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Today's Topics:

  1. Re: Silly Python riddle (asaf greenberg)
  2. Re: Silly Python riddle (Omer Zak)
  3. Re: Silly Python riddle (Ram Rachum)
  4. Re: Silly Python riddle (Shai Berger)
  5. Re: Silly Python riddle (Ram Rachum)
  6. Re: Silly Python riddle (Ram Rachum)
  7. Re: Silly Python riddle (Alon Levy)


----------------------------------------------------------------------

Message: 1
Date: Sat, 17 Dec 2011 16:27:24 +0200
From: asaf greenberg <asafgreenberg <at> gmail.com>
Subject: Re: [Python-il] Silly Python riddle
To: python-il <at> hamakor.org.il
Message-ID: <4EECA6CC.8050700 <at> gmail.com>
Content-Type: text/plain; charset="us-ascii"

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Message: 2
Date: Sat, 17 Dec 2011 16:38:42 +0200
From: Omer Zak <w1 <at> zak.co.il>
Subject: Re: [Python-il] Silly Python riddle
To: Python User Group <python-il <at> hamakor.org.il>
Message-ID: <1324132722.3846.18.camel <at> c4>
Content-Type: text/plain; charset="UTF-8"

Another attempt at golfing down the solution:
>>> f=lambda:g(0);g=(id)
>>> f()
10482512

The solution is '0);g=(id', 8 characters long.
If there is a built-in function whose name is 1-character long, then the
solution can be made to be 7 characters long.


On Sat, 2011-12-17 at 14:55 +0200, Omer Zak wrote:
> I do not have Python 2.7, so I did the following on Python 2.6.6.
> >>> f = lambda: g(); g = lambda:()
> >>> f()
> ()
> >>>
>
> In other words, '???' is replaced by '); g = lambda:('
>
> --- Omer
>
>
> On Sat, 2011-12-17 at 14:14 +0200, Ram Rachum wrote:
> > Here's a silly Python riddle for you.
> >
> >
> > Today I opened up a Python 2.7 shell, and ran two commands in it.
> >
> >
> >         >>> f = lambda: g(???)
> >         >>> f()
> >
> > (Note that these are the only commands that I ran. You're not allowed
> > to run any other commands before them.)
> >
> >
> > The riddle: What's the shortest thing you can put instead of ??? so my
> > second command would not raise an exception?
>
>

--
$ python
>>> type(type(type))
<type 'type'>          My own blog is at http://www.zak.co.il/tddpirate/
My opinions, as expressed in this E-mail message, are mine alone.
They do not represent the official policy of any organization with which
I may be affiliated in any way.
WARNING TO SPAMMERS:  at http://www.zak.co.il/spamwarning.html



------------------------------

Message: 3
Date: Sat, 17 Dec 2011 16:41:14 +0200
From: Ram Rachum <ram <at> rachum.com>
Subject: Re: [Python-il] Silly Python riddle
To: Omer Zak <w1 <at> zak.co.il>
Cc: Python User Group <python-il <at> hamakor.org.il>
Message-ID:
       <CANXboVYLqt8UVrhYGP=C-18P=xTTnWjrbu6JJFfKO4C2=5rdzw <at> mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"

Haha, I didn't think of all those creative answers when I asked this
riddle! I'm sure happy that none of them (so far!) have reached 7
characters, which is the length of my solution.

On Sat, Dec 17, 2011 at 4:38 PM, Omer Zak <w1 <at> zak.co.il> wrote:

> Another attempt at golfing down the solution:
> >>> f=lambda:g(0);g=(id)
> >>> f()
> 10482512
>
> The solution is '0);g=(id', 8 characters long.
> If there is a built-in function whose name is 1-character long, then the
> solution can be made to be 7 characters long.
>
>
> On Sat, 2011-12-17 at 14:55 +0200, Omer Zak wrote:
> > I do not have Python 2.7, so I did the following on Python 2.6.6.
> > >>> f = lambda: g(); g = lambda:()
> > >>> f()
> > ()
> > >>>
> >
> > In other words, '???' is replaced by '); g = lambda:('
> >
> > --- Omer
> >
> >
> > On Sat, 2011-12-17 at 14:14 +0200, Ram Rachum wrote:
> > > Here's a silly Python riddle for you.
> > >
> > >
> > > Today I opened up a Python 2.7 shell, and ran two commands in it.
> > >
> > >
> > >         >>> f = lambda: g(???)
> > >         >>> f()
> > >
> > > (Note that these are the only commands that I ran. You're not allowed
> > > to run any other commands before them.)
> > >
> > >
> > > The riddle: What's the shortest thing you can put instead of ??? so my
> > > second command would not raise an exception?
> >
> >
>
> --
> $ python
> >>> type(type(type))
> <type 'type'>          My own blog is at http://www.zak.co.il/tddpirate/
> My opinions, as expressed in this E-mail message, are mine alone.
> They do not represent the official policy of any organization with which
> I may be affiliated in any way.
> WARNING TO SPAMMERS:  at http://www.zak.co.il/spamwarning.html
>
> _______________________________________________
> Python-il mailing list
> Python-il <at> hamakor.org.il
> http://hamakor.org.il/cgi-bin/mailman/listinfo/python-il
>
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Message: 4
Date: Sat, 17 Dec 2011 17:01:29 +0200
From: Shai Berger <shai <at> platonix.com>
Subject: Re: [Python-il] Silly Python riddle
To: python-il <at> hamakor.org.il
Message-ID: <201112171701.29250.shai <at> platonix.com>
Content-Type: Text/Plain; CHARSET=US-ASCII

On Saturday 17 December 2011, Ram Rachum wrote:
> Here's a silly Python riddle for you.
>
> Today I opened up a Python 2.7 shell, and ran two commands in it.
>
> >>> f = lambda: g(???)
> >>> f()
>
> (Note that these are the only commands that I ran. You're not allowed to
> run any other commands before them.)
>
> The riddle: What's the shortest thing you can put instead of *???* so my
> second command would not raise an exception?
>
I have a cheating solution that goes down to 3: """

It's cheating because your second line is not a command (and in fact, it makes
the whole thing not yet a commnd):

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

But still, at this point, no exception has been raised.


------------------------------

Message: 5
Date: Sat, 17 Dec 2011 17:03:55 +0200
From: Ram Rachum <ram <at> rachum.com>
Subject: Re: [Python-il] Silly Python riddle
To: Shai Berger <shai <at> platonix.com>
Cc: python-il <at> hamakor.org.il
Message-ID:
       <CANXboVaAgUyrhoHtvdVXckBM25QnYQhjRqzKrCRvPfvbveh8Ew <at> mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"

On Sat, Dec 17, 2011 at 5:01 PM, Shai Berger <shai <at> platonix.com> wrote:

> On Saturday 17 December 2011, Ram Rachum wrote:
> > Here's a silly Python riddle for you.
> >
> > Today I opened up a Python 2.7 shell, and ran two commands in it.
> >
> > >>> f = lambda: g(???)
> > >>> f()
> >
> > (Note that these are the only commands that I ran. You're not allowed to
> > run any other commands before them.)
> >
> > The riddle: What's the shortest thing you can put instead of *???* so my
> > second command would not raise an exception?
> >
> I have a cheating solution that goes down to 3: """
>
> It's cheating because your second line is not a command (and in fact, it
> makes
> the whole thing not yet a commnd):
>
> >>> f = lambda: g(""")
> ... f()
> ...
>
> But still, at this point, no exception has been raised.
>
>
Yeah, that's a good cheat :)
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Message: 6
Date: Sat, 17 Dec 2011 18:27:38 +0200
From: Ram Rachum <ram <at> rachum.com>
Subject: Re: [Python-il] Silly Python riddle
To: Shai Berger <shai <at> platonix.com>, python-il
       <python-il <at> hamakor.org.il>
Message-ID:
       <CANXboVbypSFHwp=XyGSbs1R9ArZCp+eUCtP945bffeM6LfcLJg <at> mail.gmail.com>
Content-Type: text/plain; charset="iso-8859-1"

On Sat, Dec 17, 2011 at 6:14 PM, Shai Berger <shai <at> platonix.com> wrote:
>
>  >
> > (Note that these are the only commands that I ran. You're not allowed to
> > run any other commands before them.)
> >
> > The riddle: What's the shortest thing you can put instead of *???* so my
> > second command would not raise an exception?
> >
> >
> ???= (yield)
>
> right?
>
> (mailed privately, to avoid ruining the fun...)
>

Yep!!! I just almost finished writing the email to tell everyone that when
I got your answer.

