( …continues from previous week. )
Welcome to the Perl review for Week 113 of the Weekly Challenge! Here we will take the time to discuss the submissions offered by the team, factor out the common methodologies that came up in those solutions, and highlight some of the unique approaches and unusual code created.
Why do we do these challenges?
I suppose any answers to that would be as wide ranging and varied as the people who choose to join the team. One thing is clear: it’s not a competition, and there are no judges, even if there is a “prize” of sorts. About that – I think of it more as an honorarium periodically awarded to acknowledge the efforts we make towards this strange goal. So there’s no determination to find the fastest, or the shortest, or even, in some abstract way, the best way to go about things, although I’m certain the individuals have their own aspirations and personal drives. As Perl is such a wonderfully expressive language, this provides quite a bit of fodder to the core idea of TMTOWTDI, producing a gamut of wonderfully varied techniques and solutions.
Even the tasks themselves are often open to a certain amount of discretionary interpretation. What we end up with is a situation where each participant is producing something in the manner they find the most interesting or satisfying. Some team members focus on carefully crafted complete applications that thoroughly vet input data and handle every use case they can think up. Others chose to apply themselves to the logic of the underlying puzzle and making it work in the most elegant way they can. Some eschew modules they would ordinarily reach for, others embrace them, bringing to light wheels perhaps invented years ago that happen to exactly solve the problem in front of them today.
I’ve been considering this question for some time and have found one binding commonality between all of us solving these challenges each week, in that however we normally live our lives, the task in front of us more than likely has nothing to do with any of that . And I think this has great value. We all do what we do, out in the real world, and hopefully we do it well. The Weekly Challenge provides a opportunity to do something germane to that life yet distinctly different; if we only do the things we already know how to do we only do the same things over and over. This is where the “challenge” aspect comes into play.
So we can consider the Weekly Challenge as providing a problem space outside of our comfort zone, as far out from comfort as we wish to take things. From those reaches we can gather and learn things and bring what we want back into our lives. Personally, I think that’s its greatest value of all.
Every week there is an enormous global collective effort made by the team, analyzing and creatively coding the submissions, and that effort deserves credit due. And that’s why I’m here, to try and figure out how to do that.
Let’s have a look and see what we can find.
For context before we begin, you may wish to revisit either of the pages for the original tasks or the summary recap of the challenge. But don’t worry, the challenges themselves will be briefly summarized, presented below as we progress from task by task. Oh, and one more thing before we get started:
Getting in Touch with Us
Email › Please email me (Colin) with any feedback, notes, clarifications or whatnot about this review.
GitHub › Submit a pull request to us for any issues you may find with this page.
Twitter › Join the discussion on Twitter!
I’m always curious as to what the people think of these efforts. Everyone here at the PWC would like to hear any feedback you’d like to give.
So finally, without further ado…
• Task 1 • Task 2 • BLOGS •
TASK 1
Represent Integer
Submitted by: Mohammad S Anwar
You are given a positive integer $N and a digit $D.
Write a script to check if $N can be represented as a sum of positive integers having $D at least once. If check passes print 1 otherwise 0.
Example
Input: $N = 25, $D = 7
Output: 0
As there are 2 numbers between 1 and 25 having the digit 7 i.e. 7 and 17. If we add up both we don’t get 25.
Input: $N = 24, $D = 7
Output: 1
about the solutions
Abigail, Adam Russell, Arne Sommer, CheokYin Fung, Christian Jaeger, Colin Crain, Dave Jacoby, David Schwartz, Duncan C. White, E. Choroba, Flavio Poletti, Jaldhar H. Vyas, James Smith, Jorg Sommrey, Niels van Dijke, Pete Houston, Roger Bell_West, Simon Green, Stuart Little, Ulrich Rieke, and W. Luis Mochan
There were 21 submissions for the first task this past week. Right off the bat, the first question that struck me doing the overview was: “What question are they answering?” Some times this was clear, others decidedly less so. If there’s one thing that has become clear to me on these review pages, it’s that given any ambiguity in a text, given enough people and enough time every possible position will eventually be witnessed. It’s just a given to me now.
When the dust settled, however, three main, largely incompatible interpretations emerged from the chaos. These were whether the number could be considered a sum of every smaller integer containing the given digit, or alternately the sum of some unique combination of individual elements from the list of valid smaller numbers, or more broadly some combination of any of the valid numbers repeated any number of times. The first approach requires all of the valid numbers to be used once; the second any subset of the numbers; the third any combination allowing repetition. Each interpretation led to its own set of wildly different strategies and algorithms.
use ALL THE NUMBERS
Adam Russell, Dave Jacoby, Duncan C. White, Jaldhar H. Vyas, and Niels van Dijke
To see whether a number qualifies under this criteria, we need only determine the list of numbers smaller than the target that contain the selected digit and sum them. If they sum to the target number we have a winner.
Dave will start us off with an example. As you can see, this interpretation allows a very compact solution.
sub represent_int ( $n, $d ) {
my $s = 0;
for my $i ( 1 .. $n ) {
$s += $i if $i =~ /$d/;
}
return $n == $s ? 1 : 0;
}
Duncan takes the simplification one step further, hardwiring the digit selection into the program as 7
. On the other hand, he takes this choice as an opportunity to call his short script lucky7
, of which I thoroughly approve.
die "Usage: lucky7 N D\n" unless @ARGV==2;
my $total = sum( grep { /7/ } 1..$n );
my $correct = ($total == $n) ? 1 : 0;
additional languages: Prolog
As you can see there isn’t a lot of room in this type of solution for variation, but Adam still did his best. He makes a creative decision to restrict his toolset, disallowing regular expressions to match out the digit, substituting instead a pair of grep
statements and an equality . He also uses unpack
to perform his sums, after first using pack
to shove them into 8bit character octets. Note the size limitation, though, so if for some reason you wanted to employ this technique in a more generalpurpose manner something like an unsigned integer i*
type would probably be a better choice. Read all about in the Perl pack tutorial.
sub is_represented{
my($n, $d) = @_;
my @contains = grep { grep { $_ == $d } split(//) } (1 .. $n);
return $n == unpack("%32C*", pack("C*", @contains));
}
use any subset
Arne Sommer, James Smith, Roger Bell_West, Ulrich Rieke, and W. Luis Mochan
To consider the possibilities for combining elements from the list of valid digit numbers, we can consider those numbers as a set, and select various subsets of all sizes. As a set, each element is unique and is either present or not. Thus there are at most 2^{m} subsets to search, with m being the number of members of the set of possible summands containing the selected digit.
