¿Resolver "Quién posee la cebra" mediante programación?

Editar: este rompecabezas también se conoce como "Einstein's Riddle"

El Quién posee la cebra (puede probar la versión en línea aquí ) es un ejemplo de un conjunto clásico de rompecabezas y apuesto a que la mayoría de las personas en Stack Overflow pueden resolverlo con lápiz y papel. Pero, ¿cómo sería una solución programática?

Según las pistas que se enumeran a continuación ...

• Hay cinco casas.
• Cada casa tiene su propio color único.
• Todos los dueños de casa son de diferentes nacionalidades.
• Todos tienen diferentes mascotas.
• Todos beben diferentes bebidas.
• Todos fuman cigarrillos diferentes.
• El inglés vive en la casa roja.
• El sueco tiene un perro.
• El danés bebe té.
• La casa verde está en el lado izquierdo de la casa blanca.
• Beben café en la casa verde.
• El hombre que fuma Pall Mall tiene pájaros.
• En la casa amarilla fuman Dunhill.
• En la casa del medio beben leche.
• El noruego vive en la primera casa.
• El hombre que fuma Blend vive en la casa de al lado con gatos.
• En la casa al lado de la casa donde tienen un caballo, fuman Dunhill.
• El hombre que fuma Blue Master bebe cerveza.
• El alemán fuma Príncipe.
• El noruego vive al lado de la casa azul.
• Beben agua en la casa al lado de la casa donde fuman Blend.

... ¿quién es el dueño de la cebra?

Aquí hay una solución en Python basada en la restricción de programación:

from constraint import AllDifferentConstraint, InSetConstraint, Problem

# variables
colors        = "blue red green white yellow".split()
nationalities = "Norwegian German Dane Swede English".split()
pets          = "birds dog cats horse zebra".split()
drinks        = "tea coffee milk beer water".split()
cigarettes    = "Blend, Prince, Blue Master, Dunhill, Pall Mall".split(", ")

# There are five houses.
minn, maxn = 1, 5
problem = Problem()
# value of a variable is the number of a house with corresponding property
variables = colors + nationalities + pets + drinks + cigarettes
problem.addVariables(variables, range(minn, maxn+1))

# Each house has its own unique color.
# All house owners are of different nationalities.
# They all have different pets.
# They all drink different drinks.
# They all smoke different cigarettes.
for vars_ in (colors, nationalities, pets, drinks, cigarettes):
    problem.addConstraint(AllDifferentConstraint(), vars_)

# In the middle house they drink milk.
#NOTE: interpret "middle" in a numerical sense (not geometrical)
problem.addConstraint(InSetConstraint([(minn + maxn) // 2]), ["milk"])
# The Norwegian lives in the first house.
#NOTE: interpret "the first" as a house number
problem.addConstraint(InSetConstraint([minn]), ["Norwegian"])
# The green house is on the left side of the white house.
#XXX: what is "the left side"? (linear, circular, two sides, 2D house arrangment)
#NOTE: interpret it as 'green house number' + 1 == 'white house number'
problem.addConstraint(lambda a,b: a+1 == b, ["green", "white"])

def add_constraints(constraint, statements, variables=variables, problem=problem):
    for stmt in (line for line in statements if line.strip()):
        problem.addConstraint(constraint, [v for v in variables if v in stmt])

and_statements = """
They drink coffee in the green house.
The man who smokes Pall Mall has birds.
The English man lives in the red house.
The Dane drinks tea.
In the yellow house they smoke Dunhill.
The man who smokes Blue Master drinks beer.
The German smokes Prince.
The Swede has a dog.
""".split("\n")
add_constraints(lambda a,b: a == b, and_statements)

nextto_statements = """
The man who smokes Blend lives in the house next to the house with cats.
In the house next to the house where they have a horse, they smoke Dunhill.
The Norwegian lives next to the blue house.
They drink water in the house next to the house where they smoke Blend.
""".split("\n")
#XXX: what is "next to"? (linear, circular, two sides, 2D house arrangment)
add_constraints(lambda a,b: abs(a - b) == 1, nextto_statements)

def solve(variables=variables, problem=problem):
    from itertools  import groupby
    from operator   import itemgetter

    # find & print solutions
    for solution in problem.getSolutionIter():
        for key, group in groupby(sorted(solution.iteritems(), key=itemgetter(1)), key=itemgetter(1)):
            print key, 
            for v in sorted(dict(group).keys(), key=variables.index):
                print v.ljust(9),
            print

if __name__ == '__main__':
    solve()

Salida:

1 yellow    Norwegian cats      water     Dunhill  
2 blue      Dane      horse     tea       Blend    
3 red       English   birds     milk      Pall Mall
4 green     German    zebra     coffee    Prince   
5 white     Swede     dog       beer      Blue Master

Se necesitan 0.6 segundos (CPU 1.5GHz) para encontrar la solución.
La respuesta es "el alemán posee cebra".

