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Level River St

 Discussion between l4m2 and Karl

Imported from a comment discussion on codegolf.stackexchange.c...
Level River St
Dec 11, 2024 00:45
Oops - just found a possible hole in my own proof. I just realised that there are 24 possible permutations of the 4 faces of the S, only one of which is the 2 fold rotational symmetry operation. Still I think it should hold up.
Level River St
Dec 11, 2024 00:40
I will revise my code accordingly but it may takes some time and it won't be as short. Going to bed now.
Level River St
Dec 11, 2024 00:39
Therefore if I write a function that returns true only for the S in all 30 possible orientations, it will return the same result for 60 permutations of the inputs: that is 30 possible orientations x 2 permutations of the inputs that returned the same result for the fixed orientation
Level River St
Dec 11, 2024 00:36
Hi Karl, I think the following proof satisfies me for the 4 faces in an S: If I write a function that returns true only for the S in one particular orientation, it will return the same result for exactly one permutation of the inputs, corresponding to rotation about the symmetry axis of the S.
Level River St
Dec 10, 2024 23:33
There are definitely fully asymmetric configurations with 5 faces selected. 2 fold rotational symmetry doesn't seem to be an issue. I think you would need a central face in such a configuration, and because each face has an odd number of sides, 2 fold rotational symmetry seems impossible
Level River St
Dec 10, 2024 23:29
The trouble with these is that they all have 2 fold rotational symmetry about an edge so only have 30 images under symmetry. So a test would only return TRUE for 30 inputs. I'm not sure if that meets the symmetry requirements of the challenge.
Level River St
Dec 10, 2024 23:27
They can be summarized as follows: select north pole, south pole and one other side in the northen hemisphere at longitude 0. You can now select any of the following longitudes from the southern hemisphere +36, -36, +108, -108. You can' select 180 as that is achiral.
Level River St
Dec 10, 2024 23:26
Hi, there is a full list of 2-colourings of the dodecahedron at baumanneduard.ch/2FarbDode.htm Ones with 4 faces selected are in green. There are 4 chiral configurations (in 2 pairs) with 4 faces selected, in a block at top right.
 

 Discussion on answer by Kilisi: Is wo…

Imported from a comment discussion on workplace.stackexchange....
Level River St
Oct 20, 2023 20:34
This is exactly what butchers do. Dominant hand is bare and wields the meat cleaver. Non-dominant hand holds and turns the meat and wears a chainmail glove.
 

