Challenge 153 Task #1 - Factorials left, factorials right, factorials everywhere!

Task #1 - Left Factorials


Original Description

Submitted by: Mohammad S Anwar

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


Full Source

On the linked OEIS site we can find an example for calculating a left factorial:

!5 = 0! + 1! + 2! + 3! + 4! = 1 + 1 + 2 + 6 + 24 = 34.

From this we can see that to calculate a left factorial !n for a number n we need to sum all factorials from 0 up to n-1.

To do that we will first have to define a way to calculate a factorial. The factorial n! of a number n is the product of all numbers form 1 up to n

For the summing and multiplying of ranges I will use sum0 and product from List::Util.

For a factorial we generate the range 1 .. $n and pass it to the imported product routine. For the left factorials we first generate a list of the factorials that we need and then sum them up.

sub fac($n) {
    product( 1 .. $n );

sub left_factorial($n) {
    return sum0( map { fac() } 0 .. ( $n - 1 ) );

Finally, some wrapper code to generate and print the requested range of left_factorials

sub run() {
    say join( ', ', left_factorials( 1, 10 ) );

sub left_factorials ( $from, $to ) {
    return map { left_factorial() } $from .. $to;

Haskell Solution

Full Source

In Haskell we can make use of lazy lists to define lists of all factorials and all left factorials.

To do this we will define a starting value in both cases and then use zipWith to combine the list itself with another list.

For the factorials the starting value will be 1 (as 0! = 1). And then we zip the facs list with a list off all integers > 1.

So the first element of this list will be 1 as we have defined. The second will be 1 * 1 = 1 - the first element from this list (1) multiplied with the first integer (1). The third element will be 1 * 2 = 2 - the second element from this list (1) multiplied with the second integer (2). This process will be repeated up to the maximum index that will be requested from this list.

facs :: [Integer]
facs = 1 : zipWith (*) facs [1 ..]

For the left factorials we do something similar. The starting value is a 0 this time, the operator will be + instead of *, and the list we will be zipping with will be the list of factorials we just defined.

leftFactorials :: [Integer]
leftFactorials = 0 : zipWith (+) leftFactorials facs

And again some driver code to print the !1 .. !10.

main :: IO ()
main = drop 1 leftFactorials & take 10 & print

Raku Solution

Full Source

After solving it in Haskell I got curious if I could do the same in raku. There are lazy lists/sequences in raku, so the only obstacle should be my lack of skill in that language. I’ve never really tried raku out much and couldn’t figure out how to do it from reading the docs. A blog post by Andrew Shitov about how to define the factorials helped me out. I still don’t get all of it, and I’m not able to explain, but I was able to adapt the solution to make it work for left factorials

# I don't know how to write raku/perl6. With help from this blog post by Andrew
# Shitov I could figure out, how to generate a lazy, inifinite sequence of all
# factorials and adapt to the left factorial problem

my @facs = 1, * * ++$ ... *;
my @left_factorials = 0, 1, * + @facs[ ++$ ] ... *;

say @left_factorials[1..10];

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