Another way to look at this is all combinations of k elements considered for the range k = 1 to m. The limit to the number of summands is of course all of those available. Several solutions brought in a combinatorics module to provide the combinations.
additional languages: Misc, Raku
As you may have noticed, grep
was the goto tool to obtain a list of numbers smaller than the target composed with at least one instance of the given digit. After filtering to find his list of numbers, Arne uses Algorithm::Combinatorics
to produce all combinations from this list of lengths from 1 to all of them. Summing the combinations as we go, if we find a match we can report and exit immediately.
use Algorithm::Combinatorics 'combinations';
my @candidates = grep { /$D/ } (1 .. $N);
for my $size (1 .. @candidates)
{
for my $comb (combinations(\@candidates, $size))
{
say ": Considering " . join(' + ', @$comb) if $verbose;
if (sum(@$comb) == $N)
{
say 1;
exit;
}
}
}
say 0;
additional languages: Python, Raku, Ruby, Rust
One artful way to generate 2^{n}1 total combinations from an array of n elements is to create a list of numbers and examine their binary bits, with each bit mapped to a specific member in the set. If a bit is set, we include that element from the list of all valid digitnumbers in the sum.
Here Roger uses bitshifting to generate his powers of 2:
sub ri {
my ( $n, $d )=@_;
my @e = grep /$d/, ( 1..$n );
foreach my $i ( 1..( 1<<( scalar @e ) )1 ) {
my $s = 0;
foreach my $ii ( 0..$#e ) {
if ( 1<<$ii & $i ) {
$s += $e[$ii];
}
}
if ( $s == $n ) {
return 1;
}
}
return 0;
}
As mentioned before, every element in the set of valid number options is either present or not. By assigning each number to a bit position, we can use a list of binary numbers to represent the various distinct subsets.
Luis explores three ways along an increasingly optimized path to find his solution. In the first method he sets up a binary bitmask, much like Roger did, but found it excessively slow. Optimizing, he came up with a recursive algorithm that bails out when the sum goes past the target, limiting whole chains of possibilities from ever being evaluated.
The third solution improves the recursion by noticing that any target larger than 10 • d can be be constructed from some number 10 • d + k, which will contain d, and some number that contains d in the ones place. If that proves to be that case, the solution shortcircuits without further evaluation.
say( "Inputs: N = $N D = $D: Output: 1 as $N>=", $D * 10 ), exit if $N >= $D * 10;
my @set = reverse grep { m / $D/ } ( 1..$N ); # ordered set from large to small.
my @answer = find( $N, @set );
say( "Inputs: N = $N D = $D: Output: ",
@answer? "1 as $N = ". join( "+", @answer ):"0" );
sub find {
my ( $goal, @set )=@_;
while( defined ( my $current = shift @set ) ){
next if $current > $goal;
return ( $current ) if $current == $goal;
my @answer = find( $goal  $current, @set );
return ( $current, @answer ) if @answer;
}
return ( );
}
About that shortcircuit: James will talk over and go into a more complete analysis:
James also shares the understanding that a given number can only be be used once in the sum, and makes a detailed analysis of the possible valid solutions, in the end drastically limiting the search field.
From his notes;

If $d is equal to 0
 Any number between 100 & 109 can be represented by itself
 For numbers over 109 we can represent these as 100109 + a number ending in 0…
 e.g. 534 / 0 = 104 + 430
So if $d is equal to 0 then all numbers > 100 are possible

If $n is between 10$d and 10$d+9 then it can be represented as $d$x
 For numbers > than this we can do a similar trick to above
 We can reprent them as a number ending in $d and a number where $d is the penultimate digit
 e.g. 107 & 9 = 9 + 98
 e.g. 450 & 8 = 68 + 382
 e.g. 435 & 2 = 12 + 423
So if $d is not equal to 0 then all numbers greater than 10x$d are possible
Interesting… this establishes a hard upper bound for failing to produce a sum.
Next:
 Finally we get to the list of numbers less than this  as the only digit that can contain $d is the last one we just try to see if we can find a sum of numbers ending in $d which have the same last digit as $n and less than or equal to $n. Note as we have already removed the numbers greater than 100 we now we only need to loop up to 3  as the next number would be 100 + 4$d….
Now for the resulting code:
## Return 1 if both conditions hold true...
$n >= ( $t += $_ * 10 + $d ) &&
( $n % 10 == $t % 10 ) && return 1 for 0..3;
## Return 0 if no solution is possible...
0;
It seems that under a certain limit there are only a few possible ways to come to a solution.
use ANY COMBINATION of ANY AMOUNT
Abigail, CheokYin Fung, Colin Crain, David Schwartz, E. Choroba, Flavio Poletti, Jorg Sommrey, Pete Houston, Simon Green, and Stuart Little
The most common interpretation of the task was to find some enumeration of values from the set of valid numbers that when summed yield the target. Each number is available for inclusion in the target sum any number of times. Allowing this repetition opens up the upper bound for the number of summands in the solution to the target divided by the smallest element, which is the single digit, or 10 in the case of 0.
RECURSION was MADE FOR THIS
Colin Crain, David Schwartz, E. Choroba, Flavio Poletti, Pete Houston, and Simon Green
The somewhat openended nature of the length of the solution suggests a recursive method. Given a list of sorted options, we can iterate through them, subtracting the selected value from the total and passing the remaining quantity. Hitting the target or exceeding it are the base cases, returning success or failure for that attempt.
David gives a straightforward brute force approach. Given enough time it will find a solution if there is one. On success the positive result is immediately propagated up the chain, stopping further attempts.
sub brute {
my ($sum, $n, @vals) = @_;
# Base cases:
return 1 if $sum == $n;
return 0 if $sum > $n;
# Try to add a number and see what happens
for (@vals){
return 1 if brute ($sum + $_, $n, @vals);
}
return 0;
}
Simon tightens the restrictions on recursion by passing only the values less than the remainder, filtering out pointless cycles.
sub _find_numbers {
my ( $remainder, $numbers, $all_numbers ) = @_;
# Find out what digits remain (must be <= remainder)
my @can_use = grep { $_ <= $remainder } @$all_numbers;
foreach my $number (@can_use) {
if ( $remainder == $number ) {
# We have a solution, return that
return [ @$numbers, $number ];
}
# Recurse this function
my $solution = _find_numbers( $remainder  $number, [ @$numbers, $number ], $all_numbers );
if ($solution) {
return $solution;
}
}
# There is no solution.
return;
}
additional languages: Raku
For my own solution, I wanted to see the number chains I was creating. This complicated matters as I needed to keep partial lists as I progressed. I also constrained the list of potential values further both by filtering for values equal to or greater than the current selection and for values, that when added to the running sum would not exceed the target. This last distinction culled the recursions significantly. Further analysis showed me that above a certain value for each digit all numbers could be fitted, which would allow a shortcircuit should the target be above a calculable figure. I wrote a routine to do this but never wired it in because, as I said, I enjoyed looking at the numbers. Even without the short circuit the method will find a solution in a reasonable time even for large targets.