Para instalar el constraintmódulo a través de pip: pip install python-restricint

Para instalar manualmente:

• descargar:

  $ wget https://pypi.python.org/packages/source/p/python-constraint/python-constraint-1.2.tar.bz2#md5=d58de49c85992493db53fcb59b9a0a45

• extracto (Linux / Mac / BSD):

  $ bzip2 -cd python-restrictint-1.2.tar.bz2 | tar xvf -

• extracto (Windows, con 7zip ):

  > 7z e python-restricint-1.2.tar.bz2
  > 7z e python-restrictint-1.2.tar

• Instalar en pc:

  $ cd python-restricint-1.2
  $ python setup.py install

En Prolog, podemos instanciar el dominio simplemente seleccionando elementos de él :) (haciendo elecciones mutuamente excluyentes , para mayor eficiencia). Usando SWI-Prolog,

select([A|As],S):- select(A,S,S1),select(As,S1).
select([],_). 

left_of(A,B,C):- append(_,[A,B|_],C).  
next_to(A,B,C):- left_of(A,B,C) ; left_of(B,A,C).

zebra(Owns, HS):-     % house: color,nation,pet,drink,smokes
  HS   = [ h(_,norwegian,_,_,_),    h(blue,_,_,_,_),   h(_,_,_,milk,_), _, _], 
  select([ h(red,brit,_,_,_),       h(_,swede,dog,_,_), 
           h(_,dane,_,tea,_),       h(_,german,_,_,prince)], HS),
  select([ h(_,_,birds,_,pallmall), h(yellow,_,_,_,dunhill),
           h(_,_,_,beer,bluemaster)],                        HS), 
  left_of( h(green,_,_,coffee,_),   h(white,_,_,_,_),        HS),
  next_to( h(_,_,_,_,dunhill),      h(_,_,horse,_,_),        HS),
  next_to( h(_,_,_,_,blend),        h(_,_,cats, _,_),        HS),
  next_to( h(_,_,_,_,blend),        h(_,_,_,water,_),        HS),
  member(  h(_,Owns,zebra,_,_),                              HS).

Se ejecuta al instante:

?- time( (zebra(Who,HS), writeln(Who), nl, maplist(writeln,HS), nl, false 
          ; writeln('no more solutions!') )).
german

h( yellow, norwegian, cats,   water,  dunhill   )
h( blue,   dane,      horse,  tea,    blend     )
h( red,    brit,      birds,  milk,   pallmall  )
h( green,  german,    zebra,  coffee, prince    )     % formatted by hand
h( white,  swede,     dog,    beer,   bluemaster)

no more solutions!
% 1,706 inferences, 0.000 CPU in 0.070 seconds (0% CPU, Infinite Lips)
true.

Un póster ya mencionó que Prolog es una solución potencial. Esto es cierto, y es la solución que usaría. En términos más generales, este es un problema perfecto para un sistema de inferencia automatizado. Prolog es un lenguaje de programación lógica (y un intérprete asociado) que forman dicho sistema. Básicamente permite la conclusión de hechos de declaraciones hechas usando First Order Logic . FOL es básicamente una forma más avanzada de lógica proposicional. Si decide que no desea usar Prolog, puede usar un sistema similar de su propia creación utilizando una técnica como modus ponens para realizar el saque de conclusiones.

Por supuesto, deberá agregar algunas reglas sobre las cebras, ya que no se menciona en ninguna parte ... Creo que la intención es que pueda descubrir las otras 4 mascotas y deducir que la última es la cebra. Querrás agregar reglas que indiquen que una cebra es una de las mascotas, y cada casa solo puede tener una mascota. Obtener este tipo de conocimiento de "sentido común" en un sistema de inferencia es el principal obstáculo para usar la técnica como una verdadera IA. Hay algunos proyectos de investigación, como Cyc, que intentan dar un conocimiento tan común a través de la fuerza bruta. Se han encontrado con una cantidad interesante de éxito.

Compatible con SWI-Prolog:

% NOTE - This may or may not be more efficent. A bit verbose, though.
left_side(L, R, [L, R, _, _, _]).
left_side(L, R, [_, L, R, _, _]).
left_side(L, R, [_, _, L, R, _]).
left_side(L, R, [_, _, _, L, R]).

next_to(X, Y, Street) :- left_side(X, Y, Street).
next_to(X, Y, Street) :- left_side(Y, X, Street).

m(X, Y) :- member(X, Y).

get_zebra(Street, Who) :- 
    Street = [[C1, N1, P1, D1, S1],
              [C2, N2, P2, D2, S2],
              [C3, N3, P3, D3, S3],
              [C4, N4, P4, D4, S4],
              [C5, N5, P5, D5, S5]],
    m([red, english, _, _, _], Street),
    m([_, swede, dog, _, _], Street),
    m([_, dane, _, tea, _], Street),
    left_side([green, _, _, _, _], [white, _, _, _, _], Street),
    m([green, _, _, coffee, _], Street),
    m([_, _, birds, _, pallmall], Street),
    m([yellow, _, _, _, dunhill], Street),
    D3 = milk,
    N1 = norwegian,
    next_to([_, _, _, _, blend], [_, _, cats, _, _], Street),
    next_to([_, _, horse, _, _], [_, _, _, _, dunhill], Street),
    m([_, _, _, beer, bluemaster], Street),
    m([_, german, _, _, prince], Street),
    next_to([_, norwegian, _, _, _], [blue, _, _, _, _], Street),
    next_to([_, _, _, water, _], [_, _, _, _, blend], Street),
    m([_, Who, zebra, _, _], Street).

