TL;DR

On with TASK #2 from the Perl Weekly Challenge #117. Enjoy!

# The challenge

You are given size of a triangle.

Write a script to find all possible paths from top to the bottom right corner.

In each step, we can either move horizontally to the right (H), or move downwards to the left (L) or right (R).

BONUS: Try if it can handle triangle of size 10 or 20.

Example 1:

Input: $N = 2 S / \ / _ \ /\ /\ /__\ /__\ E Output: RR, LHR, LHLH, LLHH, RLH, LRH  Example 2: Input:$N = 1

S
/ \
/ _ \ E

Output: R, LH


# The questions

Just to be very nitpicky, the size of a triangle is a positive integer size. This becomes very evident very quickly, so yes itâ€™s nitpicking.

Kudos for the bonus challenge of handling triangles of size 10 or 20.

Iâ€™m assuing that any order will do.

Iâ€™m also assuming that itâ€™s fine to print out each sequence on its own line, instead of on a single line with comma separation. This is particularly important as $N growsâ€¦ # The solution Buckle up, there will be a lot to discuss! ## The initial solution I started with a solution in Raku that used SetHash. Then I translated that solution into Perl, which does not have SetHash, but of course it has the good olâ€™ hash. Then I figured (in Perl) that I could do without the hash by using arrays only. So I backported that solution in Raku too, which provided a performance boost. This is the solution in Raku at that point: #!/usr/bin/env raku use v6; sub find-possible-paths ($N) {
my @solution = [''],;
for 1 .. $N ->$i {
my @new_iteration = [],;
for 0 ..^ @solution.elems -> $j { my$previous = @solution[$j]; my$left     = @new_iteration[$j]; my$right    = @new_iteration[$j + 1] = [];$left.push:  (@solution[$j].flat X~ 'L').Slip;$right.push: (@solution[$j].flat X~ 'R').Slip;$right.push: (@new_iteration[$j].flat X~ 'H').Slip; } @solution = @new_iteration;$i.note;
}
return @solution[*-1].flat;
}

sub MAIN ($N = 2) { find-possible-paths($N).join(', ').put; }


I still have to get the hang of when to use .flat and when to use .Slip. I hope to fully understand it some dayâ€¦

The idea is to use dynamic programming for computing all possible paths to all nodes layer by layer, starting from the top down to the bottom. In the last iteration, our solution is in the last element of the bottom right of the pyramid.

As anticipated, itâ€™s done layer by layer. From a layer, itâ€™s possible to compute all ways to get to any node in the layer below. For this reason, we donâ€™t need to keep the whole pyramid in memory, only the previous layer and the current one.

(Now that Iâ€™m writing, I think we could even optimize this by getting incrementally rid of the previous layer as we consume it, but letâ€™s not digress).

It worksâ€¦ up to about 11 or 12, where it starts to eat a lot of memory. Actually, in my 4 GB VM, the process was killed for going out of memory while computing the solution for 12 ðŸ˜…

The corresponding solution in Perl:

sub find_possible_paths ($N) { my @solution = (['']); for my$i (1 .. $N) { my @new_iteration = []; for my$j (0 .. $#solution) { my$previous = $solution[$j];
my $left =$new_iteration[$j]; my$right    = $new_iteration[$j + 1] = [];
for my $p ($previous->@*) {
push $left->@*,$p . 'L';
push $right->@*,$p . 'R';
}
for my $p ($left->@*) {
push $right->@*,$p . 'H';
}
}
@solution = @new_iteration;
}
return $solution[-1]->@*; }  It is quite faster, but shares the same memory problem. Soâ€¦ weâ€™re at about 11 in my Linux VM, with exponential/factorial expansions as I can guess. E. Choroba extended the challenge to 20. OK, we need to change approach. ## Back to the drawing board At this point, we can observe that we can address this as a string manipulation problem. Letâ€™s start from the solution for $N = 2 (rearranged):

LLHH
LRH
LHLH
LHR
RLH
RR


We can observe that, for every path, we can generally find a simpler path by substituting a LH sequence with a shortcut R.

