Wolfram Mathematica

12:00 AM

@drN (1) The interpolating function will always match at the specified points, because this is what interpolation is all about. (2) What happens in between is specified by the interpolation scheme. Usually, a smooth function is desired.

drN

@halirutan The Interpol. func. is not a smooth function then? Isn't it impossible to plot continuous data? We always just plot discrete data right, so I can never tell how bad the interpol. func. is?

halirutan

@drN If you would know the complete underlying function of your data, then you would see that the int. fun. is not always perfect.

Example:

data = Table[{x, Sin[5 x]}, {x, 0, 2 Pi, 2 Pi/6.}];
Show[
 Plot[{Interpolation[data][x], Sin[5 x]}, {x, 0, 2 Pi}],
 ListPlot[data, PlotStyle -> {PointSize[.03], Red}]
 ]

drN

@halirutan So the interpolation order here is quite poor, I take?

halirutan

@drN This has nothing to do with the order.

drN

@halirutan In that case, what is it?

halirutan

12:05 AM

The InterpolationOrder just defines the degree of the underlying polynomial which is used for interpolation.

drN

@halirutan Right. So higher "degree" polynomials are better aren't they?

halirutan

@drN I assume you never heard of the sampling theorem?

@drN Not necessarily, no.

drN

@halirutan I understand what it is from the name but I have not heard it.

halirutan

Ok, what you see in the image is the following:

Our underlying function (red) has a high frequency, but we used only 7 sampling points.

This is not enough to catch all details of the function and therefore, your interpolation does not represent the underlying function, although it matches perfectly on the sampling points.

drN

@halirutan Aaah... that makes sense.

halirutan

12:09 AM

This is called aliasing

Aliasing

In signal processing and related disciplines, aliasing refers to an effect that causes different signals to become indistinguishable (or aliases of one another) when sampled. It also refers to the distortion or artifact that results when the signal reconstructed from samples is different from the original continuous signal. Aliasing can occur in signals sampled in time, for instance digital audio, and is referred to as temporal aliasing. Aliasing can also occur in spatially sampled signals, for instance digital images. Aliasing in spatially sampled signals is called spatial aliasing...

drN

@halirutan Rings a bell, though said bell is tolling very far away right now ;)

halirutan

@drN In essence, your first sentence was right.

drN

@halirutan Yes, but only for the sampled points... ?

halirutan

Since you only have your sampling points of your (measured) data and you almost never know what happens in between, the interp. matches perfectly.

@drN Yes.

@drN You can do the following: Take your data, but only every 3rd point. Interpolate and draw the curve with all points and see whether it still fits ;-)

drN

It makes sense..

J. M.

12:30 AM

@halirutan Hah, aliasing...

@drN "Higher order does not necessarily mean higher accuracy."

Mr.Wizard

2:58 AM

Anyone here?

rm -rf

@Mr.Wizard Hello

Mr.Wizard

@rm-rf Hello. :-) I believe this is a duplicate of this. Could you confirm my understanding?

rm -rf

@Mr.Wizard No, I don't think so... very related, but not a dupe

Mr.Wizard

@rm-rf Reasoning?

rm -rf

@Mr.Wizard the older question asks for ways to generate all N possible partitions of sublist length K (total sublists M). There is no requirement or mention of duplicates. The new one, on the other hand, asks for choosing K elements from N total in groups of M, such that there are no duplicates

Mr.Wizard

3:11 AM

sorry, brb

okay, let me take another look

@rm-rf I see nothing in the new question that makes me think that not all partitions should be generated. Partitions are by definition duplicate-free, assuming that the original set is.

rm -rf

For instance, with Range@10 and sublist size 2, the old question would have solutions that look like {{{1, 2}, {3, 4}, {5, 6}, {7, 8}, {9, 10}}, {{1, 2}, {3, 4}, {5, 6}, {7, 9}, {8, 10}}, ... }, i.e., each sublist length is fixed at 5 because of the choice of 2

However, in the new question, you're free to choose the number of sublists (he calls groups). So if I choose to take them 3 at a time, a solution would be {{{1,2},{3,4},{5,6}}, {{1,2}, {3,4},{7,8}},...}

The solution set here is much larger than in the other one

Mr.Wizard

@rm-rf I don't read it that way (I'm not saying your wrong). I thought "groups" refers only to the number of elements in each partition.

rm -rf

The older question is a specific case of the new one when the number of "groups" is fixed at length_of_original/sublist_length

Mr.Wizard

@rm-rf Wait, you're saying the answer is the powerset of the partitions from the older question?

