# Dinesman's multiple-dwelling problem

Dinesman's multiple-dwelling problem
You are encouraged to solve this task according to the task description, using any language you may know.

The task is to 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

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```

## BBC BASIC

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(Smith-Fletcher)<>1"       REM "Fletcher does not live on a floor adjacent to Cooper's":      stmt7\$ = "ABS(Fletcher-Cooper)<>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

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 UTF-8 encoded Unicode characters, but falls back to ISO 8859-1 if a string cannot be parsed as UTF-8. 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

`#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.

## Clojure

This solution uses the contributed package clojure.core.logic, a miniKanren-based 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 one-to-one 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/symbol-macrolet [_ (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

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.length-1,        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:

[This output is incorrect: it has Fletcher on the bottom floor, Baker on the top, and Cooper and Fletcher adjacent.]

`[[Fletcher, Cooper, Miller, Smith, Baker]]`

### Simpler Version

`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.

## Elixir

Translation of: Ruby

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: [x|y])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

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

`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(Smith-Fletcher)<>1"       ! "Fletcher does not live on a floor adjacent to Cooper's":      stmt7\$="ABS(Fletcher-Cooper)<>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(Smith-Fletcher)<>1 AND ABS(Fletcher-Cooper)<>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 !  BakerEND 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

All rules are encoded in the ``meets-constraints?`` word. Any variations to the rules requires modifying ``meets-constraints?``

`USING: kernel    combinators.short-circuit    math math.combinatorics math.ranges     sequences    qw prettyprint ;IN: rosetta.dinesman : /= ( x y -- ? ) = not ;: fifth ( seq -- elt ) 4 swap nth ; : meets-constraints? ( 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] all-permutations [ meets-constraints? ] filter ; : >names ( seq -- seq )    [ 1 - qw{ baker cooper fletcher miller smith } nth ] map ; : dinesman ( -- )    solutions [ >names . ] each ;`
Output:
`{ "fletcher" "cooper" "miller" "smith" "baker" }`

## Forth

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 A-Z are used as digits. That will work up to 36 floors.

Although this is not ANS Forth, one should have little trouble converting it.

