}
to
if(topFloor(s, "B"){
}
*/
if (topFloor(s, "B")) {//B is on Top Floor
return false;
}
if (bottomFloor(s, "C")) {//C is on Bottom Floor
return false;
}
if (topFloor(s, "F")  bottomFloor(s, "F")) {// F is on top or bottom floor
return false;
}
if (!livesAbove(s, "M", "C")) {// M does not live above C
return false;
}
if (adjacent(s, "S", "F")) { //S lives adjacent to F
return false;
}
return !adjacent(s, "F", "C"); //F does not live adjacent to C
}
public static void main(String[] args) {
Set<String> set = new HashSet<String>();
generatePermutations(new String[]{"B", "C", "F", "M", "S"}, set, ""); //Generates Permutations
for (Iterator<String> iterator = set.iterator(); iterator.hasNext();) {//Loops through iterator
String permutation = iterator.next();
if (!isPossible(permutation)) {//checks to see if permutation is false if so it removes it
iterator.remove();
}
}
for (String s : set) {
System.out.println("Possible arrangement: " + s);
/*
Prints out possible arranagement...changes depending on what you change in the "isPossible method"
*/
}
}
}
 Output:
Possible arrangement: SCBFM
JavaScript[edit]
ES6[edit]
More flexibility[edit]
(Full occupancy and no cohabitation included in the predicate)
The generality of nesting concatMap, and returning values enclosed in a list (empty where the test fails, populated otherwise), is the same as that of a using a list comprehension, to which it is formally equivalent. (concatMap is the bind operator for the list monad, and (a > [a]) is the type of the 'return' function for a list monad. The effect is to define a cartesian product, and apply a predicate to each member of that product. Any empty lists returned where a predicate yields false are eliminated by the concatenation component of concatMap.
The predicates here can be varied, and the depth of concatMap nestings can be adjusted to match the number of unknowns in play, with each concatMap binding one name, and defining the list of its possible values.
(() => {
'use strict';
// concatMap :: (a > [b]) > [a] > [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
// range :: Int > Int > [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n  m) + 1
}, (_, i) => m + i);
// and :: [Bool] > Bool
const and = xs => {
let i = xs.length;
while (i)
if (!xs[i]) return false;
return true;
}
// nubBy :: (a > a > Bool) > [a] > [a]
const nubBy = (p, xs) => {
const x = xs.length ? xs[0] : undefined;
return x !== undefined ? [x].concat(
nubBy(p, xs.slice(1)
.filter(y => !p(x, y)))
) : [];
}
// PROBLEM DECLARATION
const floors = range(1, 5);
return concatMap(b =>
concatMap(c =>
concatMap(f =>
concatMap(m =>
concatMap(s =>
and([ // CONDITIONS
nubBy((a, b) => a === b, [b, c, f, m, s]) // all floors singly occupied
.length === 5,
b !== 5, c !== 1, f !== 1, f !== 5,
m > c, Math.abs(s  f) > 1, Math.abs(c  f) > 1
]) ? [{
Baker: b,
Cooper: c,
Fletcher: f,
Miller: m,
Smith: s
}] : [],
floors), floors), floors), floors), floors);
// > [{"Baker":3, "Cooper":2, "Fletcher":4, "Miller":5, "Smith":1}]
})();
 Output:
[{"Baker":3, "Cooper":2, "Fletcher":4, "Miller":5, "Smith":1}]
Less flexibility[edit]
For a different tradeoff between efficiency and generality, we can take full occupancy and no cohabitation out of the predicate, and assume them in the shape of the search space.
In the version above, with nested applications of concatMap, the requirement that all apartments are occupied by one person only is included in the test conditions. Alternatively, we can remove any flexibility about such civic virtues from the predicate, and restrict the universe of conceivable living arrangements, by using concatMap just once, and applying it only to the various permutations of full and distinct occupancy.
ES6 splat assignment allows us to bind all five names in a single application of concatMap. We now also need a permutations function of some kind.
(() => {
'use strict';
// concatMap :: (a > [b]) > [a] > [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
// range :: Int > Int > [Int]
const range = (m, n) =>
Array.from({
length: Math.floor(n  m) + 1
}, (_, i) => m + i);
// and :: [Bool] > Bool
const and = xs => {
let i = xs.length;
while (i)
if (!xs[i]) return false;
return true;
}
// permutations :: [a] > [[a]]
const permutations = xs =>
xs.length ? concatMap(x => concatMap(ys => [
[x].concat(ys)
],
permutations(delete_(x, xs))), xs) : [
[]
];
// delete :: a > [a] > [a]
const delete_ = (x, xs) =>
deleteBy((a, b) => a === b, x, xs);
// deleteBy :: (a > a > Bool) > a > [a] > [a]
const deleteBy = (f, x, xs) =>
xs.reduce((a, y) => f(x, y) ? a : a.concat(y), []);
// PROBLEM DECLARATION
const floors = range(1, 5);
return concatMap(([c, b, f, m, s]) =>
and([ // CONDITIONS (assuming full occupancy, no cohabitation)
b !== 5, c !== 1, f !== 1, f !== 5,
m > c, Math.abs(s  f) > 1, Math.abs(c  f) > 1
]) ? [{
Baker: b,
Cooper: c,
Fletcher: f,
Miller: m,
Smith: s
}] : [], permutations(floors));
// > [{"Baker":3, "Cooper":2, "Fletcher":4, "Miller":5, "Smith":1}]
})();
[{"Baker":3, "Cooper":2, "Fletcher":4, "Miller":5, "Smith":1}]
jq[edit]
Since we are told that "Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors", we can represent the apartment house as a JSON array, the first element of which names the occupant of the 1st floor, etc.
The solution presented here does not blindly generate all permutations. It can be characterized as a constraintoriented approach.
# Input: an array representing the apartment house, with null at a
# particular position signifying that the identity of the occupant
# there has not yet been determined.
# Output: an elaboration of the input array but including person, and
# satisfying cond, where . in cond refers to the placement of person
def resides(person; cond):
range(0;5) as $n
 if (.[$n] == null or .[$n] == person) and ($ncond) then .[$n] = person
else empty # no elaboration is possible
end ;
# English:
def top: 4;
def bottom: 0;
def higher(j): . > j;
def adjacent(j): (.  j)  (. == 1 or . == 1);
Solution:
[]
 resides("Baker"; . != top) # Baker does not live on the top floor
 resides("Cooper"; . != bottom) # Cooper does not live on the bottom floor
 resides("Fletcher"; . != top and . != bottom) # Fletcher does not live on either the top or the bottom floor.
 index("Cooper") as $Cooper
 resides("Miller"; higher( $Cooper) ) # Miller lives on a higher floor than does Cooper
 index("Fletcher") as $Fletcher
 resides("Smith"; adjacent($Fletcher)  not) # Smith does not live on a floor adjacent to Fletcher's.
 select( $Fletcher  adjacent( $Cooper )  not ) # Fletcher does not live on a floor adjacent to Cooper's.
Out:
$ jq n f Dinesman.jq
[
"Smith",
"Cooper",
"Baker",
"Fletcher",
"Miller"
]
Julia[edit]
using Combinatorics
function solve(n::Vector{<:AbstractString}, pred::Vector{<:Function})
rst = Vector{typeof(n)}(0)
for candidate in permutations(n)
if all(p(candidate) for p in predicates)
push!(rst, candidate)
end
end
return rst
end
Names = ["Baker", "Cooper", "Fletcher", "Miller", "Smith"]
predicates = [
(s) > last(s) != "Baker",
(s) > first(s) != "Cooper",
(s) > first(s) != "Fletcher" && last(s) != "Fletcher",
(s) > findfirst(s, "Miller") > findfirst(s, "Cooper"),
(s) > abs(findfirst(s, "Smith")  findfirst(s, "Fletcher")) != 1,
(s) > abs(findfirst(s, "Cooper")  findfirst(s, "Fletcher")) != 1]
solutions = solve(Names, predicates)
foreach(x > println(join(x, ", ")), solutions)
 Output:
Smith, Cooper, Baker, Fletcher, Miller
Kotlin[edit]
// version 1.1.3
typealias Predicate = (List<String>) > Boolean
fun <T> permute(input: List<T>): List<List<T>> {
if (input.size == 1) return listOf(input)
val perms = mutableListOf<List<T>>()
val toInsert = input[0]
for (perm in permute(input.drop(1))) {
for (i in 0..perm.size) {
val newPerm = perm.toMutableList()
newPerm.add(i, toInsert)
perms.add(newPerm)
}
}
return perms
}
/* looks for for all possible solutions, not just the first */
fun dinesman(occupants: List<String>, predicates: List<Predicate>) =
permute(occupants).filter { perm > predicates.all { pred > pred(perm) } }
fun main(args: Array<String>) {
val occupants = listOf("Baker", "Cooper", "Fletcher", "Miller", "Smith")
val predicates = listOf<Predicate>(
{ it.last() != "Baker" },
{ it.first() != "Cooper" },
{ it.last() != "Fletcher" && it.first() != "Fletcher" },
{ it.indexOf("Miller") > it.indexOf("Cooper") },
{ Math.abs(it.indexOf("Smith")  it.indexOf("Fletcher")) > 1 },
{ Math.abs(it.indexOf("Fletcher")  it.indexOf("Cooper")) > 1 }
)
val solutions = dinesman(occupants, predicates)
val size = solutions.size
if (size == 0) {
println("No solutions found")
}
else {
val plural = if (size == 1) "" else "s"
println("$size solution$plural found, namely:\n")
for (solution in solutions) {
for ((i, name) in solution.withIndex()) {
println("Floor ${i + 1} > $name")
}
println()
}
}
}
 Output:
1 solution found, namely: Floor 1 > Smith Floor 2 > Cooper Floor 3 > Baker Floor 4 > Fletcher Floor 5 > Miller
Lua[edit]
local wrap, yield = coroutine.wrap, coroutine.yield
local function perm(n)
local r = {}
for i=1,n do r[i]=i end
return wrap(function()
local function swap(m)
if m==0 then
yield(r)
else
for i=m,1,1 do
r[i],r[m]=r[m],r[i]
swap(m1)
r[i],r[m]=r[m],r[i]
end
end
end
swap(n)
end)
end
local function iden(...)return ... end
local function imap(t,f)
local r,fn = {m=imap, c=table.concat, u=table.unpack}, f or iden
for i=1,#t do r[i]=fn(t[i])end
return r
end
local tenants = {'Baker', 'Cooper', 'Fletcher', 'Miller', 'Smith'}
local conds = {
'Baker ~= TOP',
'Cooper ~= BOTTOM',
'Fletcher ~= TOP and Fletcher~= BOTTOM',
'Miller > Cooper',
'Smith + 1 ~= Fletcher and Smith  1 ~= Fletcher',
'Cooper + 1 ~= Fletcher and Cooper  1 ~= Fletcher',
}
local function makePredicate(conds, tenants)
return load('return function('..imap(tenants):c','..
