Logs on 2024-10-09 (liberachat/#haskell)
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| 03:49:23 | <Axman6> | monochrom: '"<file cmd" is the same as "cmd <file"' you just blew my mind, how handy! |
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| 04:03:03 | <haskellbridge> | <mauke> Re: sh, both {foo and {;foo are errors |
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| 04:04:09 | <haskellbridge> | <mauke> {foo tries to run a command called '{foo' and {; is a syntax error |
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| 04:06:16 | <haskellbridge> | <mauke> a consequence of { not being special to the lexer, but also being a "keyword" with special syntax |
| 04:13:01 | <Axman6> | >:( |
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| 04:26:25 | <jackdk> | I was writing a shell script to use with `git bisect run` today; I ended up completing the bisection before I got the script to work right. Task failed successfully. |
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| 09:48:05 | <kuribas> | I wish there was more data science on haskell. The time I spend debugging pandas issues with missing indices etc would not have been there with a typed dataframes. |
| 09:49:00 | <kuribas> | We could easily port the polars dataframe library to haskell, with some typesafe wrapper on top. |
| 09:49:07 | <kuribas> | Or even to idris2. |
| 09:50:01 | <kuribas> | Of course the chance to get anyone on board with using idris or haskell in my company would be close to 0% |
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| 10:00:48 | <kuribas> | But I'd say for a long running service, haskell would be a better choice than Python. |
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| 10:08:07 | <tomsmeding> | kuribas: I am not at all familiar with that space, but given that there are (external) type checkers for python, would someone not have contrived some external typing annotations for pandas or something? |
| 10:08:18 | <tomsmeding> | or is the type system not expressive enough for that |
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| 10:13:45 | <kuribas> | tomsmeding: it's not expressive enough. There are actually type annotations for pandas. They often don't work well. |
| 10:14:08 | <tomsmeding> | I see |
| 10:14:11 | <kuribas> | And they cannot capture the fact "I have a column with name 'foobar' and type 'int'" |
| 10:14:23 | <kuribas> | Which would be possible in haskell or idris. |
| 10:14:32 | <tomsmeding> | I would categorise that under "not expressive enough" |
| 10:14:48 | <tomsmeding> | but I guess it's also impossible to properly describe python's fancy s(p)licing syntac |
| 10:14:51 | <tomsmeding> | *syntax |
| 10:15:26 | <kuribas> | splicing is just a special dunder method I think. |
| 10:15:36 | <tomsmeding> | ah |
| 10:16:13 | <kuribas> | The syntax creates a splice object. So in the end it's just sugar. |
| 10:16:29 | <kuribas> | And it's passed to __getitem__ |
| 10:16:35 | <tomsmeding> | I see |
| 10:18:25 | <kuribas> | The debugability of python is nice though, but very slow... |
| 10:18:49 | <Rembane> | dir() <3 |
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| 11:52:33 | <lieuwex> | tomsmeding: hi |
| 11:55:27 | <kuribas> | tomsmeding: Are you going to munihac? |
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| 11:57:41 | <leah2> | i am :3 |
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| 12:02:00 | <tomsmeding> | lieuwex: hi |
| 12:02:07 | <tomsmeding> | I have a suspicion how you ended up here |
| 12:02:11 | <lieuwex> | yes |
| 12:02:12 | <tomsmeding> | kuribas: nope, sorry, but have fun! |
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| 12:10:26 | <dminuoso> | 10:14:23 kuribas │ Which would be possible in haskell or idris. |
| 12:10:28 | <dminuoso> | Hardly in Haskell. |
| 12:11:22 | <dminuoso> | I mean yes, you can shoehorn anything in the type system. But you can also shoehorn runtime annotations into python and do type checking via symbolic execution. |