Well, unless thereâ€™s no LH sequence, of course.

Letâ€™s consider the first one LLHH. It contains one LH group (in the middle), so another solution is LRH (second line). At this point, this other solution cannot be simplified any more.

Then letâ€™s consider LHLH (third line). It has two LH groups, which we can substitute individually or both at the same time, to find LHR (fourth line), RLH (fifth line), and RR (sixth line). Again, no more simplifications are possible past this point.

Now we can observe that this technique allows us to:

• concentrate only on finding out valid sequences without an R, because we can find the ones with R by the simplification process, and more importantly
• avoid keeping track of possible duplicates, because different (valid) sequences with Ls and Hs will always yield different simplified strings.

Hence, our problem is now broken down into two parts:

• code an algorithm to find all valid sequences of Ls and Hs, in the same number, that can represent a path from the top of the triangle down to the bottom-right;
• code an algorithm to find all simplifications of a sequence of Ls and Hs only.

Additionally, we will code this with the constraint of consuming as little memory as possible, in an iterative way; this will let us start getting results immediately, and avoid filling up the memory. Although it will probably take a lot of time!

### Finding valid sequences with Ls and Hs only

Any path from start to end MUST be compound of an equal number of L and H characters. For a triangle whose side has length $N, there will be $N of both characters in sequences that do not comprise Rs.

How to find them all? I opted for a brutish force approach.

We can image that the $N characters L and $N characters H are arranged in an array that is 2 * $N long. For this reason, all candidate sequences can be found by finding all possible ways of getting $N positions in the array, assuming that those positions are filled with Ls and the rest with Hs.

To find all combinations, Iâ€™ve adapted a function from a previous post:

sub combinations_iterator ($k, @items) { my @indexes = (0 .. ($k - 1));
my $n = @items; return sub { return unless @indexes; my (@combination, @remaining); my$j = 0;
for my $i (0 .. ($n - 1)) {
if ($j <$k && $i ==$indexes[$j]) { push @combination,$items[$i]; ++$j;
}
else {
push @remaining, $items[$i];
}
}
for my $incc (reverse(-1, 0 .. ($k - 1))) {
if ($incc < 0) { @indexes = (); # finished! } elsif ((my$v = $indexes[$incc]) < $incc -$k + $n) {$indexes[$_] = ++$v for $incc .. ($k - 1);
last;
}
}
return \@combination;
}
}


This function returns an iterator that allows us to generate all possible distinct sequences with $N Ls and $N Hs.

Which, of course, is not what we need, right?

Not all the sequences generated in this way will comply with the rules. We are bound to move inside the triangle, so for example a sequence like HLLH cannot be admitted. This means that we have to check each candidate and reject those that make us fall outside of the triangle.

This is how we use the combinations_iterator to generate the candidate sequences and reject the ones that are not good for us:

sub basic_case_iterator ($N) { my$N2 = 2 * $N; my$cs;
return sub {
$cs //= combinations_iterator($N, 0 .. $N2 - 1); CANDIDATE: while (my$Ls = $cs->()) { my @sequence = ('H') x$N2;
@sequence[$Ls->@*] = ('L') x$N;
my $count = 0; for my$item (@sequence) {
$count +=$item eq 'L' ? 1 : -1;
next CANDIDATE if $count < 0; } return join '', @sequence; } return; }; }  After generating a @sequence, we check that we donâ€™t fall out of the triangle. To do this, we check that the number of H characters does not overcome the number of L characters at any time. Actually, we can trim some time from this function by observing that all valid sequences will always start with an L and end with an H, so thereâ€™s no point in considering anything different: sub basic_case_iterator ($N) {
--$N; my$N2 = 2 * $N; my$cs;
return sub {
$cs //= combinations_iterator($N, 0 .. $N2 - 1); CANDIDATE: while (my$Ls = $cs->()) { my @sequence = ('H') x$N2;
@sequence[$Ls->@*] = ('L') x$N;
my $count = 1; # we will force starting with an L for my$item (@sequence) {
$count +=$item eq 'L' ? 1 : -1;
next CANDIDATE if $count < 0; } return join '', 'L', @sequence, 'H'; } return; }; }  This means that we have to choose 1 less position for L from a pool of 2 less possible positions (hence the --$N at the beginning), our $count starts from 1 (because our real candidate always starts with L) and our return value has to account for an initial L and a final H. Boring optimization, after all! ### Finding all alternatives for a sequence Now that we have all seed sequences that only have L and H characters, itâ€™s time to generate all possible variants with the simplification we discussed earlier. Hereâ€™s the code for the impatients: sub expand_with_Rs_iterator ($sequence) {
my $indexes; my @parts; my ($i, $n,$max);
return sub {
if (! $indexes) { # initialize @parts = grep {length} split m{(LH)}mxs,$sequence;
$indexes = [grep {$parts[$_] eq 'LH'} 0 ..$#parts];
$n =$indexes->@*;
$max = 0;$max = ($max << 1) | 1 for 1 ..$n;
$i = 0; return$sequence;
}
return if $i >=$max;
++$i; my @Rs = split m{}mxs, sprintf "%0${n}b", $i; my @copy = @parts; for my$j (0 .. $#Rs) { next unless$Rs[$j];$copy[$indexes->[$j]] = 'R';
}
return join '', @copy;
};
}


The function takes a string $sequence and provides an iterator to go through all its alternative simplified forms, starting from the whole $sequence itself (i.e. without simplification).

The key insight here is to consider that a generic $sequence will be comprised of some LH groups and other stuff. The simplification can only happen on LH groups, so we divide the $sequence into @parts isolating all LH groups, like this:

  LL (LH) H (LH) HL (LH) H


If we find out that there are $k$ such groups, it means that each of them can be either in its longer form LH (letâ€™s call this state 0) or in its simplified form R (letâ€™s call this state 1).

Each LH group is independent of the others. So we have a sequence of $k$ things, each of which can independently take one or another valueâ€¦ If only we had some way to generate all possible statesâ€¦

Wait a minute! This is just counting in binary with $k$ bits!

For this reason, we keep a counter $i and we turn it into a binary form as we increment it along the way. Each binary representation will tell us which groups we have to leave alone and which we have to turn into Rsâ€¦ and weâ€™re done! ### Putting all together Now we have to put things together: sub find_possible_paths_iterator ($N) {
my ($basic_it,$fit);
return sub {
$basic_it //= basic_case_iterator($N);
while ('necessary') {
$fit //= expand_with_Rs_iterator($basic_it->() // return);
if (my $item =$fit->()) { return $item }$fit = undef;
}
};
}


We return an iterator, keeping track of two iterators inside:

• $basic_it is our iterator through all possible valid sequences with Ls and Hs only; • $fit is our iterator through all possible variants/simplifications of a starting sequence from the previous iterator.

The function just makes sure to draw one more L/H sequence when needed (think of it as a sort of outer loop) and get all possible variants from it (this would be a sort of inner loop).

We now just have to consume it:

my $n = shift // 2; my$it = find_possible_paths_iterator($n); while (my$c = $it->()) { say$c }


Weâ€™re done at last!

## The final Perl solution

The whole program in Perl is the following. It uses the faster (but memory taxing) approach for values of $N up to 10 included, then switches to the iterator-based approach for bigger input values. #!/usr/bin/env perl use 5.024; use warnings; use experimental qw< postderef signatures >; no warnings qw< experimental::postderef experimental::signatures >; use constant THRESHOLD => 10; sub find_possible_paths ($N) {
my @solution = (['']);
for my $i (1 ..$N) {
my @new_iteration = [];
for my $j (0 ..$#solution) {
my $previous =$solution[$j]; my$left     = $new_iteration[$j];
my $right =$new_iteration[$j + 1] = []; for my$p ($previous->@*) { push$left->@*, $p . 'L'; push$right->@*, $p . 'R'; } for my$p ($left->@*) { push$right->@*, $p . 'H'; } } @solution = @new_iteration; } return$solution[-1]->@*;
}