') return ' ..
imap(conds,function(c)
return string.format("(%s)",c)
end):c"and "..
" end ",'',nil,{TOP=5, BOTTOM=1})()
end
local function solve (conds, tenants)
local try, pred, upk = perm(#tenants), makePredicate(conds, tenants), table.unpack
local answer = try()
while answer and not pred(upk(answer)) do answer = try()end
if answer then
local floor = 0
return imap(answer, function(person)
floor=floor+1;
return string.format(" %s lives on floor %d",tenants[floor],person)
end):c"\n"
else
return nil, 'no solution'
end
end
print(solve (conds, tenants))
 Output:
Baker lives on floor 3 Cooper lives on floor 2 Fletcher lives on floor 4 Miller lives on floor 5 Smith lives on floor 1
Mathematica / Wolfram Language[edit]
Loads all names into memory as variables, then asserts various restrictions on them before trying to resolve them by assuming that they're integers. This works by assuming that the names are the floors the people are on. This method is slow but direct.
{Baker, Cooper, Fletcher, Miller, Smith};
(Unequal @@ %) && (And @@ (0 < # < 6 & /@ %)) &&
Baker < 5 &&
Cooper > 1 &&
1 < Fletcher < 5 &&
Miller > Cooper &&
Abs[Smith  Fletcher] > 1 &&
Abs[Cooper  Fletcher] > 1 //
Reduce[#, %, Integers] &
 Output:
Baker == 3 && Cooper == 2 && Fletcher == 4 && Miller == 5 && Smith == 1
Alternate Version[edit]
A much quicker and traditional method. This generates all permutations of a list containing the five names as strings. The list of permutations is then filtered using the restrictions given in the problem until only one permutation is left.
p = Position[#1, #2][[1, 1]] &;
Permutations[{"Baker", "Cooper", "Fletcher", "Miller", "Smith"}, {5}];
Select[%, #[[5]] != "Baker" &];
Select[%, #[[1]] != "Cooper" &];
Select[%, #[[1]] != "Fletcher" && #[[5]] != "Fletcher" &];
Select[%, #~p~"Miller" > #~p~"Cooper" &];
Select[%, Abs[#~p~"Smith"  #~p~"Fletcher"] > 1 &];
Select[%, Abs[#~p~"Cooper"  #~p~"Fletcher"] > 1 &]
 Output:
{{"Smith", "Cooper", "Baker", "Fletcher", "Miller"}}
Perl[edit]
A solution that parses a structured version of the problem text, translates it into a Perl expression, and uses it for a bruteforce search:
Setup
use strict;
use warnings;
use feature qw(state say);
use List::Util 1.33 qw(pairmap);
use Algorithm::Permute qw(permute);
our %predicates = (
#  object  sprintf format for Perl expression 
# +++
'on bottom' => [ '' , '$f[%s] == 1' ],
'on top' => [ '' , '$f[%s] == @f' ],
'lower than' => [ 'person' , '$f[%s] < $f[%s]' ],
'higher than' => [ 'person' , '$f[%s] > $f[%s]' ],
'directly below' => [ 'person' , '$f[%s] == $f[%s]  1' ],
'directly above' => [ 'person' , '$f[%s] == $f[%s] + 1' ],
'adjacent to' => [ 'person' , 'abs($f[%s]  $f[%s]) == 1' ],
'on' => [ 'ordinal' , '$f[%s] == \'%s\'' ],
);
our %nouns = (
'person' => qr/[az]+/i,
'ordinal' => qr/1st  2nd  3rd  \d+th/x,
);
sub parse_and_solve {
my @facts = @_;
state $parser = qr/^(?<subj>$nouns{person}) (?<not>not )?(?@{[
join '', pairmap {
"(?<pred>$a)" .
($b>[0] ? " (?<obj>$nouns{$b>[0]})" : '')
} %predicates
]})$/;
my (@expressions, %ids, $i);
my $id = sub { defined $_[0] ? $ids{$_[0]} //= $i++ : () };
foreach (@facts) {
/$parser/ or die "Cannot parse '$_'\n";
my $pred = $predicates{$+{pred}};
my $expr = '(' . sprintf($pred>[1], $id>($+{subj}),
$pred>[0] eq 'person' ? $id>($+{obj}) : $+{obj}). ')';
$expr = '!' . $expr if $+{not};
push @expressions, $expr;
}
my @f = 1..$i;
eval 'no warnings "numeric";
permute {
say join(", ", pairmap { "$f[$b] $a" } %ids)
if ('.join(' && ', @expressions).');
} @f;';
}
Note that it can easily be extended by modifying the %predicates
and %nouns
hashes at the top.
Problem statement
Since trying to extract information from freeform text feels a little too flaky, the problem statement is instead expected as structured text with one fact per line, each of them having one of these two forms:

<name> <position>

<name> not <position>
...where <position>
can be any of:

on bottom

on top

lower than <name>

higher than <name>

directly below <name>

directly above <name>

adjacent to <name>

on <numeral>
(e.g. 1st, 2nd, etc.)
It is assumed that there are as many floors as there are different names.
Thus, the problem statement from the task description translates to:
parse_and_solve(<DATA>);
__DATA__
Baker not on top
Cooper not on bottom
Fletcher not on top
Fletcher not on bottom
Miller higher than Cooper
Smith not adjacent to Fletcher
Fletcher not adjacent to Cooper
 Output:
2 Cooper, 5 Miller, 3 Baker, 1 Smith, 4 Fletcher
When there are multiple matching configurations, it lists them all (on separate lines).
Perl 6[edit]
By parsing the problem[edit]
use MONKEYSEENOEVAL;
sub parse_and_solve ($text) {
my %ids;
my $expr = (grammar {
state $c = 0;
rule TOP { <fact>+ { make join ' && ', $<fact>>>.made } }
rule fact { <name> (not)? <position>
{ make sprintf $<position>.made.fmt($0 ?? "!(%s)" !! "%s"),
$<name>.made }
}
rule position {
 on bottom { make "\@f[%s] == 1" }
 on top { make "\@f[%s] == +\@f" }
 lower than <name> { make "\@f[%s] < \@f[{$<name>.made}]" }
 higher than <name> { make "\@f[%s] > \@f[{$<name>.made}]" }
 directly below <name> { make "\@f[%s] == \@f[{$<name>.made}]  1" }
 directly above <name> { make "\@f[%s] == \@f[{$<name>.made}] + 1" }
 adjacent to <name> { make "\@f[%s] == \@f[{$<name>.made}] + (11)" }
 on <ordinal> { make "\@f[%s] == {$<ordinal>.made}" }
 { note "Failed to parse line " ~ +$/.prematch.comb(/^^/); exit 1; }
}
token name { :i <[a..z]>+ { make %ids{~$/} //= $c++ } }
token ordinal { [1st  2nd  3rd  \d+th] { make +$/.match(/(\d+)/)[0] } }
}).parse($text).made;
EVAL 'for [1..%ids.elems].permutations > @f {
say %ids.kv.map({ "$^[email protected][$^b]" }) if (' ~ $expr ~ ');
}'
}
parse_and_solve Q:to/END/;
Baker not on top
Cooper not on bottom
Fletcher not on top
Fletcher not on bottom
Miller higher than Cooper
Smith not adjacent to Fletcher
Fletcher not adjacent to Cooper
END
Supports the same grammar for the problem statement, as the Perl solution.
 Output:
Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1
Simple solution[edit]
# Contains only five floors. 5! = 120 permutations.
for (flat (1..5).permutations) > $b, $c, $f, $m, $s {
say "Baker=$b Cooper=$c Fletcher=$f Miller=$m Smith=$s"
if $b != 5 # Baker !live on top floor.
and $c != 1 # Cooper !live on bottom floor.
and $f != 15 # Fletcher !live on top or the bottom floor.
and $m > $c # Miller lives on a higher floor than Cooper.
and $s != $f1$f+1 # Smith !live adjacent to Fletcher
and $f != $c1$c+1 # Fletcher !live adjacent to Cooper
;
}
Adding more people and floors requires changing the list that's being used for the permutations, adding a variable for the new person, a piece of output in the string and finally to adjust all mentions of the "top" floor. Adjusting to different rules requires changing the multiline if statement in the loop.