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| 12:13:52 | <dminuoso> | What I want is something like tags in go structs in Haskell, recoverable via generics or Data.Data.Data |
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| 12:14:46 | <dminuoso> | It beats having to either use type level tricks like in servant to attach actionable information to say handlers, or ramming it all into TH like in yesod. |
| 12:15:27 | <dminuoso> | The more information you embed into Haskell types, the more akward diagnostics become and quirkier your interfaces get. |
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| 12:18:55 | <hseg> | going through the botan devlog, I see reference to a crypto-schemes repo, but can't find it anywhere atm. Anyone know where it vanished off to? Also, I suppose the "cryptographic typeclasses" refered to are in the various Botan.*.Class modules? |
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| 12:45:25 | <kuribas> | dminuoso: I don't mean it's very ergonomic, but it's possible with type level lists. |
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| 12:48:02 | <dminuoso> | And those type level lists will propagate through your entire code, or you have to erase them. |
| 12:48:35 | <kuribas> | Yeah, but that's the idea. |
| 12:48:50 | <dminuoso> | I'm slowly starting to wonder whether some of those "lets encode it into the type system for safety" is solving problems as an excuse to just explore some funky ideas, rather than it stemming from a series of production outages due to some impedance mismatching. |
| 12:48:54 | <kuribas> | Or you have a constraint (HasColumn "foobar"). |
| 12:49:17 | <kuribas> | Not production outages, just a lot of work debugging code. |
| 12:49:27 | <dminuoso> | Like I said, I think if we were able to tag/add metadata to data types and constructors, I think that would be much more useful and easier to work with. |
| 12:49:30 | <kuribas> | In an adhoc way I mean. |
| 12:49:51 | <kuribas> | dminuoso: sure, but how is that different from a dependent type. |
| 12:49:59 | <kuribas> | ? |
| 12:50:22 | <dminuoso> | You dont need a dependent type system, which Haskell does not even have. |
| 12:50:40 | <kuribas> | It emulates it. |
| 12:50:48 | <kuribas> | Type level strings, lists, Nats... |
| 12:51:02 | <dminuoso> | And yet people rather pick up Idris when they want dependent types. :-) |
| 12:51:10 | <kuribas> | they should :) |
| 12:51:19 | <kuribas> | It's so much more ergonomic on idris. |
| 12:51:29 | <dminuoso> | Sure they should. This is not an argument about whether dependent types are good, but in Haskell they just overcomplicate interfaces. |
| 12:51:39 | <dminuoso> | To the point where I think its very hard to justify it. |
| 12:51:55 | <kuribas> | Well, one justification is the maturity of the compiler and the ecosystem. |
| 12:52:06 | <kuribas> | It's much less of a risk in a production environment. |
| 12:52:27 | <dminuoso> | All that perceived time saved in debugging, is wasted again troubleshooting illegible type errors, or fighting strange type mismatches because there's an extra type tag here and there. |
| 12:52:49 | <dminuoso> | In Haskell anyway. |
| 12:52:49 | <kuribas> | I am glad to take that trade-off. |
| 12:53:01 | <kuribas> | But not everyone does, I know... |
| 12:53:07 | <dminuoso> | You trade time spend in debugging.. for time spend in debugging. |
| 12:53:08 | <dminuoso> | :-) |
| 12:53:33 | <kuribas> | I am happier with a type error, that I can disentangle, that a big stacktrace, which may not even have the answer. |
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| 13:24:45 | <kuribas> | And in idris you have nice ways to program with types, like typed holes. |
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| 13:45:27 | <kuribas> | complex types don't necessarily mean a lot of type level debugging IMO. |
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| 13:45:41 | <kuribas> | But you don't want to have too much complexity in the types. |