En el intérprete:

?- get_zebra(Street, Who).
Street = ...
Who = german

Así es como lo haría. Primero generaría todas las n-tuplas ordenadas

(housenumber, color, nationality, pet, drink, smoke)

5 ^ 6 de esos, 15625, fácilmente manejables. Luego filtraría las condiciones booleanas simples. hay diez de ellos, y cada uno de ellos esperaría filtrar 8/25 de las condiciones (1/25 de las condiciones contienen un sueco con un perro, 16/25 contienen un no sueco con un no perro) . Por supuesto, no son independientes, pero después de filtrarlos, no deberían quedar muchos.

Después de eso, tienes un buen problema gráfico. Cree un gráfico con cada nodo que represente una de las n-tuplas restantes. Agregue bordes al gráfico si los dos extremos contienen duplicados en alguna posición de n-tupla o violan cualquier restricción 'posicional' (hay cinco de esos). Desde allí, ya casi está en casa, busque en el gráfico un conjunto independiente de cinco nodos (sin que ninguno de los nodos esté conectado por bordes). Si no hay demasiados, posiblemente podría generar exhaustivamente todas las 5 tuplas de n-tuplas y simplemente filtrarlas nuevamente.

Este podría ser un buen candidato para el código de golf. Alguien probablemente pueda resolverlo en una línea con algo como haskell :)

idea de último momento: el paso de filtro inicial también puede eliminar información de las restricciones posicionales. No mucho (1/25), pero sigue siendo significativo.

Otra solución de Python, esta vez usando PyKE de Python (Python Knowledge Engine). De acuerdo, es más detallado que usar el módulo "restricción" de Python en la solución de @JFSebastian, pero proporciona una comparación interesante para cualquiera que esté buscando un motor de conocimiento en bruto para este tipo de problema.

pistas.kfb

categories( POSITION, 1, 2, 3, 4, 5 )                                   # There are five houses.
categories( HOUSE_COLOR, blue, red, green, white, yellow )              # Each house has its own unique color.
categories( NATIONALITY, Norwegian, German, Dane, Swede, English )      # All house owners are of different nationalities.
categories( PET, birds, dog, cats, horse, zebra )                       # They all have different pets.
categories( DRINK, tea, coffee, milk, beer, water )                     # They all drink different drinks.
categories( SMOKE, Blend, Prince, 'Blue Master', Dunhill, 'Pall Mall' ) # They all smoke different cigarettes.

related( NATIONALITY, English, HOUSE_COLOR, red )    # The English man lives in the red house.
related( NATIONALITY, Swede, PET, dog )              # The Swede has a dog.
related( NATIONALITY, Dane, DRINK, tea )             # The Dane drinks tea.
left_of( HOUSE_COLOR, green, HOUSE_COLOR, white )    # The green house is on the left side of the white house.
related( DRINK, coffee, HOUSE_COLOR, green )         # They drink coffee in the green house.
related( SMOKE, 'Pall Mall', PET, birds )            # The man who smokes Pall Mall has birds.
related( SMOKE, Dunhill, HOUSE_COLOR, yellow )       # In the yellow house they smoke Dunhill.
related( POSITION, 3, DRINK, milk )                  # In the middle house they drink milk.
related( NATIONALITY, Norwegian, POSITION, 1 )       # The Norwegian lives in the first house.
next_to( SMOKE, Blend, PET, cats )                   # The man who smokes Blend lives in the house next to the house with cats.
next_to( SMOKE, Dunhill, PET, horse )                # In the house next to the house where they have a horse, they smoke Dunhill.
related( SMOKE, 'Blue Master', DRINK, beer )         # The man who smokes Blue Master drinks beer.
related( NATIONALITY, German, SMOKE, Prince )        # The German smokes Prince.
next_to( NATIONALITY, Norwegian, HOUSE_COLOR, blue ) # The Norwegian lives next to the blue house.
next_to( DRINK, water, SMOKE, Blend )                # They drink water in the house next to the house where they smoke Blend.

relaciones.krb

#############
# Categories

# Foreach set of categories, assert each type
categories
    foreach
        clues.categories($category, $thing1, $thing2, $thing3, $thing4, $thing5)
    assert
        clues.is_category($category, $thing1)
        clues.is_category($category, $thing2)
        clues.is_category($category, $thing3)
        clues.is_category($category, $thing4)
        clues.is_category($category, $thing5)