 Output:
Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1
Phix[edit]
Simple static/hardcoded solution (brute force search)
enum Baker, Cooper, Fletcher, Miller, Smith
constant names={"Baker","Cooper","Fletcher","Miller","Smith"}
procedure test(sequence flats)
if flats[Baker]!=5
and flats[Cooper]!=1
and not find(flats[Fletcher],{1,5})
and flats[Miller]>flats[Cooper]
and abs(flats[Smith]flats[Fletcher])!=1
and abs(flats[Fletcher]flats[Cooper])!=1 then
for i=1 to 5 do
?{names[i],flats[i]}
end for
end if
end procedure
for i=1 to factorial(5) do
test(permute(i,tagset(5)))
end for
 Output:
{"Baker",3} {"Cooper",2} {"Fletcher",4} {"Miller",5} {"Smith",1}
Something more flexible. The nested rules worked just as well, and of course the code will cope with various content in names/rules.
sequence names = {"Baker","Cooper","Fletcher","Miller","Smith"},
rules = {{"!=","Baker",length(names)},
{"!=","Cooper",1},
{"!=","Fletcher",1},
{"!=","Fletcher",length(names)},
{">","Miller","Cooper"},
 {"!=",{"abs","Smith","Fletcher"},1},
{"nadj","Smith","Fletcher"},
 {"!=",{"abs","Fletcher","Cooper"},1},
{"nadj","Fletcher","Cooper"}}
function eval(sequence rule, sequence flats)
{string operand, object op1, object op2} = rule
if string(op1) then
op1 = flats[find(op1,names)]
 elsif sequence(op1) then
 op1 = eval(op1,flats)
end if
if string(op2) then
op2 = flats[find(op2,names)]
 elsif sequence(op2) then
 op2 = eval(op2,flats)
end if
switch operand do
case "!=": return op1!=op2
case ">": return op1>op2
 case "abs": return abs(op1op2)
case "nadj": return abs(op1op2)!=1
end switch
return 9/0
end function
procedure test(sequence flats)
for i=1 to length(rules) do
if not eval(rules[i],flats) then return end if
end for
for i=1 to length(names) do
?{names[i],flats[i]}
end for
end procedure
for i=1 to factorial(length(names)) do
test(permute(i,tagset(length(names))))
end for
Same output
PicoLisp[edit]
Using Pilog (PicoLisp Prolog). The problem can be modified by changing just the 'dwelling' rule (the "Problem statement"). This might involve the names and number of dwellers (the list in the first line), and statements about who does (or does not) live on the top floor (using the 'topFloor' predicate), the bottom floor (using the 'bottomFloor' predicate), on a higher floor (using the 'higherFloor' predicate) or on an adjacent floor (using the 'adjacentFloor' predicate). The logic follows an implied AND, and statements may be arbitrarily combined using OR and NOT (using the 'or' and 'not' predicates), or any other Pilog (Prolog in picoLisp) builtin predicates. If the problem statement has several solutions, they will be all generated.
# Problem statement
(be dwelling (@Tenants)
(permute (Baker Cooper Fletcher Miller Smith) @Tenants)
(not (topFloor Baker @Tenants))
(not (bottomFloor Cooper @Tenants))
(not (or ((topFloor Fletcher @Tenants)) ((bottomFloor Fletcher @Tenants))))
(higherFloor Miller Cooper @Tenants)
(not (adjacentFloor Smith Fletcher @Tenants))
(not (adjacentFloor Fletcher Cooper @Tenants)) )
# Utility rules
(be topFloor (@Tenant @Lst)
(equal (@ @ @ @ @Tenant) @Lst) )
(be bottomFloor (@Tenant @Lst)
(equal (@Tenant @ @ @ @) @Lst) )
(be higherFloor (@Tenant1 @Tenant2 @Lst)
(append @ @Rest @Lst)
(equal (@Tenant2 . @Higher) @Rest)
(member @Tenant1 @Higher) )
(be adjacentFloor (@Tenant1 @Tenant2 @Lst)
(append @ @Rest @Lst)
(or
((equal (@Tenant1 @Tenant2 . @) @Rest))
((equal (@Tenant2 @Tenant1 . @) @Rest)) ) )
 Output:
: (? (dwelling @Result)) @Result=(Smith Cooper Baker Fletcher Miller) # Only one solution > NIL
PowerShell[edit]
# Floors are numbered 1 (ground) to 5 (top)
# Baker, Cooper, Fletcher, Miller, and Smith live on different floors:
$statement1 = '$baker ne $cooper and $baker ne $fletcher and $baker ne $miller and
$baker ne $smith and $cooper ne $fletcher and $cooper ne $miller and
$cooper ne $smith and $fletcher ne $miller and $fletcher ne $smith and
$miller ne $smith'
# Baker does not live on the top floor:
$statement2 = '$baker ne 5'
# Cooper does not live on the bottom floor:
$statement3 = '$cooper ne 1'
# Fletcher does not live on either the top or the bottom floor:
$statement4 = '$fletcher ne 1 and $fletcher ne 5'
# Miller lives on a higher floor than does Cooper:
$statement5 = '$miller gt $cooper'
# Smith does not live on a floor adjacent to Fletcher's:
$statement6 = '[Math]::Abs($smith  $fletcher) ne 1'
# Fletcher does not live on a floor adjacent to Cooper's:
$statement7 = '[Math]::Abs($fletcher  $cooper) ne 1'
for ($baker = 1; $baker lt 6; $baker++)
{
for ($cooper = 1; $cooper lt 6; $cooper++)
{
for ($fletcher = 1; $fletcher lt 6; $fletcher++)
{
for ($miller = 1; $miller lt 6; $miller++)
{
for ($smith = 1; $smith lt 6; $smith++)
{
if (InvokeExpression $statement2)
{
if (InvokeExpression $statement3)
{
if (InvokeExpression $statement5)
{
if (InvokeExpression $statement4)
{
if (InvokeExpression $statement6)
{
if (InvokeExpression $statement7)
{
if (InvokeExpression $statement1)
{
$multipleDwellings = @()
$multipleDwellings+= [PSCustomObject]@{Name = "Baker" ; Floor = $baker}
$multipleDwellings+= [PSCustomObject]@{Name = "Cooper" ; Floor = $cooper}
$multipleDwellings+= [PSCustomObject]@{Name = "Fletcher"; Floor = $fletcher}
$multipleDwellings+= [PSCustomObject]@{Name = "Miller" ; Floor = $miller}
$multipleDwellings+= [PSCustomObject]@{Name = "Smith" ; Floor = $smith}
}
}
}
}
}
}
}
}
}
}
}
}
The solution sorted by name:
$multipleDwellings
 Output:
Name Floor   Baker 3 Cooper 2 Fletcher 4 Miller 5 Smith 1
The solution sorted by floor:
$multipleDwellings  SortObject Property Floor Descending
 Output:
Name Floor   Miller 5 Fletcher 4 Baker 3 Cooper 2 Smith 1
Prolog[edit]
Using CLPFD[edit]
Works with SWIProlog and library(clpfd) written by Markus Triska.
: use_module(library(clpfd)).
: dynamic top/1, bottom/1.
% Baker does not live on the top floor
rule1(L) :
member((baker, F), L),
top(Top),
F #\= Top.
% Cooper does not live on the bottom floor.
rule2(L) :
member((cooper, F), L),
bottom(Bottom),
F #\= Bottom.
% Fletcher does not live on either the top or the bottom floor.
rule3(L) :
member((fletcher, F), L),
top(Top),
bottom(Bottom),
F #\= Top,
F #\= Bottom.
% Miller lives on a higher floor than does Cooper.
rule4(L) :
member((miller, Fm), L),
member((cooper, Fc), L),
Fm #> Fc.
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L) :
member((smith, Fs), L),
member((fletcher, Ff), L),
abs(FsFf) #> 1.
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L) :
member((cooper, Fc), L),
member((fletcher, Ff), L),
abs(FcFf) #> 1.
init(L) :
% we need to define top and bottom
assert(bottom(1)),
length(L, Top),
assert(top(Top)),
% we say that they are all in differents floors
bagof(F, X^member((X, F), L), LF),
LF ins 1..Top,
all_different(LF),
% Baker does not live on the top floor
rule1(L),
% Cooper does not live on the bottom floor.
rule2(L),
% Fletcher does not live on either the top or the bottom floor.
rule3(L),
% Miller lives on a higher floor than does Cooper.
rule4(L),
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L),
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L).
solve(L) :
bagof(F, X^member((X, F), L), LF),
label(LF).
dinners :
retractall(top(_)), retractall(bottom(_)),
L = [(baker, _Fb), (cooper, _Fc), (fletcher, _Ff), (miller, _Fm), (smith, _Fs)],
init(L),
solve(L),
maplist(writeln, L).
 Output:
? dinners. baker,3 cooper,2 fletcher,4 miller,5 smith,1 true ; false.
true ==> predicate succeeded.
false ==> no other solution.
About flexibility : each name is associated with a floor, (contiguous floors differs from 1).
