Can someone with version 11.2 test, if it is still possible to open Mathematica version 9 with that activation key?

I'm currently on a different license and not on my university license, but a colleague pointed out that this does not seem to work.

@halirutan The way our license/passkey works has changed, too. I noticed some months ago I cannot open V9 or earlier

@MichaelE2 I was afraid of that. This is surely annoying.

@halirutan Yeah, it's annoying, but it's their property, I guess.

@MichaelE2 For me it is crucial. I'm using function versioning in the IntelliJ plugin so that I can let the user decide for which Mathematica version he is writing code. If he uses a symbol that is only available in a later version, I can mark it as an error. @Szabolcs is heavily relying on this because he tries to keep IGraphM compatible with different versions. In addition, this means that he or any other package developer can not ensure his package runs on the main versions.

@halirutan Maybe they'll give you keys?

@MichaelE2 Yes, I'll have free keys for the Plugin and this is awesome. And I'm sure it would not be a problem to provide keys for the earlier versions. The bigger picture here is that many researchers and groups still use version 9 and this restriction makes it (again) harder for package devs.

@halirutan I get the problem. And I've seen complaints about WRI not always supporting outside developers in the past....not sure what else to say, really

@Edmund yode found the same thing, but the cloudplatform docs he linked to: chat.stackexchange.com/transcript/message/39892852#39892852 aren't working for 11.3 functions yet it seems

@halirutan I'm not sure I understand the problem. How was your copy of version 9 activated? Did you use MathLM? You should be able to store multiple activations in the same Mathpass file?

@Searke No, I guess the thing is that we are using university licenses that are renewed every year and we have premier support and always get the latest versions.

Yes. And you may need a new password for those newer versions

but you can keep the old password in your mathpass file

this might be a bit annoying though.

and probably not that well documented.

@Searke The activation keys always expire after a year.

Oh and your universities don't intentionally buy version 9 for you.

@Searke Yes. So I always have an up-to-date key for the latest version but until recently, I could still open older versions.

That's is a problem. I'm not sure I've heard it before.

Definitely worth writing a suggestion to support about.

But as I said, I'm currently on a pre-release key and I haven't checked myself. I just heard from a colleague who contacted his distributor and the distributor gave him a special v9 key.

That made me wonder, why I should not be allowed to open v9 when I have a key for 11.2

@halirutan @b3m2a1 any ideas? ^

@Kuba seems to work fine on my old Mac. It could be Windows specific--the last time I used Mathematica on Windows I tried to give someone a Dynamic-heavy framework and it just made the front-end hang without end. My guess is they use a battle-tested 3D graphics library, but rolled parts of their own 2D stuff and this isn't playing nice with Windows.

@b3m2a1 ok, it is definitely OS/soft specific because it works well on my VM MacOS

on the same hardware where it jumps in win10

Hi @Kuba, @halirutan!

Does anyone know why RegionQ@Interval[{-Infinity, Infinity}] returns False?

Lemme check the doc

While e.g. RegionQ@InfiniteLine[{0, 0}, {0, 1}] returns True.

It could be specific to v11.1, I don't have the latest.

Interval doesn't return a region, AFAIK

@ItamarG3 Nevertheless, RegionQ@Interval[{-1, 1}] returns True.

huh. weird.

I fail to see the rationale behind this if it is a feature by design and not a bug.

This is so weird.

Maybe because Infinity is treated a bit differently than an equivalent using InfiniteLine?

@IstvánZachar Seems like a regression to me. Try it in an earlier version:

In[6]:= $Version
RegionQ@Interval[{-Infinity, Infinity}]

Out[6]= "11.0.1 for Linux x86 (64-bit) (September 21, 2016)"

Out[7]= True

I would say it is a bug because other infinite regions return true like

In[3]:= RegionQ[ImplicitRegion[x^2 + y^2 > 1, {x, y}]]

Out[3]= True

@ItamarG3 I saw your suggested edit to the tag wiki of the fullsimplify tag. Did you also see the tag simplifying-expressions? Its tag info mentions FullSimplify. Perhaps making fullsimplify a synonym would be a better alternative to introducing the new tag fullsimplify.

