RabbitFarm

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

Part 1

You are given a list of integers, @list. Write a script to find the total count of Good airs.

Solution

good_pair(Numbers, Pair):-
    length(Numbers, L),
    fd_domain(I, 1, L),
    fd_domain(J, 1, L),
    I #<# J,
    fd_labeling([I, J]), 
    fd_element(I, Numbers, Ith),  
    fd_element(J, Numbers, Jth), 
    Ith #= Jth,
    Pair = [I, J].   

Sample Run

$ gprolog --consult-file prolog/ch-1.p
| ?- good_pair([1, 2, 3, 1, 1, 3], Pair).

Pair = [1,4] ? a

Pair = [1,5]

Pair = [3,6]

Pair = [4,5]

no
| ?- good_pair([1, 2, 3], Pair).         

no
| ?- good_pair([1, 1, 1, 1], Pair).

Pair = [1,2] ? a

Pair = [1,3]

Pair = [1,4]

Pair = [2,3]

Pair = [2,4]

Pair = [3,4]

yes
| ?- 

Notes

I take a clpfd approach to this problem and the next. This allows a pretty concise solution. Here we get the length of the list of numbers, constrain the indices for the pair and then specify the additional conditions of a Good Pair.

Part 2

You are given an array of integers, @array and three integers $x,$y,$z. Write a script to find out total Good Triplets in the given array.

Solution

good_triplet(Numbers, X, Y, Z, Triplet):-
    length(Numbers, I),
    fd_domain(S, 1, I),
    fd_domain(T, 1, I),
    fd_domain(U, 1, I),
    S #<# T, T #<# U,   
    fd_labeling([S, T, U]),   
    fd_element(S, Numbers, Sth),  
    fd_element(T, Numbers, Tth),  
    fd_element(U, Numbers, Uth), 
    Ast is abs(Sth - Tth), Ast #=<# X,     
    Atu is abs(Tth - Uth), Atu #=<# Y,     
    Asu is abs(Sth - Uth), Asu #=<# Z, 
    Triplet = [Sth, Tth, Uth].   

Sample Run

$ gprolog --consult-file prolog/ch-2.p
| ?- good_triplet([3, 0, 1, 1, 9, 7], 7, 2, 3, Triplet).

Triplet = [3,0,1] ? a

Triplet = [3,0,1]

Triplet = [3,1,1]

Triplet = [0,1,1]

no
| ?- good_triplet([1, 1, 2, 2, 3], 0, 0, 1, Triplet).   

no
| ?-

Notes

Again for part 2 a clpfd solution ends up being fairly concise. In fact, the approach here is virtually identical to part 1. The differences are only that we are looking for a triple, not a pair, and slightly different criteria.

References

Challenge 199