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| 14:21:49 | <vektor> | What's the usual suspects for when equality fails, but the string repr of the two data points looks identical? I've checked that there should be no custom Show instance in play that hides information, and all aspects of the data should be displayable. |
| 14:21:49 | <vektor> | In my concrete case, a quickcheck "===" fails, but the counterexample is a more complex version of "a /= a". |
| 14:23:38 | <tomsmeding> | vektor: what types are involved? |
| 14:23:53 | <Lears> | vektor: If there's a Float/Double in there, it could be a NaN. |
| 14:25:39 | <vektor> | tomsmeding is "too many to count" a good answer? It's my DSL with lots of annotations, so it's a lot. Nothing crazy though. There's a optional field that contains a function here, which shouldn't be relevant in this case. |
| 14:26:13 | <vektor> | Lears the datatype can contain floats, but there's no floats in this example that's failing on me here. |
| 14:26:20 | <tomsmeding> | floating-point numbers are nr 1. candidate |
| 14:26:22 | <tomsmeding> | oh |
| 14:26:39 | <tomsmeding> | any Data.Semigroup(Arg)? :p |
| 14:27:03 | <tomsmeding> | are all Eq instances auto-generated? |
| 14:28:02 | <tomsmeding> | if the Show instances try to pretty-print stuff, perhaps you didn't properly print parentheses in all cases and there is some mismatch in association order? i.e. (a + b) + c /= a + (b + c) but maybe they pretty-print the same |
| 14:29:24 | <vektor> | Custom Eq is a good shout, let me look. |
| 14:29:25 | <vektor> | Data.Semigroup shouldn't be, but I'll check too. |
| 14:29:25 | <vektor> | Shows are all custom derived, except the one that contains a function in a field, but as said that constructor shouldn't even be in play. |
| 14:29:36 | <vektor> | * Shows are auto-derived except that one. |
| 14:30:04 | <tomsmeding> | (`Arg` has an Eq instance that ignores some fields) |
| 14:32:24 | <vektor> | Hmm, there's one custom Eq that looks very dodgy. |
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| 14:35:30 | <vektor> | I've removed that custom Eq for a derived one, any good advice on how to find where that Eq instance was used? I suspect there was a method to my previous madness in defining that custom Eq and I'd like to make sure. |
| 14:37:36 | <vektor> | Removing the Eq altogether and checking all the places that break I suppose, right? |
| 14:38:55 | <tomsmeding> | indeed :p |
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| 16:55:42 | <seydar> | i'm so sorry but i have a linear algebra question and i don't know who else to turn to |
| 16:56:19 | <seydar> | i have two 2d points, a and b, and i'm trying to translate a third point, c, by magnitude m in the direction orthogonal to <a, b> |
| 16:57:51 | <seydar> | is it just m * (a x b) * c ? am i just misusing my libraries? |
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| 16:58:43 | <dolio> | That doesn't seem right. |
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| 17:00:49 | <seydar> | turns out there's no cross product in R^2, which appears to be my first issue |
| 17:03:47 | <ncf> | there's a ##math channel |
| 17:04:19 | <ncf> | oh you're there already |
| 17:04:38 | <seydar> | i panicked when i didn't get an immediate response, i'm sorry |
| 17:04:44 | <seydar> | jumped here |
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| 17:34:56 | <darkling> | Direction orthogonal to AB is (b2-a2, a1-b1), IIRC. Normalise that, and multiply by m. Add to c. |
| 17:35:30 | <seydar> | why am i making this so hard |
| 17:35:31 | <seydar> | thank you |
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| 18:15:10 | <tomsmeding> | seydar: a way to remember darkling's formula is to recall that two vectors are orthogonal iff their inner product is zero |
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| 18:16:05 | <tomsmeding> | the vector a -> b is (b1-a1, b2-a2); what has inner product zero with that? Well, exchange the two components (so that both products are equal) then negate one of the components (so that they cancel) |
| 18:17:32 | <darkling> | I just worked out the first vector and multiplied by the matrix to rotate through 90° :) |