Congrats for solving the riddle Shai.

So as Shai said, the solution is:


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


Funny, isn't it? I was surprised to see that the `yield` keyword can be
used in a lambda function.

So when you type `f()`, it just returns a generator. If you'll try to
exhaust it, an exception will be raised because `g` doesn't exist, but
that's a new line :)

It's funny that in this case, Python seems to throw away the value of the
lambda function! As we know, the `yield` keyword actually forms a statement
whose value is `None`, unless you used the generator's `.send` instead of
`.next`. So you could also use `.send` to send in whatever value you want
into the lambda function, and Python will just throw it away. Unless I'm
missing something.

So that's the only case I can think of where Python completely throws away
the value of a lambda function.


Another funny thing that I learned from this riddle is that when you do a
function invocation in Python, Python accesses the function *before* it
looks at the arguments.

So if were to do:

adfgadgof(1 / 0)


Python will complain about the non-existent function before it even sees
the division-by-zero.
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Message: 7
Date: Sat, 17 Dec 2011 18:34:11 +0200
From: Alon Levy <alonlevy1 <at> gmail.com>
Subject: Re: [Python-il] Silly Python riddle
To: Ram Rachum <ram <at> rachum.com>
Cc: python-il <python-il <at> hamakor.org.il>, Shai Berger
       <shai <at> platonix.com>
Message-ID:
       <CALijEpYiZu7z2dJCtvTeexi1OjgoLZkHTf7tzwos_PZ12MJekQ <at> mail.gmail.com>
Content-Type: text/plain; charset=UTF-8

cute and interesting.

On Sat, Dec 17, 2011 at 6:27 PM, Ram Rachum <ram <at> rachum.com> wrote:
> On Sat, Dec 17, 2011 at 6:14 PM, Shai Berger <shai <at> platonix.com> wrote:
>>
>> >
>> > (Note that these are the only commands that I ran. You're not allowed to
>> > run any other commands before them.)
>> >
>> > The riddle: What's the shortest thing you can put instead of *???* so my
>>
>> > second command would not raise an exception?
>> >
>> >
>> ???= (yield)
>>
>> right?
>>
>> (mailed privately, to avoid ruining the fun...)
>
>
> Yep!!! I just almost finished writing the email to tell everyone that when I
> got your answer.
>
> Congrats for solving the riddle Shai.
>
> So as Shai said, the solution is:
>
>>
>> >>> f = lambda: g((yield))
>> >>> f()
>
>
> Funny, isn't it? I was surprised to see that the `yield` keyword can be used
> in a lambda function.
>
> So when you type `f()`, it just returns a generator. If you'll try to
> exhaust it, an exception will be raised because `g` doesn't exist, but
> that's a new line :)
>
> It's funny that in this case, Python seems to throw away the value of the
> lambda function! As we know, the `yield` keyword actually forms a statement
> whose value is `None`, unless you used the generator's `.send` instead of
> `.next`. So you could also use `.send` to send in whatever value you want
> into the lambda function, and Python will just throw it away. Unless I'm
> missing something.
>
> So that's the only case I can think of where Python completely throws away
> the value of a lambda function.
>
>
> Another funny thing that I learned from this riddle is that when you do a
> function invocation in Python, Python accesses the function before?it looks
> at the arguments.
>
> So if were to do:
>
>> adfgadgof(1 / 0)
>
>
> Python will complain about the non-existent function before it even sees the
> division-by-zero.
>
> _______________________________________________
> Python-il mailing list
> Python-il <at> hamakor.org.il
> http://hamakor.org.il/cgi-bin/mailman/listinfo/python-il
>



--
Alon Levy


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