Bottom is always 1 but Top is defined from the number of names.
Each statement of the problem is translated in a Prolog rule, (a constraint on the floors), we can add so much of rules that we want, and a modification of one statement only modified one rule.
To solve the problem, library clpfd does the job.
Plain Prolog version[edit]
select([AAs],S): select(A,S,S1),select(As,S1).
select([],_).
dinesmans(X) :
%% Baker, Cooper, Fletcher, Miller, and Smith live on different floors
%% of an apartment house that contains only five floors.
select([Baker,Cooper,Fletcher,Miller,Smith],[1,2,3,4,5]),
%% Baker does not live on the top floor.
Baker =\= 5,
%% Cooper does not live on the bottom floor.
Cooper =\= 1,
%% Fletcher does not live on either the top or the bottom floor.
Fletcher =\= 1, Fletcher =\= 5,
%% Miller lives on a higher floor than does Cooper.
Miller > Cooper,
%% Smith does not live on a floor adjacent to Fletcher's.
1 =\= abs(Smith  Fletcher),
%% Fletcher does not live on a floor adjacent to Cooper's.
1 =\= abs(Fletcher  Cooper),
%% Where does everyone live?
X = ['Baker'(Baker), 'Cooper'(Cooper), 'Fletcher'(Fletcher),
'Miller'(Miller), 'Smith'(Smith)].
main : bagof( X, dinesmans(X), L )
> maplist( writeln, L), nl, write('No more solutions.')
; write('No solutions.').
Ease of change (flexibility) is arguably evident in the code. The output:
[Baker(3), Cooper(2), Fletcher(4), Miller(5), Smith(1)] No more solutions.
Testing as soon as possible[edit]
dinesmans(X) :
%% 1. Baker, Cooper, Fletcher, Miller, and Smith live on different floors
%% of an apartment house that contains only five floors.
Domain = [1,2,3,4,5],
%% 2. Baker does not live on the top floor.
select(Baker,Domain,D1), Baker =\= 5,
%% 3. Cooper does not live on the bottom floor.
select(Cooper,D1,D2), Cooper =\= 1,
%% 4. Fletcher does not live on either the top or the bottom floor.
select(Fletcher,D2,D3), Fletcher =\= 1, Fletcher =\= 5,
%% 5. Miller lives on a higher floor than does Cooper.
select(Miller,D3,D4), Miller > Cooper,
%% 6. Smith does not live on a floor adjacent to Fletcher's.
select(Smith,D4,_), 1 =\= abs(Smith  Fletcher),
%% 7. Fletcher does not live on a floor adjacent to Cooper's.
1 =\= abs(Fletcher  Cooper),
%% Where does everyone live?
X = ['Baker'(Baker), 'Cooper'(Cooper), 'Fletcher'(Fletcher),
'Miller'(Miller), 'Smith'(Smith)].
Running it produces the same output, but more efficiently. Separate testing in SWI shows 1,328 inferences for the former, 379 inferences for the latter version. Moving rule 7. up below rule 4. brings it down to 295 inferences.
PureBasic[edit]
Prototype cond(Array t(1))
Enumeration #Null
#Baker
#Cooper
#Fletcher
#Miller
#Smith
EndEnumeration
Procedure checkTenands(Array tenants(1), Array Condions.cond(1))
Protected i, j
Protected.cond *f
j=ArraySize(Condions())
For i=0 To j
*f=Condions(i) ; load the function pointer to the current condition
If *f(tenants()) = #False
ProcedureReturn #False
EndIf
Next
ProcedureReturn #True
EndProcedure
Procedure C1(Array t(1))
If Int(Abs(t(#Fletcher)t(#Cooper)))<>1
ProcedureReturn #True
EndIf
EndProcedure
Procedure C2(Array t(1))
If t(#Baker)<>5
ProcedureReturn #True
EndIf
EndProcedure
Procedure C3(Array t(1))
If t(#Cooper)<>1
ProcedureReturn #True
EndIf
EndProcedure
Procedure C4(Array t(1))
If t(#Miller) >= t(#Cooper)
ProcedureReturn #True
EndIf
EndProcedure
Procedure C5(Array t(1))
If t(#Fletcher)<>1 And t(#Fletcher)<>5
ProcedureReturn #True
EndIf
EndProcedure
Procedure C6(Array t(1))
If Int(Abs(t(#Smith)t(#Fletcher)))<>1
ProcedureReturn #True
EndIf
EndProcedure
If OpenConsole()
Dim People(4)
Dim Conditions(5)
Define a, b, c, d, e, i
;
; Load all conditions
Conditions(i)=@C1(): i+1
Conditions(i)=@C2(): i+1
Conditions(i)=@C3(): i+1
Conditions(i)=@C4(): i+1
Conditions(i)=@C5(): i+1
Conditions(i)=@C6()
;
; generate and the all legal combinations
For a=1 To 5
For b=1 To 5
If a=b: Continue: EndIf
For c=1 To 5
If a=c Or b=c: Continue: EndIf
For d=1 To 5
If d=a Or d=b Or d=c : Continue: EndIf
For e=1 To 5
If e=a Or e=b Or e=c Or e=d: Continue: EndIf
People(#Baker)=a
People(#Cooper)=b
People(#Fletcher)=c
People(#Miller)=d
People(#Smith)=e
If checkTenands(People(), Conditions())
PrintN("Solution found;")
PrintN("Baker="+Str(a)+#CRLF$+"Cooper="+Str(b)+#CRLF$+"Fletcher="+Str(c))
PrintN("Miller="+Str(d)+#CRLF$+"Smith="+Str(e)+#CRLF$)
EndIf
Next
Next
Next
Next
Next
Print("Press ENTER to exit"): Input()
EndIf
Solution found; Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1
Python[edit]
By parsing the problem statement[edit]
This example parses the statement of the problem as given and allows some variability such as the number of people, floors and constraints can be varied although the type of constraints allowed and the sentence structure is limited
 Setup
Parsing is done with the aid of the multiline regular expression at the head of the program.
import re
from itertools import product
problem_re = re.compile(r"""(?msx)(?:
# Multiple names of form n1, n2, n3, ... , and nK
(?P<namelist> [azAZ]+ (?: , \s+ [azAZ]+)* (?: ,? \s+ and) \s+ [azAZ]+ )
# Flexible floor count (2 to 10 floors)
 (?: .* house \s+ that \s+ contains \s+ only \s+
(?P<floorcount> twothreefourfivesixseveneightnineten ) \s+ floors \s* \.)
# Constraint: "does not live on the n'th floor"
(?: (?P<not_live> \b [azAZ]+ \s+ does \s+ not \s+ live \s+ on \s+ the \s+
(?: topbottomfirstsecondthirdfourthfifthsixthseventheighthninthtenth) \s+ floor \s* \. ))
# Constraint: "does not live on either the I'th or the J'th [ or the K'th ...] floor
(?P<not_either> \b [azAZ]+ \s+ does \s+ not \s+ live \s+ on \s+ either
(?: \s+ (?: or \s+)? the \s+
(?: topbottomfirstsecondthirdfourthfifthsixthseventheighthninthtenth))+ \s+ floor \s* \. )
# Constraint: "P1 lives on a higher/lower floor than P2 does"
(?P<hi_lower> \b [azAZ]+ \s+ lives \s+ on \s+ a \s (?: higherlower)
\s+ floor \s+ than (?: \s+ does) \s+ [azAZ]+ \s* \. )
# Constraint: "P1 does/does not live on a floor adjacent to P2's"
(?P<adjacency> \b [azAZ]+ \s+ does (?:\s+ not)? \s+ live \s+ on \s+ a \s+
floor \s+ adjacent \s+ to \s+ [azAZ]+ (?: 's )? \s* \. )
# Ask for the solution
(?P<question> Where \s+ does \s+ everyone \s+ live \s* \?)
)
""")
names, lennames = None, None
floors = None
constraint_expr = 'len(set(alloc)) == lennames' # Start with all people on different floors
def do_namelist(txt):
" E.g. 'Baker, Cooper, Fletcher, Miller, and Smith'"
global names, lennames
names = txt.replace(' and ', ' ').split(', ')
lennames = len(names)
def do_floorcount(txt):
" E.g. 'five'"
global floors
floors = 'twothreefourfivesixseveneightnineten'.split('').index(txt)
def do_not_live(txt):
" E.g. 'Baker does not live on the top floor.'"
global constraint_expr
t = txt.strip().split()
who, floor = t[0], t[2]
w, f = (names.index(who),
('firstsecondthirdfourthfifthsixth' +
'seventheighthninthtenthtopbottom').split('').index(floor)
)
if f == 11: f = floors
if f == 12: f = 1
constraint_expr += ' and alloc[%i] != %i' % (w, f)
def do_not_either(txt):
" E.g. 'Fletcher does not live on either the top or the bottom floor.'"
global constraint_expr
t = txt.replace(' or ', ' ').replace(' the ', ' ').strip().split()
who, floor = t[0], t[6:1]
w, fl = (names.index(who),
[('firstsecondthirdfourthfifthsixth' +
'seventheighthninthtenthtopbottom').split('').index(f)
for f in floor]
)
for f in fl:
if f == 11: f = floors
if f == 12: f = 1
constraint_expr += ' and alloc[%i] != %i' % (w, f)
def do_hi_lower(txt):
" E.g. 'Miller lives on a higher floor than does Cooper.'"
