Advice on solving zebra type puzzle

Question

I need your help in solving the following problem:

There're 3 girls (Ann,Susan,Alice) who need to choose what color shoes and dress to wear. There're 3 possible colors for shoes and dresses: white, blue and green.

Main conditions:

Ann hates white.
Susan wears shoes and dress of same color.
Alice has white shoes.
Alice and Ann's shoes and dress have different colors.

My code satisfies only 2 conditions; I'm kinda having hard times meeting conditions on the same color for Susan and whereas other girls need to have different color garments.

Here's what I come up with:

PREDICATES
   girl(symbol)
   shoes(symbol,symbol)
   skirt(symbol,symbol)
   hates(symbol,symbol)
   will_wear(symbol, symbol, symbol)


CLAUSES
   will_wear(X,Y,Z):-
      girl(X),
      shoes(X,Y),
      skirt(X,Z),
      not(hates(X,Y)),
      not(hates(X,Z)).

   girl(ann).
   girl(susan).
   girl(alice).

   hates(ann,white).

   skirt(_,white).
   skirt(_,blue).
   skirt(_,green).

   shoes(alice,white).
   shoes(_,blue).
   shoes(_,green).

GOAL
   will_wear(Name,Shoes,Dress).

Code above works fine, but gives too many solutions. Plus, I couldn't come up with any logical solution for the condition for Susan to wear shoes and dress of same color.

Thanks.

Is this homework? It's ok if it is, just let us know and tell us what you've tried so far.... — Matthew, Oct 06 '11 at 16:34
It actually is. The code above is the thing I came up with myself. Teacher gave us only conditions. I think main problem is the logic of solution. I've already been thinking on this for 2 hours. Kinda makes me wanna laugh at myself for that :) — Luke, Oct 06 '11 at 16:37
Don't feel bad. I spent way more than 2 hours on my first 'simple' prolog assignment. Keep at it. It's a different way of thinking. I'd love to help you, but all that knowledge has left my brain years ago... :) — i_am_jorf, Oct 06 '11 at 16:45

score 1 · Answer 1 · answered Oct 06 '11 at 17:29

1

Off the top of my head, I'm thinking something along these lines:

only_wears(Girl,Color):-
    shoes(Girl, Color),
    skirt(Girl, Color).

different_shoes(F, S):-
    shoes(F,F_color),
    shoes(S,S_color),
    not(equals(F_color,S_color)).

different_skirts(F, S):-
    skirt(F,F_color),
    skirt(S,S_color),
    not(equals(F_color,S_color)).

I do wonder if there's a way to pass clauses to other clauses, because different_shoes and different_skirts are identical in structure.

You would initialize it like so:

only_wears(ann, white).
different_shoes(alice, ann).
different_skirt(alice, ann).

answered Oct 06 '11 at 17:29

Ehtesh Choudhury

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`not(equals(A,B))` should really be `A \= B`, I think – configurator Oct 06 '11 at 17:55
And about passing clauses, it's possible but not pretty. – configurator Oct 06 '11 at 17:55
i had the same idea on clauses, but couldn't find answer anywhere. Your version still gives 19 solutions, but i need only 3. – Luke Oct 06 '11 at 17:57
Do you mean there should only be 3 solutions? – Ehtesh Choudhury Oct 06 '11 at 18:00
yep, other solutions are just duplicates or it's impossible (girls can't wear same shoes or dress) – Luke Oct 06 '11 at 18:03

score 1 · Accepted Answer · answered Oct 06 '11 at 17:58

1

If I understand the conditions correctly, they're not what Shurane answered.

This will make sure that a girl wears a dress and shoes with the same colour:

same_color(Girl) :-
    shoes(Girl, Color),
    dress(Girl, Color).

I'll leave the different colour one as an exercise, but hint that to say two things are not the same you say A \= B. Please leave a comment if you have a hard time with different_color - and tell me what you've tried.

answered Oct 06 '11 at 17:58

configurator

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I found a solution for similar problem online. Like I said logic there is a bit different. Thank you all for your help. Will try to develop my version of solution though :) – Luke Oct 06 '11 at 18:05
idea is that for similar colors we use correspond clause correspond(X,Y,Z):-girl(X),shoes(Y),dress(Z), X=susan,Y=green,Y<>Z. And that should be customized for each of the girls. Then there's solution clause: solution(X1,Y1,Z1,X2,Y2,Z2,X3,Y3,Z3):- X1=susan,correspond(X1,Y1,Z1), X2=alice, correspond(X2,Y2,Z2), X3=ann, correspond(X3,Y3,Z3), Y1<>Y2, Y2<>Y3, Y1<>Y3, Z1<>Z2, Z2<>Z3, Z1<>Z3. – Luke Oct 06 '11 at 18:07
Isn't that the same as `only_wears` in my code? Also, `dress()` ought to be `skirt()`. – Ehtesh Choudhury Oct 06 '11 at 18:07
I think what configurator meant was that clause's logic has to be bind to "color" attribute – Luke Oct 06 '11 at 18:10
Ooh, I see. Although, that still seems a lot like `only_wears(Girl, _)` – Ehtesh Choudhury Oct 06 '11 at 18:40
@Shurane: Yes, it _does_ the same thing. But it doesn't _mean_ the same thing. The important thing in software development in general and Prolog in particular is that the meaning are all correct. Everything else follows from that. – configurator Oct 06 '11 at 21:35

Advice on solving zebra type puzzle

2 Answers2