Anyway I got to go

hm. somehow I missed that. Thanks for pointing it out

@IstvánZachar Nice to see you. I see you already received feedback about the Interval, don't have anything to add really.

@JacobAkkerboom @ItamarG3 yes, I think we don;t need this tag as it look like a function documentation :) and there is one which is broader.

@Kuba I see. Thanks for the explanation :D

@IstvánZachar I have reported it. If you like to make a question out of this, I could point user21 to it who should be able to give further insight.

It's always fun to start joining another SE community.

@halirutan Exactly, this was my idea with HalfLine, Plane, etc. which are also unbounded regions. Moreover, I remember my code working in earlier versions. That could be my failing memory though, of course.

@ItamarG3 welcome to MMA.SE :)

:D

@halirutan Thanks for reporting it, I'm not sure my licence is valid anymore to report anything.

I had forgotten the struggle of low-rep users. Nice to be reminded of it XD

@halirutan Unfortunately I don't have the time to post a question, but if anyone wants to, please feel free to do so!

@IstvánZachar Then let's wait what the support says. In the end it might be on user21's desk anyway.

@halirutan @IstvánZachar A more direct comparison: RegionQ@ImplicitRegion[-Infinity < x < Infinity, {x}] returns True.

And it affects region-related calls like Integrate[Exp[-x^2], {x} \[Element] Interval[{-Infinity, Infinity}]]

@ItamarG3 I think general idea has been that as far as built-in symbols/functions go, a tag should organize multiple ones, especially in Mathematica, which tends to have several functions related to the same task.

@MichaelE2 such as graphics-something instead of having one for graphics, one for show etc.?

#dailyfun

@ItamarG3 graphics is not just for Graphics, though. It's for the topic of graphics in M. graphics3d is perhaps a better example of a seemingly narrow focus. OTOH, visualization in 3D has been a strength of M from the beginning. -- Or you can look at it this way: Tags are created by a weakly organized community of human beings, who are bound to be inconsistent from time to time. -- Or maybe I misunderstood your question....

@Kuba That's a tough one. If you are on Windows, you could use GPUZ to see if the left example really uses more GPU. I doubt that but I'm not sure. You could create a small 3D plot with many polygons and observe if the GPU kicks in when you rotate it because this is a task for the GPU.

If it does, you can try to find out if the resizing of your graphics does anything similar.

What I would rather try is to decrease ticks and tickmarks and see if it changes. What happens for example if you use Frame -> True but turn off frame ticks?

For instance, does this make it worse?

ticks = Table[x, {x, -1, 1, .05}];
g1 = Graphics[Circle[], Frame -> True, ImageSize -> Full,
   AspectRatio -> Full,
   FrameTicks -> {{ticks, ticks}, {ticks, ticks}}];
g2 = Graphics[Circle[], Frame -> False, ImageSize -> Full,
   AspectRatio -> Full];
resizeArea =
  Framed@Pane[#, AppearanceElements -> "ResizeArea",
     ImageSize -> 100 {1, 1/GoldenRatio}, FrameMargins -> 15] &;

{resizeArea[g1], resizeArea[g2]}

What happens if you use AspectRatio->1?

@halirutan yes, it is because of ticks.

@halirutan Frame -> True, FrameTicks -> None is smooth

@Kuba And does the increased number of ticks make it worse?

Moreover, the issue for the simple case (not the real app case) disappeared after clearing ini.m files (after I failed to find any reasonable trigger)

But now with your code I am able to reproduce it

So the number of ticks clearly magnifies the problem

@Kuba That is bad. Even with that many ticks it runs smoothly. I have no jumps like you experience.