| 18:17:44 | <Lears> | You can also embed R^2 in R^3 and project a cross product with a normal vector back into the plane. |
| 18:18:03 | <tomsmeding> | incidentally, (a1, a2, 1) x (b1, b2, 1) = (a2-b2, b1-a1, _) |
| 18:18:08 | <Lears> | That said, there are /two/ orthogonal directions. I hope they guess the right one! |
| 18:18:25 | <tomsmeding> | (and indeed this yields the other, lol) |
| 18:19:19 | <tomsmeding> | darkling: but then you actually have to _remember_ stuff! |
| 18:19:31 | tomsmeding | would have to re-derive that matrix first |
| 18:20:00 | <tomsmeding> | which is admittedly not terribly difficult, but solving the dot product is about the same amount of work :p |
| 18:20:03 | <darkling> | That's one's simple enough, and I've done enough of them, I can remember it by now. :) |
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| 18:20:52 | <tomsmeding> | {{0,1},{-1,0}} |
| 18:20:57 | <tomsmeding> | right |
| 18:21:24 | <dolio> | Why (a1, a2, 1) etc.? |
| 18:21:33 | <tomsmeding> | that matrix is quite literally "swap the components and negate one of them" :p |
| 18:21:42 | <darkling> | (Or just visualise an angled vector, then its 90° rotation, and it's pretty obvious you swap the ordinates and negate one) |
| 18:22:36 | <tomsmeding> | dolio: visualise it in 3D with two points a and b that are a bit away from the origin, but fairly close together |
| 18:22:55 | <tomsmeding> | er, hm |
| 18:22:58 | <Lears> | Coming in late: "just multiply by i" |
| 18:23:07 | <darkling> | That works, too. |
| 18:23:17 | <tomsmeding> | dolio: I don't know, it works |
| 18:23:39 | <dolio> | I guess my question is: why 1? |
| 18:24:12 | <tomsmeding> | (because I recalled there was some trick kind of like this, and I forgot what the third component needed to be, put "1" there, it worked, and I posted) |
| 18:24:22 | <dolio> | Haha. Okay. |
| 18:24:34 | <tomsmeding> | it doesn't work with 0, in any case |
| 18:24:39 | <dolio> | Hmm. |
| 18:24:48 | <dolio> | Oh yeah, that'll be all z. |
| 18:24:52 | <tomsmeding> | yes |
| 18:26:07 | <dolio> | I guess you need to give them some z so that the result has some component in the x-y plane. |
| 18:26:26 | <dolio> | Probably doesn't matter how much. |
| 18:26:38 | <tomsmeding> | ah, that's correct |
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| 18:27:23 | <tomsmeding> | (a1,a2,x) x (b1,b2,x) = (a2*x - b2*x, x*b1 - x*a1, a1*b2 - a2*b1) |
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| 18:30:12 | <tomsmeding> | hm, not sure if that third component is useful in any way |
| 18:30:21 | <dolio> | Oh, and you need it to be the same z for both so that the component is orthogonal to the line between the two. |
| 18:30:35 | <tomsmeding> | it's the determinant of {{a1,a2},{b1,b2}}, but it's not the length of any of the relevant vectors |
| 18:31:02 | <tomsmeding> | dolio: I mean, that is correct because the algebra works out, but that's the part I don't have intuition for |
| 18:31:18 | <tomsmeding> | why does the projection into the plane end up orthogonal to the line segment you started with? |
| 18:33:15 | <tomsmeding> | my moving points around in space, mentally, and visualising what the normal of the (O, A, B) triangle does, I can see that it points in roughly the right direction to be orthogonal to AB, but I wouldn't be able to tell (without working out the algebra) whether that's exact |
| 18:33:29 | <tomsmeding> | *by moving points around |
| 18:33:31 | <dolio> | Because you're sort of rotating the plane of the x-y vectors to change the 3d orthogonal vector. |
| 18:33:50 | <tomsmeding> | I mean, yes |
| 18:34:27 | <dolio> | If you add the same z component to both, you're rotating it so that the projection bisects the angle between the original vectors. |
| 18:34:43 | <tomsmeding> | "the projection"? |
| 18:35:06 | <dolio> | Of the cross product. |
| 18:35:36 | <tomsmeding> | does it bisect anything? |