There’s a detailed methodology in comments, flushed out with improved writing in the proper writeup for the challenge.
sub sum_from_list ($target, $numlist, $partsum = 0, $partial = []) {
for my $nextval ( $numlist>@* ) {
if ($partsum + $nextval == $target) {
push $partial>@*, $nextval;
return $partial;
}
my @newpart = ( $partial>@*, $nextval );
my $newpsum = $partsum + $nextval;
my @newlist = grep { $_ >= $nextval && $_ <= $target  $newpsum} $numlist>@*;
next if scalar @newlist == 0;
my $sol = sum_from_list ($target, \@newlist, $newpsum, \@newpart);
return $sol if defined $sol;
}
return undef;
}
Choroba produces his own analysis of the mathematical bounds:
For D = 1, there’s always the trivial solution: N = 1 + 1 + … + 1.
For D > 1 and any N >= D * 10, we can subtract D as many times as we get between D * 10 and D * 10 + 9. Let’s call this number M. M contains D and M + D + D + … + D = N, so it’s the solution.
For D = 0 and any N > 100, we can subtract 10 as many times as we get between 100 and 110. Let’s call this number M. M contains 0 and M + 10 + 10 + 10 + … + 10 = N, so it’s the solution.
Therefore, the only possible nonrepresentable integers are N < D * 10 for D > 0, and N < 100 for D = 0.
Building from this, he presents us with two routines that work in tandem to work through the possibilities.
sub represent_integer {
my ($integer, $digit) = @_;
return 1 if 1 == $digit
 1 < $digit && $integer >= $digit * 10
 0 == $digit && $integer > 99;
return _represent_integer($integer, $digit)
}
sub _represent_integer {
my ($integer, $digit) = @_;
return 1 if 0 == $integer;
for (my $having = $integer; $having > 0; $having) {
next unless $having =~ /$digit/;
return 1 if represent_integer($integer  $having, $digit);
}
return 0
}
One interesting detail is that to run his tests he uses a lookup table for verification, and the entirety of the table of all unsolvable targetdigit combinations is only 336 values over all 10 digits.
D • k + 10 • c
Abigail, CheokYin Fung, and Stuart Little
One very interesting solution pattern we saw popping up in various ways was that a solution could be broken down into some multiple of the digit plus a remainder that is a constant times 10. If we can find this pair, then one factor from the multiple can be added to the value ending in 10, creating a number that contains the digit in the one’s place and a list of digit values.
CY gives us two very similar techniques. In the first, she tries subtracting the digit from the target, and checking the remainder for the presence of the digit. If found then we have a summation of some multiple instances of the digit and a number containing the digit.
In the second technique she checks multiples of the digit to see whether the last digit matches the last digit in the target. If so, then the difference between that multiple and the target will be some multiple of 10. We remove one instance of the digit from the digit multiple and add it to to the tens figure and that value will now end in the digit and the list will now all sum. On top of this she adds some conditional shortcircuiting for the cases of d equal to 4, 6 and 8, further on top of clauses for a target divisible by the digit (always the case when d=1) and the target being above the upper cutoff for possible failure. She’s really working all the angles.
Marvelous.
sub last_digit {
# Example I: if N = 82, D = 9, it hints 82 = 72+10 = 9*8+10 = 9*7+19
# Example II: if N = 64, D = 7, it hints 64 = 14+50 = 7*2+50 = 7*1+57
# Example III: if N = 30, D = 8, the set {8, 18, 28} ...
# Example IV: if N = 44, D = 6, it hints 44 = 24+20 = 6*4+20 = 6*3+26
my $digit = $_[0];
my $short = $_[1];
my $last_digit_of_short = $short % 10;
my $i = 1;
while ($digit*$i < $short) {
if ($digit*$i % 10 == $last_digit_of_short ) {
return 1;
}
$i++;
}
return 0;
}
additional languages: Haskell, Node, Python, Raku
In a tight functional style, Stuart replicates the same technique. It looks different, but works out the same.
sub lastDigSumm($nr,$dig,$nrSummands) {
return (($nr  $nrSummands * $dig) % 10 == 0)
&& ($nrSummands * $dig <= $nr)
&& ($nrSummands * (($dig 1) * 10 + $dig) >= $nr);
}
sub lastDig($nr,$dig) {
return !!(grep {lastDigSumm($nr,$dig,$_)} (1..9));
}
sub sol($nr,$dig) {
$dig == 0 && return ($nr >= 101  ($nr % 10 == 0));
return (($nr >= $dig * 11)  lastDig($nr,$dig));
}
say 0+sol(@ARGV);
additional languages: Awk, Bash, C, Lua, Node, Python, Ruby
Abigail then goes above and beyond, presenting a full mathematical proof as to why this technique holds true in all cases, and narrowing the scope for the tens multiplier to less than
d / gcd(d,10)
I truly enjoy seeing this effort made. Sure it made sense before, but now we can definitively say we know it, which is always a good thing.
They also go into further detail about the upper bound definition and the special cases that surround the digit being 0, all in rigorous form. Truly excellent work.
Read their complete writeup here.
my @gcds = (0, 1, 2, 1, 2, 5, 2, 1, 2, 1);
MAIN: while (<>) {
my ($N, $D) = split;
if ($D == 0) {
say $N >= 100  $N % 10 == 0 ? 1 : 0;
next MAIN;
}
if ($N >= $D * 10) {
say 1;
next MAIN;
}
for (my $i = 0; $i < $D / $gcds [$D]; $i ++) {
my $T = $N  10 * $i  $D;
if ($T >= 0 && $T % $D == 0) {
say 1;
next MAIN;
}
}
say 0;
}
BOUNDING the SEARCH SPACE
Abigail, CheokYin Fung, Colin Crain, E. Choroba, Flavio Poletti, Jorg Sommrey, and Pete Houston
Here is some further analysis that was presented around the bounds of surrounding possible outcomes.
Some considerations:
 The task neither requires the summands to be distinct nor a solution to have more than one summand.
 Every integer n that is a multiple of the digit d can be represented as a multiple sum of the given digit.
 Every integer having d in its decimal representation is a solution with itself as the sole summand.
 For all d > 0 and 10 * d <= n < 10 * (d + 1) the number starts with the digit d and thus is a solution itself.
 For all d > 0 and 10 * (d + 1) <= n there is a number m with 10 * d <= m < 10 * (d + 1) starting with d and n  m is a multiple of d. Thus n is representable as a sum of numbers that have the digit d in their decimal representation.
 For d = 0 and 100 <= n an analogous consideration is applicable when taking d=10 instead. As leading zeros do not count, with the taken modification the second digit becomes zero.
