# ETOOBUSY ðŸš€ minimal blogging for the impatient

# PWC153 - Left Factorials

**TL;DR**

Here we are with TASK #1 from The Weekly Challenge #153. Enjoy!

# The challenge

Write a script to compute

`Left Factorials`

of`1`

to`10`

. Please refer OEIS A003422 for more information.

Expected Output:`1, 2, 4, 10, 34, 154, 874, 5914, 46234, 409114`

# The questions

Our fine host is a fox disguised as an innocent lamb!

After having endured countless petty nitpicks (by many, yours truly included) about missing stuff in formulating the challenges, we are gently redirected to a place that leaves no doubt about what weâ€™re asked to do:

\[!n = \sum_{k = 0}^{n-1} k!\]Well played indeed!

# The solution

Why provide a solution when we can provideâ€¦ *three*?

Raku goes first, with two of them that will be printed one along the other, for comparison:

```
#!/usr/bin/env raku
use v6;
sub MAIN (Int:D $min = 1, Int:D $max = 10) {
($min .. $max).map({left-factorial($_)}).join(', ').put;
($min .. $max).map({left-factorial-cached($_)}).join(', ').put;
}
```

The first takes advantage of the `multi`

mechanism, separating lower
values from higher ones.

```
multi sub left-factorial (Int:D $n where 0 <= * <= 2) { $n }
multi sub left-factorial (Int:D $n where * > 2) {
my $f = 1;
1 + (1 ..^ $n).map({$f *= $^x}).sum;
}
```

These higher ones are always calculated over
and over, so in this case weâ€™re not reusing any previous calculation,
i.e. calculating `left-factorial`

for 5 does not leverage our previous
calculation for 4.

This is a *totally* unacceptable loss of performance, of course. ðŸ™„

In this case, it makes *total* sense to keep those previous values around,
because weâ€™re requested to print a sequence of consecutive values. This
leads us to the second solution, which keeps track of previous values in
a few `state`

variables:

```
sub left-factorial-cached (Int:D $n where * >= 0) {
state $factorial = 1;
state $k = 1;
state @left-factorials = 0, 1, 2;
while $n > @left-factorials.end {
$factorial *= ++$k;
@left-factorials.push: @left-factorials[*-1] + $factorial;
}
return @left-factorials[$n];
}
```

This last solution might be readily translated into Perl, but we
take a lazy detour here and let Perl manage caching for us. This
goes at the expense of *over-caching*, because values for the
`factorial`

functions are all cached too, while this is not strictly
necessary:

```
#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';
use Memoize;
my $min = shift // 1;
my $max = shift // 10;
say join ', ', map { left_factorial($_) } $min .. $max;
memoize('left_factorial');
sub left_factorial ($n) {
return $n if $n <= 2;
return factorial($n - 1) + left_factorial($n - 1);
}
memoize('factorial');
sub factorial ($n) {
return 1 if $n < 2;
return $n * factorial($n - 1);
}
```

Readability is probably better here, though. This solution makes it clear the recursive nature of the approach, while at the same time acknowledging that some dynamic programming can help speed things up as inputs go high.

Soâ€¦ what do you prefer?

*Comments? Octodon, , GitHub, Reddit, or drop me a line!*