sub combinations_iterator ($k, @items) { my @indexes = (0 .. ($k - 1));
my $n = @items; return sub { return unless @indexes; my (@combination, @remaining); my$j = 0;
for my $i (0 .. ($n - 1)) {
if ($j <$k && $i ==$indexes[$j]) { push @combination,$items[$i]; ++$j;
}
else {
push @remaining, $items[$i];
}
}
for my $incc (reverse(-1, 0 .. ($k - 1))) {
if ($incc < 0) { @indexes = (); # finished! } elsif ((my$v = $indexes[$incc]) < $incc -$k + $n) {$indexes[$_] = ++$v for $incc .. ($k - 1);
last;
}
}
return \@combination;
}
}

sub basic_case_iterator_longer ($N) { my$N2 = 2 * $N; my$cs;
return sub {
$cs //= combinations_iterator($N, 0 .. $N2 - 1); CANDIDATE: while (my$Ls = $cs->()) { my @sequence = ('H') x$N2;
@sequence[$Ls->@*] = ('L') x$N;
my $count = 0; for my$item (@sequence) {
$count +=$item eq 'L' ? 1 : -1;
next CANDIDATE if $count < 0; } return join '', @sequence; } return; }; } sub basic_case_iterator ($N) {
--$N; my$N2 = 2 * $N; my$cs;
return sub {
$cs //= combinations_iterator($N, 0 .. $N2 - 1); CANDIDATE: while (my$Ls = $cs->()) { my @sequence = ('H') x$N2;
@sequence[$Ls->@*] = ('L') x$N;
my $count = 1; # we will force starting with an L for my$item (@sequence) {
$count +=$item eq 'L' ? 1 : -1;
next CANDIDATE if $count < 0; } return join '', 'L', @sequence, 'H'; } return; }; } sub expand_with_Rs_iterator ($sequence) {
my $indexes; my @parts; my ($i, $n,$max);
return sub {
if (! $indexes) { # initialize @parts = grep {length} split m{(LH)}mxs,$sequence;
$indexes = [grep {$parts[$_] eq 'LH'} 0 ..$#parts];
$n =$indexes->@*;
$max = 0;$max = ($max << 1) | 1 for 1 ..$n;
$i = 0; return$sequence;
}
return if $i >=$max;
++$i; my @Rs = split m{}mxs, sprintf "%0${n}b", $i; my @copy = @parts; for my$j (0 .. $#Rs) { next unless$Rs[$j];$copy[$indexes->[$j]] = 'R';
}
return join '', @copy;
};
}

sub find_possible_paths_iterator ($N) { my ($basic_it, $fit); return sub {$basic_it //= basic_case_iterator($N); while ('necessary') {$fit //= expand_with_Rs_iterator($basic_it->() // return); if (my$item = $fit->()) { return$item }
$fit = undef; } }; } my$n = shift // 2;
my $use_iterator =$ENV{USE_ITERATOR} ? 1
: defined($ENV{USE_ITERATOR}) ? 0 :$n > THRESHOLD;
if ($use_iterator) { my$it = find_possible_paths_iterator($n); while (my$c = $it->()) { say$c }
}
else {
say for find_possible_paths(\$n);
}


The two solutions are equivalent, although they provide sequences in a different orderâ€¦ we were assuming that any ordering is fine, right?!?

# Conclusion

This has been a hell of a ride! Andâ€¦ somehow tiring.

It would be great to code an iterator-based solution in Raku too, especially because I suspect that the gather/take mechanism is perfect for coding it with lazyness. Probably.

Wellâ€¦ maybe some other time!

Comments? Octodon, , GitHub, Reddit, or drop me a line!