RabbitFarm

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

Part 1

You are given an array of integers. Write a script to calculate the number of integers smaller than the integer at each index.

Solution

smaller([], _, 0).
smaller([H|Integers], X, Y):-
    smaller(Integers, X, Y0),
    ((X > H, succ(Y0, Y));
     (X =< H, Y = Y0)).

count_smaller(Integers, CountSmaller):-
    maplist(smaller(Integers), Integers, CountSmaller).

Sample Run

% gprolog --consult-file prolog/ch-1.p
| ?- count_smaller([2, 2, 2], CountSmaller). 

CountSmaller = [0,0,0]

yes
| ?- count_smaller([6, 5, 4, 8], CountSmaller).

CountSmaller = [2,1,0,3] ? 

yes
| ?- count_smaller([8, 1, 2, 2, 3], CountSmaller).

CountSmaller = [4,0,1,1,3] ? 

yes
| ?- 

Notes

Probably this is the most obvious way to count up smaller elements as required. In order to cut down on the recursion I call smaller/3 via a maplist/3.

Part 2

You are given an array of integers representing the strength. Write a script to return the sum of the powers of all possible combinations; power is defined as the square of the largest number in a sequence, multiplied by the smallest.

Solution

group_hero(Group, GroupHero):-
    findall(Hero, (
        sublist(SubList, Group),
        max_list(SubList, Maximum),
        min_list(SubList, Minimum),
        Hero #= Maximum**2 * Minimum
    ), Heroes),
    sum_list(Heroes, GroupHero). 

Sample Run

% gprolog --consult-file prolog/ch-2.p 
| ?- group_hero([2, 1, 4], GroupHero).

GroupHero = 141

yes
| ?-

Notes

The core of this problem is to enumerate all the Power Sets of the Group list. In other programming languages enumerating all sublists of a list is straightforward enough, but requires much more code. Here, with Prolog, we need only call sublist/2 with backtracking. We use a findall/3 to generate all the necessary backtracking and create the list of intermediate sums, which are then all summed for the final solution.

References

Challenge 244