 Discussion on question by Peter Kagey…

Imported from a comment discussion on codegolf.stackexchange.c...
Level River St
Jul 5, 2022 07:59
@user202729 Your opinion that this question is clear enough is based on the fact that you already have a good understanding of what homeomorphism means. To see things from another point of view, check out the original text of codegolf.stackexchange.com/q/193502/15599 which was written in music theory, and was closed (for not being self contained) until Xnor provided a more accessible explanation. Note that Xnor does not attempt to explain fully what n-tone equal temperament means, he goes directly to the case log2(3) which is the specific case required to answer this challenge.
Level River St
Jul 5, 2022 07:59
@user202729 sending users to wikipedia is to be avoided. It's a long standing guideline that challenges should be self contained. I tried to find the origin of this on Meta but when I searched for self contained I got so many more up-to-date hits I couldn't find the original. Wikipedia can change, and the article contains lots of extraneous information which is not relevant to this challenge. (I hope) I captured the essence in the grey text in my comment 6 hours ago (though I wish I'd said side of the square not edge), without needing to even define "continous bijective function."
Level River St
Jul 5, 2022 07:59
@user202729 personally I'm finding it easiest to visualise by drawing a dual graph with one vertex for each edge of the square and one vertex for each rectangle. That way I don't see the corners of the individual rectangles, which I now understand are not relevant and hence distracting.
Level River St
Jul 5, 2022 07:59
@user202729 Important thing isn't to prove, it's to give clear enough definition (within the question, not externally) that the reader can understand & prove themself. I admit it's difficult! I suggest the following based on my current understanding: Examples of homeomorphisms are: /n Slightly moving the boundaries of the rectangles, so long as rectangles that were previously adjacent to each other remain adjacent to each other and vice versa, and rectangles that were previously adjacent to an edge of a square remain adjacent to the same edge and vice versa /n Rotating or flippng the square
Level River St
Jul 5, 2022 07:59
@user202729 Challenges should be written in self contained, jargon free plain English, though I appreciate that is difficult to do. I have noted on some other questions that too much math jargon is used and risks losing people (and if I'm confident I understand the jargon, I'll dive in and edit it myself.) Similarly I've rewritten significant portions of music and chemisty codegolf challenges (2 areas in which I have knowledge) to dejargonise them (sometimes to prevent or reverse closure) always with the proviso that the OP could roll them back if they disagreed. (So far, nobody has.)
Level River St
Jul 5, 2022 07:59
@user202729 I was hoping to see some activity from Peter to correct both what I consider an inadequate explanation, and whatever issues he had seen. That hasn't happened yet, but I've gone ahead anyway and cast the 2nd reopen vote because I'd like to see a (valid) answer (and I get a hint that you may have one?) If you're convinced you're right, go ahead and add some explanation. It was the fact that the vertices of the individual rectangles don't matter that threw me - I got that those of the square do. The linked wikipedia article does say that spotting homeomorphism requires "practice."
Level River St
Jul 5, 2022 07:59
@user202729 Thanks for the explanation. Two reasons I didn't retract the close vote. One is that the explanation came from you, not Peter, and as the challenge author his interpretation matters. I see from the side conversation that Peter and you seem to agree. However the explanation should be in the body (for example, in the form the following partitions are equivalent because....) The second reason I've not retracted my close vote (or at least, have not considered issuing a reopen vote) is that Peter himself cast one of the close votes and I don't know what his reason was.
Level River St
Jul 5, 2022 07:59
@user202729 sacred in the sense of "cannot be violated." Meaning that the inter-rectangle edge that cuts the west edge of the square in image 5 can't be moved (continuously or not) to the north edge to make it look like a rotation of image 2. Whereas you appear to be OK with changing the number of edges that cut each of the 4 sides of each individual rectangle. Is that right? In the pictures for n=5 NW rectangle has a different number of edges cutting its southern edge in each picture.
Level River St
Jul 5, 2022 07:59
@user202729 re:additional property: so you're saying n=4 pictures 2 and 5 are different because the NW vertex and edges of the square are sacred, but the n=5 pictures are the same because the vertices/edges of the rectangles are not? Specifically, the southern edge of the NW rectangle bounds 2 other rectangles in one picture and one other rectangle in the other.
Level River St
Jul 5, 2022 07:59
@user202729 by "4-way corner" I mean a corner between 4 rectangles. Regarding math terms, I did glean from wikipedia that homeomorphism doesn't mean there has to be a continuous deformation (specific example from wikipedia - unknotted torus and trefoil knotted torus) but the question specifically says "slightly" moving the boundaries of the rectangles. I'd consider temporarily deforming a rectangle into a pentagon (and changing a line segment from horizontal to vertical) as more than slightly moving a boundary. Better definition of homeomorphism (as it pertains to this problem) is required.
Level River St
Jul 5, 2022 07:59
Voting to close because of the edit "as user 202729 pointed out..." It looks to me that to get from one to the other you have to slide through the point where the Northwest rectangle forms a 4-way corner with the rectangles East, South and Southeast of it. Also, if these two are equivalent, why aren't the 5th and 2nd examples for n=4 equivalent? Extend the west rectangle till it fills the whole west side, then rotate 90 deg. I think this needs clearer definition.
 

 Discussion on answer by Level River S…

Imported from a comment discussion on music.stackexchange.com/...
Level River St
Mar 28, 2021 05:07
@obscurans I've added in a bit more about commas and different meantones, which I think addresses your point, but makes the answer dwell a lot on meantone temperaments. While I agree that equal temperament is a meantone, I've also heard "meantone" used to refer specifically to quarter comma meantone (and the wikipedia article on meatones agrees with this.)
Level River St
Mar 28, 2021 05:07
@phoog Thanks. Just fifth is 701.955 cents of an equal tempered semitone, a discrepancy of under 2 cents. Just major third is 386.314 cents of an equal tempered semitone, a discrepancy of nearly 14 cents from 400. Just minor third gives even bigger discrepancy at 315.641 cents, nearly 16 cents from the equal tempered 300. So I'd say the minor third is the worst interval in equal temperament, even worse than major third. One way to arrive at this just minor third is to subtract the major third from the fifth 701.995-386.314 = 315.641 . Note that the differences add up.
 