global constraint_expr
t = txt.replace('.', '').strip().split()
name_indices = [names.index(who) for who in (t[0], t[1])]
if 'lower' in t:
name_indices = name_indices[::1]
constraint_expr += ' and alloc[%i] > alloc[%i]' % tuple(name_indices)
def do_adjacency(txt):
''' E.g. "Smith does not live on a floor adjacent to Fletcher's."'''
global constraint_expr
t = txt.replace('.', '').replace("'s", '').strip().split()
name_indices = [names.index(who) for who in (t[0], t[1])]
constraint_expr += ' and abs(alloc[%i]  alloc[%i]) > 1' % tuple(name_indices)
def do_question(txt):
global constraint_expr, names, lennames
exec_txt = '''
for alloc in product(range(1,floors+1), repeat=len(names)):
if %s:
break
else:
alloc = None
''' % constraint_expr
exec(exec_txt, globals(), locals())
a = locals()['alloc']
if a:
output= ['Floors are numbered from 1 to %i inclusive.' % floors]
for a2n in zip(a, names):
output += [' Floor %i is occupied by %s' % a2n]
output.sort(reverse=True)
print('\n'.join(output))
else:
print('No solution found.')
print()
handler = {
'namelist': do_namelist,
'floorcount': do_floorcount,
'not_live': do_not_live,
'not_either': do_not_either,
'hi_lower': do_hi_lower,
'adjacency': do_adjacency,
'question': do_question,
}
def parse_and_solve(problem):
p = re.sub(r'\s+', ' ', problem).strip()
for x in problem_re.finditer(p):
groupname, txt = [(k,v) for k,v in x.groupdict().items() if v][0]
#print ("%r, %r" % (groupname, txt))
handler[groupname](txt)
 Problem statement
This is not much more than calling a function on the text of the problem!
if __name__ == '__main__':
parse_and_solve("""
Baker, Cooper, Fletcher, Miller, and Smith
live on different floors of an apartment house that contains
only five floors. Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?""")
print('# Add another person with more constraints and more floors:')
parse_and_solve("""
Baker, Cooper, Fletcher, Miller, Guinan, and Smith
live on different floors of an apartment house that contains
only seven floors. Guinan does not live on either the top or the third or the fourth floor.
Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?""")
 Output
This shows the output from the original problem and then for another, slightly different problem to cover some of the variability asked for in the task.
Floors are numbered from 1 to 5 inclusive. Floor 5 is occupied by Miller Floor 4 is occupied by Fletcher Floor 3 is occupied by Baker Floor 2 is occupied by Cooper Floor 1 is occupied by Smith # Add another person with more constraints and more floors: Floors are numbered from 1 to 7 inclusive. Floor 7 is occupied by Smith Floor 6 is occupied by Guinan Floor 4 is occupied by Fletcher Floor 3 is occupied by Miller Floor 2 is occupied by Cooper Floor 1 is occupied by Baker
By using the Amb operator[edit]
In this example, the problem needs to be turned into valid Python code for use with the Amb operator. Setup is just to import Amb.
The second set of results corresponds to this modification to the problem statement:
Baker, Cooper, Fletcher, Miller, Guinan, and Smith live on different floors of an apartment house that contains only seven floors. Guinan does not live on either the top or the third or the fourth floor. Baker does not live on the top floor. Cooper does not live on the bottom floor. Fletcher does not live on either the top or the bottom floor. Miller lives on a higher floor than does Cooper. Smith does not live on a floor adjacent to Fletcher's. Fletcher does not live on a floor adjacent to Cooper's. Where does everyone live
from amb import Amb
if __name__ == '__main__':
amb = Amb()
maxfloors = 5
floors = range(1, maxfloors+1)
# Possible floors for each person
Baker, Cooper, Fletcher, Miller, Smith = (amb(floors) for i in range(5))
for _dummy in amb( lambda Baker, Cooper, Fletcher, Miller, Smith: (
len(set([Baker, Cooper, Fletcher, Miller, Smith])) == 5 # each to a separate floor
and Baker != maxfloors
and Cooper != 1
and Fletcher not in (maxfloors, 1)
and Miller > Cooper
and (Smith  Fletcher) not in (1, 1) # Not adjacent
and (Fletcher  Cooper) not in (1, 1) # Not adjacent
) ):
print 'Floors are numbered from 1 to %i inclusive.' % maxfloors
print '\n'.join(sorted(' Floor %i is occupied by %s'
% (globals()[name], name)
for name in 'Baker, Cooper, Fletcher, Miller, Smith'.split(', ')))
break
else:
print 'No solution found.'
print '# Add another person with more constraints and more floors:'
# The order that Guinan is added to any list of people must stay consistant
amb = Amb()
maxfloors = 7
floors = range(1, maxfloors+1)
# Possible floors for each person
Baker, Cooper, Fletcher, Miller, Guinan, Smith = (amb(floors) for i in range(6))
for _dummy in amb( lambda Baker, Cooper, Fletcher, Miller, Guinan, Smith: (
len(set([Baker, Cooper, Fletcher, Miller, Guinan, Smith])) == 6 # each to a separate floor
and Guinan not in (maxfloors, 3, 4)
and Baker != maxfloors
and Cooper != 1
and Fletcher not in (maxfloors, 1)
and Miller > Cooper
and (Smith  Fletcher) not in (1, 1) # Not adjacent
and (Fletcher  Cooper) not in (1, 1) # Not adjacent
) ):
print 'Floors are numbered from 1 to %i inclusive.' % maxfloors
print '\n'.join(sorted(' Floor %i is occupied by %s'
% (globals()[name], name)
for name in 'Baker, Cooper, Fletcher, Miller, Guinan, Smith'.split(', ')))
break
else:
print 'No solution found.'
 Output:
Floors are numbered from 1 to 5 inclusive. Floor 1 is occupied by Smith Floor 2 is occupied by Cooper Floor 3 is occupied by Baker Floor 4 is occupied by Fletcher Floor 5 is occupied by Miller # Add another person with more constraints and more floors: Floors are numbered from 1 to 7 inclusive. Floor 1 is occupied by Baker Floor 2 is occupied by Cooper Floor 3 is occupied by Miller Floor 4 is occupied by Fletcher Floor 5 is occupied by Guinan Floor 6 is occupied by Smith
Simple Solution[edit]
from itertools import permutations
class Names:
Baker, Cooper, Fletcher, Miller, Smith = range(5)
seq = [Baker, Cooper, Fletcher, Miller, Smith]
strings = "Baker Cooper Fletcher Miller Smith".split()
predicates = [
lambda s: s[Names.Baker] != len(s)1,
lambda s: s[Names.Cooper] != 0,
lambda s: s[Names.Fletcher] != 0 and s[Names.Fletcher] != len(s)1,
lambda s: s[Names.Miller] > s[Names.Cooper],
lambda s: abs(s[Names.Smith]  s[Names.Fletcher]) != 1,
lambda s: abs(s[Names.Cooper]  s[Names.Fletcher]) != 1];
for sol in permutations(Names.seq):
if all(p(sol) for p in predicates):
print " ".join(Names.strings[s] for s in sol)
 Output:
Fletcher Cooper Miller Smith Baker
Racket[edit]
This is a direct translation of the problem constraints using an amb operator to make the choices (and therefore continuations to do the search). Since it's a direct translation, pretty much all aspects of the problem can change. Note that a direct translation was preferred even though it could be made to run much faster.
#lang racket
;; A quick `amb' implementation
(define fails '())
(define (fail) (if (pair? fails) ((car fails)) (error "no more choices!")))
(define (amb xs)
(let/cc k (set! fails (cons k fails)))
(if (pair? xs) (begin0 (car xs) (set! xs (cdr xs)))
(begin (set! fails (cdr fails)) (fail))))
(define (assert . conditions) (when (memq #f conditions) (fail)))
;; Convenient macro for definining problem items
(definesyntaxrule (with: all (name ...) #:in choices body ...)
(let* ([cs choices] [name (amb cs)] ... [all `([,name name] ...)]) body ...))
;; ===== problem translation starts here =====
;; Baker, Cooper, Fletcher, Miller, and Smith live on different floors
;; of an apartment house that contains only five floors.
(with: residents [Baker Cooper Fletcher Miller Smith] #:in (range 1 6)
;; Some helpers
(define (ontop x) (for/and ([y residents]) (x . >= . (car y))))
(define (onbottom x) (for/and ([y residents]) (x . <= . (car y))))
(define (adjacent x y) (= 1 (abs ( x y))))
(assert
;; ... live on different floors ...
(assert (= 5 (length (removeduplicates (map car residents)))))
;; Baker does not live on the top floor.
(not (ontop Baker))
;; Cooper does not live on the bottom floor.
(not (onbottom Cooper))
;; Fletcher does not live on either the top or the bottom floor.
(not (ontop Fletcher))
(not (onbottom Fletcher))
;; Miller lives on a higher floor than does Cooper.
(> Miller Cooper)
;; Smith does not live on a floor adjacent to Fletcher's.
(not (adjacent Smith Fletcher))
;; Fletcher does not live on a floor adjacent to Cooper's.
(assert (not (adjacent Fletcher Cooper))))
;; Where does everyone live?
(printf "Solution:\n")
(for ([x (sort residents > #:key car)]) (apply printf " ~a. ~a\n" x)))
 Output:
Solution: 5. Miller 4. Fletcher 3. Baker 2. Cooper 1. Smith
REXX[edit]
This REXX version tries to keep the rules as simple as possible, with easytoread if statements.
