TL;DR

On with TASK #2 from the Perl Weekly Challenge #119. Enjoy!

# The challenge

Write a script to generate sequence starting at 1. Consider the increasing sequence of integers which contain only 1’s, 2’s and 3’s, and do not have any doublets of 1’s like below. Please accept a positive integer $N and print the $Nth term in the generated sequence.

1, 2, 3, 12, 13, 21, 22, 23, 31, 32, 33, 121, 122, 123, 131, …

Example

Input: $N = 5 Output: 13 Input:$N = 10
Output: 32

Input: $N = 60 Output: 2223  # The questions As a meta-question, I wonder if there’s a hidden scheme in pairing this challenge with the one on nibbles (i.e. “different aggregations of consecutive bits”). Anyway. The description of the sequence and how to build it seems fine: we’re assuming that we will consider any positional representation of integers, in a base that contains at least symbols 1, 2, and 3; this provides an ordering for any two candidates and we’re fine. We’re also told to start counting from 1, which is the value corresponding to an input $N equal to 1.

The definition of doublet is a bit fuzzy to me, but English is definitely not my mother language. Anyway, seeing that 121 and 131 are in the list but 11, 111, and 112 are not, I guess we’re dealing with two consecutive 1 digits.

# The solution

This is my first stab at the problem, coded in Raku. I’m sticking to a representation of the “numbers” that is completely held as strings. It’s basically a way of counting in base 3, but not really:

#!/usr/bin/env raku
use v6;

sub sequence-without-one-on-one (Int:D $N is copy where 0 < *) { my$candidate = '1';
while ($N > 1) {$candidate = succ-of($candidate); --$N if $candidate !~~ /11/; } return$candidate;
}

sub succ-of (Str:D $x) { my ($carry, @succ) = (True, $x.comb.reverse>>.Int.Slip); for @succ ->$item is rw {
($item,$carry) = ($item + 1, False) if$carry;
($item,$carry) = (1        , True)  if $item > 3; last unless$carry;
}
@succ.push: 1 if $carry; @succ.reverse.join: ''; } my @inputs = @*ARGS ?? @*ARGS !! qw< 5 10 60 >; sequence-without-one-on-one(+$_).put for @inputs;


The succ-of function takes an input and computes its successor in the 1/2/3 restricted world. It does not care about the doublets, they are filtered out in sequence-without-..., that also does the counting. Choices.

As I was saying, it’s almost like counting in base 3, but not really. We might map 1 to 0, then 2 to 1, then 3 to 1 and have a valid base-3 representation for each valid sequence, including those with doublets. But that would make it hard to generate 23 and 123 by simple counting, because they would map to 12 and 012 respectively, which are the same number in base 3. Ouch.

But this does not mean that we cannot use a counting approach that relies on native integers, instead of rolling our own like in the example above.

In particular, any base from 4 on contains all the symbols that we need; needless to say, base 4 is the one that leads to the least waste because it only has one additional digit 0 that we cannot admit, so in general there will be less skipping in this base. This brings us to this alternative implementation of succ-for:

sub succ-of (Str:D $x) { my$X = $x.parse-base(4); loop { my$candidate = (++$X).base(4); return$candidate if $candidate !~~ /0/; } }  I don’t know if it’s more efficient but I think it makes it easier to see what’s going on (the initial parse-base(4) is just there for efficiency, having my$X = +$x would work as well, although in this case I would need to change the regular expression test in the loop). Let’s now turn to Perl. We can code pretty much the same solution, only we need to provide an implementation for the change of base between 4 and 10: #!/usr/bin/env perl use 5.024; use warnings; use experimental qw< postderef signatures >; no warnings qw< experimental::postderef experimental::signatures >; sub sequence_without_1_on_1 ($N) {
my $candidate = 1; while ($N > 1) {
$candidate = succ_of($candidate);
--$N if$candidate !~ m{11}mxs;
}
return $candidate; } sub succ_of ($x) {
$x = base_4_to_10($x);
while ('necessary') {
my $candidate = base_10_to_4(++$x);
return $candidate if$candidate !~ m{0}mxs;
}
}

sub base_4_to_10 ($x) { my$X = 0;
for my $digit (split m{}mxs,$x) {
$X = ($X << 2) + $digit; } return$X;
}

sub base_10_to_4 ($x) { my @digits; while ($x) {
push @digits, $x & 0b11;$x >>= 2;
}
return join '', @digits ? reverse @digits : 0;
}

my @inputs = @ARGV ? @ARGV : qw< 5 10 60 >;
say sequence_without_1_on_1($_) for @inputs;  I’m taking advantage of Perl’s laxer approach to variables, e.g. by reusing variable $x in succ_of instead of taking a new one. No big deal, actually; this goes at the expense of readability, so I’m not sure it’s a win overall.

The conversion functions take advantage of the fact that 4 is a multiple of 2, hence it’s easy to use bit fiddling to do the conversions. This is also why I have the meta-question in the previous section - both solutions rely on bit fiddling in the end!

Well… I guess it’s everything at this point! Stay safe and have fun!