 Discussion on question by CourageousP…

Imported from a comment discussion on space.stackexchange.com/...
Level River St
Mar 22, 2021 13:40
Rate of sublimation of ice is limited by heat transfer - and remember there is no convection in space. It takes about 2000kJ to sublime a kg of ice. The solar radiation is about 1.3kW/m2, so a 1m2 exposed ice area would lose a kg of ice in about half an hour from solar radiation (plus more from conduction if in contact with the lunar surface which gets very hot in the day.) The point is, I think you've got a lot more than 2 minutes to transport your ice - but you want to avoid direct contact with the lunar surface if possible.
 

 Discussion on answer by DavidSupports…

Imported from a comment discussion on travel.stackexchange.com...
Level River St
Dec 18, 2020 22:20
@user2299523 National Express bus 707 was my thought too, but they are infrequent at the best of times, and are currently not running / unbookable due to COVID 19. But anyway, I would only attempt this with a taxi, AND be well prepared to miss my second flight.
 

 Discussion on question by Ritesh Sing…

Imported from a comment discussion on physics.stackexchange.co...
Level River St
Feb 27, 2020 23:14
A billion is a pretty small number when it comes to subatomic particles. Consider that a gram of hydrogen contains 6x10^23 hydrogen atoms. To put it another way, a billionth of a billionth of a gram of hydrogen contains just over half a billion billion hydrogen atoms, each consisting of a proton and an electron.
 

 Discussion on question by msamogh: 99…

Imported from a comment discussion on codegolf.stackexchange.c...
Level River St
Apr 19, 2018 02:27
@FryAmTheEggman I think if there is doubt it should be left opened. this applies to all questions, not just this type, if someone finds a suitable approach it could be reopened This is my problem - where are they going to post their approach? Post it on meta and beg for the question to be opened so they can post it in the proper place? The fact is people are going to see the question closed and not bother. That said, I think the case for this question is weaker than work it harder and I will not be pushing for it to be reopened despite my general beliefs.
Level River St
Apr 19, 2018 02:27
@FryAmTheEggman My Meta answer on the work it harder song, advocating keeping it open got 17 votes compared to 5 against. I scored 303 with my pattern based approach in Ruby, while other Ruby answers not exploiting the pattern came in at 643, 486 and 483 bytes. I'm only 28 bytes behind Peter in Golfscript at 275, which is remarkable given the typical relative lengths of answers in these languages. You cannot claim that Peter "showed" anything until you do a like for like language comparison. Closing is a self fulfilling prophecy, as you'll never see what approaches may be tried.
Level River St
Apr 19, 2018 02:27
@FryAmTheEggman if all the answers were like daniellindie's, I would agree with you. But if you look at tfeld's answer and elpedro's comment, there are sufficient patterns in the data to employ strategies specific to this text that would not work on the rickroll. This has already been discussed here codegolf.meta.stackexchange.com/q/6956/15599 . Note also that in this case, freedom to order the list as you please gives added flexibility in solving the challenge.
 

 Discussion on question by PyRulez: St…

Imported from a comment discussion on codegolf.stackexchange.c...
Level River St
Mar 4, 2018 00:26
I'm not close voting because it may be possible to keep all edge lengths the same on higher iterations but I'm really not sure if it is, or indeed if any mathematicians have tried to prove or disprove it.
Level River St
Mar 4, 2018 00:26
I've no idea how to specify. The hexagons are not regular, there are 2 angles identical to the rhombic triacontahedron (beta on this page mathworld.wolfram.com/RhombicTriacontahedron.html) with the rest of the sum of the internal angles of the hexagon (720 degrees) being split equally between the other 4 angles. The first iteration can have all edge lengths equal but for the second this then becomes physically impossible as far as I can tell. Each iteration gives more different types of edges, hence more types of faces in the next. Maybe you need to sandbox or close this until resolved.
Level River St
Mar 4, 2018 00:26
Also you haven't specified by how much the object should be chamfered. As far as I am aware and apparently in the examples from the page you linked on chamfering, you can have anything from very slight chamfering (large pentagonal faces and narrow hexagons, almost a dodecahedron) through to very strong chamfering (tiny pentagonal faces, almost a rhombic triacontahedron.) how much should we chamfer?
Level River St
Mar 4, 2018 00:26
I think it would be worth pointing out that a chamfered dodecahedron is not the same as the familiar truncated icosahedron (soccerball.) Truncated icosahedron has 20 regular hexagon faces, surrounded by 3 pentagons. Chamfered dodecahedron has 30 stretched hexagon faces, surrounded by 2 pentagons.
 

 Discussion on question by TheFamousDi…

Imported from a comment discussion on workplace.stackexchange....
Level River St
Jun 17, 2017 12:54
It's only unprofessional if done to excess. The current president of a certain country comes to mind....
 

