The h Bar

12:01 AM

I think it really just comes back to this solid-state stuff. You would really like to be able to define two sets, first, the covectors which fetch the $n^\text{th}$ components of a vector; second, the vectors which return $1$ for exactly one of these covectors and $0$ for all the others.

those vectors are your basis vectors e_i^\bullet, i=1,2,\dots D, and those covectors are their dual elements e^i_\bullet, i = 1,2,\dots D.

doublefelix

But, if I take \partial_\mu dx^\nu, this does not actually give \frac{\partial x^\nu}{\partial x^\mu}, which is what I want in order to get a delta function. You would have to define it as a delta function, no?

Like, if defining the cotan basis as dx^\nu naturally led to the delta function under the usual way of thinking about differentials, it would be well-motivated

CR Drost

you definitely need a Kronecker delta, not sure about a delta-function...

doublefelix

oops, sorry, I meant a kronecker

CR Drost

But I guess my point is that if you try to raise the index of $e^i_\nu$ you then get some $e^{i~\mu}$ which is not necessarily related directly to $e_i^\mu$, you need some sort of conversion factor in there.

CR Drost

12:19 AM

@doublefelix I should also add, since I don't think I've stated it explicitly, that I agree that naively putting $dx^\mu$ and $\partial_\mu$ in the middle of these things is some shorthand for some much more complex process that's harder to understand, and it probably won't work with every single expression ever.

In particular $dx^\mu$ mostly comes into discussions of the metric, where $ds^2 = g_{\mu\nu}~dx^\mu~dx^\nu$ describes the distance between two very nearby points of the space.

doublefelix

ooooh, interesting

CR Drost

and again $\partial_\mu$ if you view it coordinate-independently takes a scalar field and "suspends" it for later computation on by a vector's directional derivative.

doublefelix

I think this is good rationale to think of the \partial _\mu and dx^\mu as akin to derivatives and differentials while not necessarily being literally that

the line element thing is a good analogy

CR Drost

right. I mean they kind of are literally that in that one then parameterizes a path over some other coordinate, let's say x^\mu(q), and then calculates its derivatives with respect to that coordinate $v^\mu(q),$ and then calculates$\Delta s$ by integrating $\int dq~ \sqrt{g_{\mu\nu} ~ v^\mu~v^\nu}$ just as if it were $\Delta s = \int dq~\frac{ds}{dq}.$

But whenever physicists work with differentials there is usually some hand-wave involved.

heather

12:48 AM

is there anyone in here who knows how to formally calculate algorithmic complexity?

DanielSank

Define "formally".

The way I normally see it done is by thinking about the algorithm.

E.g. "There's a loop here over all elements in the list. That's order N".

heather

Um...I guess I just want to know a. whether or not this algorithm is polynomial time and b. whether or not this algorithm is more or less efficient than another.

sorry, I guess I don't know enough to respond to that question =)

the two algorithms are the first and third here - the one I don't know the complexity of is the third one, and the first one I know is polynomial time (that was the result of the paper linked in the question).

does that make sense @DanielSank?

dmckee

1:22 AM

@BernardoMeurer Something I would add to @John's very reasonable thoughts is that generally the only thing that calls for one header to include another is if you need to be sure a struct included so that it can appear in a header.

Something like

` struct feature_t* sorterFeatureList(struct feature_t* head);`

and for those cases it is just as useful to declare the structure without defining it. Like this

struct feature_t;

instead of including the header at all.

Indeed the headers for opaque types often do exactly that so that the programmer has to look in the source file to know how the struct is arranged. It's a hint that you shouldn't care how the struct is arranged. This is the same kind of encapsulation as you find in object oriented language that say 'you only need to know the interface, don't worry about the implementation'.

heather

@dmckee I don't suppose you know/would mind helping in all of your wise professorial ways?

dmckee

@heather I can try, but I don't admit to any wisdom.

heather

=) it's just the question above about algorithmic complexity

specifically, I want to know a. whether or not this algorithm is polynomial time and b. whether or not this algorithm is more or less efficient than another.

the algorithm in question being the third here, compared to the first on the same question

dmckee

The absolute algorithmic complexity of either depends on the complexity of several of the sub-steps. Notably the is computable step, the compute step and the eliminate element step.