Names of the tenants can be easily listed, and the floors are numbered according to the American system,
that is, the ground floor is the 1^{st} floor, the next floor up is the 2^{nd} floor, etc.
The REXX program is broken up into several parts:
 preamble where names and floors are defined.
 iterating all possibilities (permutations would be faster, but with obtuse code).
 evaluation of the possibilities.
 elimination of cohabitation possibilities (tenants must live on separate floors).
 elimination of possibilities according to the rules.
 displaying the possible solution(s), if any.
 displaying the number of solutions found.
Note that the TH function has extra boilerplate to handle larger numbers.
With one more REXX statement, the tenants could be listed by the order of the floors they live on;
(currently, the tenants are listed in the order they are listed in the names variable).
The "rules" that contain == could be simplified to = for readability.
/*REXX program solves the Dinesman's multiple─dwelling problem with "natural" wording.*/
names= 'Baker Cooper Fletcher Miller Smith' /*names of multiple─dwelling tenants. */
tenants=words(names) /*the number of tenants in the building*/
floors=5; top=floors; bottom=1; #=floors; /*floor 1 is the ground (bottom) floor.*/
sols=0
do !.1=1 for #; do !.2=1 for #; do !.3=1 for #; do !.4=1 for #; do !.5=1 for #
do p=1 for tenants; _=word(names,p); upper _; call value _, !.p
end /*p*/
do j=1 for #1 /* [↓] people don't live on same floor*/
do k=j+1 to #; if !.j==!.k then iterate !.5 /*cohab?*/
end /*k*/
end /*j*/
call Waldo /* ◄══ where the rubber meets the road.*/
end; end; end; end; end /*!.5 & !.4 & !.3 & !.2 & !.1*/
say 'found' sols "solution"s(sols). /*display the number of solutions found*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
Waldo: if Baker == top then return
if Cooper == bottom then return
if Fletcher == bottom  Fletcher == top then return
if Miller \> Cooper then return
if Smith == Fletcher1  Smith == Fletcher+1 then return
if Fletcher == Cooper 1  Fletcher == Cooper +1 then return
sols=sols+1
say; do p=1 for tenants; tenant=right( word(names, p), 30)
say tenant 'lives on the' !.p  th(!.p) "floor."
end /*p*/
return /* [↑] show tenants in order in NAMES.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
s: if arg(1)=1 then return ''; return "s" /*a simple pluralizer function.*/
th: arg x; x=abs(x); return word('th st nd rd', 1 +x// 10* (x//100%10\==1)*(x//10<4))
output
Baker lives on the 3rd floor. Cooper lives on the 2nd floor. Fletcher lives on the 4th floor. Miller lives on the 5th floor. Smith lives on the 1st floor. found 1 solution.
Ring[edit]
floor1 = "return baker!=cooper and baker!=fletcher and baker!=miller and
baker!=smith and cooper!=fletcher and cooper!=miller and
cooper!=smith and fletcher!=miller and fletcher!=smith and
miller!=smith"
floor2 = "return baker!=4"
floor3 = "return cooper!=0"
floor4 = "return fletcher!=0 and fletcher!=4"
floor5 = "return miller>cooper"
floor6 = "return fabs(smithfletcher)!=1"
floor7 = "return fabs(fletchercooper)!=1"
for baker = 0 to 4
for cooper = 0 to 4
for fletcher = 0 to 4
for miller = 0 to 4
for smith = 0 to 4
if eval(floor2) if eval(floor3) if eval(floor5)
if eval(floor4) if eval(floor6) if eval(floor7)
if eval(floor1)
see "baker lives on floor " + baker + nl
see "cooper lives on floor " + cooper + nl
see "fletcher lives on floor " + fletcher + nl
see "miller lives on floor " + miller + nl
see "smith lives on floor " + smith + nl ok ok ok ok ok ok ok
next
next
next
next
next
Output:
baker lives on floor 2 cooper lives on floor 1 fletcher lives on floor 3 miller lives on floor 4 smith lives on floor 0
Ruby[edit]
By parsing the problem[edit]
Inspired by the Python version.
def solve( problem )
lines = problem.split(".")
names = lines.first.scan( /[AZ]\w*/ )
re_names = Regexp.union( names )
# Later on, search for these keywords (the word "not" is handled separately).
words = %w(first second third fourth fifth sixth seventh eighth ninth tenth
bottom top higher lower adjacent)
re_keywords = Regexp.union( words )
predicates = lines[1..2].flat_map do line #build an array of lambda's
keywords = line.scan( re_keywords )
name1, name2 = line.scan( re_names )
keywords.map do keyword
l = case keyword
when "bottom" then >(c){ c.first == name1 }
when "top" then >(c){ c.last == name1 }
when "higher" then >(c){ c.index( name1 ) > c.index( name2 ) }
when "lower" then >(c){ c.index( name1 ) < c.index( name2 ) }
when "adjacent" then >(c){ (c.index( name1 )  c.index( name2 )).abs == 1 }
else >(c){ c[words.index(keyword)] == name1 }
end
line =~ /\bnot\b/ ? >(c){not l.call(c) } : l # handle "not"
end
end
names.permutation.detect{candidate predicates.all?{predicate predicate.(candidate)}}
end
The program operates under these assumptions:
 Sentences end with a ".".
 Every capitalized word in the first sentence is a name, the rest is ignored.
 There are as many floors as there are names.
 The only relevant words beside the names are: first, second, third,.., tenth, bottom, top, higher, lower, adjacent,(and) not. The rest, including the last sentence, is ignored.
Program invocation:
#Direct positional words like top, bottom, first, second etc. can be combined; they refer to one name.
#The relative positional words higher, lower and adjacent can be combined; they need two names, not positions.
demo1 = "Abe Ben Charlie David. Abe not second top. not adjacent Ben Charlie.
David Abe adjacent. David adjacent Ben. Last line."
demo2 = "A B C D. A not adjacent D. not B adjacent higher C. C lower D. Last line"
problem1 = "Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that
contains only five floors. Baker does not live on the top floor. Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor. Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's. Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?"
# from the Python version:
problem2 = "Baker, Cooper, Fletcher, Miller, Guinan, and Smith
live on different floors of an apartment house that contains
only seven floors. Guinan does not live on either the top or the third or the fourth floor.
Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?"
[demo1, demo2, problem1, problem2].each{problem puts solve( problem ) ;puts }
 Output:
Ben David Abe Charlie B A C D Smith Cooper Baker Fletcher Miller Baker Cooper Miller Fletcher Guinan Smith
Simple solution[edit]
names = %i( Baker Cooper Fletcher Miller Smith )
predicates = [>(c){ :Baker != c.last },
>(c){ :Cooper != c.first },
>(c){ :Fletcher != c.first && :Fletcher != c.last },
>(c){ c.index(:Miller) > c.index(:Cooper) },
>(c){ (c.index(:Smith)  c.index(:Fletcher)).abs != 1 },
>(c){ (c.index(:Cooper)  c.index(:Fletcher)).abs != 1 }]
puts names.permutation.detect{candidate predicates.all?{predicate predicate.call(candidate)}}
 Output:
Smith Cooper Baker Fletcher Miller
Run BASIC[edit]
This program simply iterates by looking at each room available for each person. It then looks to see if it meets the requirements for each person by looking at the results of the iteration. It makes sure the room numbers add up to 15 which is the requirement of adding the floors in 1 + 2 + 3 + 4 + 5 = 15.
for baler = 1 to 4 ' can not be in room 5
for cooper = 2 to 5 ' can not be in room 1
for fletcher = 2 to 4 ' can not be in room 1 or 5
for miller = 1 to 5 ' can be in any room
for smith = 1 to 5 ' can be in any room
if baler <> cooper and fletcher <> miller and miller > cooper and abs(smith  fletcher) > 1 and abs(fletcher  cooper) > 1 then
if baler + cooper + fletcher + miller + smith = 15 then ' that is 1 + 2 + 3 + 4 + 5
rooms$ = baler;cooper;fletcher;miller;smith
print "baler: ";baler;" copper: ";cooper;" fletcher: ";fletcher;" miller: ";miller;" smith: ";smith
end
end if
end if
next smith
next miller
next fletcher
next cooper
next baler
print "Can't assign rooms" ' print this if it can not find a solution
baler: 3 copper: 2 fletcher: 4 miller: 5 smith: 1
Scala[edit]
import scala.math.abs
object Dinesman3 extends App {
val tenants = List("Baker", "Cooper2", "Fletcher4", "Miller", "Smith")
val (groundFloor, topFloor) = (1, tenants.size)
/** Rules with related tenants and restrictions*/
val exclusions =
List((suggestedFloor0: Map[String, Int]) => suggestedFloor0("Baker") != topFloor,
(suggestedFloor1: Map[String, Int]) => suggestedFloor1("Cooper2") != groundFloor,
(suggestedFloor2: Map[String, Int]) => !List(groundFloor, topFloor).contains(suggestedFloor2("Fletcher4")),
(suggestedFloor3: Map[String, Int]) => suggestedFloor3("Miller") > suggestedFloor3("Cooper2"),
(suggestedFloor4: Map[String, Int]) => abs(suggestedFloor4("Smith")  suggestedFloor4("Fletcher4")) != 1,
(suggestedFloor5: Map[String, Int]) => abs(suggestedFloor5("Fletcher4")  suggestedFloor5("Cooper2")) != 1)
tenants.permutations.map(_ zip (groundFloor to topFloor)).
