*This post is part of a series on Mohammad Anwar’s excellent Perl Weekly Challenge, where Perl and Raku hackers submit solutions to two different challenges every week. (It’s a lot of fun, if you’re into that sort of thing.)*

For challenge #2 this week, the task is to start with $1, and by either adding $1 or doubling the amount, reach $200 in the smallest possible number of steps.

## “Greedy never works”

In a 75-minute lecture some decades ago, my Advanced Algorithms professor said, over and over, “greedy never works,” while all the while showing us exceptions to that mantra. Greedy algorithms operate by always making the locally optimal choice, which might be the biggest/smallest, closest to the goal, etc., depending on the problem. However, for a great many problems, the locally optimal choice is not always the best. For example, suppose you’re trying to climb the following mountain range (The ASCIIHorn):

B /\ /\A / \ / \/ \ / C \

Let’s say you’re at position (A), and your goal is to find the highest point. A greedy algorithm would *always* choose to go higher, never lower (even temporarily), so you’d reach the smaller left peak, and be stuck there. This is known as a hill climbing problem, for which greedy doesn’t work, because you have to go down through the valley (C) before you go up to the goal (B).

But in the case of *this* challenge, greedy does work, but only if we go in the opposite direction: given $200, you can divide by 2 or subtract $1, and your goal is now to get to $1 in the least number of steps. So how does it work? If we *can* divide by two (i.e., if our amount is an even number), then divide by two. Otherwise, subtract 1. This is greedy, because dividing by two will *always* result in the biggest deduction (or at least a tie, in the case you had $2). We can easily work this out by hand:

$200 even, divide by 2 100 even, divide by 2 50 even, divide by 2 25 odd, subtract 1 12 even, divide by 2 6 even, divide by 2 3 odd, subtract 1 2 even, divide by 2 (subtract 1 is equivalent) 1 done.

Thus, the solution is achieved in 8 moves. The moves are in the opposite order to what we want, but making a reversed list is trivial in Perl and Raku.

## Greedy Solution

#### Perl

sub double_or_plus( $start, $end ) { my @path = $end; while ($end != $start) { $end = $end % 2 ? $end - 1 : $end / 2; unshift @path, $end; } @path; }

#### Raku

sub double-or-plus( Int $start, Int $end is copy ) { my @path = $end; while ($end != $start) { $end = $end %% 2 ?? ($end / 2).Int !! $end -1; @path.unshift($end); } @path; }

Note the use of `unshift`

in both cases, to produce a reversed array on the fly.

## Exhaustive Solution

If you don’t yet trust the greedy solution, it’s easy to write an efficient exhaustive validator (shown here in Perl). This one takes the `$start`

and `$end`

, but it also takes a `$max_path`

value, which is the length of the path from `double_or_plus()`

. That allows us to avoid pointlessly searching paths that are already longer than the one we found.

sub exhaustive( $start, $end, $max_path ) { my @shortest = (0) x ($max_path + 1); sub ( $start, $end, @path ) { push @path, $start; return if @path > @shortest or $start > $end; @shortest = @path and return if $start == $end; __SUB__->($start + 1, $end, @path); __SUB__->($start * 2, $end, @path); }->($start, $end); @shortest; }

This uses some newer Perl features: `signatures`

, and `current_sub`

. These made my closure a little bit cleaner looking, but are absolutely not required.

With this sub, we can purchase a bit of confidence that our greedy result really is the optimal one. I ran this in a loop, increasing the `$end`

as high as I was patient enough to wait. (It’s not necessary to increase `$start`

, because we will have already validated `1..$start`

.) The greedy solution was optimal up to an end of 100,000, which is about when I hit **Ctrl-C**.

## Why Greedy Works

A full proof is beyond the scope of this blog post. However, conceptually, you can imagine that dividing by two will *always* get you at least as close as subtracting $1. (That’s the greedy part.) It’s optimal in this case, because we are always *only* in one of the following states:

- Even number: Can divide by 2, which gets us closer more quickly
- Odd number: Subtracting is the only option, but guarantees next number is even

Thus, dividing when we can gets us to lower numbers faster, and there is no penalty for our greed, since when we do have to subtract, we just get another even number. To prove otherwise, you would have to find an even number such that subtracting leads to a shorter path than dividing. Try to find that counterexample, and you might just prove our greedy solution is in fact optimal.