RogerBW's Blog

Perl Weekly Challenge 113: Represent-Recreate 19 May 2021

I’ve been doing the Perl Weekly Challenges. The latest involved finding constrained integer decompositions and more binary tree furkling. (Note that this is open until 23 May 2021.)

TASK #1 › Represent Integer

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.

This might better be expressed as: "as a sum of positive integers, each of which contains the digit $D". It's implied that each integer can only be used once, and I chose to assume this because it made things more fun.

So there's a two-part solution that I find reasonably workable. First, get a list of all the valid integers. (So if $D is 7, that would be 7, 17, 27, etc.) Then use a bitmask to determine and sum all possible combinations of those integers: 7, 17, 7+17, 27, 27+7, etc.

(Some of the languages have automatic combination generators, but this was straightforward enough that it didn't seem worth using that level of sophistication.)

sub ri {
  my ($n,$d)=@_;

Here's my list of valid integers (using Perl's "treat a thing as a string if it's in a stringy sort of context" approach).

  my @e=grep /$d/,(1..$n);

So if I have four valid integers in the list, I'll run the mask from 1 to 2⁴-1=15.

  foreach my $i (1..(1<<(scalar @e))-1) {
    my $s=0;

Add up any entries that match the current mask.

    foreach my $ii (0..$#e) {
      if (1<<$ii & $i) {
    if ($s==$n) {
      return 1;
  return 0;

There might be some scope for optimisation of the order in which things appear, but this gets the job done and is easy to implement everywhere. Raku replaces << with +< and & with +& for no obvious reason. In Python, Ruby and Rust I compile a regexp and then match it in their various different ways.

  e=[i for i in range(1,n+1) if re.match(dd,str(i))]
  e=1.upto(n).find_all {|i| i.to_s =~ dd}

  let dd=Regex::new(&d.to_string()).unwrap();
  let e=(1..=n).filter(|i| dd.is_match(&i.to_string())).collect::<Vec<u64>>();

(Note that you have to filter the iterator before you collect it into a vector.)

And because I was interested, I plotted out the values for all digits and $N in the range 1-100. ("X" indicates a true return.)

  1 .X........
  2 ..X.......
  3 ...X......
  4 ....X.....
  5 .....X....
  6 ......X...
  7 .......X..
  8 ........X.
  9 .........X
 10 XX........
 11 .X........
 12 .XX.......
 13 .X.X......
 14 .XX.X.....
 15 .X...X....
 16 .X.X..X...
 17 .X.....X..
 18 .X..X...X.
 19 .X.......X
 20 XXX..X....
 21 .XX.......
 22 .XX...X...
 23 .XXX......
 24 .XX.X..X..
 25 .XX..X....
 26 .XXX..X.X.
 27 .XX....X..
 28 .XX.X...XX
 29 .XX......X
 30 XXXX.X....
 31 .XXX......
 32 .XXX..X...
 33 .XXX......
 34 .XXXX..X..
 35 .XXX.X....
 36 .XXX..X.X.
 37 .XXX...X..
 38 .XXXX...XX
 39 .XXX.....X
 40 XXXXXX....
 41 .XXXX.....
 42 .XXXX.X...
 43 .XXXX.....
 44 .XXXX..X..
 45 .XXXXX....
 46 .XXXX.X.X.
 47 .XXXX..X..
 48 .XXXX.X.XX
 49 .XXXX....X
 50 XXXXXX....
 51 .XXXXX.X..
 52 .XXXXXX...
 53 .XXXXX....
 54 .XXXXX.XX.
 55 .XXXXX....
 56 .XXXXXX.X.
 57 .XXXXX.X.X
 59 .XXXXX...X
 60 XXXXXXX...
 61 .XXXXXXX..
 62 .XXXXXX...
 63 .XXXXXX...
 65 .XXXXXX...
 66 .XXXXXX.X.
 69 .XXXXXX..X
 71 .XXXXXXX..
 72 .XXXXXXX..
 73 .XXXXXXX..
 75 .XXXXXXX..

TASK #2 › Recreate Binary Tree

You are given a Binary Tree.

Write a script to replace each node of the tree with the sum of all the remaining nodes.

I.e. each node's value should become (sum of all old node values) minus (old value of this node). I really can't be bothered to write a parser for this weird ASCII format, so I start with the flat array representation I've used before. This makes life much simpler. In this case I used -1 as the null value; which means it's simply a matter of passing over the list once to get a sum of all valid entries, then passing over it again to work out the new values. Raku:

sub rbt(@ti) {
  my $s=0;
  for @ti -> $n {
    if ($n>=0) {
  my @to;
  for @ti -> $n {
    if ($n>=0) {
      push @to,$s-$n;
    } else {
      push @to,$n;
  return @to;

and all the rest are basically the same.

Full code on github.

  1. Posted by RogerBW at 07:33pm on 24 May 2021

    In part 1, some people assumed no number repetition, others didn't, and some assumed that it's only valid if all the D-containing numbers sum to N – which is also a valid interpretation of the problem statement. All of them produce interesting results, but I'd call that an underspecified problem.

    In part 2, the binary tree part is of course irrelevant; there's nothing about the problem that requires a tree structure, and otherwise it's just a sum and a map. One could use an actual tree representation, but I can't see that being any faster or more efficient.

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