TL;DR

On with TASK #2 from The Weekly Challenge #176. Enjoy!

# The challenge

Write a script to find out all Reversible Numbers below 100.

A number is said to be a reversible if sum of the number and its reverse had only odd digits.

For example,

36 is reversible number as 36 + 63 = 99 i.e. all digits are odd.
17 is not reversible as 17 + 71 = 88, none of the digits are odd.


Output

10, 12, 14, 16, 18, 21, 23, 25, 27,
30, 32, 34, 36, 41, 43, 45, 50, 52,
54, 61, 63, 70, 72, 81, 90


# The questions

I’m assuming that we’re talking about positive integer numbers here, right?

Also, the concept of reverse is defined not in matematical terms (at least that I can recognize, but YMMV) but in textual terms (take the digits and read them in reverse order).

# The solution

• Interestingly, the reversing operation is not invertible. All integers with a trailing 0 map onto an integer that has fewer digits, but the reverse does not happen.
• single-digit number don’t cut it, because they happen to be their own reverse, which means doubling when doing the sum, which means an even number (multiple of 2).
• two-digits number cannot have their sum go beyond 99, which means that the two digits must have a sum that is at most 9.

The last bullet bears an explanation. If we have a number $h \cdot 10 + l$, we end up with the following sum:

$S = (h + l) \cdot 10 + (h + l)$

Now if $h + l > 9$, it can only be a number below 19 (the maximum possible value with summing two digits is $9 + 9 = 18 < 19$), i.e. it’s a number of type $10 + o$ where $o < 9$ . This leads us to:

$S = (10 + o) \cdot 10 + 10 + o \\ S = 1 \cdot 100 + (o + 1) \cdot 10 + o$

The second digit is actually $o + 1$ because $o < 9 \Rightarrow o + 1 < 10$. This can’t possibly have all odd digits, because it contains both $o + 1$ and $o$. Only one of them can be odd, right?

So we’re left with two-digit integers $h \cdot 10 + l$ where $h + l < 10$, and these are the integers we will look for, starting from Perl:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

say join ', ', reversible_numbers();

sub reversible_numbers {
my @retval;
for my $lo (0 .. 4) { for (my$hi = $lo + 1;$hi <= 9 - $lo;$hi += 2) {
push @retval, $lo * 10 +$hi if $lo; push @retval,$hi * 10 + $lo; } } return sort {$a <=> $b } @retval; }  Almost every time we find a good number, the reverse is good as well. The only exception is when the good number ends with a 0, in which case we ignore the reverse. This accounts for the if$lo in the first push line.

We’re actually going bottom to top here; the first digit is constrained to be at most 4, because it’s the lowest first digit that allows us finding out all target numbers considering the double finding (i.e. 45 is a solution as well as 54 and from that point on we will only find integers that we already found thanks to this reversing rule).

For the same reason, the other digit starts from the first plus 1, so that they have opposite oddness (yielding an odd sum) and we don’t count candidates twice. This is another time where I found that the C-style for hit the nail right in the head.

The resulting array of collected items is not sorted, so we might just return it as we’re not requested to do the sorting. Anyway it does not cost too much and it’s nice to see the result in ascending order, so why not?

The Raku counterpart is mostly the same, only taking advantage of gather/take, which I love:

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

reversible-numbers().join(', ').put;

sub reversible-numbers {
return gather {
for 0 .. 4 -> $lo { my$hi = $lo + 1; while$hi <= 9 - $lo { take$lo * 10 + $hi if$lo;
take $hi * 10 +$lo;
\$hi += 2;
}
}
}.sort
}


And this is all for today, stay safe and hydrated!