 The remaining cases are n < 10 * d with the modified d. Further analysis can be applied to these, e.g checking the special cases where d is one, even or five or is already occurring in n. However, skipping any refinements and performing a brute force approach on this small solution space instead.
sub rep_int ($n, $d) {
$d = 10;
return 1 if $n >= $d * 10;
# keys are strings, using the numeric values.
my %sum = (0 => 0);
# All numbers containing the digit $d.
for (my $num = $d; $num <= $n; $num += 10) {
# All sums found so far.
for my $sum (values %sum) {
# New sums arise from the current sum plus multiples of the
# current number.
for (my $new = $sum + $num; $new <= $n; $new += $num) {
return 1 if $new == $n;
$sum{$new} = $new;
}
}
}
# Not found.
0;
}
additional languages: Raku
Transcribed from his blog writeup:
If a value N is such that N>10⋅D, it means that it can be expressed as the following sum:
N = 10 • D + K
Now we can consider that K can be expressed in terms of its integer division by D like follows:
K = q • D + r
with 0 ≤ r < D ≤ 9
Hence, we can write N as follows:
N = 10 • D + q • D + r N = q • D + (10 • D + r)
Now, q⋅D is the same as summing D to itself q times, so it can be represented in terms of “sum of positive integers having D at least once”.
On the other hand, considering the restrictions on D and r, the value 10 ⋅ D + r is the twodigits number where the first digit is D and the last digit is r, hence it contains digit D and complies with the rule.
As a result, N is the sum of two compliant addentds and can thus be decomposed according to the rules.
After this he concludes a simple recursion will suffice:
sub represent_integer ($n, $d) {
return 0 if $n < $d; # no point in checking this
return 1 if $n >= 10 * $d; # n * d + (10 * d + i) (i < 9)
return 1 if $n =~ m{$d}mxs; # match one digit
$n = $d;
while ($n > 0) {
return 1 if represent_integer($n, $d);
$n = 10;
}
return 0;
}
a UNIQUELY FUNCTIONAL APPROACH
I don’t know where to begin on Christian’s awesome contribution. No, seriously, I don’t know where to start.
What Christian has done here is not just used functional programming techniques to solve the problems, but indeed appears to have constructed a whole functional programming framework of modules to sit on top of Perl, allowing Perl to work as a functional language.
It’s a lot to take in. So we’re going to start slow, familiarizing ourselves a little bit at a time. Rather than jumping to the conclusive end of a series of optimizations as I normally would when looking over someone’s multiple submissions, this time we will give instead the most basic brute force form presented as an opening example:
sub maybe_choose_brute ($N, $ns) {
__ 'Choose a combination of numbers from $ns (repetitions allowed)
that added together equal $N; undef if not possible. This
solution is brute force in that it is picking additional
numbers from the left end of $ns, one after another,
depthfirst.';
sub ($chosen) {
my $check = __SUB__;
warn "check (brute): " . show($chosen) if $verbose;
my $sum = $chosen>sum;
if ($sum == $N) {
$chosen
} elsif ($sum > $N) {
undef
} else {
$ns>any(
sub ($n) {
$check>(cons($n, $chosen))
}
)
}
}
>(null)
}
This returns an anonymous recursive routine that uses __SUB__
to acquire a reference to itself, which tells you the weirdness is just getting started. A trio of these are actually presented, growing more optimized as they go. We can identify some familiar patterns we’ve een in other recursive solutions, and this one is definitely on the simpler side, with no rarefaction of the options list at all, and bases cases of meeting or exceeding the target.
I encourage everyone to go and read his extensive writeup and introduction to the framework.
Additional Submissions in Guest Languages
additional languages: Raku
additional languages: Raku
TASK 2
Recreate Binary Tree
Submitted by: Mohammad S Anwar
You are given a Binary Tree.
Write a script to replace each node of the tree with the sum of all the remaining nodes.
Example
Input Binary Tree
1
/ \
2 3
/ / \
4 5 6
\
7
Output Binary Tree
27
/ \
26 25
/ / \
24 23 22
\
21
about the solutions
Abigail, Adam Russell, Arne Sommer, CheokYin Fung, Christian Jaeger, Colin Crain, Dave Jacoby, David Schwartz, Duncan C. White, E. Choroba, Flavio Poletti, Jaldhar H. Vyas, James Smith, Jorg Sommrey, Laurent Rosenfeld, Niels van Dijke, Roger Bell_West, Simon Green, Stuart Little, and W. Luis Mochan
There were 20 submissions for the second task this past week.
Binary tree problems always involve two giant, related complications: input and output. The tree structure itself, once built up through a system of nodes of some sort, is a fascinating, longstanding integral part of computer science. But parent and child nodes are a data structure that really only comfortably fits in two forms: either as an abstract collection of pointers and values in a computer memory or as a physically drawn graph on a piece of paper. Intermediate states, describing the tree rather than displaying it, certainly have been devised, but these seem to exist in that worst of all possible worlds where neither computer nor human can easily understand what’s being presented.
This task starts with a simple statement: “You are given a binary tree…” and we have immediately lost consensus the idea. “How?", “What?", “Where?” — all of these are very good questions that will need some decisions before the task can even be started, and there is no obvious answer to any of them. “Who?", at least is fortunately already answered: this is you, the willing participant, and as for “Why?", well that will just have to wait for another day.
Do not misunderstand me; generally I find ambiguity in the task definitions to be a good thing. The added requirement to assist in formulating the question indisputably adds creativity to the solutions and broadens the genepool, so to speak. It’s just that in this particular problem every solution that I’ve come across comes with a set of associated faults — it’s one of those frustrating tradeoffs where nobody walks away happy, but what can you do?
the I/O problem
Consider the example tree:
1
/ \
2 3
/ / \
4 6 7
Before we even start notice that this tree lacks a leaf at the “5” node, allowing it to be easily drawn. Binary trees do not, generally, look like this, but this is also a perfectly valid tree.
To read in our data and output the results of our computation we need to present not just the values of our tree but also the information of the interconnections between those values, in a one or twodimensional form. These expressions can be in turn broken onto two ideas: a linear, serialized datastream, or some sort of graphical stab at adding additional information in a second dimension. For linear data in a string, we can draw on traditional set notation, recursively defined by portraying an individual node as {left, value, right}, with the placeholders left and right replaced with either additional nodes as described or the null set, ∅. This format builds out in both directions from the center in a manner mimicking a graphical representation, but leaves the root node buried in the center and hard to spot:
{{{∅,4,∅},2,∅},1,{{∅,6,∅},3,{∅,7,∅}}}
Alternately, we could rearrange our nodes to {value, left, right}, which might improve the readability:
{1,{2,{4,∅,∅},∅},{3,{6,∅,∅},{7,∅,∅}}}
Oh, now it makes sense! Thanks for that! And yes, that was sarcasm should that not have been absolutely clear. In full disclosure I believe I’ve used both of these encodings before in these challenges.
