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rm -rf
rm -rf
12:00 AM
@halirutan Installing TTF + restart worked for me.
 
halirutan
halirutan
@rm-rf OK.
 
rm -rf
rm -rf
Thanks for testing! This was driving me nuts for a few days
 
halirutan
halirutan
@rm-rf No problem. Glad I could help.
 
 
10 hours later…
Vitaliy Kaurov
Vitaliy Kaurov
9:43 AM
Hey guys someone perhaps already posted here a link to the famous Mathematica Sierpinski Triangle blog that took the world by storm. I made a summary of some data here if you are curious:

http://community.wolfram.com/groups/-/m/t/138400
2
 
Yves Klett
Yves Klett
10:02 AM
@VitaliyKaurov wow! I was also very happy to see the cow shine once more :-)
 
Vitaliy Kaurov
Vitaliy Kaurov
10:43 AM
user image
2
@YvesKlett that's another hidden image from that blog. And yes indeed the cow shines and he also wrote poetic verses there - **specifically for Mathematica cow**. Quoting ;-)

"This cow does not cower. Infinity cannot bully this bull, cannot bloviate this bovine. By all appearances this cow is wearing infinity on its mane. Its horns are probably made of ℵℵ⋱ down 4 or 5 levels, an immutability surpassed only by that of the tusks of the Alephant. Our cow isn't staring into infinity. It's looking down at infinity, observing infinity with detached understanding. If our cow were not so enlightin…
 
 
2 hours later…
Yves Klett
Yves Klett
12:51 PM
@VitaliyKaurov cow! once more.
 
 
3 hours later…
Zibadawa
Zibadawa
3:44 PM
Sometimes what makes one piece of code finish in 15 minutes versus another for an identical task takes 15 hours is mystifying. I expected the 15 minute code to be the slower of the two, due to substantially more iterations. Neither one uses up a meaningful amount of memory.
And I swear the first time I ran the 15 hour code it was done in 14 minutes. Go figure.
 
rcollyer
rcollyer
@Zibadawa if the code isn't to long, post it here, and we might be able to explain why.
@VitaliyKaurov Mooo!
 
Zibadawa
Zibadawa
@rcollyer I think it is a bit too long. And I'm not sure how to reduce it to a minimal state, since I have little idea exactly what is causing it to go so slow. The basic idea is I have a matrix, and I want look through the matrix for 0 entries, and then want to change that entry and several others according to group multiplications (the matrix is thought of as indexed by the elements of the group, as enumerated by mathematica).
The 15-minute version iterated over all matrix entries and used an If statement to check for a 0 and then altered the matrix. The 15-hour version used a while statement and Position to find the next zero entry, so that it wouldn't have to look at non-zero entries ever.
I'm considering a third idea that involves partitioning up the indices according to which ones would get modified simultaneously. The size of the matrix is relatively small; 56x56 in this case, there's just several hundred of them to iterate over, each using slightly different rules for which set of entries to modify when a 0 is found, plus computing the group multiplications when relevant. Haven't tried that, yet.
 
rcollyer
rcollyer
4:09 PM
@Zibadawa So, do you loop over the positions returned from Position, or do you call Position repeatedly?
 
Zibadawa
Zibadawa
@rcollyer I was calling it repeatedly, which definitely wasn't the most efficient use of that function call.
 
rcollyer
rcollyer
@Zibadawa no definitely, not. Pull it out of the loop, and see what happens.
 
Zibadawa
Zibadawa
I'd have to collect the indices I'm modifying each time and delete them from the position list. Which is do-able, just not something I wanted to implement right then.
or otherwise make some sort of running tally of entries already adjusted and take a complement
 
rcollyer
rcollyer
That doesn't make sense to me. Are you looking for all zero positions?
 
Zibadawa
Zibadawa
@rcollyer Well, let me more accurately describe what I'm truly doing. Perhaps there is an entirely different and smarter way to do it.
 
rcollyer
rcollyer
4:13 PM
@Zibadawa Sure.
 
Zibadawa
Zibadawa
I want to construct a matrix of variables, from which I will then build up some equations to then solve. Some entries in the matrix need to correspond to the same variable, depending on various relations coming from group multiplications (which are essentially already computed before this stage).
 
rcollyer
rcollyer
Is the matrix a member of the group, or otherwise conform to the group symmetry?
 