#########################
# Inverse Relationships

# Foreach A=1, assert 1=A
inverse_relationship_positive
    foreach
        clues.related($category1, $thing1, $category2, $thing2)
    assert
        clues.related($category2, $thing2, $category1, $thing1)

# Foreach A!1, assert 1!A
inverse_relationship_negative
    foreach
        clues.not_related($category1, $thing1, $category2, $thing2)
    assert
        clues.not_related($category2, $thing2, $category1, $thing1)

# Foreach "A beside B", assert "B beside A"
inverse_relationship_beside
    foreach
        clues.next_to($category1, $thing1, $category2, $thing2)
    assert
        clues.next_to($category2, $thing2, $category1, $thing1)


###########################
# Transitive Relationships

# Foreach A=1 and 1=a, assert A=a
transitive_positive
    foreach
        clues.related($category1, $thing1, $category2, $thing2)
        clues.related($category2, $thing2, $category3, $thing3)

        check unique($thing1, $thing2, $thing3) \
          and unique($category1, $category2, $category3)
    assert
        clues.related($category1, $thing1, $category3, $thing3)

# Foreach A=1 and 1!a, assert A!a
transitive_negative
    foreach
        clues.related($category1, $thing1, $category2, $thing2)
        clues.not_related($category2, $thing2, $category3, $thing3)

        check unique($thing1, $thing2, $thing3) \
          and unique($category1, $category2, $category3)
    assert
        clues.not_related($category1, $thing1, $category3, $thing3)


##########################
# Exclusive Relationships

# Foreach A=1, assert A!2 and A!3 and A!4 and A!5
if_one_related_then_others_unrelated
    foreach
        clues.related($category, $thing, $category_other, $thing_other)
        check unique($category, $category_other)

        clues.is_category($category_other, $thing_not_other)
        check unique($thing, $thing_other, $thing_not_other)
    assert
        clues.not_related($category, $thing, $category_other, $thing_not_other)

# Foreach A!1 and A!2 and A!3 and A!4, assert A=5
if_four_unrelated_then_other_is_related
    foreach
        clues.not_related($category, $thing, $category_other, $thingA)
        clues.not_related($category, $thing, $category_other, $thingB)
        check unique($thingA, $thingB)

        clues.not_related($category, $thing, $category_other, $thingC)
        check unique($thingA, $thingB, $thingC)

        clues.not_related($category, $thing, $category_other, $thingD)
        check unique($thingA, $thingB, $thingC, $thingD)

        # Find the fifth variation of category_other.
        clues.is_category($category_other, $thingE)
        check unique($thingA, $thingB, $thingC, $thingD, $thingE)
    assert
        clues.related($category, $thing, $category_other, $thingE)


###################
# Neighbors: Basic

# Foreach "A left of 1", assert "A beside 1"
expanded_relationship_beside_left
    foreach
        clues.left_of($category1, $thing1, $category2, $thing2)
    assert
        clues.next_to($category1, $thing1, $category2, $thing2)

# Foreach "A beside 1", assert A!1
unrelated_to_beside
    foreach
        clues.next_to($category1, $thing1, $category2, $thing2)
        check unique($category1, $category2)
    assert
        clues.not_related($category1, $thing1, $category2, $thing2)


###################################
# Neighbors: Spatial Relationships

# Foreach "A beside B" and "A=(at-edge)", assert "B=(near-edge)"
check_next_to_either_edge
    foreach
        clues.related(POSITION, $position_known, $category, $thing)
        check is_edge($position_known)

        clues.next_to($category, $thing, $category_other, $thing_other)

        clues.is_category(POSITION, $position_other)
        check is_beside($position_known, $position_other)
    assert
        clues.related(POSITION, $position_other, $category_other, $thing_other)

# Foreach "A beside B" and "A!(near-edge)" and "B!(near-edge)", assert "A!(at-edge)"
check_too_close_to_edge
    foreach
        clues.next_to($category, $thing, $category_other, $thing_other)

        clues.is_category(POSITION, $position_edge)
        clues.is_category(POSITION, $position_near_edge)
        check is_edge($position_edge) and is_beside($position_edge, $position_near_edge)

        clues.not_related(POSITION, $position_near_edge, $category, $thing)
        clues.not_related(POSITION, $position_near_edge, $category_other, $thing_other)
    assert
        clues.not_related(POSITION, $position_edge, $category, $thing)

# Foreach "A beside B" and "A!(one-side)", assert "A=(other-side)"
check_next_to_with_other_side_impossible
    foreach
        clues.next_to($category, $thing, $category_other, $thing_other)

        clues.related(POSITION, $position_known, $category_other, $thing_other)
        check not is_edge($position_known)

        clues.not_related($category, $thing, POSITION, $position_one_side)
        check is_beside($position_known, $position_one_side)

        clues.is_category(POSITION, $position_other_side)
        check is_beside($position_known, $position_other_side) \
          and unique($position_known, $position_one_side, $position_other_side)
    assert
        clues.related($category, $thing, POSITION, $position_other_side)