A popular alternate idea is to fit the data into an appropriatelysized template of a full tree and present the tree and a list of values separated by a defined delimiter, possibly with a specifically defined nullset character as well that will not otherwise appear in the data. Numbering the nodes in a breadthfirst traversal, each node, filled or null, gets a spot in the list, so the list position immediately defines the node location. This format has the advantage of maintaining the parentchild relationships mathematically in the indexing. The children for a parent located at index n will always be found at index 2n+1 and 2n+2. Because of this mathematical relationship the data can considered a real tree drawn flat: each array index is a node, able to be traversed in the same manner as a more traditional form. Only the practical particulars differ.
Some examples would be in order to help clarify. Be aware there is no formal specification for this format, so users must decide for themselves what works for them.
1,2,3,4,,6,7
1234*67
111 222 333 444 ∅ 666 777
I like this format, but acknowledge for larger structures it gets unwieldy quickly, as each additional level to the tree adds 2^{n} more elements to be accounted for, empty or not. For a sparse tree looking more like a linked list, with 6 nodes on 6 levels, we still require 63 pieces of data. Perhaps we can leave out some implicit nulls after the last value on the last level, but that’s the best we can do to minimize things. I find this format fairly humanreadable for smaller trees, but for anything big the bookkeeping quickly becomes ridiculous.
One immediate improvement we can make to readability in this format is to break each level of the tree into its own line in a file. This can at least hold off the deluge of data to make it visually parsable for another level or two:
1
2 3
4 X 6 7
This presents the level information nicely, but the parentchild relationships still get lost without a direct line displaying the connection.
So why don’t we draw the lines too? Well this brings us full circle to our opening example, and I will try and sum up the problem as succinctly as possible:
Because a textbased solution is just freaking terrible at that job.
I chose my trivial example carefully because it nicely encapsulates the problems with the idea. The tree only works in a clean and readable way because node 5 is missing. If we need that data node, then the tradeoffs begin:
1
/ \
2 3
/ \ / \
4 5 6 7
The spacing is all dependent on keeping the elements in the last row separated, which leads to increasingly larger gaps on the levels above. Further, our toy example has only singlecharacter values. What about this:
1
/ \
22222 3
/ \ / \
4 5 666 7
/ \ / \ /
888 11111 12 13 14
We’re really losing the plot now. We’re looking less like a tree and more like a polycyclic aromatic hydrocarbon. I’ve made some judgement calls on where to place the connecting lines, surely, to make it as pretty as possible, and I really can’t see any alternative patterns that make the outcome fundamentally clearer. This is as good as it gets. Add another level and of course everything gets worse.
Even if you wanted to draw this, calculating the variable spacing to get things to line up gracefully, well lets say this a nontrivial problem. I grappled with it myself and settled on basing everything off the longest value in the tree. I’m pleased with the results, albeit the trees are sometimes wider than I’d like. That’s my tradeoff. Generously, I’ll state that the problem is complex. Off the record I’d use stronger language.
Some brave souls, in the past, have even gone as far as to parse these drawings for input. Although I’d like to make unequivocally clear that I admire the will to make such an quixotic effort — without these beautiful crazy schemes life would be so much less interesting — I believe the underlying format, not the idea of a parsing implementation, has fundamental flaws. It just doesn’t work well in a generalpurpose way to define a tree to a computer program without some serious work.
Of course anything is possible. But adaptive spacing and keeping track of missing children and their children in turn would be a real handful. We could count forward and backward slashes, sure, but assigning the linkages to the correct children in a sparse tree, without fixed positioning would necessitate some seriously fancy trickery of one kind or another.
so WHAT DID WE END UP DOING?
We saw hardwired data, because we said it was given, without being too particular as to how. “Given” doesn’t actually require external input, depending on how you think about it.
We also saw serialized input from the command line, parsed, watered and grown into trees. Alternately, we also had serialized input that was transformed in place. Some sort of serialization scheme seemed to be a goto plan.
We had a wide variety of concepts deployed for a recursive data structure defining a tree, from arrays of arrays mimicking the set definition to node and tree objects, to onestop module imports we haven’t previously seen here.
There were only a few attempts to output the tree in a graphical representation. Mostly the altered value from the nodes were presented as a list, but there were some more ambitious efforts. Having finished early I dusted off a routine I had made for PWC 057 and did that refactoring I had said I would get to, so later I’ll present it in its new artdeco glory.
so do we BUILD THIS TREE OR NOT?
Abigail, Arne Sommer, Duncan C. White, Laurent Rosenfeld, Roger Bell_West, and Simon Green
One emergent consequence of serializing the input was that with all of the values right there in a line, it wasn’t necessary, strictly speaking, to reconfigure the data into a tree to perform the transformation. The argument might be made that it was unsporting, but summing all of the numbers found and replacing them with new values are operations independant of whatever linear format is employed. As the underlying tree structure is unchanged, if the format described a tree going in, it will describe a tree going out. One might reasonably consider what happens in the middle to be a black box, and that if we accomplish our goal, it doesn’t matter the means we use to get there. This proved to be a quite popular way to think about it.
Simon wants a serialized string from the command line.
Two regular expressions first match, gathering and summing every number found, then with a second pass substitute each match found for the difference of the sum and the value. It’s remarkably concise. I admire the elegance.
sub main {
my $string = shift;
# Calculate the sum of all digits
my $sum = sum( $string =~ /(\d+)/g );
$string =~ s/(\d+)/$sum$1/eg;
say $string;
}
additional languages: Python, Raku, Ruby, Rust
In Roger’s world he acts on data already in array form. In this case two iterations get the job done, first to sum the digits and then to apply the difference equation. Any null nodes are indicated by the value 1
.
sub rbt {
my $ti=shift;
my $s=0;
foreach my $n (@{$ti}) {
if ($n>=0) {
$s+=$n;
}
}
my @to;
foreach my $n (@{$ti}) {
if ($n>=0) {
push @to,$s$n;
} else {
push @to,$n;
}
}
return \@to;
}
additional languages: Misc, Raku
Arne expects a string of values delineated with spaces; pipes separate layers and stars indicate null data but ultimately the format is inconsequential, as all of this infrastructure is passed through when we transform the numbers. First a regular expression matches out and then sums the digits. Then, after splitting the string into and array on whitespace a loop is set up and a new array is built; if the value is a number it is swapped with the difference from the sum, and if it’s not it’s left unmolested. The new array, joined up, becomes the recreated tree.
my $tree = shift(@$ARGV) // '1  2 3  4 * 5 6  * 7';
my $sum = sum(grep(/\d/, split(/\s+/, $tree)));
say ": Sum: $sum" if $verbose;
my @elems;
for my $elem (split(/\s+/, $tree))
{
$elem =~ /\d/
? push(@elems, $sum  $elem)
: push(@elems, $elem);
}
say join(" ", @elems);
additional languages: Raku
This is an odd case in a task full of odd cases, starting work as an array and moving to a tree for display. Laurent starts with a toy “tree” in linear form, being an array of the digits 1 through 7, encoding a full tree of three levels and seven nodes. The processing for the task, summing and replacing with the difference, it then effected on this array in two simple lines.
