RabbitFarm

The examples used here are from the weekly challenge problem statement and demonstrate the working solution.

Part 2

You are given a list of positive numbers, @n, having at least 3 numbers. Write a script to find the triplets (a, b, c) from the given list that satisfies a set of rules.

Solution

magical_triple_sum(Numbers, Triple, TripleSum):-
    sublist([A, B, C], Numbers),
    A + B > C,
    B + C > A,
    A + C > B,
    Triple = [A, B, C],
    sum_list(Triple, TripleSum).

magical_triple(Numbers, Triple):-
    fd_maximize(magical_triple_sum(Numbers, Triple, TripleSum), TripleSum). 

Sample Run

$ gprolog --consult-file prolog/ch-2.p
| ?- magical_triple([4, 2, 3], Triple).

Triple = [4,2,3] ? 

(1 ms) yes
| ?- magical_triple([1, 3, 2], Triple). 

no
| ?- magical_triple([1, 1, 3, 2], Triple).

no
| ?- magical_triple([1, 2, 3, 2], Triple).

Triple = [2,3,2] ? 

yes

Notes

The "magical" rules, if not clear from the above code are:

• a + b > c

• b + c > a

• a + c > b

• a + b + c is maximum.

I don't routinely do a lot of constraint programming, but I don't think a smaller solution for this can be written! Indeed, the code here is largely a rewriting of the given rules in slightly modified Prolog form.

References

Challenge 187