# RabbitFarm

### 2023-11-05

The Weekly Challenge 241 (Prolog Solutions)

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

## Part 1

*You are given an array (3 or more members) of integers in increasing order and a positive
integer. Write a script to find out the number of unique Arithmetic Triplets satisfying
the given rules.*

### Solution

```
arithmetic_triplets(Numbers, Difference, TripletCount):-
length(Triplet, 3),
member(I, Triplet),
member(J, Triplet),
member(K, Triplet),
fd_domain(Triplet, Numbers),
fd_all_different(Triplet),
Difference #= J - I,
Difference #= K - J,
I #< J,
J #< K,
findall(Triplet, fd_labeling(Triplet), Triplets),
length(Triplets, TripletCount).
```

### Sample Run

```
% gprolog --consult-file prolog/ch-1.p
| ?- arithmetic_triplets([0, 1, 4, 6, 7, 10], 3, TripletCount).
TripletCount = 2 ?
yes
| ?- arithmetic_triplets([4, 5, 6, 7, 8, 9], 2, TripletCount).
TripletCount = 2 ?
yes
| ?-
```

### Notes

The rules for *arithmetic triples* are a) i < j < k b) nums[j] - nums[i] == diff and
c) nums[k] - nums[j] == diff, where diff is a provided parameter. The code above
implements these rules directly, letting Prolog do all the work for us!

## Part 2

*You are given an array of unique positive integers greater than 2. Write a script to sort
them in ascending order of the count of their prime factors, tie-breaking by ascending
value.*

### Solution

```
prime_factors(N, L):-
N > 0,
prime_factors(N, L, 2).
prime_factors(1, [], _):-
!.
prime_factors(N, [F|L], F):-
R is N // F,
N =:= R * F,
!,
prime_factors(R, L, F).
prime_factors(N, L, F):-
next_factor(N, F, NF),
prime_factors(N, L, NF).
next_factor(_, 2, 3):-
!.
next_factor(N, F, NF):-
F * F < N,
!,
NF is F + 2.
next_factor(N, _, N).
kvf_insert_sort(List,Sorted):-
i_sort(List,[],Sorted).
i_sort([],Acc,Acc).
i_sort([H|T],Acc,Sorted):-
kvf_insert(H,Acc,NAcc),
i_sort(T,NAcc,Sorted).
kvf_insert(K0-V0,[K1-V1|T],[K1-V1|NT]):-
V0 > V1,
kvf_insert(K0-V0,T,NT).
kvf_insert(K0-V0,[K1-V1|T],[K0-V0,K1-V1|T]):-
V0 < V1.
kvf_insert(K0-V0,[K1-V1|T],[K1-V1|NT]):-
V0 = V1,
K0 > K1,
kvf_insert(K0-V0,T,NT).
kvf_insert(K0-V0,[K1-V1|T],[K0-V0,K1-V1|T]):-
V0 = V1,
K0 < K1.
kvf_insert(K0-V0, [], [K0-V0]).
write_factor_sorted([K-_|[]]):-
write(K),
nl.
write_factor_sorted([K-_|T]):-
write(K),
write(', '),
write_factor_sorted(T).
factor_counter(Number, Number-FactorCount):-
prime_factors(Number, Factors),
length(Factors, FactorCount).
factor_sorter(Numbers, FactorsSorted):-
maplist(factor_counter, Numbers, Factors),
kvf_insert_sort(Factors, FactorsSorted).
```

### Sample Run

```
% gprolog --consult-file prolog/ch-2.p
| ?- factor_sorter([11, 8, 27, 4], FactorsSorted), write_factor_sorted(FactorsSorted).
11, 4, 8, 27
FactorsSorted = [11-1,4-2,8-3,27-3] ?
yes
| ?-
```

### Notes

This code is build mainly from pieces from previous challenges. The prime factorization code is something I've used several times and the modified Insertion Sort is a minor modification of code from TWC 233.

## References

posted at: 23:23 by: Adam Russell | path: /prolog | permanent link to this entry

#### Recursive Loops and Code Re-Use

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

## Part 1

*You are given an array (3 or more members) of integers in increasing order and a positive
integer. Write a script to find out the number of unique Arithmetic Triplets satisfying
the given rules.*

### Solution

```
use v5.38;
sub arithmetic_triplets{
my $counter = 0;
my $difference = shift;
arithmetic_triplets_r($difference, \$counter, [@_[0 .. @_ -1]], [@_[1 .. @_ -1]], [@_[2 .. @_ -1]]);
return $counter;
}
sub arithmetic_triplets_r{
my $difference = $_[0];
my $counter = $_[1];
my @i = @{$_[2]};
my @j = @{$_[3]};
my @k = @{$_[4]};
if(@i > 0 && @j > 0 && @k > 0){
$$counter++ if $j[0] - $i[0] == $difference && $k[0] - $j[0] == $difference;
arithmetic_triplets_r($difference, $counter, [@i], [@j], [@k[1 .. @k - 1]]);
}
elsif(@i > 0 && @k == 0 && @j > 0){
arithmetic_triplets_r($difference, $counter, [@i], [@j[1 .. @j - 1]], [@j[2 .. @j - 1]]);
}
elsif(@i > 0 && @k == 0 && @j == 0){
arithmetic_triplets_r($difference, $counter, [@i[1 .. @i - 1]], [@i[2 .. @i - 1]], [@i[3 .. @i - 1]]);
}
}
MAIN:{
my $difference;
$difference = 3;
say arithmetic_triplets $difference, 0, 1, 4, 6, 7, 10;
$difference = 2;
say arithmetic_triplets $difference, 4, 5, 6, 7, 8, 9;
}
```

### Sample Run

```
$ perl perl/ch-1.pl
2
2
```

### Notes

The rules for *arithmetic triples* are a) i < j < k b) nums[j] - nums[i] == diff and
c) nums[k] - nums[j] == diff, where diff is a provided parameter. The code above
implements these rules somewhat in the obvious way, looping thricely over the list, but
recursively.

## Part 2

*You are given an array of unique positive integers greater than 2. Write a script to sort
them in ascending order of the count of their prime factors, tie-breaking by ascending
value.*

### Solution

```
use v5.38;
sub prime_factor{
my $x = shift(@_);
my @factors;
for (my $y = 2; $y <= $x; $y++){
next if $x % $y;
$x /= $y;
push @factors, $y;
redo;
}
return @factors;
}
sub prime_order{
my %factor_i = map{($_, 0 + prime_factor($_))} @_;
my $factor_sorter = sub{
my $c = $factor_i{$a} <=> $factor_i{$b};
return $c unless !$c;
return $a <=> $b;
};
return sort $factor_sorter @_;
}
MAIN:{
say join q/, /, prime_order 11, 8, 27, 4;
}
```

### Sample Run

```
$ perl perl/ch-2.pl
11, 4, 8, 27
```

### Notes

This code borrows from two previous challenges: The prime factor code has been used
several times, but in this case I referred to the Attractive Number challenge from TWC
041. The sorting is a variant of the *frequency sort* from TWC 233. If you write enough
code you don't need GitHub Copilot, you can just re-use your own work!

## References

posted at: 18:19 by: Adam Russell | path: /perl | permanent link to this entry