# Foreach "A left of B"...
#   ... and "C=(position1)" and "D=(position2)" and "E=(position3)"
# ~> assert "A=(other-position)" and "B=(other-position)+1"
left_of_and_only_two_slots_remaining
    foreach
        clues.left_of($category_left, $thing_left, $category_right, $thing_right)

        clues.related($category_left, $thing_left_other1, POSITION, $position1)
        clues.related($category_left, $thing_left_other2, POSITION, $position2)
        clues.related($category_left, $thing_left_other3, POSITION, $position3)
        check unique($thing_left, $thing_left_other1, $thing_left_other2, $thing_left_other3)

        clues.related($category_right, $thing_right_other1, POSITION, $position1)
        clues.related($category_right, $thing_right_other2, POSITION, $position2)
        clues.related($category_right, $thing_right_other3, POSITION, $position3)
        check unique($thing_right, $thing_right_other1, $thing_right_other2, $thing_right_other3)

        clues.is_category(POSITION, $position4)
        clues.is_category(POSITION, $position5)

        check is_left_right($position4, $position5) \
          and unique($position1, $position2, $position3, $position4, $position5)
    assert
        clues.related(POSITION, $position4, $category_left, $thing_left)
        clues.related(POSITION, $position5, $category_right, $thing_right)


#########################

fc_extras

    def unique(*args):
        return len(args) == len(set(args))

    def is_edge(pos):
        return (pos == 1) or (pos == 5)

    def is_beside(pos1, pos2):
        diff = (pos1 - pos2)
        return (diff == 1) or (diff == -1)

    def is_left_right(pos_left, pos_right):
        return (pos_right - pos_left == 1)

driver.py (en realidad más grande, pero esta es la esencia)

from pyke import knowledge_engine

engine = knowledge_engine.engine(__file__)
engine.activate('relations')

try:
    natl = engine.prove_1_goal('clues.related(PET, zebra, NATIONALITY, $nationality)')[0].get('nationality')
except Exception, e:
    natl = "Unknown"
print "== Who owns the zebra? %s ==" % natl

Salida de muestra:

$ python driver.py

== Who owns the zebra? German ==

#   Color    Nationality    Pet    Drink       Smoke    
=======================================================
1   yellow   Norwegian     cats    water    Dunhill     
2   blue     Dane          horse   tea      Blend       
3   red      English       birds   milk     Pall Mall   
4   green    German        zebra   coffee   Prince      
5   white    Swede         dog     beer     Blue Master 

Calculated in 1.19 seconds.

Fuente: https://github.com/DreadPirateShawn/pyke-who-owns-zebra

Aquí hay un extracto de la solución completa usando NSolver , publicado en Einstein's Riddle en C # :

// The green house's owner drinks coffee
Post(greenHouse.Eq(coffee));
// The person who smokes Pall Mall rears birds 
Post(pallMall.Eq(birds));
// The owner of the yellow house smokes Dunhill 
Post(yellowHouse.Eq(dunhill));

Aquí hay una solución sencilla en CLP (FD) (ver también clpfd):

:- use_module(library(clpfd)).

solve(ZebraOwner) :-
    maplist( init_dom(1..5), 
        [[British,  Swedish,  Danish,  Norwegian, German],     % Nationalities
         [Red,      Green,    Blue,    White,     Yellow],     % Houses
         [Tea,      Coffee,   Milk,    Beer,      Water],      % Beverages
         [PallMall, Blend,    Prince,  Dunhill,   BlueMaster], % Cigarettes
         [Dog,      Birds,    Cats,    Horse,     Zebra]]),    % Pets
    British #= Red,        % Hint 1
    Swedish #= Dog,        % Hint 2
    Danish #= Tea,         % Hint 3
    Green #= White - 1 ,   % Hint 4
    Green #= Coffee,       % Hint 5
    PallMall #= Birds,     % Hint 6
    Yellow #= Dunhill,     % Hint 7
    Milk #= 3,             % Hint 8
    Norwegian #= 1,        % Hint 9
    neighbor(Blend, Cats),     % Hint 10
    neighbor(Horse, Dunhill),  % Hint 11
    BlueMaster #= Beer,        % Hint 12
    German #= Prince,          % Hint 13
    neighbor(Norwegian, Blue), % Hint 14
    neighbor(Blend, Water),    % Hint 15
    memberchk(Zebra-ZebraOwner, [British-british, Swedish-swedish, Danish-danish,
                                 Norwegian-norwegian, German-german]).

init_dom(R, L) :-
    all_distinct(L),
    L ins R.

neighbor(X, Y) :-
    (X #= (Y - 1)) #\/ (X #= (Y + 1)).

Ejecutándolo, produce:

3? - tiempo (resolver (Z)).
% 111,798 inferencias, 0.016 CPU en 0.020 segundos (78% CPU, 7166493 Labios)
Z = alemán.

En PAIP (Capítulo 11), Norvig resuelve el acertijo de cebra utilizando un Prólogo incrustado en Lisp .