 Discussion on question by Peter Lenke…

Imported from a comment discussion on codegolf.stackexchange.c...
Level River St
Jun 9, 2017 07:22
@PeterLenkefi ok, Iso is best avoided as it's confusing. Isooctyl is 6-methylheptyl (similar to isobutyl being 2-methylpropyl) but isooctANE has a completely different structure, 2,2,4 trimethylpentane. The reason for this anomaly is that isooctane is synthesized by dimerising isobutene, which is a convenient way of making a high performance ("high-octane") gasoline component.
Level River St
Jun 9, 2017 07:22
The rules mention iso, sec and tert. Are these permitted/required? For example is C4H9 CH (CH(CH3)2) C4H9 called 5(isopropyl)nonane or 5(1methylethyl)nonane or both? The first method works great in some cases, but the second method always works. Also, are parenthesis for clarity required?
 

 Discussion on answer by Lilienthal: "…

Imported from a comment discussion on workplace.stackexchange....
Level River St
May 23, 2016 12:46
@Martijn +1. I can`t resist adding: The right to vote for whoever you want gives you the obligation to think about who you vote for.
 

 The Nineteenth Byte

The Nineteenth Byte: General discussion for codegolf.stackexc...
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Level River St
May 15, 2016 18:15
@trichoplax I would like to see your algorithm, I hope you post it.
Level River St
May 15, 2016 18:06
@trichoplax no problem, stroke-width=3.464102 it is. So long as it's the same for everyone. As you can see the SVG is a real mess and it was important to clarify that before golfing. I reckon it will still be under 600 bytes when I've finished.
Level River St
May 15, 2016 17:54
what algorithm did you use?
Level River St
May 15, 2016 17:53
the algorithm I use is tiny. It took me a while to pick it though. The hardest thing was drawing the fragments, actually, especially in svg as the arcs aren't defined by their centres.
Level River St
May 15, 2016 17:52
@trichoplax Hi Trichoplax, I'm hopeless at using this Chat. I assume the above comment refers to my answer to your sierpinski knot question
steveverrill
Aug 8, 2015 14:26
@Doorknob, your data is very interesting. If you feel chat is the best place I won't argue, but I encourage you to post it as an answer to that question (and my preference was to leave it where it is)
steveverrill
Aug 8, 2015 14:24
@Orlp: "I'm more concerned about the fact that whatever answer's given will be completely outdated once Pyth, CJam continue development" I see absolutely no problem with that, provided the answer is dated, which it will be. I think it would be interesting to come back in a few years time and see if it needs revision.
 

 Discussion on question by smci: Which…

Imported from a comment discussion on travel.stackexchange.com...
steveverrill
Sep 2, 2015 09:47
@smci A shit travel agent can do worse than confuse airport codes. I got a call from my friend saying her husband's work had flown him from Colombia to Aberdeen (ABZ) in the north of Scotland and asking for my help. He was supposed to be going to the small town of Abingdon, which doesn't have an airport. The closest one is LHR, london heathrow.
 

 Discussion on question by Mark Lakata…

Imported from a comment discussion on codegolf.stackexchange.c...
steveverrill
Jun 13, 2015 21:18
Also, tag should be code-challenge, not code-golf
steveverrill
Jun 13, 2015 21:18
Nice idea, but could do with a bit more introduction. To be clear, is any compiled language / compiler allowed? (even gcc has variations between versions, not to mention that it's not the only C compiler around.) And if so, how will you check? (I'm close voting for this reason. If you want a readily avilable standard maybe Ideone.com would do, but it would invalidate your existing answer.) I assume interpreted languages are out because it should be possible to have an infinite loop that generates errors.
 

 Discussion between F. Hauri and steve…

Imported from a comment discussion on codegolf.stackexchange.c...
steveverrill
Mar 26, 2015 17:42
Ok take care bye.
steveverrill
Mar 26, 2015 17:41
Cool yeah i've seen it, hopefully you will get some answers soon. I will answer if I have time, but I'm pretty busy.
steveverrill
Mar 26, 2015 17:38
Check out my edit here, for example. codegolf.stackexchange.com/q/45415/15599 The linked wikipedia page was as clear as mud.
steveverrill
Mar 26, 2015 17:37
Yeah thanks for the info. Maybe you should link the first wikipedia page to the question. I generally don't edit other peoples questions for clarity unless they are about music, because classically trained musicians are unable to give up on their terminology.