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you
15:22
Reposted fromjapande japande
you
18:12
you
18:04
you
17:55
8713_34af
CHOMP! — cowboyo
Reposted fromjamiew jamiew
you
17:54
3496_5b66
Reposted fromjamiew jamiew
you
17:48
you
17:46
you
17:40
0047_b8b3

focsa:

Check it if your credit card number has been stolen on the internet?! via wdl.lug.ro
Reposted fromjamiew jamiew
you
17:40
4967_164c
Always Be Coding. (via Joey deVilladancroak)
Reposted fromjamiew jamiew
you
17:38
0527_49d7_500
Reposted fromwalterra walterra
you
17:37
haben will! (nur ohne emo bandl)
you
17:37
you
17:35
writing code is like writing a cookbook.
for a four year old.
in japanese.
— the perfect analogy
you
17:33
Design is not just what it looks like and feels like. Design is how it works.”

Steve Jobs (via eyebo)


you
08:54

Halting Poem

Scooping the Loop Snooper

an elementary proof of the undecidability of the halting problem

No program can say what another will do.
Now, I won't just assert that, I'll prove it to you:
I will prove that although you might work til you drop,
you can't predict whether a program will stop.

Imagine we have a procedure called P
that will snoop in the source code of programs to see
there aren't infinite loops that go round and around;
and P prints the word "Fine!" if no looping is found.

You feed in your code, and the input it needs,
and then P takes them both and it studies and reads
and computes whether things will all end as the should
(as opposed to going loopy the way that they could).

Well, the truth is that P cannot possibly be,
because if you wrote it and gave it to me,
I could use it to set up a logical bind
that would shatter your reason and scramble your mind.

Here's the trick I would use - and it's simple to do.
I'd define a procedure - we'll name the thing Q -
that would take and program and call P (of course!)
to tell if it looped, by reading the source;

And if so, Q would simply print "Loop!" and then stop;
but if no, Q would go right back to the top,
and start off again, looping endlessly back,
til the universe dies and is frozen and black.

And this program called Q wouldn't stay on the shelf;
I would run it, and (fiendishly) feed it itself.
What behaviour results when I do this with Q?
When it reads its own source, just what will it do?

If P warns of loops, Q will print "Loop!" and quit;
yet P is supposed to speak truly of it.
So if Q's going to quit, then P should say, "Fine!" -
which will make Q go back to its very first line!

No matter what P would have done, Q will scoop it:
Q uses P's output to make P look stupid.
If P gets things right then it lies in its tooth;
and if it speaks falsely, it's telling the truth!

I've created a paradox, neat as can be -
and simply by using your putative P.
When you assumed P you stepped into a snare;
Your assumptions have led you right into my lair.

So, how to escape from this logical mess?
I don't have to tell you; I'm sure you can guess.
By reductio, there cannot possibly be
a procedure that acts like the mythical P.

You can never discover mechanical means
for predicting the acts of computing machines.
It's something that cannot be done. So we users
must find our own bugs; our computers are losers!

by Geoffrey K. Pullum
Stevenson College
University of California
Santa Cruz, CA 95064

From Mathematics magazine VOL73. No. 4, Oct 2000 319-320

you
08:18
Geepster (Geek hipster) is a person who derives his or her identity by mastering complex, but lesser-known, technologies
— Larry O’Brien @ SD Times Magz
you
08:05

http://snurl.com/1yrfz

Ini dia buat yang pusing nyari judul TA :)
Mudah-mudahan membantu
you
07:54
9200_8ec5_500
Gambar ini menerangkan bagaimana seorang programmer berhadapan dengan kode.
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