Silly Python riddle

http://blog.garlicsim.org/post/14506657016/silly-python-riddle

by cool-RR • 15 years ago

77 43 15 years ago

blog.garlicsim.org

teaspoon• 15 years ago

"""

cool-RR7• 15 years ago

Doesn't count... If only for the fact that you won't get a >>> prompt after it.

jgeralnik• 15 years ago

Something like

);f=(set

Leading to the whole line reading:

f = lambda: g();f=(set)

8 characters

aroberge• 15 years ago

Alternatively

);f=set#

makeramen• 15 years ago

0);g=(id

8 chars

VMG• 15 years ago

the winning solutions seems to be 7 chars

ul52556• 15 years ago

>>> f = lambda: g((yield))

>>> f()

cool-RR7• 15 years ago

Yep, that's the one! Congrats!

ul52556• 15 years ago

lurking around here for years ... this riddle finally made me register an account with HN ...

tocomment• 15 years ago

Can someone explain. It would seem like g still isn't defined. What's special about 'yield' that makes g valid?

martin_k7• 15 years ago

You're right, it isn't. The presence of yield makes the return value of f() a generator object. The function body is not executed initially. When you call .next() on the generator object execution starts and runs until it hits the first yield, raising a NameError because g is not defined.

edit: typo

molbioguy• 15 years ago

Sorry for the dumb question (I'm still inexperienced with python), but can someone explain why the yield needs to be enclosed in ()?

slashcom• 15 years ago

Not a dumb question at all.

Short answer: it throws a syntax error otherwise.

Long answer: I don't know.

ul52556• 15 years ago

Whatever goes inside a lambda must be an expression. The parentheses around yield make it a yield expression. Otherwise yield would be a statement which is a different thing as far as Python is concerned.

You can also use it inside generators:

  >>> def f():
         while True:
             xyz = (yield)
             print xyz
  >>> g = f()
  >>> g.next()
  # nothing happens as xyz == None
  >>> g.send('huhu')
  huhu

obtu• 15 years ago

yield was first introduced as a statement similar to return. In Python not every statement is an expression; another example of non-expression statement is assignment, which was banned from expressions so that it couldn't be confused with ==.

Since then someone proposed (PEP 342) that yield could be used not just to yield values from coroutines, but to send one back in. yield becomes an expression, mostly used as the right side of an assignment. The parentheses are there because the grammar seems to handle the yield statement and yield expressions differently (the PEP mentions that "yield 12, 42" would have become confusing).

phzbOx2• 15 years ago

I had this (with 8 characters):

  );g=(int

giving:

  f = lambda: g();g=(int)
  f() -> 0

phzbOx2• 15 years ago

There is also this: f = lambda: g();g=(f)

Which produce a 'RuntimeError: maximum recursion depth exceeded'; But that's only because my computer is not powerful enough :)

rplnt• 15 years ago

And it's even shorter than the yield one.

sapphirecat• 15 years ago

I was going to attempt this, but the solution was posted inline in the blog entry, and I read it by accident. My window was just big enough that it was the last line displayed. =(

cool-RR7• 15 years ago

Sorry, I couldn't find a way to blacken the text with Tumblr.

I now did a (Scroll down...) thing to "hide" the solution.

martin_k7• 15 years ago

I'm not sure, why you think the value of the lambda function is thrown away. The lambda function returns a generator object. The function body of a generator is generally not executed until you call .next() on it, so that's why you don't get a NameError instantly. Also the value, that you .send() into the generator is not thrown away either. In your example the generator just runs into an exception before it goes into a state in which it would accept a .send() call. Just consider this modified example:

  >>> f=lambda: (yield)
  >>> gen=f()
  >>> gen.next()
  >>> gen.send('foobar')
  'foobar'

cool-RR7• 15 years ago

I think you misunderstood which value gets thrown away.

Consider this:

    >>> g = lambda x: 7
    >>> f = lambda: g((yield))
    >>> list(f())
    [None]

Can you tell me where the 7 went to? As far as I see, it went into oblivion.

martin_k7• 15 years ago

As i said, f() returns a generator object in this case. So you're calling list on a generator that yields None once, the result of which is a list that contains None. If you change the `yield` to `yield "foobar"`, list(f()) will get you `["foobar"]`.

Edit: Perhaps, to clear things up more, about where the value 7 went- your snippet can be expressed without using lambdas as

  >>> def g(x):
  ...  return 7
  ... 
  >>> def f():
  ...  v = (yield)
  ...  g(v)
  ... 
  >>> list(f())
  [None]

Note that there's no return before g(v) because generators (at least in 2.7) can only yield, not return a value.

ul52556• 15 years ago

Strange. With Python 2.6 I get [None, 7] while with Python 2.7 it is [None]. Looking at the OP codes the 2.7 one's look weird.

Python 2.6:

  >>> dis.dis(f)
    1           0 LOAD_GLOBAL              0 (g)
                3 LOAD_CONST               0 (None)
                6 YIELD_VALUE
                7 CALL_FUNCTION            1
               10 RETURN_VALUE

Python 2.7:

  >>> dis.dis(f)
    1           0 LOAD_GLOBAL              0 (g)
                3 LOAD_CONST               0 (None)
                6 YIELD_VALUE
                7 CALL_FUNCTION            1
               10 POP_TOP
               11 LOAD_CONST               0 (None)
               14 RETURN_VALUE

In essence the Python 2.7 compiler decides to throw away the result of the g() function call and replaces it with None. Does anyone understand why?

nyellin• 15 years ago

I haven't been following the thread, but Ram posted the same riddle on Python-IL mailing list a few days ago. You can see the discussion there.

http://hamakor.org.il/pipermail/python-il/2011-December/thre...

coppero1237• 15 years ago

f = lambda g:(0) f() <function <lambda> at 0x113478>.

jmathes• 15 years ago

f = lambda: g(""")

f()

technically not two commands, and it will look like:

>>> f = lambda: g(""")

. . . f()

ngkabra• 15 years ago

At 12 characters, this is definitely not the shortest, but it is a totally different approach than the other solutions:

    >>> f = lambda: g()if 0 else(1)
    >>> f()
    1

jerfelix• 15 years ago

Creative solution! You can knock it down a character by removing the "1" and simply returning an empty tuple.

    >>> f = lambda: g()if 0 else()
    >>> f()
    ()

Silly Python riddle

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