RabbitFarm
2025-04-27
The Weekly Challenge 318 (Prolog Solutions)
The examples used here are from the weekly challenge problem statement and demonstrate the working solution.
Part 1: Group Position
You are given a string of lowercase letters. Write a script to find the position of all groups in the given string. Three or more consecutive letters form a group. Return “” if none found.
We can do this in a single predicate which uses maplist to get the groupings with a small utility predicate.
-
group(Letters, Letter, Group):-
length(Letters, LengthLetters),
delete(Letters, Letter, Deleted),
length(Deleted, LengthDeleted),
Difference is LengthLetters - LengthDeleted,
Difference >= 3,
length(G1, Difference),
maplist(=(Letter), G1),
append(G1, _, G2),
append(_, G2, Letters),
atom_codes(Group, G1).
group(_, _, nil).
◇
-
Fragment referenced in 3.
-
groupings(Word, Groupings):-
sort(Word, UniqueLetters),
maplist(group(Word), UniqueLetters, Groups),
delete(Groups, nil, Groupings).
◇
-
Fragment referenced in 3.
The rest of the code just wraps this single predicate into a file.
Sample Run
$ gprolog --consult-file prolog/ch-1.p | ?- groupings("abccccd", Groupings). Groupings = [cccc] ? yes | ?- groupings("aaabcddddeefff", Groupings). Groupings = [aaa,dddd,fff] ? yes | ?- groupings("abcdd", Groupings). Groupings = [] yes | ?-
Part 2: Reverse Equals
You are given two arrays of integers, each containing the same elements as the other. Write a script to return true if one array can be made to equal the other by reversing exactly one contiguous subarray.
This is going to be a quick one, but there’s going to be a few pieces we need to take care of. First we will check that we can subtract/3 the two words (character code lists) and obtain an empty list. Then we’ll check in which places the words differ. If they differ in one place or more then we’re done. Otherwise we’ll test the reversal of the sublist.
-
length(DifferenceIndices, NumberDifferences),
NumberDifferences > 0,
nth(1, DifferenceIndices, FirstIndex),
last(DifferenceIndices, LastIndex),
findall(E, (
between(FirstIndex, LastIndex, I),
nth(I, List1, E)
), SubList1),
findall(E, (
between(FirstIndex, LastIndex, I),
nth(I, List2, E)
), SubList2),
◇
-
reverse(SubList1, Reverse1),
reverse(SubList2, Reverse2),
append(SubList1, Suffix1, S1),
append(SubList2, Suffix2, S2),
append(Reverse1, Suffix1, S3),
append(Reverse2, Suffix2, S4),
append(Prefix1, S1, List1),
append(Prefix2, S2, List2),
(append(Prefix1, S3, List2); append(Prefix2, S4, List1))
◇
All these pieces will be assembled into reverse_equals/2.
Finally, let’s assemble our completed code into a single file.
Sample Run
$ gprolog --consult-file prolog/ch-2.p | ?- reverse_equals([3, 2, 1, 4], [1, 2, 3, 4]). true ? yes | ?- reverse_equals([1, 3, 4], [4, 1, 3]). no | ?- reverse_equals([2], [2]). yes | ?-
References
posted at: 23:10 by: Adam Russell | path: /prolog | permanent link to this entry
Group Position Reversals
The examples used here are from the weekly challenge problem statement and demonstrate the working solution.
Part 1: Group Position
You are given a string of lowercase letters. Write a script to find the position of all groups in the given string. Three or more consecutive letters form a group. Return “” if none found.
Here’s our one subroutine, this problem requires very little code.
-
sub groupings{
my($s) =
@_;
my
@groups;
my
@group;
my($current, $previous);
my
@letters = split //, $s;
$previous = shift
@letters;
group = ($previous);
do {
$current = $_;
if($previous eq $current){
push
@group, $current;
}
if($previous ne $current){
if(
@group >= 3){
push
@groups, [
@group];
}
group = ($current);
}
$previous = $current;
} for
@letters;
if(
@group >= 3){
push
@groups, [
@group];
}
my
@r = map {q/"/␣.␣join(q//,␣
@{$_}) . q/"/␣}␣
@groups;
return join(q/, /,
@r) || q/""/;
}
◇
-
Fragment referenced in 2.
Putting it all together...
The rest of the code just runs some basic tests.
-
MAIN:{
say groupings q/abccccd/;
say groupings q/aaabcddddeefff/;
say groupings q/abcdd/;
}
◇
-
Fragment referenced in 2.
Sample Run
$ perl perl/ch-1.pl "cccc" "aaa", "dddd", "fff" ""
Part 2: Reverse Equals
You are given two arrays of integers, each containing the same elements as the other. Write a script to return true if one array can be made to equal the other by reversing exactly one contiguous subarray.
Here’s the process we’re going to follow.
- scan both arrays and check where and how often they differ
- if they differ in zero places return true!
- if they differ in one or more places check to see if the reversal makes the two arrays equal
Now let’s check and see how many differences were found.
-
return 1 if
@{$indices_different} == 0;
$indices_different = [sort {$a <=> $b}
@{$indices_different}];
my $last_i = $indices_different->[
@{$indices_different} - 1];
my $length = 1 + $last_i - $indices_different->[0];
my
@u_ = reverse
@{$u}[$indices_different->[0] .. $last_i];
my
@v_ = reverse
@{$v}[$indices_different->[0] .. $last_i];
splice
@{$u}, $indices_different->[0], $length,
@u_;
splice
@{$v}, $indices_different->[0], $length,
@v_;
return 1 if join(q/,/,
@{$u}) eq join(q/,/,
@{$t});
return 1 if join(q/,/,
@{$v}) eq join(q/,/,
@{$s});
return 0;
◇
The rest of the code combines the previous steps and drives some tests.
-
MAIN:{
say reverse_equals [3, 2, 1, 4], [1, 2, 3, 4];
say reverse_equals [1, 3, 4], [4, 1, 3];
say reverse_equals [2], [2];
}
◇
-
Fragment referenced in 7.
Sample Run
$ perl perl/ch-2.pl 1 0 1
References
posted at: 19:21 by: Adam Russell | path: /perl | permanent link to this entry