$sum += $_ for @$tree;
my $new_tree = [ map $sum  $_, @$tree ];
Then, after the fact, this list is transformed first into an arrayofarrays, broken down into the levels of a tree, and then this array is printed out given padding within each level in an inverse ratio to its depth, with the root padded the most and the leaves the least.
ASSEMBLE an ADHOC DATA STRUCTURE
CheokYin Fung, David Schwartz, E. Choroba, Niels van Dijke, and W. Luis Mochan
The essential quality of the settheory definition for a binary tree is the recursive structure: each three member node contains two child nodes (or null sets) holding the same definition, with the pattern repeated inward as far as it goes. We don’t need any further abstraction to implement this than an array of 3element arrays of a hash of 3key hashes. Either will do the job perfectly well.
Input: [1,[2,[4,[],[7,[],[]]],[]],[3,[5,[],[]],[6,[],[]]]]
Output: [27,[26,[24,[],[21,[],[]]],[]],[25,[23,[],[]],[22,[],[]]]]
Luis makes a nice segue from the previous section as his serialized data is already an array of arrays. Once input the string is given a suspicious onceover, and once cleared is then passed to an eval
where it realizes its true nature. Once there a pair of recursive functions first perform the summation, then replace the values.
sub sum_tree { #sum and do some rough validation
my $node=shift;
die "Wrong format" unless ref($node) eq "ARRAY";
return 0 if @$node==0;
return $node>[0]
+sum0 map {sum_tree($node>[$_])} (1,2) if @$node==3;
die "Wrong format";
}
sub subtract_tree {
my ($node, $from)=@_;
return [] if @$node==0;
return [$from$node>[0],
map {subtract_tree($node>[$_], $from)} (1,2)];
}
A third recursive routine converts the tree back into the string form it came in:
sub stringify_tree {
my $node=shift;
return "[]" if @$node==0;
return sprintf("[%s,%s,%s]", $node>[0],
map {stringify_tree($node>[$_])} (1,2));
}
Niels also keeps his tree as an arrayofarrays, in [value, left, right]
form:
'Test 1' => [
[ 1,[ 2,[ 4,[undef, 7]]],[ 3,[ 5, 6]]],
[27,[26,[24,[undef,21]]],[25,[23,22]]]
],
These arrays can then be traversed in two steps to first sum and then update the tree.
sub sumOfTree {
my ($arT) = @_;
my $sum;
foreach my $n (@$arT) {
if (ref($n) eq 'ARRAY') {
$sum += sumOfTree($n);
} elsif (defined $n) {
$sum += $n;
}
}
return $sum;
}
sub updateTree {
my ($arT, $sum) = @_;
foreach my $n (@$arT) {
if (ref($n) eq 'ARRAY') {
updateTree($n,$sum);
} elsif (defined $n) {
$n = $sum  $n;
}
}
}
David, and Choroba later, are the first to bring us a new revelation to the processing in the task. You may have by now observed that we need to make two passes through the data: once to gather the sum of the nodes, then again to alter their values. As an array these passes are trivial, but given a proper tree structure it becomes a little more complex to perform a nodewise traversal to access the values, and this action need to be repeated for each processing step.
So what if we were to apply a little functional programming methodology and refactor out just the traversal into its own routine, which we could then give a coderef as an argument to effect the processing? That. of course, is exactly what we have been given here:
sub preorder (&@) {
my ($expr, $node) = @_;
$_ = $node>{"val"};
$node>{"val"} = &$expr;
# the & forces it to run without the prototype
# important because we switch from asking for an anonblock
# to a code ref (which encapsulates that block)
&preorder ($expr, $node>{"left"}) if $node>{"left"};
&preorder ($expr, $node>{"right"}) if $node>{"right"};
}
preorder {$sum += $_; $_;} $test;
preorder {$sum  $_} $test;
This routine is the real meat of the demonstration, with I/O relegated to ancillary status: the input is hardwired, and the output is just the list of altered values. But it’s not about that anyways. A proper tree is constructed from a hash of hashes, and walked with this routine to get the results.
I like that he took the trouble to add a subroutine signature, allowing him to call it as
preorder {BLOCK} $tree;
It’s a small thing, but a nice detail.
build a NODE, build an OBJECT, grow a TREE
Colin Crain, Dave Jacoby, E. Choroba, Jaldhar H. Vyas, James Smith, and Jorg Sommrey
The classic Object Oriented way to build a binary tree is to abstract the structure into a Node object that knows how to be a node, and perhaps a Tree object that knows all about stringing these nodes together. Keeping the mechanics encapsulated makes adding functionality a little easier to keep straight, with everything kept within its proper frame of reference.
We mentioned Choroba in passing previously, and here he demonstrates how easily a traversal method can be brought into a simple object. It’s compact and elegant and I like it. We could swap the anonymous coderefs as proper methods if we wanted to, but here they remain the oneoff transformations that they are, easily configurable to do exactly what we want.
{ package Node;
use Moo;
has value => (is => 'rw', required => 1);
has leftchild => (is => 'rw', predicate => 1);
has rightchild => (is => 'rw', predicate => 1);
sub walk {
my ($self, $sub) = @_;
$self>leftchild>walk($sub) if $self>has_leftchild;
$self>rightchild>walk($sub) if $self>has_rightchild;
$sub>($self);
}
}
$one>walk(sub { $sum += $_[0]>value});
$one>walk(sub { $_[0]>value($sum  $_[0]>value) });
James has decided to flush out the work he started in PWC 094, bringing us today two objects, a Tree
, and a BinaryTree
.
In this expanding of the featureset of his packages, he joins the club of solutions that abstract the tree traversal into its own method, walk
, which takes a function as input and optionally a few variables to apply to each node as it goes along. He then takes the idea of “recreating” the tree to a completely different place, by not effecting the changes on the input tree, but rather by using his walk
function to make a deep clone of the original structure first and changing that instead. In the end we have two trees: the original, unaltered, and a “recreated” version with the new data.