Works with: 4tH version 3.6.20
`  0 enum baker                         \ enumeration of all tenants    enum cooper    enum fletcher    enum millerconstant smith create names                           \ names of all the tenants  ," Baker"  ," Cooper"  ," Fletcher"  ," Miller"  ," Smith"                            \ get name, type itdoes> 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 : num@ c@ [char] 0 - ;                 ( a -- n): floor chars over + num@ ;            ( 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 num@ 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 + c@ 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
```

 This example is incomplete. Examples should state what changes to the problem text are allowed. Please ensure that it meets all task requirements and remove this message.

The List monad is perfect for this kind of problem. One can express the problem statements in a very natural and concise way:

Works with: GHC version 6.10+
`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)main = print [ (b,c,f,m,s) | [b,c,f,m,s] <- permutations [1..5], b/=5,c/=1,f/=1,f/=5,m>c,abs(s-f)>1,abs(c-f)>1] `

## Icon and Unicon

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 allglobal 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 solutionsend procedure showsolution()       # show the soluton   write("The solution is:")   every write("   ",n := !nameL, " lives in ", nameT[n])   returnend  procedure eval(n)              # evaluate a rule   r := copy(rules[n-top()])   every r[i := 2 to *r] := rv(r[i])   if get(r)!r then suspendend procedure rv(x)                # return referenced value if it existsreturn \nameT[x] | xend 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   returnend procedure distinct(a,b)        # ensure each name is distinct   if nameT[n := !nameL] = nameT[n ~== key(nameT)] then fail   suspendend procedure notadjacent(n1,n2)   # ensure n1,2 are not adjacent   if not adjacent(n1,n2) then suspendend procedure adjacent(n1,n2)      # ensure n1,2 are adjacent   if abs(n1 - n2) = 1 then suspendend procedure bottom()             # return bottom    return if *nameL > 0 then 1 else 0end procedure top()                # return top   return *nameLend`
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

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 floorpossible=: (#~ 'C' ~: {."1) possible         NB. Cooper not on bottom floorpossible=: (#~ 'F' ~: {:"1) possible         NB. Fletcher not on top floorpossible=: (#~ 'F' ~: {."1) possible         NB. Fletcher not on bottom floorpossible=: (#~ </@i."1&'CM') possible        NB. Miller on higher floor than Cooperpossible=: (#~ 0 = +/@E."1~&'SF') possible   NB. Smith not immediately below Fletcherpossible=: (#~ 0 = +/@E."1~&'FS') possible   NB. Fletcher not immediately below Smithpossible=: (#~ 0 = +/@E."1~&'CF') possible   NB. Cooper not immediately below Fletcherpossible=: (#~ 0 = +/@E."1~&'FC') possible   NB. Fletcher not immediately below Cooper`

`   possibleSCBFM`

(bottom floor) Smith, Cooper, Baker, Fletcher, Miller (top floor)

## Java

 This example is incomplete. Examples should state what changes to the problem text are allowed. Please ensure that it meets all task requirements and remove this message.

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);      }    }    return;  }   private static boolean topFloor(String permutation, String person)  {  return permutation.endsWith(person);  }   private static boolean bottomFloor(String permutation, String person)  {  return permutation.startsWith(person);  }   public static boolean livesAbove(String permutation, String upperPerson, String lowerPerson)  {  return permutation.indexOf(upperPerson) > permutation.indexOf(lowerPerson);  }   public static boolean adjacent(String permutation, String person1, String person2)  {  return (Math.abs(permutation.indexOf(person1) - permutation.indexOf(person2)) == 1);  }   private static boolean isPossible(String s)  {    // Conditions here    if (topFloor(s, "B"))      return false;    if (bottomFloor(s, "C"))      return false;    if (topFloor(s, "F") || bottomFloor(s, "F"))      return false;    if (!livesAbove(s, "M", "C"))      return false;    if (adjacent(s, "S", "F"))      return false;    if (adjacent(s, "F", "C"))      return false;    return true;  }   public static void main(String[] args)  {    Set<String> set = new HashSet<String>();    generatePermutations(new String[] { "B", "C", "F", "M", "S" }, set, "");    for (Iterator<String> iterator = set.iterator(); iterator.hasNext(); )    {      String permutation = iterator.next();      if (!isPossible(permutation))        iterator.remove();    }    for (String s : set)      System.out.println("Possible arrangement: " + s);  }}`
Output:
`Possible arrangement: SCBFM`

## jq

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 constraint-oriented 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 persondef resides(person; cond):  range(0;5) as \$n  | if (.[\$n] == null or .[\$n] == person) and (\$n|cond) 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"]`

## Mathematica / Wolfram Language

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

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

A solution that parses a structured version of the problem text, translates it into a Perl expression, and uses it for a brute-force 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/[a-z]+/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 free-form 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 topCooper not on bottomFletcher not on topFletcher not on bottomMiller higher than CooperSmith not adjacent to FletcherFletcher 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

###  By parsing the problem

Translation of: Perl
`sub parse_and_solve (\$text) {    my %ids;    my \$expr = (grammar {        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}] + (-1|1)" }            || on <ordinal>          { make "\@f[%s] == {\$<ordinal>.made}"            }            || { note "Failed to parse line " ~ +\$/.prematch.comb(/^^/); exit 1; }        }         token name    { :i <[a..z]>+              { make %ids{~\$/} //= (state \$)++ } }        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({ "\$^a=@f[\$^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

Works with: rakudo version 2015-11-15
`# 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 != 1|5       # Fletcher !live  on top or the bottom floor.        and \$m  > \$c        # Miller    lives on a higher floor than Cooper.        and \$s != \$f-1|\$f+1 # Smith    !live  adjacent to Fletcher        and \$f != \$c-1|\$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 multi-line if statement in the loop.

