35
1
Your task is to write some code in Python 2 or 3 such that this expression:
(a+b)(c+d) == a*c + b*c + a*d + b*d
will evaluate to True without raising any exceptions.
To clarify, I will copy your code into a file, then from the file import *. Then I will type the expression into the console and verify that it is True.
This is code-golf, so the answer with the shortest length (in bytes) wins.
Esolanging Fruit
Posted 2016-10-28T22:53:24.467
Reputation: 13 542
20
Removed 4 bytes thanks to Dennis.
The latest version is inspired by the "some reason" in xnor's answer.
class t(int):__add__=type
a=b=t()
c=d=0
jimmy23013
Posted 2016-10-28T22:53:24.467
Reputation: 34 042
10
class m(int):__call__=__add__=lambda*x:m()
a=b=c=d=m()
Make an object that inherits from int, except adding or calling just returns a copy of itself.
Same length:
class m(int):__call__=__add__=lambda a,b:a
a=b=c=d=m()
I thought min or {}.get would work in place of lambda a,b:a, but for some reason they act only on the second argument.
xnor
Posted 2016-10-28T22:53:24.467
Reputation: 115 687
1(it's [tag:code-golf]) – Addison Crump – 2016-10-28T23:10:24.187
1Oops, I only saw [tag:programming-puzzle], will golf. – xnor – 2016-10-28T23:11:06.130
3That was quite a reduction o-o – Addison Crump – 2016-10-28T23:19:06.077
@xnor It doesn't work because min already has a __self__ attribute, so the class skips binding its own self. Why min has __self__ is another question... – matsjoyce – 2016-10-29T19:00:42.260
@matsjoyce: No, it doesn't have anything to do with the fact that min has a __self__. min.__self__ is just an artifact of how built-in functions and built-in methods are implemented as the same type. min doesn't work here because unlike functions written in Python, built-in functions don't support the descriptor protocol, which is responsible for binding the first argument. – user2357112 supports Monica – 2016-10-30T01:07:16.123
@user2357112 Ah, OK, I thought I got it working with sum yesterday, guess I was looking at wrong thing then. – matsjoyce – 2016-10-30T14:43:43.343
3
class e:__mul__=lambda*o:0;__add__=lambda*o:lambda x:0
a=b=c=d=e()
Jonathan Allan
Posted 2016-10-28T22:53:24.467
Reputation: 67 804
1
While it cannot really compete with the existing answers, this one actually performs the calculation in question:
from sympy.abc import*
type(a+b).__call__=lambda x,y:(x*y).expand()
Explanation:
sympy.abc contains all single-letter symbols, in particular ones named a, b, c, and d.a+b is an Add object, which represents a general sum.type(a+b).__call__= […] monkey-patches the Add class to give it evaluation capabilities, in this case enabling it to work like a multiplication of caller and callee.expand is necessary to make the expressions actually equal (since SymPy only performs thorough equality checks on demand). 
Wrzlprmft
Posted 2016-10-28T22:53:24.467
Reputation: 2 772
Nice! There's
0 .__mul__forlambda y:0but it's the same length. – xnor – 2016-10-28T23:28:53.010x.countsaves a byte. – Dennis – 2016-10-28T23:36:31.490With
a=t();b=c=d=0, you can useclass t(int):__add__=lambda*x:tto save three bytes. – Dennis – 2016-10-29T00:24:40.770@Dennis Thanks. Good idea... – jimmy23013 – 2016-10-29T00:32:37.737
1I don't get it...
type(t(), t())ort().type(t())throws an exception, so what is happening when you dot() + t()? – feersum – 2016-10-29T03:56:20.757@feersum just
type(t()). The LHS becomes(type(b))(0)ort(0), which is equal to0. – Jonathan Allan – 2016-10-29T09:59:46.080@JonathanAllan But
+is a binary operator. Why would__add__be called with 1 argument? – feersum – 2016-10-29T10:10:09.4971@feersum
__add__is called with two, but the first is interpreted asself, onlyotheris passed totype. Weird, yes. – Jonathan Allan – 2016-10-29T10:12:28.543@feersum The difference is, I believe, that inbuilts use method-wrappers (py3) or unbound methods (py2) (rather than methods or bound methods respectively, like
MySpam.__myEggs__would usually be). I could well be wrong though, interested to know... – Jonathan Allan – 2016-10-29T10:23:39.453@JonathanAllan But try this:
numargs = lambda *a: len(a); class t(int): __add__ = numargst() + t()results in2! – feersum – 2016-10-29T11:03:56.4701
@feersum:
– user2357112 supports Monica – 2016-10-29T18:25:17.497a + bfirst triesa.__add__(b).a.__add__istype, so that becomestype(b). The key difference between this case and the usual case for methods is that usually,a.__add__would be a different object from the thing you set__add__to in the class definition, due to the descriptor protocol, which ordinary function objects implement. (There are also a few other tricky bits that aren't relevant here.)