# RabbitFarm

### 2022-10-30

The Weekly Challenge 188 (Prolog Solutions)

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

## Part 1

*You are given list of integers @list of size $n and divisor $k. Write a script to
find out count of pairs in the given list that satisfies a set of rules.*

### Solution

```
divisible_pair(Numbers, K, Pair):-
length(Numbers, NumbersLength),
between(1, NumbersLength, I),
succ(I, INext),
between(INext, NumbersLength, J),
nth(I, Numbers, Ith),
nth(J, Numbers, Jth),
IJModK is (Ith + Jth) mod K,
IJModK == 0,
Pair = [I, J].
divisible_pairs(Numbers, K, Pairs):-
findall(Pair, divisible_pair(Numbers, K, Pair), Pairs).
```

### Sample Run

```
$ gprolog --consult-file prolog/ch-1.p
| ?- divisible_pairs([4, 5, 1, 6], 2, Pairs), length(Pairs, NumberPairs).
NumberPairs = 2
Pairs = [[1,4],[2,3]]
(1 ms) yes
| ?- divisible_pairs([1, 2, 3, 4], 2, Pairs), length(Pairs, NumberPairs).
NumberPairs = 2
Pairs = [[1,3],[2,4]]
(1 ms) yes
| ?- divisible_pairs([1, 3, 4, 5], 3, Pairs), length(Pairs, NumberPairs).
NumberPairs = 2
Pairs = [[1,4],[3,4]]
yes
| ?- divisible_pairs([5, 1, 2, 3], 4, Pairs), length(Pairs, NumberPairs).
NumberPairs = 2
Pairs = [[1,4],[2,4]]
yes
| ?- divisible_pairs([7, 2, 4, 5], 4, Pairs), length(Pairs, NumberPairs).
NumberPairs = 1
Pairs = [[1,4]]
yes
```

### Notes

The rules, if not clear from the above code are : the pair (i, j) is eligible if and only if

0 <= i < j < len(list)

list[i] + list[j] is divisible by k

There really is not too much here beyond translating the rules into Prolog. The first
condition on i and j are handled by default using `between/3`

. Once we define how to find
one such pair we use `findall/3`

to obtain them all.

## Part 2

*You are given two positive integers $x and $y. Write a script to find out the number of
operations needed to make both ZERO.*

### Solution

```
count_zero(X, Y, Count):-
count_zero(X, Y, 0, Count), !.
count_zero(0, 0, Count, Count).
count_zero(X, Y, CountAccum, Count):-
X > Y,
XNew is X - Y,
succ(CountAccum, CountAccumSucc),
count_zero(XNew, Y, CountAccumSucc, Count).
count_zero(X, Y, CountAccum, Count):-
Y > X,
YNew is Y - X,
succ(CountAccum, CountAccumSucc),
count_zero(X, YNew, CountAccumSucc, Count).
count_zero(X, Y, CountAccum, Count):-
X == Y,
XNew is X - Y,
YNew is Y - X,
succ(CountAccum, CountAccumSucc),
count_zero(XNew, YNew, CountAccumSucc, Count).
```

### Sample Run

```
$ gprolog --consult-file prolog/ch-2.p
| ?- count_zero(5, 4, Count).
Count = 5
(1 ms) yes
| ?- count_zero(4, 6, Count).
Count = 3
yes
| ?- count_zero(2, 5, Count).
Count = 4
yes
| ?- count_zero(3, 1, Count).
Count = 3
yes
| ?- count_zero(7, 4, Count).
Count = 5
yes
```

### Notes

The operations are dictated by these rules:

`$x = $x - $y if $x >= $y`

or

`$y = $y - $x if $y >= $x (using the original value of $x)`

Be carefully examining the rules we can see that we can arrange `count_zero/4`

predicates
in a somewhat concise way. I find this preferable to Prolog's if/else syntax which
absolutely could have been used here. I would argue that the slightly longer form here is
worthwhile in that it is much more readable.

One thing I found especially convenient was that due to the immutable nature of Prolog
variables we don;t have to do any extra accounting for the possibly changed value of `X`

when computing an updated `Y`

. The
Perl solution to this
requires a temporary variable, for example.

## References

posted at: 18:55 by: Adam Russell | path: /prolog | permanent link to this entry