Just saw a nice post with solutions to Euler problem #2 in several different languages.

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

Find the sum of all the even-valued terms in the sequence which do not exceed four million.

This one is so easy in Perl 6 it’s not worth writing a script, I just did it as a one-liner in the REPL:

> say [+] (1, 1, *+* ... 4000000).grep(* !% 2) 4613732

Breaking it down: `1, 1, *+* ... *`

is the standard way to generate the infinite Fibonacci sequence in p6. It creates a list of numbers using the `...`

“series” operator. The list starts with `1, 1`

, and each additional number is created by adding the two previous numbers (that’s the `*+*`

bit). The final `*`

means the series has no limit. By switching the limit to 4000000 instead, we explicitly say “stop the series as soon as the number 4000000 is exceeded.”

> say ~(1, 1, *+* ... 4000000) 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946 17711 28657 46368 75025 121393 196418 317811 514229 832040 1346269 2178309 3524578

Then `grep`

filters on the test `* !% 2`

. `!%`

is the “divisible by” operator; `* !% 2`

generates a closure that checks its one input for divisibility by 2.

> say ~(1, 1, *+* ... 4000000).grep(* !% 2) 2 8 34 144 610 2584 10946 46368 196418 832040 3524578

Finally `[+]`

is the reduce metaop for addition; it sums the list that follows.

I know that when people see one-liners, frequently they think the result has been golfed down. But this is really the natural way to express this problem in Perl 6.

**Update**: Very shortly after I wrote this, the “is divisible by” operator was changed to `%%`

. At the moment, both versions of the operator work in Rakudo, but I would expect `!%`

to go away eventually.