I read the question again and I'm just not seeing this (again, not saying you're wrong). He writes:

rm -rf

@Mr.Wizard Ah, ok... I can see how the new one can be read that way as well.

Mr.Wizard

3:18 AM

So I am expecting result to be below combinations for group of 2,

{A, B}, {C, D}

{A, C}, {B, D},

{A, D}, {B, C},

rm -rf

@Mr.Wizard For some reason I just took that as an unfortunate side effect of his having chosen only 4 elements for his example. But I agree with your interpretation too...

Mr.Wizard

If it follows from the specification of groups of two that this is the result, and e.g. {A, B} is not on a line by itself as a separate return, doesn't that mean that all elements are to be used in each case?

rm -rf

it seems like whuber read it the same way as I did

Mr.Wizard

Okay, I guess we'll have to let the OP clarify this.

I posted a comment requesting clarification. Thanks R.M, and see you later.

rm -rf

@Mr.Wizard Ok. Btw, I think your interpretation is most likely what the OP meant, since he explicitly mentioned the equation for the number of combinations (that is possible only with your reading)...

On my first read, it just seemed like he meant "groups" as in groups of pairs or triplets, etc.

drN

7:56 AM

Why am I unable to do this: `dataD = Import["ExampleData/OriginOfSpecies", "Text"]`? It throws an error at me which says `Import::nffil: File not found during Import. >>`.
This works: `Evaluate ExampleData[{"Text","OriginOfSpecies"}]` but this doesn't import data does it?

Matariki

8:09 AM

@drN This will give you the text: dataD = ExampleData[{"Text", "OriginOfSpecies"}]; Not certain what you mean by import.

drN

@Matariki Import is the command in mathematica to "import" data is what I thought...

Matariki

@drN But no need to import here. The text will be loaded into the variable

drN

@Matariki I would like to perform word counts etc. on the Origin of Species... so won't I need to "Import" it rather than just display it?

Matariki

@drN if you execute the ExampleData[... you'll end up with the text which you can manipulate as usual

drN

@Matariki Oh... will give that a go! tnx@!

Matariki

8:13 AM

@drn I am not certain that this will include the complete Origin Of Species. It might be just an excerpt!

It is after all just an example

drN

@Matariki Oh is that all is available then? I guess thats good! Wouldn't want to deal with Darwin's entire discourse! ;)

halirutan

8:37 AM

Can someone shed some light into what the hell bitcoin code is??

0

Q: Visualizing the bitcoin code?

I'm a graphic designer from Switzerland and I'm blown away by some of the graphical results that Mathematica is able to produce. However I have no clue of mathematics or coding myself. But I'm asking myself: Since you are able to visualize code with Mathematica, would it be possible to visualize ...

graphics3d

I'm somewhat confused of the results google is giving me.

Oleksandr R.

8:49 AM

@halirutan unclear. I guess you know what Bitcoin is, but what aspect this poster wants to look at is rather obscure.

halirutan

@OleksandrR. I googled some digital currency and that's why I'm so confused.

Oleksandr R.

@halirutan yes, Bitcoin is a (rather complex) cryptosystem that implements a digital currency. The code is open source, but I suppose the code by itself is not very enlightening as to how Bitcoin works in real life.

halirutan

But since he says he has no clue about programming, I assume he mean something else. Why would a programming-noob want to visualize source code?

Oleksandr R.

@halirutan the OP claims not to have a clue about math either, so visualizing a cryptosystem seems perhaps impractical as well. I suspect actually they do not know exactly what they want, and the post will be closed as NaRQ. But, on the off chance they can describe something concrete, let's leave it for the time being.

kirma

Bitcoin transactions produce a traceable graph, that might be the "code" he's referring to. Or not.

eprint.iacr.org/2012/584.pdf

Oleksandr R.

9:02 AM

@kirma true; but this graph is enormous (many gigabytes for the compressed representation). Visualising it meaningfully will be difficult.

Oleksandr R.

9:22 AM

@kirma very interesting paper

kirma

9:39 AM

@OleksandrR. of course. :-|

Rojo

1:18 PM

Hello

Any image processing enthusiasts around?

rm -rf

2:06 PM

@Rojo If you only want an "enthusiast", what's up? ;)

Rojo

@rm-rf Hehe

acl

I feel that this answer deserves a lot more upvotes than it has.

Rojo

@rm-rf Is the email address that is (or was) on your profile where I once sent you the munk notebook still available?

rm -rf

yeah

btw, you use MATLAB, right?