From BinaryTree
:
sub walk {
my( $self, $fn, $global, $local, $dir ) = @_;
$local = $fn>( $self, $global, $local, $dir'' );
$self>left>walk( $fn, $global, $local, 'left' ) if $self>has_left;
$self>right>walk( $fn, $global, $local, 'right' ) if $self>has_right;
return;
}
sub clone {
my( $self, $clone_fn ) = @_;
$clone_fn = sub { $_[0] };
my $clone = {};
$self>walk( sub { my( $node, $global, $local, $dir ) = @_;
if(exists $global>{'tree'} ) {
my $child = BinaryTree>new( $clone_fn>( $node>value ) );
$dir eq 'left' ? $local>add_child_left( $child ) : $local>add_child_right( $child );
return $child;
}
$global>{'tree'} = BinaryTree>new( $clone_fn>( $node>value ) );
return $global>{'tree'};
}, $clone );
return $clone>{'tree'};
}
Jorg brings us a similarly augmented BinaryTree
package, with an unusual twist on updating the node values. He provides a generic traversal routine much like those we’ve seen, but on the first traversal, to gather and sum the values, he also collects references to the values themselves within the nodes. Then once he has the sum, he only needs to iterate through this array of scalar references and replace the value referenced with the transformed difference. Sneaky!
From BinaryTree
:
# Depthfirst NLR traversal of the binary tree starting from its root.
# The code ref is called for every node with $_ set to the current node.
sub traverse ($self, $code) {
# Recursively process the tree in NLR order. Nodes are
# not blessed and thus have no methods.
do {local $_ = $self; $code>()};
traverse($_, $code) for grep $_, $self>@[1 .. $#$self];
}
And in the main body of the program:
sub recreate_tree ($tree) {
# Get the sum of all node values and collect references to them.
my $sum;
my @nodes;
$tree>traverse(sub {
$sum += $_>[0];
push @nodes, \$_>[0];
});
# Adjust the nodes' values as the sum minus the old value.
$$_ = $sum  $$_ for @nodes;
}
talk to a MODULE
Adam Russell, and Stuart Little
There were packages all over the place in this challenge, however almost all of these were classes made by the submitters, crafted for the job at hand. There were a few outside imports that showed up, in each case some of the more uncommon modules we don’t often see.
additional languages: Prolog
modules:
Graph;
Graph::Easy::Parser;
Adam states that he likes the consistency of having the Graph
module to work with, enabling his to build up a reusable library of routines to draw on when solving these problems.
The Graph::Easy
boxes are a little easier to see as a tree if you tilt your head a little to the left.
++ ++ ++ ++
 27  ==>  26  ==>  24  ==>  21 
++ ++ ++ ++
H
H
v
++ ++
 25  ==>  22 
++ ++
H
H
v
++
 23 
++
The Graph
framework gives easy, immediate access to the node values, here stored in vertices
:
sub sum_remaining{
my($graph, $visited) = @_;
my $sum = 0;
for my $vertex ($graph>vertices()){
$sum += $vertex if $vertex != $visited;
}
return $sum;
}
Additional routines provided are used to transmute the data into a new graph for output:
sub dfs_update{
my($graph, $vertex, $graph_updated, $previous) = @_;
my @successors = $graph>successors($vertex);
for my $successor (@successors){
my $sum_remaining = sum_remaining($graph, $vertex);
$graph_updated>add_edge($previous, $sum_remaining) if $previous;
dfs_update($graph, $successor, $graph_updated, $sum_remaining);
}
$graph_updated>add_edge($previous, sum_remaining($graph, $vertex)) if !@successors;
}
additional languages: Haskell, Node, Python, Raku
**modules:**
Tree::DAG_Node
Simon also brings us a module we haven’t seen in quite some time, Tree::DAG_Node
.
27
 26
  24
  
   21
  
  
 
 25
 23
 
 
 22


I have to say I’m a little sketchy on some of the notation, but the values are correct. The input is taken as a serialized tree array in string format, spaceseparated with a single “.
” for null value. The transformation is quickly effected in a few lines to a new array, and then the array is converted to a hash using treeList2Hash()
. After that a second function, mkDAG()
converts the hash to describe a Directed Acyclic Graph for the module.
sub treeList2Hash($t) {
(! scalar @{$t}  $t>[0] eq '.') && return {};
my @rest = @{$t}[1..scalar @{$t}1];
my $ix = ixSplit(\@rest);
my @left = @rest[0..$ix];
my @right = @rest[$ix+1..$#rest];
return {
name => $t>[0],
left => treeList2Hash(\@left),
right => treeList2Hash(\@right),
};
}
sub mkDAG($h) {
(! scalar keys %{$h}) && return Tree::DAG_Node>new({ name => "" });
my $root = Tree::DAG_Node>new({ name => $h>{name} });
my %left = %{$h>{left}};
$root>add_daughter(mkDAG(\%left));
my %right = %{$h>{right}};
$root>add_daughter(mkDAG(\%right));
return $root;
}
a new FUNCTIONAL PARADIGM, REDUX
Again, where to start with this magnificent undertaking? Everything looks a little different, yet mostly similar. We have an example of a Node
object, adapted with a system of typing and given a method to traverse and apply a function. Ok, we’ve seen this before. That was functional, but this is another level:
package PFLANZE::Node {
use FP::Predicates qw(is_string maybe instance_of);
*is_maybe_Node = maybe instance_of("PFLANZE::Node");
use FP::Struct [
[\&is_maybe_Node, "left"],
[\&is_string, "value"],
[\&is_maybe_Node, "right"]
] => ("FP::Struct::Show", "FP::Struct::Equal");
sub map ($self, $fn) {
my $l = $self>left;
my $r = $self>right;
my $l2 = defined($l) ? $l>map($fn) : undef;
my $r2 = defined($r) ? $r>map($fn) : undef;
$fn>($l2, $self>value, $r2)
}
_END_
}
PFLANZE::Node::constructors>import;
Later on we see a couple more routines we should recognize:
sub tree_sum($t) {
$t>map(
sub ($l, $value, $r) {
$value + ($l // 0) + ($r // 0)
}
)
}
TEST { tree_sum $in } 28;
sub tree_recreate($t) {
my $sum = tree_sum($t);
$t>map(
sub ($l, $value, $r) {
Node($l, $sum  $value, $r)
}
)
}
It all has an intense, rigorous style, which is to be expected. Every value or lack thereof always needs to be accounted for. Tests are continuous and pervasive. We’ll be looking further at this fascinating work as we explore Christian’s submissions in the weeks to come.
Again, Christian is providing an admirably detailed companion blog that provides extensive documentation not only for the specific solutions but also the framework he is constructing. I look forward to the next installment.
a PRETTY DISPLAY for the PEOPLE
additional languages: Raku
As I’ve stated previously, presenting a binary tree for terminal or text output in a pleasing manner is a nontrivial task — not too hard to get okay but quite challenging to get unambiguously right. Most submissions shied away from this aspect of the task this time, and I can hardly blame them. There has been quite a lot of work by the team in the past on the subject, and the aforementioned difficulties leave it as a fairly unsatisfying project to tackle.
