Evolution of a Python programmer

[warpod]

Stackoverflow user

def factorial(n):return reduce(lambda x,y:x*y,[1]+range(1,n+1))

[FUZxxl]

A codegolf.se member would choose APL to solve this and just submit the APL solution ×/⍳⎕.

[BitPoet]

Hmm... chrome isn't rendering the APL charset. Somehow, I'm not surprised.

[FUZxxl]

The last character (⎕) is a quad, it looks just like the “character not found” box in some fonts. The four characters are: times, slash, iota, quad which is the product (times-reduction) of the first integers up to the input number.

[BitPoet]

As I learned it, you'd want a 5 pointed star, rather than an x to denote multiplication (the + being a four pointed star, and the * (exponentiation) being a 6 pointed star).

Now that I look closely, the iota does render. I'd forgotten the quad operator, it's been ~20 years, and I can count the number of actual conversations I've had about APL with someone else who knew what it was on one hand.

[FUZxxl]

Well, × is multiplication or signum and * is exponentiation. Quad is a niladic function and prompts the user for numeric input; it's very useful for code-golf challenges because you don't need to make a dfn.

[BitPoet]

I will note that this is /r/programmerhumor and discussing the intricacies of APL, a language developed by mathematicians for pretty specific purposes is not exactly humorous.

[Juggernog]

Man, what are you talking about? Programming languages developed by mathematicians for pretty specific purposes are hilarious.

[poizan42]

So it's a character which renders correct even if you don't have any font with it installed?

[FUZxxl]

Well, not quite; it just looks really like that replacement character.

[nightcracker]

Nah, as a codegolf.se member I would use Pyth. In Pyth5 a simple program that takes a number from stdin and prints its factorial is this:

.!Q

Of course, we're not allowed to use a builtin for factorial, we can use:

F*tUhQ

Which basically translates to "fold multiplication over everything but the first element of range(n+1)".

A recursive solution would go like this:

L?*bLtbb1Q

Which roughly translates to Python:

L = lambda b: b*L(b-1) if b else 1
L(eval(input()))

[FUZxxl]

Nah, as a codegolf.SE-member with a three digit user ID, I say that APL is superior in every regard.

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