Output:
`Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1`

## PicoLisp

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) built-in 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```

## Prolog

### Using CLPFD

Works with SWI-Prolog and library(clpfd) written by Markus Triska.

`:- use_module(library(clpfd)). :- dynamic top/1, bottom/1. % Baker does not live on the top floorrule1(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(Fs-Ff) #> 1. % Fletcher does not live on a floor adjacent to Cooper's.rule6(L) :-	member((cooper, Fc), L),	member((fletcher, Ff), L),	abs(Fc-Ff) #> 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

`select([A|As],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

`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

 This example is incomplete. Examples should state what changes to the problem text are allowed. Please ensure that it meets all task requirements and remove this message.
`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 #TrueEndProcedure 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

### By parsing the problem statement

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 multi-line regular expression at the head of the program.

`import refrom itertools import product problem_re = re.compile(r"""(?msx)(?: # Multiple names of form n1, n2, n3, ... , and nK(?P<namelist> [a-zA-Z]+ (?: , \s+ [a-zA-Z]+)* (?: ,? \s+ and) \s+ [a-zA-Z]+ ) # Flexible floor count (2 to 10 floors)| (?:  .* house \s+ that \s+ contains \s+ only \s+  (?P<floorcount> two|three|four|five|six|seven|eight|nine|ten ) \s+ floors \s* \.) # Constraint: "does not live on the n'th floor" |(?: (?P<not_live>  \b [a-zA-Z]+ \s+ does \s+ not \s+ live \s+ on \s+ the \s+  (?: top|bottom|first|second|third|fourth|fifth|sixth|seventh|eighth|ninth|tenth) \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 [a-zA-Z]+ \s+ does \s+ not \s+ live \s+ on \s+ either  (?: \s+ (?: or \s+)? the \s+           (?: top|bottom|first|second|third|fourth|fifth|sixth|seventh|eighth|ninth|tenth))+ \s+ floor \s* \. ) # Constraint: "P1 lives on a higher/lower floor than P2 does"|(?P<hi_lower> \b  [a-zA-Z]+ \s+ lives \s+ on \s+ a \s (?: higher|lower)   \s+ floor \s+ than (?: \s+ does)  \s+  [a-zA-Z]+ \s* \. ) # Constraint: "P1 does/does not live on a floor adjacent to P2's"|(?P<adjacency>  \b [a-zA-Z]+ \s+ does (?:\s+ not)? \s+ live \s+ on \s+ a \s+   floor \s+ adjacent \s+ to \s+  [a-zA-Z]+ (?: 's )? \s* \. ) # Ask for the solution|(?P<question> Where \s+ does \s+ everyone \s+ live \s* \?) )""") names, lennames = None, Nonefloors = Noneconstraint_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 = '||two|three|four|five|six|seven|eight|nine|ten'.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),            ('|first|second|third|fourth|fifth|sixth|' +             'seventh|eighth|ninth|tenth|top|bottom|').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),             [('|first|second|third|fourth|fifth|sixth|' +               'seventh|eighth|ninth|tenth|top|bottom|').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:        breakelse:    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

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      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.'    print `
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

Translation of: D
`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

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(define-syntax-rule (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 (on-top    x) (for/and ([y residents]) (x . >= . (car y))))  (define (on-bottom x) (for/and ([y residents]) (x . <= . (car y))))  (define (adjacent x y) (= 1 (abs (- x y))))  (assert   ;; ... live on different floors ...   (assert (= 5 (length (remove-duplicates (map car residents)))))   ;; Baker does not live on the top floor.   (not (on-top Baker))   ;; Cooper does not live on the bottom floor.   (not (on-bottom Cooper))   ;; Fletcher does not live on either the top or the bottom floor.   (not (on-top Fletcher))   (not (on-bottom 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

This REXX version tries to keep the rules as simple as possible, with easy-to-read   if   statments.
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 1st floor, the next floor up is the 2nd floor, etc.

The REXX program is broken up into several parts:

• preamble where names and floors are defined.
• iterating all possibilities (permutations would be better and faster).
• evaluation of the possibilities.
• elimination of possibilities because of cohabitation (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   subroutine 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 pgm: Dinesman's multiple-dwelling problem with "natural" wording.*/names= 'Baker Cooper Fletcher Miller Smith'      /*names of the tenants.*/floors=5;     top=floors;     bottom=1;      #=floors;         sols=0                                       /*floor 1 is the ground floor.   */     do !.1=1 for #;do !.2=1 for #;do !.3=1 for #;do !.4=1 for #;do !.5=1 for #       do p=1 for words(names); _=word(names,p); upper _; call value _,!.p       end   /*p*/                                       /*  [â†“] don't live on same floor.*/      do j=1 for #-1; do k=j+1 to #; if !.j==!.k then iterate !.5; end;end      call Waldo  /* â—„â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€where the rubber meets the road.*/     end /*!.5*/;    end /*!.4*/;    end /*!.3*/;    end /*!.2*/;   end /*!.1*/ say;        say  'found'  sols  "solution"s(sols)'.'exit                                   /*stick a fork in it, we're done.*//*â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€Waldo subroutineâ”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€*/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    == Fletcher-1  |  Smith    == Fletcher+1  then return         if Fletcher == Cooper-1    |  Fletcher == Cooper+1    then return say;  sols=sols+1                      /*list tenants in order in list. */                      do p=1  for words(names);        _=word(names,p)                      say right(_,20) 'lives on the' !.p||th(!.p) "floor."                      end   /*p*/return/*â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€one-liner subroutinesâ”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€â”€*/s: if arg(1)=1 then return ''; return 's'   /*a simple pluralizer funct.*/th:procedure;parse 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 3rd floor.
Cooper lives on 2nd floor.
Fletcher lives on 4th floor.
Miller lives on 5th floor.
Smith lives on 1st floor.

found 1 solution.
```