Last year, I spent way too much time trying to figure this out, and with a little extra time this week decide to revisit that code and do the refactoring I always wanted to get around to. Inspired in part by Simon Proctor and his Raku submission for PWC 057, I also swapped my slashes for Unicode boxdrawing characters, giving what I consider a quite pleasing result. It makes me think of artdeco robots for some reason, which is, obviously, also pleasing to me.
Input:
┏━━━━━━┫6┣━━━━━━┓
┃ ┃
┏━━┫8┃ ┏━━┫6┣━━┓
┃ ┃ ┃
┃2┣┓ ┏┫3┣┓ ┃9┣┓
┃ ┃ ┃ ┃
┃3┃ ┃5┃ ┃1┃ ┃1┃
Output:
┏━━━━━━━━━━━━━┫38┣━━━━━━━━━━━━━┓
┃ ┃
┏━━━━━┫36┃ ┏━━━━━┫38┣━━━━━┓
┃ ┃ ┃
┃42┣━┓ ┏━┫41┣━┓ ┃35┣━┓
┃ ┃ ┃ ┃
┃41┃ ┃39┃ ┃43┃ ┃43┃
I even went and rewrote the whole thing in Raku, which lead to more refactoring, some of which I backported to this version. Other bits, like replacing some of the simpler routines with constants, could still be done. All this prettyprinting is of course outside the immediate scope of the challenge; this just takes a serialized array and outputs it.
For the challenge itself I made a pair of Node
and BTree
classes with methods to do the transformation, which was a nice exercise. We take a serial list to load a BTree object, then when the work on it is done another method transforms in back into an array. I’m proud of the work, but it’s nothing we haven’t seen already. This, on the other hand, is different:
## predeclare
sub space;
sub dash;
sub vert;
sub rtee;
sub ltee;
sub downr;
sub downl;
sub print_tree (@tree) {
## originally created for PWC 0571 "invertsugar"
## updated for box drawing elements and cleaned up for PWC 113
my $value_width = get_max_value_width(@tree);
## magic trick here, as we get longer values we pretend we're at the top of
## a larger tree to keep from running out of space between adjacent values
## between two parent nodes on the lowest level
my $num_levels = get_level(scalar @tree  1 ) + int($value_width/2);
my $index = 0;
while ($index < scalar @tree) {
my $level = get_level($index);
my $spacer = 2**($num_levels  $level + 1);
my $white = ($spacer/2 + 1 + $value_width) > $spacer
? $spacer
: $spacer/2 + 1 + $value_width;
my $dashes = $spacer  $white;
my $level_node_count = 2 ** $level;
my $node_line;
my $vert_line;
## draw the nodes of each level and any connecting lines to the next
for (1..$level_node_count) {
## if the node is defined draw it in
if (defined $tree[$index]) {
## centers value in a slot $value_width wide, leaning right for odd fits
my $this_width = length($tree[$index]);
my $right_pad_count = int(($value_width$this_width)/2);
my $right_pad = space($right_pad_count);
my $left_pad = space($value_width  $this_width  $right_pad_count);
my $value_format = "${left_pad}%${this_width}s${right_pad}";
my $node = sprintf $value_format, $tree[$index];
## draw connecting lines if children present, or whitespace if not
my $left_branch = defined @tree[2 * $index + 1]
? space($white2) . downr . dash($dashes) . ltee
: space($spacer1). vert;
my $right_branch = defined $tree[2 * $index + 2]
? rtee . dash($dashes) . downl . space($white$value_width2)
: vert . space($spacer$value_width1);
$node_line .= $left_branch . $node . $right_branch;
## construct the vert connector line
my $left_vert = defined $tree[2 * $index + 1]
? space($spacer/2+$value_width1) . vert . space($dashes+1)
: space($spacer);
my $right_vert = defined $tree[2 * $index + 2]
? space($dashes+$value_width+1) . vert . space($spacer/21)
: space($spacer);
$vert_line .= $left_vert . $right_vert;
}
## else insert equivalent whitespace
else {
$node_line .= space(2 * $spacer);
$vert_line .= space($spacer + 2 + $dashes*2 + $value_width*2);
}
$index++;
}
say $node_line;
say $vert_line;
}
}
sub space ($val) { return q( ) x $val }
sub dash ($val) { return q(━) x $val }
sub vert { return q(┃) }
sub rtee { return q(┣) }
sub ltee { return q(┫) }
sub downr { return q(┏) }
sub downl { return q(┓) }
sub get_level ($n) {
## determines the 0based level of a node from its index
return $n > 0 ? int log($n+1)/log(2)
: 0;
}
sub get_max_value_width (@tree) {
## finds the widest string representation in the array and returns the width
my $max = 0;
$_ > $max and $max = $_ for map { scalar split // } grep defined, @tree;
return $max;
}
Additional Submissions in Guest Languages
additional languages: Awk, Bash, C, Lua, Node, Python, Ruby
additional languages: Raku
additional languages: Raku
BLOGS
That66s it for me this week, people! Warped by the rain, driven by the snow, resolute and unbroken by the torrential influx, I somehow continue to maintain my bearings. Looking forward to next wave, the perfect wave, I am: your humble servant.
But if Your THIRST for KNOWLEDGE is not SLAKED,
then RUN (dont walk!) to the WATERING HOLE
and READ these BLOG LINKS:
( don’t think, trust your training, you’re in the zone, just do it … )
Aaron Smith
Abigail
 Perl Weekly Challenge: Represent Integer ( Perl )
 Perl Weekly Challenge 113: Recreate Binary Tree ( Perl )
Adam Russell
 Weekly Challenge 113 — Perl — RabbitFarm ( Perl )
 Weekly Challenge 113 — Prolog — RabbitFarm ( Prolog )
Arne Sommer
 Re Re Raku (and Perl) ( Perl & Raku )
CheokYin Fung
Christian Jaeger
 The Perl Weekly Challenges, #113 ( Perl )
Colin Crain
Flavio Poletti
 PWC113  Represent Integer  ETOOBUSY ( Perl & Raku )
 PWC113  Recreate Binary Tree  ETOOBUSY ( Perl & Raku )
Jaldhar H. Vyas
 Perl Weekly Challenge: Week 113 ( Perl & Raku )
James Smith
 Perl Weekly Challenge #113 ( Perl )
Laurent Rosenfeld
Luca Ferrari
 Perl Weekly Challenge 113: sums and trees — Luca Ferrari — Open Source advocate, human being ( Raku )
Roger Bell_West
 RogerBW’s Blog: Perl Weekly Challenge 113: RepresentRecreate ( Perl & Raku )
Simon Green
 Weekly Challenge 113 ( Perl )
W. Luis Mochan