Yes it is strange. I stress the FE a lot so maybe I'm to blame but I work equally often on 2 machines + one laptop + Rolf's machine and this is the only one with the problem

@Kuba My best guess is that the FE uses a library for calculating the new positions. It might be that it lies in some OpenGL functionality but I'm honestly not sure. Have you compared the graphics drivers on both machines?

@halirutan what exactly to compare? Because drivers are different for sure as gpu is.

This is the one where problems happens

the same renderer with apple's drivers works well

@Kuba If you have the same version. These days, NVidia uses the same drivers for GT 630 and e.g. TI 1080

But I see you are already on the latest one.. hmm

@halirutan yep, I've updated it today via Win updater so I tried today on 3 versions ( the updater didn't fetch the latest so I did it manually)

And the issue was always present

@Kuba Damn..

If you need desperately to find out what is going on, you could (a) use a tool to find out which dlls are used by Mathematica and look out for differences on both machines.

(b) you could use a thing like this to monitor dll calls while you do nothing else than resizing. This is surely a huge pain and I don't know if you should consider doing this.

I have never done this on Windows, so I cannot be of any further help. But when the system is both Win10 with the same Mathematica installation, I cannot think of a different reason for the slowdown.

@halirutan I don't use v9 much anymore (also, just arrived and my harddrives are still boxed up...), but as I remember, it was possible to have more than one activation key in the mathpass file. They might provide you a v9-specific activation if you ask nicely, and if they can see that you originally had v9

Is there any obvious reason why StringCases["1213235", __?PalindromeQ, Overlaps -> All] doesn't work (the PatternTest seems to be ignored), but StringCases["1213235", x__ /; PalindromeQ@x, Overlaps -> All] does? Otherwise, I might ask on main.

@MartinEnder tutorial/StringPatterns says of ?: "a pattern for which test yields True for each character" e.g. perhaps it isn't applying PalindromeQ to the sequence, but to each character individually.

oh okay. that's quite unintuitive, but is probably useful in general

If anyone has the time, I would appreciate help at mathematica.stackexchange.com/questions/165754/…

Hmmh. I reported a bug on Sunday (well, 50+ hours ago), and haven't seen a reply to it yet. I think that's a bit unusual.

@halirutan Sorry, I see you already discussed this in detail with Searke

Damn I can't seem to reshape my linear layer to fit a ConvolutionLayer, to create a bottleneck autoencoder

With[{w = 64,
  channels = 16},
 NetChain[{LinearLayer[channels*w*w*3],
   ReshapeLayer[{3, channels, w, w}],
   ConvolutionLayer[channels, w, "Dimensionality" -> 2]},
  "Input" -> 8]]

this gives me the following error: NetChain::ninctyp2: Incompatible types for output of layer 2, ReshapeLayer[[Ellipsis]], and input to layer 3, ConvolutionLayer[16,[Ellipsis]]; a 3*16*64*64 tensor is not compatible with a rank-3 tensor, respectively.

I tried different shapes, like channels, w, w, 3 or channels, 3, w, w etc.., but no dice

ok, you can just drop the 3, it will figure it out automatically. it's a bit counterintuitive.

also i feel i am doing something wrong, my training doesn't seem to leverage the gpu at all, yet it is superbly slow. it must be busy passing data from / to the gpu?

@MartinEnder the general trick I do is to use /; to test entire string pattern blocks and ? to test individual character.

You can sort-of see how this works in this:

sp = (s : DigitCharacter ..?(Divisible[ToExpression[#], 2] &) /;
     StringTake[Echo@s, 1] == "2") ~~ __;

StringCases["asd 2424.", sp]

2424

{"2424."}

StringCases["asd 2814.", sp]

28

{"2814."}

StringCases["asd 4422.", sp]

4422

442

44

4

422

42

4

22

{"22."}

In the first case the ? eats up all of the characters before passing to /;. In the second it only eats up some. In the third it keeps trying until it finally gets a case that works for /;.