filter(p => exclusions.forall(_(p.toMap))).toList match {
case Nil => println("No solution")
case xss => {
println(s"Solutions: ${xss.size}")
xss.foreach { l =>
println("possible solution:")
l.foreach(p => println(f"${p._1}%11s lives on floor number ${p._2}"))
}
}
}
}
 Output:
Solutions: 1 possible solution: Smith lives on floor number 1 Cooper2 lives on floor number 2 Baker lives on floor number 3 Fletcher4 lives on floor number 4 Miller lives on floor number 5
Extended task[edit]
We can extend this problem by adding a tenant resp. adding conditions:
import scala.math.abs
object Dinesman3 extends App {
val tenants = List("Baker", "Cooper2", "Fletcher4", "Miller", "Rollo5", "Smith")
val (groundFloor, topFloor) = (1, tenants.size)
/** Rules with related tenants and restrictions*/
val exclusions =
List((suggestedFloor0: Map[String, Int]) => suggestedFloor0("Baker") != topFloor,
(suggestedFloor1: Map[String, Int]) => suggestedFloor1("Cooper2") != groundFloor,
(suggestedFloor2: Map[String, Int]) => !List(groundFloor, topFloor).contains(suggestedFloor2("Fletcher4")),
(suggestedFloor3: Map[String, Int]) => suggestedFloor3("Miller") > suggestedFloor3("Cooper2"),
(suggestedFloor4: Map[String, Int]) => abs(suggestedFloor4("Smith")  suggestedFloor4("Fletcher4")) != 1,
(suggestedFloor5: Map[String, Int]) => abs(suggestedFloor5("Fletcher4")  suggestedFloor5("Cooper2")) != 1,
(suggestedFloor6: Map[String, Int]) => !List(3, 4, topFloor).contains(suggestedFloor6("Rollo5")),
(suggestedFloor7: Map[String, Int]) => suggestedFloor7("Rollo5") < suggestedFloor7("Smith"),
(suggestedFloor8: Map[String, Int]) => suggestedFloor8("Rollo5") > suggestedFloor8("Fletcher4"))
tenants.permutations.map(_ zip (groundFloor to topFloor)).
filter(p => exclusions.forall(_(p.toMap))).toList match {
case Nil => println("No solution")
case xss => {
println(s"Solutions: ${xss.size}")
xss.foreach { l =>
println("possible solution:")
l.foreach(p => println(f"${p._1}%11s lives on floor number ${p._2}"))
}
}
}
}
 Output:
Solutions: 1 possible solution: Baker lives on floor number 1 Cooper2 lives on floor number 2 Miller lives on floor number 3 Fletcher4 lives on floor number 4 Rollo5 lives on floor number 5 Smith lives on floor number 6
Enhanced Solution[edit]
Combine the rules with the person names and separated the original task with an extension.
import scala.math.abs
object Dinesman2 extends App {
val groundFloor = 1
abstract class Rule(val person: String) { val exclusion: Map[String, Int] => Boolean }
/** Rules with related tenants and restrictions*/
def rulesDef(topFloor: Int) = List(
new Rule("Baker") { val exclusion = (_: Map[String, Int])(person) != topFloor },
new Rule("Cooper2") { val exclusion = (_: Map[String, Int])(person) != groundFloor },
new Rule("Fletcher4") {
val exclusion = (suggestedFloor2: Map[String, Int]) => !List(groundFloor, topFloor).contains(suggestedFloor2(person))
}, new Rule("Miller") {
val exclusion = (suggestedFloor3: Map[String, Int]) => suggestedFloor3(person) > suggestedFloor3("Cooper2")
}, new Rule("Smith") {
val exclusion = (suggestedFloor4: Map[String, Int]) => abs(suggestedFloor4(person)  suggestedFloor4("Fletcher4")) != 1
}, new Rule("Fletcher4") {
val exclusion = (suggestedFloor5: Map[String, Int]) => abs(suggestedFloor5(person)  suggestedFloor5("Cooper2")) != 1
})
def extensionDef(topFloor: Int) = List(new Rule("Rollo5") {
val exclusion = (suggestedFloor6: Map[String, Int]) => !List(3, 4, topFloor).contains((suggestedFloor6: Map[String, Int])(person))
}, new Rule("Rollo5") {
val exclusion = (suggestedFloor7: Map[String, Int]) => suggestedFloor7(person) < suggestedFloor7("Smith")
}, new Rule("Rollo5") {
val exclusion = (suggestedFloor8: Map[String, Int]) => suggestedFloor8(person) > suggestedFloor8("Fletcher4")
})
def allRulesDef(topFloor: Int) = rulesDef(topFloor) ++ extensionDef(topFloor)
val tenants = allRulesDef(0).map(_.person).distinct // Pilot balloon to get # of tenants
val topFloor = tenants.size
val exclusions = allRulesDef(topFloor).map(_.exclusion)
tenants.permutations.map(_ zip (groundFloor to topFloor)).
filter(p => exclusions.forall(_(p.toMap))).toList match {
case Nil => println("No solution")
case xss => {
println(s"Solutions: ${xss.size}")
xss.foreach { l =>
println("possible solution:")
l.foreach(p => println(f"${p._1}%11s lives on floor number ${p._2}"))
}
}
}
}
Sidef[edit]
By parsing the problem[edit]
func dinesman(problem) {
var lines = problem.split('.')
var names = lines.first.scan(/\b[AZ]\w*/)
var re_names = Regex(names.join(''))
# Later on, search for these keywords (the word "not" is handled separately).
var words = %w(first second third fourth fifth sixth seventh eighth ninth tenth
bottom top higher lower adjacent)
var re_keywords = Regex(words.join(''))
# Build an array of lambda's
var predicates = lines.ft(1, lines.end1).map{ line
var keywords = line.scan(re_keywords)
var (name1, name2) = line.scan(re_names)...
keywords.map{ keyword
var l = do {
given(keyword) {
when ("bottom") { >(c) { c.first == name1 } }
when ("top") { >(c) { c.last == name1 } }
when ("higher") { >(c) { c.index(name1) > c.index(name2) } }
when ("lower") { >(c) { c.index(name1) < c.index(name2) } }
when ("adjacent") { >(c) { c.index(name1)  c.index(name2) > abs == 1 } }
default { >(c) { c[words.index(keyword)] == name1 } }
}
}
line ~~ /\bnot\b/ ? func(c) { l(c) > not } : l; # handle "not"
}
}.flat
names.permutations { *candidate
predicates.all { predicate predicate(candidate) } && return candidate
}
}
Function invocation:
var demo1 = "Abe Ben Charlie David. Abe not second top. not adjacent Ben Charlie.
David Abe adjacent. David adjacent Ben. Last line."
var demo2 = "A B C D. A not adjacent D. not B adjacent higher C. C lower D. Last line"
var problem1 = "Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that
contains only five floors. Baker does not live on the top floor. Cooper does not live on the bottom floor.
Fletcher does not live on either the top or the bottom floor. Miller lives on a higher floor than does Cooper.
Smith does not live on a floor adjacent to Fletcher's. Fletcher does not live on a floor adjacent to Cooper's.
Where does everyone live?"
var problem2 = "Baker, Cooper, Fletcher, Miller, Guinan, and Smith
live on different floors of an apartment house that contains
only seven floors. Guinan does not live on either the top or the third or the fourth floor.
Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?"
[demo1, demo2, problem1, problem2].each{problem say dinesman(problem).join("\n"); say '' }
 Output:
Ben David Abe Charlie B A C D Smith Cooper Baker Fletcher Miller Baker Cooper Miller Fletcher Guinan Smith
Simple solution[edit]
var names = %w(Baker Cooper Fletcher Miller Smith)
var predicates = [
>(c){ :Baker != c.last },
>(c){ :Cooper != c.first },
>(c){ (:Fletcher != c.first) && (:Fletcher != c.last) },
>(c){ c.index(:Miller) > c.index(:Cooper) },
>(c){ (c.index(:Smith)  c.index(:Fletcher)).abs != 1 },
>(c){ (c.index(:Cooper)  c.index(:Fletcher)).abs != 1 },
]
names.permutations { *candidate
if (predicates.all {predicate predicate(candidate) }) {
say candidate.join("\n")
break
}
}
 Output:
Smith Cooper Baker Fletcher Miller
Tcl[edit]
It's trivial to extend this problem to deal with more floors and people and more constraints; the main internallygenerated constraint is that the names of people should begin with an upper case character so that they are distinct from internal variables. This code also relies on the caller encoding the conditions as expressions that produce a value that is/can be interpreted as a boolean.