Solución ES6 (Javascript)

Con muchos generadores ES6 y un poco de lodash . Necesitarás a Babel para ejecutar esto.

var _ = require('lodash');

function canBe(house, criteria) {
    for (const key of Object.keys(criteria))
        if (house[key] && house[key] !== criteria[key])
            return false;
    return true;
}

function* thereShouldBe(criteria, street) {
    for (const i of _.range(street.length))
        yield* thereShouldBeAtIndex(criteria, i, street);
}

function* thereShouldBeAtIndex(criteria, index, street) {
    if (canBe(street[index], criteria)) {
        const newStreet = _.cloneDeep(street);
        newStreet[index] = _.assign({}, street[index], criteria);
        yield newStreet;
    }
}

function* leftOf(critA, critB, street) {
    for (const i of _.range(street.length - 1)) {
        if (canBe(street[i], critA) && canBe(street[i+1], critB)) {
            const newStreet = _.cloneDeep(street);
            newStreet[i  ] = _.assign({}, street[i  ], critA);
            newStreet[i+1] = _.assign({}, street[i+1], critB);
            yield newStreet;
        }
    }
}
function* nextTo(critA, critB, street) {
    yield* leftOf(critA, critB, street);
    yield* leftOf(critB, critA, street);
}

const street = [{}, {}, {}, {}, {}]; // five houses

// Btw: it turns out we don't need uniqueness constraint.

const constraints = [
    s => thereShouldBe({nation: 'English', color: 'red'}, s),
    s => thereShouldBe({nation: 'Swede', animal: 'dog'}, s),
    s => thereShouldBe({nation: 'Dane', drink: 'tea'}, s),
    s => leftOf({color: 'green'}, {color: 'white'}, s),
    s => thereShouldBe({drink: 'coffee', color: 'green'}, s),
    s => thereShouldBe({cigarettes: 'PallMall', animal: 'birds'}, s),
    s => thereShouldBe({color: 'yellow', cigarettes: 'Dunhill'}, s),
    s => thereShouldBeAtIndex({drink: 'milk'}, 2, s),
    s => thereShouldBeAtIndex({nation: 'Norwegian'}, 0, s),
    s => nextTo({cigarettes: 'Blend'}, {animal: 'cats'}, s),
    s => nextTo({animal: 'horse'}, {cigarettes: 'Dunhill'}, s),
    s => thereShouldBe({cigarettes: 'BlueMaster', drink: 'beer'}, s),
    s => thereShouldBe({nation: 'German', cigarettes: 'Prince'}, s),
    s => nextTo({nation: 'Norwegian'}, {color: 'blue'}, s),
    s => nextTo({drink: 'water'}, {cigarettes: 'Blend'}, s),

    s => thereShouldBe({animal: 'zebra'}, s), // should be somewhere
];

function* findSolution(remainingConstraints, street) {
    if (remainingConstraints.length === 0)
        yield street;
    else
        for (const newStreet of _.head(remainingConstraints)(street))
            yield* findSolution(_.tail(remainingConstraints), newStreet);
}

for (const streetSolution of findSolution(constraints, street)) {
    console.log(streetSolution);
}

Resultado:

[ { color: 'yellow',
    cigarettes: 'Dunhill',
    nation: 'Norwegian',
    animal: 'cats',
    drink: 'water' },
  { nation: 'Dane',
    drink: 'tea',
    cigarettes: 'Blend',
    animal: 'horse',
    color: 'blue' },
  { nation: 'English',
    color: 'red',
    cigarettes: 'PallMall',
    animal: 'birds',
    drink: 'milk' },
  { color: 'green',
    drink: 'coffee',
    nation: 'German',
    cigarettes: 'Prince',
    animal: 'zebra' },
  { nation: 'Swede',
    animal: 'dog',
    color: 'white',
    cigarettes: 'BlueMaster',
    drink: 'beer' } ]

El tiempo de ejecución es de alrededor de 2.5 segundos para mí, pero esto se puede mejorar mucho cambiando el orden de las reglas. Decidí mantener el orden original para mayor claridad.

¡Gracias, este fue un gran desafío!

Esto es realmente un problema de resolución de restricciones. Puede hacerlo con un tipo generalizado de propagación de restricciones en programación lógica como los lenguajes. Tenemos una demostración específica para el problema de Zebra en el sistema ALE (motor de lógica de atributos):

http://www.cs.toronto.edu/~gpenn/ale.html

Aquí está el enlace a la codificación de un rompecabezas Zebra simplificado:

http://www.cs.toronto.edu/~gpenn/ale/files/grammars/baby.pl

Hacer esto de manera eficiente es otra cuestión.

La forma más fácil de resolver estos problemas mediante programación es usar bucles anidados en todas las permutaciones y verificar si el resultado satisface los predicados en la pregunta. Muchos de los predicados se pueden izar del bucle interno a los externos para reducir drásticamente la complejidad computacional hasta que la respuesta se pueda calcular en un tiempo razonable.

