Advent Calendar 2019

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The gift is presented by Scimon Proctor. Today he is talking about his solutions to Task #2: Pascal Triangle of “The Weekly Challenge - 003”.

Create a script that generates Pascal Triangle. Accept number of rows from the command line. The Pascal Triangle should have at least 3 rows. For more information about Pascal Triangle, check this wikipedia page.

So Pascal’s Triangle is pretty and I decided I wanted to get it nicely laid out like so:

                   1
                 1   1
               1   2   1
             1   3   3   1
           1   4   6   4   1
         1   5   10  10  5   1
       1   6   15  20  15  6   1
     1   7   21  35  35  21  7   1
   1   8   28  56  70  56  28  8   1
 1   9   36  84 126 126  84  36  9   1

So this challenge has two parts. Generating the numbers for the triangle and then laying it out. In this case I didn’t read the article too much but had a neat idea. For a given row (eg [1,2,1] you can generate the next row by implementing doing the following):

# Here's our row
my @row = [1,2,1];
# Make a copy with a 0 at the start
my @row1 = [0, |@row]; # [0,1,2,1] (| breaks the array down so it doesn't get added as a single object in this case)
# Make a copy with a 0 at the end
my @row2 = [|@row, 0]; # [1,2,1,0]

# For each item in row1 add it to the same item in row2
my @row3 = [
    @row1[0] + @row2[0], # 0 + 1 = 1
    @row1[1] + @row2[1], # 1 + 2 = 3
    @row1[2] + @row2[2], # 2 + 1 = 3
    @row1[3] + @row2[3], # 1 + 0 = 1
];

So… that’s a thing but it’s a bit unwieldy if only there was some kind of way to simplify this. Oh yes, the Zip metaoperator Z. Place a Z between two lists and it will default to applying the , operator to each item in turn to make a new list of lists :

(0, 1, 2, 1) Z (1, 2, 1, 0) => ((0,1), (1,2), (2,1), (1,0))

But you can give Z and inline operator to apply list for instance + giving you :

0,1,2,1) Z+ (1,2,1,0) => (1, 3, 3, 1)

Which looks very nice. So if we just wanted our script to out lists of lists that make up Pascal’s Triangle we could to this :

 Note the challenge says we need at least 3 lines so we catch that here.
multi sub MAIN( Int() $lines is copy where * > 2 ) {
    # Set up the initial state.
    # Decrement lines.
    my @row = (1);
    my @out = [ [1], ]; # Note we need a trailing comma or the array will be flattened
    $lines--;

    # Did I mention this earlier? Repeat / While is like the Perl5 do / while pair
    repeat {
        @row = (0, |(@row)) Z+ (|(@row), 0); # Here's the Zip magic
        @out.push( @row.clone ); # We need to clone our row or will get clobbered in the next loop
        $lines--;
    } while ( $lines );

    @out.say;
}

And that works but… well I wanted it pretty. The simple thought would be to join each row with spaces and then join the lot with newlines but just doing that you end up with :

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1

Which is…. a bit meh.

Of course you could make those lines up then find the length of the last line and left pad all the other by half that amount.

@out = @out.map( { $_.join(" ") } );

my $len = @out[*-1].codes;

@out.map( { ( " " x ( ($len - $_.codes) div 2) ) ~ "$_" } ).join("\n").say;

And that’s a bit nicer :

                   1
                 1   1
               1   2   1
             1   3   3   1
           1   4   6   4   1
         1   5   10  10  5   1
       1   6   15  20  15  6   1
     1   7   21  35  35  21  7   1
   1   8   28  56  70  56  28  8   1
 1   9   36  84 126 126  84  36  9   1

But it falls apart when larger numbers get added. So we need to pad our output based on the length of the biggers number. Here’s what I came up with :

sub pad( Str $val, Int $len ) {
    my $diff = $len - $val.codes;
    my $rpad = " " x ( ( $diff div 2 ) + 1 );
    my $lpad = " " x ( ( $diff div 2 ) + ( $diff % 2 ) );
    return "{$lpad}{$val}{$rpad}";
}

Pass in the number to pad and the length of the largest number and it returns the number all laiud out nicely. With that here’s my final code :

sub pad( Str $val, Int $len ) {
    my $diff = $len - $val.codes;
    my $rpad = " " x ( ( $diff div 2 ) + 1 );
    my $lpad = " " x ( ( $diff div 2 ) + ( $diff % 2 ) );
    return "{$lpad}{$val}{$rpad}";
}

multi sub MAIN( Int() $lines is copy where * > 2 ) {
    my @row = (1);
    my @out = [ [1], ];
    $lines--;
    my $max = 1;

    repeat {
        @row = (0, |(@row)) Z+ (|(@row), 0);
        @out.push( @row.clone );
        $max = @row.max;
        $lines--;
    } while ( $lines );

    my $len = $max.codes;

    @out = @out.map( -> @list { @list.map( { pad($_.Str,$len) } ).join("") } );

    $len = @out[*-1].codes;

    @out.map( { ( " " x ( ($len - $_.codes) div 2) ) ~ "$_" } ).join("\n").say;
}

multi sub MAIN($) {
    note "Please input a number of lines that must be at least 3";
}

And that seems to do the trick which is fun. (This is the code used to make the triangle at the top).

If you have any suggestion then please do share with us perlweeklychallenge@yahoo.com.

Advent Calendar 2019