Drive It In Reverse, Then. Done.

Wherein whether we read it backwards or forward, the result is still very very odd…

THE WEEKLY CHALLENGE – PERL & RAKU #176 Task 2


“Darn the wheel of the world! Why must it continually turn over? Where is the reverse gear?”

— Jack London


Reversible Numbers

Submitted by: Mohammad S Anwar

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

ANALYSIS

We could do this the easy way, or the hard way. Or maybe the hard way, and then the easy way, depending on when you start counting the effort. We’re going to front-load the work here, and logically think this thing through, which I suspect will yield an easy solution in the end. With considerably more mental effort, of course.

Or we could just test the numbers from 1 to 100 and select any that fit. That’s super easy, but also disappointingly boring.

The main drive behind my febrile over-analysis is trying to figure out in the first place why these numbers are called reversible. There are several examples of so-called reversible numbers in the literature, but none with this definition that also gives any inkling as to why they are called this. What on Earth do odd digits have to do with reversibility? I’m still wondering about that one. At this point I’m inclined to think its something made-up once for a programming problem and nothing more.

We do, of course, reverse the number at one stage. Then we add the number to its reverse, and see if the digits are odd. Not that the number is odd, which of course would also be true if it were made from odd digits, but that the all of the actual digits themselves are odd.

So let’s get on with it.

How do you make an odd number? You add an even number and an odd one. Zero is be considered even here, a premise not in serious dispute even as it occupies a unique role among the whole numbers. In any case, for whatever lingering doubts are to whether or not zero is even, none of these arguments convincingly assert it to be odd either. I’ll grant the idea that it suspiciously looks to be neither, but the fact is that mathematically it does all the things and even number should do.

However, we also need to consider we are looking for numbers less than 100. These will have only two digits, so if leading zeros are not allowed, then by nature of the reversal step all zeros in either position are out. However from the examples given we list 30 as a solution, so it appears they are allowed in some cases. Zeros, then, are in — if only temporarily when the numbers are reversed.

This application seems… inconsistent.

But then I don’t think there is a strict mathematical definition to reversing the digits of a number. Seems kind of a free-for-all, to be honest. It’s not a mathematical process, but textual, operating on the glyphs.

Let’s get back to it and keep going on this less-than-one-hundred angle. These numbers have 1 or 2 digits. Single digit numbers reverse to the same digit, and the sum of any digit with itself is 2 times the value, and hence even.

So single digit values are out.

We are left with two-digit values with one even and one odd digit.

But what about if we carry? If we carry a 1 and form a three-digit number then the leading 1 is odd, so that might work, right? No one said we couldn’t add a digit, only that all the digits that are present in the final sum are odd. The thing is that we need even-odd pairs that when swapped create odd-even pairs, or vice versa. If the rightmost sum in the sum is odd, then the left sum, comprised of the same digits, will also be odd. But carrying over a 1 from the right sum will make the left sum — which because it also will carry becomes the center digit in the final sum — adds 1 to that odd number and becomes even.

So all carrying is excluded, at least with 2-digit numbers.

Do we have enough information now? I believe we do. We need a list of two-digit numbers such that one digit is even and the other odd, where the two digits summed together do not exceed 9, and the first digit is not zero. Any number that satisfies these criteria should be so-called reversible. I’d rather call them something more descriptive, like odd-digit-reversible, but whatever. Perhaps I shall. It seems to be a bit of a free-for-all out there.

Three-Digit Numbers

For 3-digit numbers the rules are a little different. In these, the center digit will always reverse to itself, and hence sum to an even number. So the only way for a 3-digit number to be reversible is for the ones position to carry to the tens. The reverse-sum of these numbers will always be of the form 1xxx then, where x is some odd digit.

In these number the first and third position must therefore sum to an odd number greater than 9, and the middle position can be any number that will not carry with 1 added, which is to say 0 through 4. It’s easier to construct these out-of-order, modifying the loops of our 2-digit solution and adding a third interior loop to fit the center digit, and then sorting the output.

Four Digits

For four digits we get some number mnpq, with positions 1234, such that

m + q is odd

n + p is odd

if m + q carries, then the interior pair n + p must be even so it becomes odd when the carry is added at position 3 . However this would also need to carry to make position 2 odd as well. Position 2, in turn , would then carry to postion 1, which like postion 4 must be odd already. So m + q cannot carry. Internally, n + p is restrained by the same logic as the 2-digit numbers — if they at postion 2 then position 3, which was odd, will become even. So this pair cannot carry either. The constrints on carrying for the four digit numbers are the same as taking the set of two digit numbers and inserting one member of the set inside another, as these are the only allowed pairings.