(lots? some? very little? none?)

Rojo

@rm-rf yes, somewhat between some and very little

rm -rf

2:16 PM

Hmm.. ok. Would you be willing to beta test a MATLAB <--> Mma link?

Rojo

@rm-rf I would

rm -rf

@Rojo Great! Expect an email in the next few days :)

Rojo

:)

@rm-rf Sent an email, forgot the attachment

rm -rf

lol

Rojo

There it is

Any tips or ideas that come to mind, feel free to say

drN

2:56 PM

Does Findroot[..] have (undocumented) options that allows for multiple root calculation? Right now I can either do a FindRoot[eq1 == eq2, {x,x0,x1,x2,...,xn}] which is rather untidy..

J. M.

@drN No, it just finds a single root from a single seed.

If you need more than one root, there are various methods.

drN

@J.M. Yes, I noticed FindallCrossings2D but didn't quite want to go down that road.

@J.M. So generally as an advanced mma user (you), do you think it is rather crass to use Table[FindRoot[eqn1 == eqn2, {x, xn}], {xn,n}] where n is pre-decided?

J. M.

@drN I wouldn't say "crass", but if you're in the business of solving equations, you usually need to do more effort than that, unless you're absolutely sure that your function's roots are very nearly integers...

drN

@J.M. Makes sense.

halirutan

3:11 PM

@drN The problem you have is, that you cannot be sure you missed some roots.

J. M.

That's where the "effort" I speak of comes in: you'll want to find nice brackets for your roots, assuming they're odd-order roots. If your function has even-order roots, then that's quite troublesome...

halirutan

Especially, when you don't know your function explicitly, e.g. when you try to solve f1[x,y,..]==0 where f1 is a black box.

drN

@halirutan Yes. I understand what you are saying. However, I am creating a couple of interpolating functions from discrete data points. So I proceeded with two methods: 1) I used a curve fit for these data points and then Roots[] which seemed rather terrible. 2) Then I used an interpolating function with FindRoot[...] where I could go from {x,xn} since I know xn....

Table[FindRoot[eqn1 == eqn2, {x, xn}], {xn,n}]1.. I know xn is what I am saying...

J. M.

@drN Where did these "data points" come from?

drN

@J.M. These data points just represent curves. Conveniently these curves interesect.

J. M.

3:21 PM

@drN Yes, but you did not answer my question. Where did you get these curves, or points, or whatever?

drN

Oh, its a bit of a homework problem. I don't want to just ask for the solution to it. I thought I should figure out the nitty gritty of it. I know that intersecting curves have the same roots so I was trying to figure out how best to use FindRoot to do this.

Rojo

4:09 PM

How's it going @YvesKlett

Yves Klett

4:58 PM

@Rojo good :-) browser in absentia...

Now leaving to hunt for the weekend rations.

acl

5:14 PM

@halirutan this is always going to be true though (except perhaps for polynomials)

@YvesKlett do not forget the beer

Rojo

5:52 PM

@SjoerdC.deVries is that question still open?

Pick doesn't work on level one. To my experience, this is how it works I think:

Pick[expr, struct, patt]

if struct matches patt, it returns expr

If it doesn't, if struct has depth=1, it returns nothing

Sequence[]

Same if expr has depth=1

However, if both have depth>1, it requires that their lengths are equal

and it maps Pick to the arguments of expr, and the args of struct

with the same pattern

wrapped in the head of expr

Humm, let's code it :D

Rojo

6:15 PM

Something along these lines perhaps?

pick[expr_, struct_] := pick[expr, struct, True];
pick[expr_, struct_, patt_] /; MatchQ[struct, patt] := expr;
pick[expr_, struct_, patt_] /;
   Depth[struct] == 1 || Depth[expr] == 1 := Sequence[];
pick[h_[args___], _[sargs___], patt_] /;
   Length@{args} == Length@{sargs} :=
  pick[#1, #2, patt] & @@@ h @@ Transpose@{{args}, {sargs}};
_pick := Throw@"Incompatible shapes, blabla"

halirutan

Does someone know when exactly the new Stackexchange day starts?

halirutan

6:36 PM

What I mean is, when does the SE clock change to 0:00 so I can earn more rep?

Rojo

I think it's about 9 pm Argentinian time

whichh is currently 15:37 pm

halirutan

@Rojo Ok, you are -3 and I am aehmm..

right +1

So it seems to be at 1:00 am here.

I guess

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