package require Tcl 8.5
package require struct::list
proc dinesmanSolve {floors people constraints} {
# Search for a possible assignment that satisfies the constraints
struct::list foreachperm p $floors {
lassign $p {*}$people
set found 1
foreach c $constraints {
if {![expr $c]} {
set found 0
break
}
}
if {$found} break
}
# Found something, or exhausted possibilities
if {!$found} {
error "no solution possible"
}
# Generate in "nice" order
foreach f $floors {
foreach person $people {
if {[set $person] == $f} {
lappend result $f $person
break
}
}
}
return $result
}
Solve the particular problem:
set soln [dinesmanSolve {1 2 3 4 5} {Baker Cooper Fletcher Miller Smith} {
{$Baker != 5}
{$Cooper != 1}
{$Fletcher != 1 && $Fletcher != 5}
{$Miller > $Cooper}
{abs($Smith$Fletcher) != 1}
{abs($Fletcher$Cooper) != 1}
}]
puts "Solution found:"
foreach {where who} $soln {puts " Floor ${where}: $who"}
 Output:
Solution found: Floor 1: Smith Floor 2: Cooper Floor 3: Baker Floor 4: Fletcher Floor 5: Miller
uBasic/4tH[edit]
REM Floors are numbered 0 (ground) to 4 (top)
FOR B = 0 TO 4
FOR C = 0 TO 4
FOR F = 0 TO 4
FOR M = 0 TO 4
FOR S = 0 TO 4
GOSUB 100 : IF POP() THEN
GOSUB 110 : IF POP() THEN
GOSUB 120 : IF POP() THEN
GOSUB 130 : IF POP() THEN
GOSUB 140 : IF POP() THEN
GOSUB 150 : IF POP() THEN
GOSUB 160 : IF POP() THEN
PRINT "Baker lives on floor " ; B + 1
PRINT "Cooper lives on floor " ; C + 1
PRINT "Fletcher lives on floor " ; F + 1
PRINT "Miller lives on floor " ; M + 1
PRINT "Smith lives on floor " ; S + 1
ENDIF
ENDIF
ENDIF
ENDIF
ENDIF
ENDIF
ENDIF
NEXT S
NEXT M
NEXT F
NEXT C
NEXT B
END
REM "Baker, Cooper, Fletcher, Miller, and Smith live on different floors"
100 PUSH (B#C)*(B#F)*(B#M)*(B#S)*(C#F)*(C#M)*(C#S)*(F#M)*(F#S)*(M#S)
RETURN
REM "Baker does not live on the top floor"
110 PUSH B#4
RETURN
REM "Cooper does not live on the bottom floor"
120 PUSH C#0
RETURN
REM "Fletcher does not live on either the top or the bottom floor"
130 PUSH (F#0)*(F#4)
RETURN
REM "Miller lives on a higher floor than does Cooper"
140 PUSH M>C
RETURN
REM "Smith does not live on a floor adjacent to Fletcher's"
150 PUSH ABS(SF)#1
RETURN
REM "Fletcher does not live on a floor adjacent to Cooper's"
160 PUSH ABS(FC)#1
RETURN
Output:
Baker lives on floor 3 Cooper lives on floor 2 Fletcher lives on floor 4 Miller lives on floor 5 Smith lives on floor 1 0 OK, 0:1442
UNIX Shell[edit]
#!/bin/bash
# NAMES is a list of names. It can be changed as needed. It can be more than five names, or less.
NAMES=(Baker Cooper Fletcher Miller Smith)
# CRITERIA are the rules imposed on who lives where. Each criterion must be a valid bash expression
# that will be evaluated. TOP is the top floor; BOTTOM is the bottom floor.
# The CRITERIA can be changed to create different rules.
CRITERIA=(
'Baker != TOP' # Baker does not live on the top floor
'Cooper != BOTTOM' # Cooper does not live on the bottom floor
'Fletcher != TOP' # Fletcher does not live on the top floor
'Fletcher != BOTTOM' # and Fletch also does not live on the bottom floor
'Miller > Cooper' # Miller lives above Cooper
'$(abs $(( Smith  Fletcher )) ) > 1' # Smith and Fletcher are not on adjacent floors
'$(abs $(( Fletcher  Cooper )) ) > 1' # Fletcher and Cooper are not on adjacent floors
)
# Code below here shouldn't need to change to vary parameters
let BOTTOM=0
let TOP=${#NAMES[@]}1
# Not available as a builtin
abs() { local n=$(( 10#$1 )) ; echo $(( n < 0 ? n : n )) ; }
# Algorithm we use to iterate over the permutations
# requires that we start with the array sorted lexically
NAMES=($(printf "%s\n" "${NAMES[@]}"  sort))
while true; do
# set each name to its position in the array
for (( i=BOTTOM; i<=TOP; ++i )); do
eval "${NAMES[i]}=$i"
done
# check to see if we've solved the problem
let solved=1
for criterion in "${CRITERIA[@]}"; do
if ! eval "(( $criterion ))"; then
let solved=0
break
fi
done
if (( solved )); then
echo "From bottom to top: ${NAMES[@]}"
break
fi
# Bump the names list to the next permutation
let j=TOP1
while (( j >= BOTTOM )) && ! [[ "${NAMES[j]}" < "${NAMES[j+1]}" ]]; do
let j=1
done
if (( j < BOTTOM )); then break; fi
let k=TOP
while (( k > j )) && [[ "${NAMES[k]}" < "${NAMES[j]}" ]]; do
let k=1
done
if (( k <= j )); then break; fi
t="${NAMES[j]}"
NAMES[j]="${NAMES[k]}"
NAMES[k]="$t"
for (( k=1; k<=(TOPj); ++k )); do
a=BOTTOM+j+k
b=TOPk+1
if (( a < b )); then
t="${NAMES[a]}"
NAMES[a]="${NAMES[b]}"
NAMES[b]="$t"
fi
done
done
Sample output:
From bottom to top: Smith Cooper Baker Fletcher Miller
XPL0[edit]
include c:\cxpl\codes;
int B, C, F, M, S;
for B:= 1 to 4 do \Baker does not live on top (5th) floor
for C:= 2 to 5 do \Cooper does not live on bottom floor
if C#B then \Cooper & Baker live on different floors
for F:= 2 to 4 do \Fletcher doesn't live on top or bottom
if F#B & F#C & F#C1 & F#C+1 then \ and she's not adjacent to Cooper
for M:= 1 to 5 do
if M#F & M#B & M>C then \Miller lives above Cooper
for S:= 1 to 5 do \Smith is not adjacent to Fletcher
if S#M & S#F & S#C & S#B & S#F1 & S#F+1 then \show
[Text(0, "Baker "); IntOut(0, B); CrLf(0); \all
Text(0, "Cooper "); IntOut(0, C); CrLf(0); \possible
Text(0, "Fletcher "); IntOut(0, F); CrLf(0); \solutions
Text(0, "Miller "); IntOut(0, M); CrLf(0);
Text(0, "Smith "); IntOut(0, S); CrLf(0);
]
Output:
Baker 3 Cooper 2 Fletcher 4 Miller 5 Smith 1
zkl[edit]
This solution uses reflection to make the problem statement more "natural". Each dinesman is a variable, whose name is the dinesman name and value is the floor the dinesman lives on. The constraints are then a straight translation of the problem statement. The "each live on a different floors" is implicit by using permutation of the floors.
A floor plan is generated, shoved into the variables and the constraints are run. If they are all true, we have a solution. The first false constraint stops the running of the rest of them (conditional and).
This could be generalized even more by putting the variables and constraint functions in a class, then reflection could be used to automagically get the variables, variable names and constraint functions.
var Baker, Cooper, Fletcher, Miller, Smith; // value == floor
const bottom=1,top=5; // floors: 1..5
// All live on different floors, enforced by using permutations of floors
//fcn c0{ (Baker!=Cooper!=Fletcher) and (Fletcher!=Miller!=Smith) }
fcn c1{ Baker!=top }
fcn c2{ Cooper!=bottom }
fcn c3{ bottom!=Fletcher!=top }
fcn c4{ Miller>Cooper }
fcn c5{ (Fletcher  Smith).abs() !=1 }
fcn c6{ (Fletcher  Cooper).abs()!=1 }
filters:=T(c1,c2,c3,c4,c5,c6);
dudes:=T("Baker","Cooper","Fletcher","Miller","Smith"); // for reflection
foreach combo in (Utils.Helpers.permuteW([bottom..top].walk())){ // lazy
dudes.zip(combo).apply2(fcn(nameValue){ setVar(nameValue.xplode()) });
if(not filters.runNFilter(False)){ // all constraints are True
vars.println(); // use reflection to print solution
break;
}
}
 Output:
L(L("Baker",3),L("Cooper",2),L("Fletcher",4),L("Miller",5),L("Smith",1))
ZX Spectrum Basic[edit]
10 REM Floors are numbered 0 (ground) to 4 (top)
20 REM "Baker, Cooper, Fletcher, Miller, and Smith live on different floors":
30 REM "Baker does not live on the top floor"
40 REM "Cooper does not live on the bottom floor"
50 REM "Fletcher does not live on either the top or the bottom floor"
60 REM "Miller lives on a higher floor than does Cooper"
70 REM "Smith does not live on a floor adjacent to Fletcher's"
80 REM "Fletcher does not live on a floor adjacent to Cooper's"
90 FOR b=0 TO 4: FOR c=0 TO 4: FOR f=0 TO 4: FOR m=0 TO 4: FOR s=0 TO 4
100 IF B<>C AND B<>F AND B<>M AND B<>S AND C<>F AND C<>M AND C<>S AND F<>M AND F<>S AND M<>S AND B<>4 AND C<>0 AND F<>0 AND F<>4 AND M>C AND ABS (SF)<>1 AND ABS (FC)<>1 THEN PRINT "Baker lives on floor ";b: PRINT "Cooper lives on floor ";c: PRINT "Fletcher lives on floor ";f: PRINT "Miller lives on floor ";m: PRINT "Smith lives on floor ";s: STOP
110 NEXT s: NEXT m: NEXT f: NEXT c: NEXT b
 Programming Tasks
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