Aquí hay una solución simple de F # derivada de un artículo en F # Journal :

let rec distribute y xs =
  match xs with
  | [] -> [[y]]
  | x::xs -> (y::x::xs)::[for xs in distribute y xs -> x::xs]

let rec permute xs =
  match xs with
  | [] | [_] as xs -> [xs]
  | x::xs -> List.collect (distribute x) (permute xs)

let find xs x = List.findIndex ((=) x) xs + 1

let eq xs x ys y = find xs x = find ys y

let nextTo xs x ys y = abs(find xs x - find ys y) = 1

let nations = ["British"; "Swedish"; "Danish"; "Norwegian"; "German"]

let houses = ["Red"; "Green"; "Blue"; "White"; "Yellow"]

let drinks = ["Milk"; "Coffee"; "Water"; "Beer"; "Tea"]

let smokes = ["Blend"; "Prince"; "Blue Master"; "Dunhill"; "Pall Mall"]

let pets = ["Dog"; "Cat"; "Zebra"; "Horse"; "Bird"]

[ for nations in permute nations do
    if find nations "Norwegian" = 1 then
      for houses in permute houses do
        if eq nations "British" houses "Red" &&
           find houses "Green" = find houses "White"-1 &&
           nextTo nations "Norwegian" houses "Blue" then
          for drinks in permute drinks do
            if eq nations "Danish" drinks "Tea" &&
               eq houses "Green" drinks "Coffee" &&
               3 = find drinks "Milk" then
              for smokes in permute smokes do
                if eq houses "Yellow" smokes "Dunhill" &&
                   eq smokes "Blue Master" drinks "Beer" &&
                   eq nations "German" smokes "Prince" &&
                   nextTo smokes "Blend" drinks "Water" then
                  for pets in permute pets do
                    if eq nations "Swedish" pets "Dog" &&
                       eq smokes "Pall Mall" pets "Bird" &&
                       nextTo pets "Cat" smokes "Blend" &&
                       nextTo pets "Horse" smokes "Dunhill" then
                      yield nations, houses, drinks, smokes, pets ]

La salida obtenida en 9 ms es:

val it :
  (string list * string list * string list * string list * string list) list =
  [(["Norwegian"; "Danish"; "British"; "German"; "Swedish"],
    ["Yellow"; "Blue"; "Red"; "Green"; "White"],
    ["Water"; "Tea"; "Milk"; "Coffee"; "Beer"],
    ["Dunhill"; "Blend"; "Pall Mall"; "Prince"; "Blue Master"],
    ["Cat"; "Horse"; "Bird"; "Zebra"; "Dog"])]

El ejemplo de Microsoft Solver Foundation de: https://msdn.microsoft.com/en-us/library/ff525831%28v=vs.93%29.aspx?f=255&MSPPError=-2147217396

delegate CspTerm NamedTerm(string name);

public static void Zebra() {
  ConstraintSystem S = ConstraintSystem.CreateSolver();
  var termList = new List<KeyValuePair<CspTerm, string>>();

  NamedTerm House = delegate(string name) {
    CspTerm x = S.CreateVariable(S.CreateIntegerInterval(1, 5), name);
    termList.Add(new KeyValuePair<CspTerm, string>(x, name));
    return x;
  };

  CspTerm English = House("English"), Spanish = House("Spanish"),
    Japanese = House("Japanese"), Italian = House("Italian"),
    Norwegian = House("Norwegian");
  CspTerm red = House("red"), green = House("green"),
    white = House("white"),
    blue = House("blue"), yellow = House("yellow");
  CspTerm dog = House("dog"), snails = House("snails"),
    fox = House("fox"),
    horse = House("horse"), zebra = House("zebra");
  CspTerm painter = House("painter"), sculptor = House("sculptor"),
    diplomat = House("diplomat"), violinist = House("violinist"),
    doctor = House("doctor");
  CspTerm tea = House("tea"), coffee = House("coffee"),
    milk = House("milk"),
    juice = House("juice"), water = House("water");

  S.AddConstraints(
    S.Unequal(English, Spanish, Japanese, Italian, Norwegian),
    S.Unequal(red, green, white, blue, yellow),
    S.Unequal(dog, snails, fox, horse, zebra),
    S.Unequal(painter, sculptor, diplomat, violinist, doctor),
    S.Unequal(tea, coffee, milk, juice, water),
    S.Equal(English, red),
    S.Equal(Spanish, dog),
    S.Equal(Japanese, painter),
    S.Equal(Italian, tea),
    S.Equal(1, Norwegian),
    S.Equal(green, coffee),
    S.Equal(1, green - white),
    S.Equal(sculptor, snails),
    S.Equal(diplomat, yellow),
    S.Equal(3, milk),
    S.Equal(1, S.Abs(Norwegian - blue)),
    S.Equal(violinist, juice),
    S.Equal(1, S.Abs(fox - doctor)),
    S.Equal(1, S.Abs(horse - diplomat))
  );
  bool unsolved = true;
  ConstraintSolverSolution soln = S.Solve();

