Einsteins Riddle Prolog

Question

I need some help with a prolog homework for my AI class. The question is to write prolog code for einstein\'s puzzle. I know how to write it down in my own but there are som

Answer 1

This puzzle (also know as the Zebra Puzzle) has been discussed many times on Stackoverflow before, see e.g.:

• Solving "Who owns the Zebra" programmatically?
• Why cant i get the answer to the zebra puzzle in prolog?
• Solving logic puzzle in Prolog
• Einstein's riddle

Answer 2

A Prolog translation can be straightforward, rule by rule, still following the paradigm of instantiating the domain by selecting from it. Here it's the domain of house attributes; in the linked answer house attributes are fixed by a human programmer and the domain is the actual inhabited houses, which allows for a very succinct encoding.

In other words the difference is in the notation: a sophisticated notation takes us half way there already, but it was a human who invented it and followed it (like the programmer having to write down norwegian in the first house's specification directly, at the appropriate argument position) -- not a computer.

Here we try to inject as little human knowledge into the code as possible, following the homework's constraints. _{(though anything is debatable of course, and the ultimate in eschewing human interference would be a computer program that takes English text as its input ... which would again be open to criticism as to how specifically tailored that program is for finding solutions to this specific puzzle, or type of puzzles, etc., etc.)}

We code it in the top-down style. Apparently, the question is missing. It should be "who drinks water? who owns the zebra?":

zebra( Z, W ,HS) :-         
    length(        HS, 5),      % nation? color? what's that? define it later...
    member(  H1,   HS),    nation( H1, eng    ),    color( H1, red    ),
    member(  H2,   HS),    nation( H2, spa    ),    owns(  H2, dog    ),            
    member(  H3,   HS),    drink(  H3, coffee ),    color( H3, green  ),         
    member(  H4,   HS),    nation( H4, ukr    ),    drink( H4, tea    ),
    right_of(B, A, HS),    color(  A , ivory  ),    color( B , green  ),
    member(  H5,   HS),    smoke(  H5, oldgold),    owns(  H5, snails ),   
    member(  H6,   HS),    smoke(  H6, kools  ),    color( H6, yellow ), 
    middle(  C,    HS),    drink(  C , milk   ),  
    first(   D,    HS),    nation( D , nor    ),
    next_to( E, F, HS),    smoke(  E , chester),    owns(  F , fox    ),
    next_to( G, H, HS),    smoke(  G , kools  ),    owns(  H , horse  ),
    member(  H7,   HS),    smoke(  H7, lucky  ),    drink( H7, orange ),
    member(  H8,   HS),    nation( H8, jpn    ),    smoke( H8, parlamt),
    next_to( I, J, HS),    nation( I , nor    ),    color( J , blue   ),
    member(  W,    HS),    drink(  W , water  ),
    member(  Z,    HS),    owns(   Z , zebra  ).

right_of( B, A, HS) :- append( _, [A, B | _], HS).
next_to( A, B, HS) :- right_of( B, A, HS) ; right_of( A, B, HS).
middle( A, [_,_,A,_,_]).
first( A, [A | _]).

nation(H, V) :-  attr( H, nation-V).
owns(  H, V) :-  attr( H, owns-V).        % select an attribute
smoke( H, V) :-  attr( H, smoke-V).       %   from an extensible record H
color( H, V) :-  attr( H, color-V).       %   of house attributes
drink( H, V) :-  attr( H, drink-V).       %   which *is* a house

attr(House, Attr-Value) :- 
    memberchk( Attr-X, House),            % unique attribute names
    X = Value.

Testing, performing exhaustive search with a failure-driven loop,

3 ?- time((zebra(Z,W,_), maplist(nation,[Z,W],R), writeln(R), false ; true)).
[jpn,nor]
% 180,974 inferences, 0.016 CPU in 0.020 seconds (78% CPU, 11600823 Lips)
true.

Here's how the houses end up being defined:

5 ?- zebra(_, _, HS), maplist( writeln, HS),
     false.
[smoke-kools,  color-yellow, nation-nor,    owns-fox,      drink-water |_G859]
[nation-ukr,   drink-tea,    smoke-chester, owns-horse,    color-blue  |_G853]
[nation-eng,   color-red,    smoke-oldgold, owns-snails,   drink-milk  |_G775]
[nation-spa,   owns-dog,     color-ivory,   smoke-lucky,   drink-orange|_G826]
[drink-coffee, color-green,  nation-jpn,    smoke-parlamt, owns-zebra  |_G865]
false.

or, with attributes lists "frozen" by fixing their length, and then sorted,

7 ?- zebra( _, _, HS), maplist( length, HS, _), !, maplist( sort, HS, S),
     maplist( writeln, S), false.
[color-yellow, drink-water,  nation-nor,  owns-fox,    smoke-kools  ]
[color-blue,   drink-tea,    nation-ukr,  owns-horse,  smoke-chester]
[color-red,    drink-milk,   nation-eng,  owns-snails, smoke-oldgold]
[color-ivory,  drink-orange, nation-spa,  owns-dog,    smoke-lucky  ]
[color-green,  drink-coffee, nation-jpn,  owns-zebra,  smoke-parlamt]
false.

It is also easy to make the attr/2 predicate accept lists of Name-Value pairs, allowing for more naturally flowing, higher-level looking coding style with kind of "extensible records" -- one might even say "objects" -- specifications, like

zebra( Z, W ,HS):-         
    length(       HS, 5), 
    member(  H1,  HS),    attr( H1,  [nation-eng,   color-red  ] ),
    member(  H2,  HS),    attr( H2,  [nation-spa,   owns-dog   ] ),
    member(  H3,  HS),    attr( H3,  [drink-coffee, color-green] ),
    ......

etc..

Answer 3

This site is devoted to solve such puzzles with CLP(FD). But the full power of CLP(FD) is overkill here: your assignment can be solved effectively searching the entire solution space when you have adequately described the constraints.

The solution will be composed of 5 houses, where each attribute satisfy all constraints imposed by description.

Be aware to use the very same symbol for each attribute (i.e. green and green_house is wrong, choose one of them).

Also next_to seems wrong: if you number from 1 to 5, this can be computed or enumerated, but refers to the immediate neighbour.

So complete the 'solution search space' data representation, something like

Problem = [
 house(1, Nationality1, Color1, Pet1, Drinks1, Smokes1),
 house(2, Nationality2, Color2, Pet2, Drinks2, Smokes2),
 ...
],
% place constraints
member(house(_, englishman, red, _, _, _), Problem),
member(house(_, spaniard, _, dog, _, _), Problem),
...

member/2 it's the simpler Prolog builtin, but in this case suffices to solve the problem: when all constraints have been posted, the variables will bind to appropriate values. The key is the ability of member to non deterministically select a member (duh) of the solution.

So when you need to express a constraint between 2 different elements, invoke 2 times member, and place the constraints between appropriate variables: i.e.

the man who smokes Chesterelds lives in the house next to the man with the fox

will be translated to

....,
member(house(N, _, _, _, _, chesterelds), Problem),
member(house(M, _, _, fox, _, _), Problem),
next_to(N, M),
...

When expressing many constraints in such way, beware to symbols identity: could be useful to code each predicate in a separate procedure, to avoid undue aliasing. But the couterpart is also true: if the same symbol is involved in more than a constraint, will be necessary to pass around the symbol, to narrow down the search.

I will let you to think about the right representation of 'geometric' predicates: next_to and right_of could be enumerated, or expressed by means of arithmetic.