## Ruby

### By parsing the problem

Inspired by the Python version.

`def solve( problem )  lines = problem.split(".")  names = lines.first.scan( /[A-Z]\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 thatcontains 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 Smithlive on different floors of an apartment house that containsonly 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. Cooperdoes not live on the bottom floor. Fletcher does not live oneither the top or the bottom floor. Miller lives on a higherfloor than does Cooper. Smith does not live on a flooradjacent to Fletcher's. Fletcher does not live on a flooradjacent 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

Translation of: D
`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

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. However there are instances where the rooms add to 15 but the room numbers are not unique. So.. it makes sure each person has a unique number. This then meets the requirements.

`people\$ = "Baler,Cooper,Fletcher,Miller,Smith" 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 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        bad   = 0        for i = 1 to 5                                         ' make sure each room is unique          rm\$ = chr\$(i + 48)          r1  = instr(rooms\$,rm\$)          r2  = instr(rooms\$,rm\$,r1+1)          if r2 <> 0 then bad = 1        next i        if bad = 0 then goto [roomAssgn]                       ' if it is not bad it is a good assignment      end if     end if     next smith   next miller  next fletcher next coopernext balerprint "Cam't assign rooms"                                     ' print this if it can not find a solutionwait [roomAssgn]Print "Room Assignment"for i = 1 to 5print mid\$(rooms\$,i,1);" ";word\$(people\$,i,",");" ";           ' print the room assignmentsnext i`
```Room Assignment
3 Baler 2 Cooper 4 Fletcher 5 Miller 1 Smith```

## Scala

`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
```

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

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

### By parsing the problem

Translation of: Ruby
`func dinesman(problem) {  var lines = problem.split('.');  var names = lines.first.scan(/\b[A-Z]\w*/);  var re_names = Regex.new(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.new(words.join('|'));   # Build an array of lambda's  var predicates = lines.ft(1, lines.end-1).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"    }  }.flatten;   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 thatcontains 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 Smithlive on different floors of an apartment house that containsonly 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. Cooperdoes not live on the bottom floor. Fletcher does not live oneither the top or the bottom floor. Miller lives on a higherfloor than does Cooper. Smith does not live on a flooradjacent to Fletcher's. Fletcher does not live on a flooradjacent 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

Translation of: Ruby
`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|    predicates.all {|predicate| predicate(candidate) }        && (say candidate.join("\n"); break);};`
Output:
```Smith
Cooper
Baker
Fletcher
Miller
```

## Tcl

It's trivial to extend this problem to deal with more floors and people and more constraints; the main internally-generated 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.

Library: Tcllib (Package: struct::list)
`package require Tcl 8.5package 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

Translation of: BBC Basic
`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 CNEXT 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(S-F)#1    RETURN REM "Fletcher does not live on a floor adjacent to Cooper's"160 PUSH ABS(F-C)#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

Works with: Bash
`#!/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 parameterslet BOTTOM=0let TOP=\${#NAMES[@]}-1 # Not available as a builtinabs() { 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 lexicallyNAMES=(\$(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=TOP-1  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<=(TOP-j); ++k )); do    a=BOTTOM+j+k    b=TOP-k+1    if (( a < b )); then      t="\${NAMES[a]}"      NAMES[a]="\${NAMES[b]}"      NAMES[b]="\$t"    fi  donedone`

Sample output:

`From bottom to top: Smith Cooper Baker Fletcher Miller`

## XPL0

 This example is incomplete. Examples should state what changes to the problem text are allowed. Please ensure that it meets all task requirements and remove this message.
`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#C-1 & 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#F-1 & 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

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 == floorconst 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 reflectionforeach 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))
```