Zibadawa
Zibadawa
The matrix is external to the group, really. I have a group (PermutationGroup, specifically) computed, and effectively view the matrix as being indexed by the group elements.
Ultimately there's a function of two variables, each in the group, and some of the outputs must be the same (according to these group multiplications I'm mentioning), and then those outputs must satisfy various polynomial relations.
the matrix is representing that function. The entries are the outputs, the indices are the inputs.
 
rcollyer
rcollyer
I'm not seeing it. Can you describe what it should look like with the Cyclic group of order 2? Maybe that will crystalize things for me.
 
Zibadawa
Zibadawa
What I'm trying to do is create a matrix of unknown outputs, making sure outputs which should be identical have the same output stored. Then I go through the business of finding the polynomial identities and solving them, but that's a separate step with an entirely different bottleneck.
 
rcollyer
rcollyer
4:22 PM
So, the matrix must have some symmetry, right?
 
Zibadawa
Zibadawa
Let us call the function F, with domain G x G, where G is our group. Let v: G -> G be a given group homomorphism. Then F must satisfy F( g*g'*g^{-1}, v(g)*h*v(g)^{-1}) = F(g', h) for all g,g',h in G.
For the cyclic group of order two that relation is trivial, since it's abelian.
 
rcollyer
rcollyer
I know. Unfortunately, I think you have to go to order 4 before you can find one that's not.
 
Zibadawa
Zibadawa
Order 6.
S_3, more specifically.
 
rcollyer
rcollyer
I was close. My group theory text is packed.
@Zibadawa what is v(g) in this case? Just a polynomial function of g?
 
Zibadawa
Zibadawa
So the basic idea is: I'm going to want to find all such functions F which satisfy some polynomial identities in its outputs, but first I need to make sure it satisfies the relation I just specified.
 
rcollyer
rcollyer
4:27 PM
Hmm. Given an hour, I probably could come up with something. But, at the moment, I'm at a loss.
 
Zibadawa
Zibadawa
v is a group homomorphism. In my code, you can think of it as a function, which given a group element spits out another group element.
I stated that relation in its "math" version, rather than its "mathematica" version, incidentally.
 
rcollyer
rcollyer
@Zibadawa right. I was just wondering if it just a polynomial, then closure is ensured.
@Zibadawa noted.
 
Zibadawa
Zibadawa
Closure is ensured, at least at the extent I need it. So if the relation says if I start with the pair (a,b) and it tells me that (c,d) must be the same, then we have reflexivity: if I started at (c,d) instead it would have then told me that (a,b) must be the same.
So I don't have to worry about conflicts or anything like that. Once an entry has been modified (set to its unknown variable), it will not be modified again.
I need to go get lunch, I'll be back in an hour so.
 
rcollyer
rcollyer
@Zibadawa okay. I'll likely let someone else tackle this. Best of luck.
 
Zibadawa
Zibadawa
Alright. It's looking like that 15 minute code has caught the 15 hour bug. The heck? Oh well, I'll look into that when I get back.
 
Yyrkoon
Yyrkoon
4:49 PM
Hi all,
i was looking for a way to evaluate symbolically a maximum. Like if i've
Assumptions = a > c && a > b && a > 0 && b > 0 && c > 0
and
Max[{a,b,c}]
gives me
a
Is there a command in mathematica that i can use to feed the assumption into Max?
Thanks a lot!!
 
Yyrkoon
Yyrkoon
5:01 PM
I think i found a solutions:
Refine[Max[{a, b, c}], {a > c && a > b && a > 0 && b > 0 && c > 0}]
if someone has comments or suggestion please let me know
 
Zibadawa
Zibadawa
5:20 PM
I think Assuming would also do the job.
 
 
3 hours later…
Szabolcs
Szabolcs
7:58 PM
@halirutan I went with this finally: mathematica.stackexchange.com/a/34018/12
 
 
2 hours later…
halirutan
halirutan
10:23 PM
@Szabolcs I don't really agree with this method because you cannot share you package with others. It is bound to your system or relies on a fixed static path for your dependency. I was giving you bad advice because I was drawing the wrong conclusions. At least in theory, there is a far better method which works through all system! Ping me when you are around.
 

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