Prolog: Solving a Puzzle

Question

I am new to prolog and I am trying to solve the following question. I am having difficulty trying to understand the logic to solve the problem. I know its similar to zebra problem but, I am unsure how to approach. Any help would be greatly appreciated.

The answers submitted by five students to a T/F quiz are as follows.

Teresa: T T F T F
Tim:    F T T T F
Tania:  T F T T F
Tom:    F T T F T
Tony:   T F T F T

Tania got more answers right than Teresa did.
Tom got more right than Tim.
Tony did not get all the answers right, nor did he get them all wrong.

Write a Prolog program quiz(Answer) that asserts Answer is the list of t and f constants that is the correct answer to the quiz..

Sounds like a fun puzzle. Why do you want to ruin it for yourself by cheating by asking for the solution here? — wvdz, Jan 29 '15 at 21:12
I am not looking for the solution, I just need help to figure out the logic for it. A flow of steps that i could potentially use. — mayak, Jan 30 '15 at 00:18
Think about what facts are being stated first (each student with their answers) and decide how you want to represent them. For example, `answers(teresa, [t,t,f,t,f]).`, etc. You could create a predicate to generate possible correct answer sets and compute each person's score against it. Backtrack when the conditions for score comparisons don't match. There really is only one result. — lurker, Jan 30 '15 at 02:28

joel76 · Accepted Answer · 2015-01-30T11:05:31.507

1

If you use SWI-Prolog, you can use library clpfd to solve the puzzle :, I get only one solution (f,f,t,f,t).

You have a solution [A,B,C,D,E]. You initialize the possibles solutions with

[A,B,C,D,E] ins 0..1,

You reify all the answers for teresa for example

teresea([1,1,0,1,0]).
A #= 1 #<==> TA
B #= 1 #<==> TB
.....

you compute the sum of Tis

sum([TA, TB, ...], #= , Steresa),

and later you will have for Tania got more answers right than Teresa did.

Stania #> Steresa

You get the solution with

label([A,B,C,D,E]).

Hope this helps

edited Jan 30 '15 at 11:05

answered Jan 30 '15 at 08:26

joel76

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Thank you for posting this. It helped me a lot in figuring out and starting the problem. Now, I am practicing more problems like this. This was my first question on stackoverflow :) – mayak Feb 02 '15 at 17:31
If you are OK with this answer please click for accept it – joel76 Feb 02 '15 at 17:52

CapelliC · Answer 2 · 2015-01-31T15:56:55.597

small puzzles like this can be solved by generate-and-test

solve(L) :-
    % generator
    length(L, 5), maplist(tf, L),

    % Tania got more answers right than Teresa did.
    matches(L, tania, Tania),
    matches(L, teresa, Teresa), Tania > Teresa,
...

tf(t).
tf(f).

teresa(t, t, f, t, f).
tim(f, t, t, t, f).
...

Of course, matches(L, tania, Tania) counts correct Tania' answers.

But, I don't find a solution. The only tuple that 'get thru' Tony, it's its exact result. So, this condition

Tony did not get all the answers right

cannot be solved...

edit I had a bug in matches/3. Of course there is a solution.

edit well, the CLP(FD) version can be very compact, while being more general...

teresa(t, t, f, t, f).
...

matches(L, P, N) :-
    call(P, A,B,C,D,E),
    foldl(eqsum, [A,B,C,D,E], L, 0, N).
eqsum(t,Ls,Acc,N) :-    N #= Acc + (° #<==> Ls #= 1).
eqsum(f,Ls,Acc,N) :-    N #= Acc + (° #<==> Ls #= 0).

solve(L) :-
    length(°L, 5) ins 0..1,
    % Tania got more answers right than Teresa did.
    matches(L, tania, °) #> matches(L, teresa, °),
    % Tom got more right than Tim.
    matches(L, tom, °) #> matches(L, tim, °),
    % Tony did not get all the answers right, nor did he get them all wrong.
    matches(L, tony, °Tony) #> 0, Tony #< 5.

I used my lifter here.

Prolog: Solving a Puzzle

2 Answers2