  while (soln.HasFoundSolution) {
    unsolved = false;
    System.Console.WriteLine("solved.");
    StringBuilder[] houses = new StringBuilder[5];
    for (int i = 0; i < 5; i++)
      houses[i] = new StringBuilder(i.ToString());
    foreach (KeyValuePair<CspTerm, string> kvp in termList) {
      string item = kvp.Value;
      object house;
      if (!soln.TryGetValue(kvp.Key, out house))
        throw new InvalidProgramException(
                    "can't find a Term in the solution: " + item);
      houses[(int)house - 1].Append(", ");
      houses[(int)house - 1].Append(item);
    }
    foreach (StringBuilder house in houses) {
      System.Console.WriteLine(house);
    }
    soln.GetNext();
  }
  if (unsolved)
    System.Console.WriteLine("No solution found.");
  else
    System.Console.WriteLine(
"Expected: the Norwegian drinking water and the Japanese with the zebra.");
}

Esta es una solución MiniZinc para el rompecabezas de cebra como se define en Wikipedia:

include "globals.mzn";

% Zebra puzzle
int: nc = 5;

% Colors
int: red = 1;
int: green = 2;
int: ivory = 3;
int: yellow = 4;
int: blue = 5;
array[1..nc] of var 1..nc:color;
constraint alldifferent([color[i] | i in 1..nc]);

% Nationalities
int: eng = 1;
int: spa = 2;
int: ukr = 3;
int: nor = 4;
int: jap = 5;
array[1..nc] of var 1..nc:nationality;
constraint alldifferent([nationality[i] | i in 1..nc]);

% Pets
int: dog = 1;
int: snail = 2;
int: fox = 3;
int: horse = 4;
int: zebra = 5;
array[1..nc] of var 1..nc:pet;
constraint alldifferent([pet[i] | i in 1..nc]);

% Drinks
int: coffee = 1;
int: tea = 2;
int: milk = 3;
int: orange = 4;
int: water = 5;
array[1..nc] of var 1..nc:drink;
constraint alldifferent([drink[i] | i in 1..nc]);

% Smokes
int: oldgold = 1;
int: kools = 2;
int: chesterfields = 3;
int: luckystrike = 4;
int: parliaments = 5;
array[1..nc] of var 1..nc:smoke;
constraint alldifferent([smoke[i] | i in 1..nc]);

% The Englishman lives in the red house.
constraint forall ([nationality[i] == eng <-> color[i] == red | i in 1..nc]);

% The Spaniard owns the dog.
constraint forall ([nationality[i] == spa <-> pet[i] == dog | i in 1..nc]);

% Coffee is drunk in the green house.
constraint forall ([color[i] == green <-> drink[i] == coffee | i in 1..nc]);

% The Ukrainian drinks tea.
constraint forall ([nationality[i] == ukr <-> drink[i] == tea | i in 1..nc]);

% The green house is immediately to the right of the ivory house.
constraint forall ([color[i] == ivory -> if i<nc then color[i+1] == green else false endif | i in 1..nc]);

% The Old Gold smoker owns snails.
constraint forall ([smoke[i] == oldgold <-> pet[i] == snail | i in 1..nc]);

% Kools are smoked in the yellow house.
constraint forall ([smoke[i] == kools <-> color[i] == yellow | i in 1..nc]);

% Milk is drunk in the middle house.
constraint drink[3] == milk;

% The Norwegian lives in the first house.
constraint nationality[1] == nor;

% The man who smokes Chesterfields lives in the house next to the man with the fox.
constraint forall ([smoke[i] == chesterfields -> (if i>1 then pet[i-1] == fox else false endif \/ if i<nc then pet[i+1] == fox else false endif) | i in 1..nc]);

% Kools are smoked in the house next to the house where the horse is kept.
constraint forall ([smoke[i] == kools -> (if i>1 then pet[i-1] == horse else false endif \/ if i<nc then pet[i+1] == horse else false endif)| i in 1..nc]);

%The Lucky Strike smoker drinks orange juice.
constraint forall ([smoke[i] == luckystrike <-> drink[i] == orange | i in 1..nc]);

% The Japanese smokes Parliaments.
constraint forall ([nationality[i] == jap <-> smoke[i] == parliaments | i in 1..nc]);

% The Norwegian lives next to the blue house.
constraint forall ([color[i] == blue -> (if i > 1 then nationality[i-1] == nor else false endif \/ if i<nc then nationality[i+1] == nor else false endif) | i in 1..nc]);

solve satisfy;

Solución:

Compiling zebra.mzn
Running zebra.mzn
color = array1d(1..5 ,[4, 5, 1, 3, 2]);
nationality = array1d(1..5 ,[4, 3, 1, 2, 5]);
pet = array1d(1..5 ,[3, 4, 2, 1, 5]);
drink = array1d(1..5 ,[5, 2, 3, 4, 1]);
smoke = array1d(1..5 ,[2, 3, 1, 4, 5]);
----------
Finished in 47msec