But if we assume they each have the same complexity in each of the examples we can at least compare the options.

Come to think of it the length check could in principle be an issue, thought because you are only checking for zero length you can probably rig it to be O(1) in all cases of interest.

heather

is there a way to make the length check more efficient?

dmckee

1:38 AM

In the first case you are going to do $N$ is computable, $M<N$ compute, and $N-M$ eliminate element.

In the last you are going to do $N$ length, $N$ is computable, $M<N$ compute, and $N$ eliminate element.

The latter is slower, but may be of the same order.

heather

hmm, but it is slower...

dmckee

@heather Depends how your list is implemented. A bad case would be using an array of booleans: to check for emptiness you'd have to walk the array each time. Yikes.

But you can fix that flaw by also storing a current count, and then length just returns that count so the trouble goes away.

heather

it's a list of vectors

dmckee

For most reasonable ways of storing the data length > 0 can be written is such a way as to be O(1).

@heather What, in c++? Then it keeps track of it's own length so it is O(1).

heather

in python 3

so i guess really a list of lists.

dmckee

1:42 AM

I don't actually know python well enough to say for sure, but I know that python people wouldn't put up with list.length() being computationally intensive.

heather

okay.

dmckee

So don't worry about it.

So, long story shot. No obvious reason to expect a different big-O for those two code snippets, but (a) they do slightly different things and (b) it looks like the latter does more work.

heather

i guess i'd prefer that the third one was at least as efficient as the first.

i'm trying to rewrite the algorithm so it's more pythonic.

dmckee

It doesn't look to me that they come to the same end-state, so the first thing to check is that they both have an acceptable ending state.

If only one is doing the right thing, then you should write the one that does the right thing and not worry about pythonic ways for now. Maybe a python guru can suggest an alternative approach later on.

heather

okay...what do you think the difference is in end state?

er...actually, you're right, the third one is wrong again...it will never complete, because there should be things in the list that aren't removed in some cases...

darn.

thank you =)

nbro

2:11 AM

dmc saving people (intentionally mistyping your initial letters)! btw, you should also join the computer science chat, since it's pretty desertic (2-3 people, most of them inactive)

heather

"dmc" - what does that mean?

oh, sorry ::facepalms::

DanielSank

3:00 AM

@heather yeah, makes sense.

heather

good night everyone

DanielSank

Night

rob

Whoa, I have a catchphrase.

DanielSank

4:13 AM

@rob yes?

rob

6 hours ago, by AccidentalFourierTransform

what would rob say?

6 hours ago, by AccidentalFourierTransform

"this is not a useful conversation."

DanielSank

The catch phrase is "This is not a useful conversation"?

rob

Apparently. I'm not quite sure how I feel about it.

DanielSank

Sounds very moderatorish. You should be proud.

rob

I had just decided on "bemused"

I can be proudly bemused.

DanielSank

4:22 AM

True.

I have just removed a rather large amount of dead callous skin from my foot. Running barefoot seems to encourage the skin to develop a sort of armor.

It seems practical, but itchy.

rob

Sounds like the process of growing/ungrowing guitarist's calluses on your fingertips, but more surface area is involved.

DanielSank

If only I could grow guitar callouses.

I played for years and my fingers were as tender as the day I started. Probably just didn't pay enough.

Made barre harder than it should have been.

rob

4:48 AM

@DanielSank I've switched to the superlight steel strings and I play the nylon-stringed guitar a lot, so my calluses aren't what they have been in the past.

But my guitar playing ebbs and flows

DanielSank

5:45 AM

@rob yeah I'm all nylon these days.

user228700

6:01 AM

Hi, everyone! :-)

user228700

Still haven't finished the third one :'-( Was dragged to bed by mum at 3 AM.

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