Quote from: GreenBits on June 19, 2016, 03:14:51 PM

Quote from: iamnotback on June 19, 2016, 03:04:26 PM

Quote from: biggbox on June 19, 2016, 02:27:04 PM

Quote from: iamnotback on June 19, 2016, 02:20:49 PM

Quote from: biggbox on June 19, 2016, 02:18:59 PM

The issue now is with DAO, not Ether.

Incorrect.

Okay. That is too technical. It is beyond my understanding. Do you mind explaining it in layman terms or, at least in bachelor's level?

Does this help?

http://ebiquity.umbc.edu/blogger/2008/01/19/how-dr-suess-would-prove-the-halting-problem-undecidable/

Scooping the Loop Snooper
an elementary proof of the undecidability of the halting problem

Geoffrey K. Pullum, University of Edinburgh

No program can say what another will do.
Now, I wont just assert that, Ill prove it to you:
I will prove that although you might work til you drop,
you cant 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 arent 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 they 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.

Heres the trick I would use and its simple to do.
Id define a procedure well name the thing Q
that would take any 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 wouldnt 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 Qs 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 Ps output to make P look stupid.
If P gets things right then it lies in its tooth;
and if it speaks falsely, its telling the truth!

Ive 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 dont have to tell you; Im 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.
Its something that cannot be done. So we users
must find our own bugs; our computers are losers!

.

Too long, did actually read; now head hurts and nose is bleeding.

But seriously, thank you, that Turing completeness shit was jibber jabber nonsense until that poem. I get it now Grin

Let me provide a simplistic explanation of the Halting problem.

We have a program P which inputs the source code of other programs and the result of P is the result of the analysis of whether the input source code loops forever or terminates (i.e. halts).

We have another program Q which inputs the source code of a program and calls P, then prints "Loop" and terminates if P reports "Loop", otherwise Q loops.

So we call Q passing the source code of Q to itself, but this is a paradox because P can't decide what the result would be. If P's result is "Loop", then Q terminates. If P's result is "terminates", then Q loops. We have an inconsistency in both cases.

This shows that it is impossible to make a compiler that acts like P, i.e. it is impossible to analyze whether programs with unbounded recursion terminate or not, because recursion is decided (unbounded) at runtime and not at compile-time. Thus at compile-time the result of the question is undecideable.

QED.