Each combination of two digit odd-digit-reversible numbers will construct a new 4-digit number, with one small modification: the addition of a new valid group of inserted values — 01, 03, 05, 07, 09 — that allow a leading 0. Thus 2015 is allowed as a valid reversible. See what I said before about disallowing leading zeros being inconsistent? A large amount of what we’re doing here isn’t math, but symbolic pairings according to sets of rules that happen to be mathematically based. If we allowed leading zeros the scaling would be perfect, instead of the outmost pairs being bound by an additional constraint. The quantity of 4-digit numbers is not then the 2-digit quantity squared, but somewhat more. However this scaling, limited or not, works for all even numbers of digits. The number is symmetrically divided down the middle, with left and right pairs that sum to an odd number less than 10.

As no carrying is ever allowed, this pattern, inserting a member of an extended group of 2-digit solutions into the center of an n-minus-2 digit solution, can be generalized outward to all even numbers of digits.

FIVE DIGITS

At first I thought the carrying pattern from the 3-digit values would extend to all odd numbers of digits, but this is not in fact the case.

Given a number mnpqr with positions 12345, then:

m + r is odd

n + q is odd

Position 3 maps to itself so must receive a carry to be odd, thus p4 is odd and carries to p3, and so p2 carries as well to p1. But p1 initially sums the same as p5, which is odd already, and this carry will make it even, which is disallowed. The position p2 must not carry, but also must, and we have a contradiction.

The working pattern of chaining carries established in 3-digit values is broken by the presence of multiple digits surrounding the central pivot. There can be no 5-digit odd-digit-reversible numbers. I believe 7-digit ODRs work, though, for reasons I will leave as an exercise to the reader.


Ok fine: given a number with positions 1234567, p4 is the pivot, so p5 must carry to it. Thus p3 carries to p2, and p2 + p6 must initially be even and not carry, as p1 + p7 must be odd to start and remain that way. However p1 + p7 must carry to p6 to mirror the action performed on by p3 on p2. So to summarize, the ruleset is:

  • p1 + p7 is odd and must carry
  • p2 + p6 is even and must not carry
  • p3 + p5 is odd and must carry
  • p4 must not carry

I conject that what we are looking at is a pattern of non-carrying digits surrounded by carrying pairs: left ~ pivot ~ right, such that each of the left and right sides can be further reduced using the same pattern. Thus the 5-digit numbers will not work, but the 7-digit ones can be viewed as 3 + 1 + 3. The next such pattern is (3 + 1 + 3) + 1 + (3 + 1 + 3) or 15 digits.

That one, and for that matter the other odd numbers of digits, I’ll leave as an exercise to the reader.

METHOD

PERL 5 SOLUTION

I ended up making little routines to construct 2-digit, 3-digit and 4-digit odd-digit-reversible numbers. They all loop through allowed digits to construct the various numbers, according to the rules we’ve come up with above about summing each set of paired digits.

Yes, to be logically grounded I should have recomputed the extended @twos to include leading zeros, which would be easy enough using a modification of the original loop structure starting the outer loop at 0 instead of 1. But I allowed myself to just add in the short list of extra values as a list of strings. It works and it’s obviously true so I went for clarity.

use warnings;
use strict;
use utf8;
use feature ":5.26";
use feature qw(signatures);
no warnings 'experimental::signatures';


## we'll construct TWO DIGIT numbers in the form $m$n

my @twos;

FIRST: for my $m ( 1..9 ) {
    for my $n ( 0..9 ) {
        next unless ($m + $n) & 1;    ## sum is odd
        next FIRST if $m + $n > 9;    ## but will not carry
        push @twos, "$m$n";
    }
}

say "@twos";

## THREE DIGIT $m$p$n

my @threes;

ONE: for my $m ( 1..9 ) {
    TWO: for my $n ( 0..9 ) {
        next unless ($m + $n) & 1;          ## sum is odd
        next unless $m + $n > 9;            ## must carry
        for my $p (0..9) {
            next TWO if 2 * $p + 1 > 9 ;    ## must not carry
            push @threes, "$m$p$n";
        }
    }
}

@threes = sort { $a<=>$b } @threes;

say "@threes";


## FOUR DIGITS

## first we get a list of two-digit ODR numbers (that's Odd-Digit-Reversible to you)
## we could duplicate the code above or just use @twos. 
## we'll just use @twos that we already computed, then.

my @fours;

## we need the 0-based pairs to extend @twos, though

my @twos_ext = (@twos, qw( 01 03 05 07 09 ) );

## insert from @twos-ext between the digits of a value from @twos
for my $outer ( @twos ) {
    for my $inner ( @twos_ext ) {
        push @fours, (substr $outer, 0, 1).$inner.(substr $outer, 1, 1);
    }
}

@fours = sort { $a<=>$b } @fours;

say "@fours";



The Perl Weekly Challenge, that idyllic glade wherein we stumble upon the holes for these sweet descents, is now known as

The Weekly Challenge – Perl and Raku

It is the creation of the lovely Mohammad Sajid Anwar and a veritable swarm of contributors from all over the world, who gather, as might be expected, weekly online to solve puzzles. Everyone is encouraged to visit, learn and contribute at

https://theweeklychallenge.org

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