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10:05 PM
@lucas have you ever thought about visiting this room to help you from being misunderstood :-)
No offense intended, of course.
 
10:26 PM
@ACuriousMind Is everyone still angry with you?
@knzhou Howdy
 
@BernardMeurer Hi!
 
@knzhou Do you know anything about theoretical turing machines?
 
I took a class about theoretical CS once.
What's the question?
 
I'm not getting the difference between deterministic turing machines and nondeterministic ones
 
Both have the same class of solvable problems
 
10:34 PM
What exactly about it?
 
Well this:
 
If you're weirded out by the nondeterminism, you might want to look at deterministic vs. nondeterministic finite state machines, they're a bit simpler.
 
> Nondeterministic Turing machines are defined in the same way as their deterministic counterparts, except that instead of a transition function associating one and only one $(p,\tau,d)$ to each $(state, symbol)$ pair $(q, \sigma)$, there is a transition relation (which we will again denote by $\Updelta$) consisting of any number of ordered 5-tuples $(q,\sigma,p,\tau,d)$.
 
Okay, what's the issue?
 
10:42 PM
1. What difference does that make?
2. Why would that make it nondeterministic
3. What difference will having a transition function have in relation to a transition relation?
 
Okay, let everything about an initial state be $x_i$ and everything that the machine can do in one step be $y_i$.
In a deterministic TM, given $x_i$, there is a definite $y_i$ given by the transition function.
In an NTM, you have a relation, i.e. a set of pairs $(x_i, y_j)$. Given $x_i$, you are allowed to transition to all $y_j$ such that $(x_i, y_j)$ in a pair in the relation.
 
What does it mean to transition to more than one thing? Well, you can't implement that on a physical Turing machine. You can think about it a few ways:
- a number of "ghost copies" of the TM split off. each of them does one of the possible things.
- the TM rolls a die to pick which transition to do.
The latter is why we call it 'nondeterministic'. But I think imagining the first is a bit easier.
 
I get it, it makes sense now!
 
No prob!
 
10:47 PM
Why would we want a NTM though?
Seems silly
 
It's just a theoretical tool.
 
Ah, well, that explains it
 
The important point is that an NTM is the same as a TM that "already knows how to get to the answer". For example, if the question is "does this maze have a solution", the TM can always take the right path. The NTM can take all paths.
So if you want to talk about P vs. NP you'll talk about NTMs.
 
So an NTM can solve an NP problem in P time?
 
Yes.
 
10:49 PM
Balling
Wait, so is an adiabatic quantum computer an NTM?
 
No, that's a totally different thing. Quantum superposition is a lot more restrictive than this NTM "taking all paths at once" thing.
NTMs work if any of the paths work. A quantum computer's wavefunction must have every path in it, and you have to figure out a clever algorithm to make all the wrong paths destructively interfere.
So quantum computers have little to do with the P vs. NP problem.
 
Hmm. But isn't the whole point of a QC to solve problems on BQP?
 
Well, we have no clue how BQP relates to NP.
 
Cool!
Well, thanks a lot @knzhou I've added you to my 'owe a beer' list ;)
 
Thanks. :P Having just turned 21 in the US, I'll gladly take you up on the offer.
 
10:55 PM
Come to Portugal, I'm still 18 so it'll be a while until I can deliver it to you over there hahaha
 
11:32 PM
@dmckee are you any good at finding weird functions?
 
@0celo7 Not really. It's been a long time since math methods.
 
Not special functions
I need to construct a function $g(s), s\ge 0$ such that $g(s)=1$ for $s$ "small", $g(s)>0$ always, and $\int_0^\infty g=1$.
 
But when my Dad went back for his Ph.D. I got to help him with his homework. He'd reduced a problem to an integral he didn't recognize. It was the complementary error function.
 
Are there Gaussians which terminate on one side but keep going on the other?
all smoothly, of course
 
Well, the Poisson distribution cuts off at one end. Maybe that's a starting point.
 
11:36 PM
Cuts off?
I'm looking at google here
Not sure what you mean
I need something that's 1 near $0$, then smoothly transitions to a decaying function so that the whole thing has area 1 and never goes negative
I'm certain it should be a smooth plateau function near the origin summed with some lopsided distribution with support $[\epsilon,\infty)$, $\epsilon>0$.
 
@JohnRennie (InRe the cyclotron spoiler above) Then what good is it?
 
and it needs to be integrable so I can normalize it.
that's why I was thinking Gaussian
 
@dmckee What's a cyclotron?
 
Wait, what are the derivatives of a Gaussian at its peak
 
@0celo7 Hmmm ... nuclear densities are sometimes modeled with a function proportional to $1/(A + r^2)$ for some constant $A$.
@0celo7 The first one is zero. Obviously.
 
11:41 PM
TIME FOR CALC 1
@dmckee I need the derivatives to be 0 to all orders to use a smooth gluing lemma
Second derivative will be nonzero because it's a max
crap
@dmckee good thing I have my calculator with me
very interesting
do you know what the derivatives of that thing are at 0
 
Now if only I hadn't change my mind about how to write it about three times I might have had a consistent notation at the beginning. SHould be right now.
 
whoops, second one won't be 0
I'm bad at calculus, when is the second derivative 0?
 
@0celo7 At an inflection point
 
I can construct this function easily
But the problem is checking integrability :/
So it has to be some $L^1$ function
or finite sum of $L^1$ functions
AH
Wait, I need ACM
 
@ACuriousMind BJOOOOOOOOORN
 
11:53 PM
I need a smooth ramp function to multiply my Gaussian with
Not sure if I need ACM, this seems quite reasonable.
@dmckee If you multiply a Gaussian with a smooth $0\to 1$ ramp function, you should get a "ramped" Gaussian, and since it's $\le$ everywhere to the original one, it's $L^1$ integrable.
Then take the plateau, integrate it along its support in $[0,\infty)$, call that $\epsilon<1$.
Then normalize the ramped Gaussian to $1-\epsilon$
Slide it over so that the end of its support on the left just touches that of the plateau
Boom
@BernardMeurer Ain't nobody got time for Germans
 
Hi, everybody
 
@DanielSank hey
Oh, I bet you would have been able to help just now
 
@DanielSank Nihao
 
@BernardMeurer Chinese. Nice.
 

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