| 18:36:20 | <tomsmeding> | if I take A = (3, 3) and B = (7, 6), then the OAB triangle is slanted sideways quite a bit, and the cross product (the normal of that triangle) points away from the whole thing, to the side |
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| 18:37:30 | <dolio> | Hmm, oh yeah. Okay. I don't know why it works, then. |
| 18:37:39 | <tomsmeding> | perhaps B = (100, 99) works better :p |
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| 18:39:48 | <dolio> | I guess the answer is that however it rotates it is the correct way to rotate it, but I don't know exactly why. |
| 18:40:17 | <tomsmeding> | depending on who you ask, the "why" will be "look at the algebra, that's why" |
| 18:40:39 | <tomsmeding> | but usually there are deeper reasons for these things in linear algebra |
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| 18:40:55 | <tomsmeding> | seydar: look what you've done |
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| 18:41:30 | <dolio> | Yeah, it's not a 'geometric reason.' I'm not very good at geometric stuff, though. |
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| 18:42:30 | <dolio> | Obviously I was only visualizing the case where the vectors to A and B have equal length. |
| 18:42:39 | <tomsmeding> | I did the same thing :) |
| 18:42:43 | <tomsmeding> | then it's quite clear |
| 18:42:47 | <dolio> | Yeah. |
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| 18:42:56 | <seydar> | tomsmeding: this only further validates that i came to the right place |
| 18:42:59 | <tomsmeding> | then I realised my mistake and came the "er, hm" |
| 18:43:16 | <tomsmeding> | seydar: :D |
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| 19:00:25 | <Lears> | tomsmeding, dolio: Distributivity. (a1, a2, z) x (b1, b2, z) = ((a1, a2, 0) + (0, 0, z)) x ((b1, b2, 0) + (0, 0, z)) = ((a1, a2, 0) + (b1, b2, 0)) x (0, 0, z) + (0, 0, <junk>) |
| 19:00:41 | <Lears> | The projection should be orthogonal to a+b, not b-a? |
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| 19:01:46 | <dolio> | tomsmeding: Maybe this is a better thing to consider. Those vectors are what you get by going between your actual points and the point (0,0,-z). And that plane actually contains the line you care about. |
| 19:05:19 | <tomsmeding> | Lears: well, it's orthogonal to b-a. :P |
| 19:06:37 | <tomsmeding> | dolio: which "those vectors"? (a1,a2,z) and (b1,b2,z)? |
| 19:06:48 | <dolio> | Yeah. |
| 19:07:17 | <tomsmeding> | and "the plane" is the one generated by {(a1,a2,0), (b1,b2,0), (0,0,-z)}? |
| 19:07:24 | <dolio> | Yes. |
| 19:07:45 | <tomsmeding> | that's just the same plane as we've been looking at before ("the triangle OAB"), only shifted down by z :p |
| 19:07:55 | <tomsmeding> | the cross product is going to be _normal_ to that plane |
| 19:08:38 | <tomsmeding> | so I'm not sure about your "contains the line you care about" |
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| 19:09:08 | <dolio> | It's normal to a plane containing the line between the a and b points. |
| 19:09:25 | <tomsmeding> | yes ... oh |
| 19:09:30 | <tomsmeding> | well it always was |
| 19:09:40 | <tomsmeding> | okay fair |
| 19:10:09 | <tomsmeding> | the cross product is normal to the plane spun up by the triangle OAB, hence it is normal to all vectors in that plane, in particular B-A |
| 19:10:12 | <tomsmeding> | \qed? |
| 19:10:44 | <dolio> | Well, you need some fact about projections back to the x-y plane. |
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| 19:11:06 | <tomsmeding> | when I did a bachelor in maths I had some fun with a fellow student and wrote a macro which, translated to English, would I guess be \maybeBoxy |
| 19:11:11 | <tomsmeding> | it was a \qed with a ? inside |
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| 19:11:35 | <tomsmeding> | right |
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| 19:12:53 | <tomsmeding> | you can do that algebraically (the cross product is orthogonal to B-A, but the z-component of B-A is zero, hence it's orthogonal regardless of the z component of the cross product, hence also at zero, hence also when projected to the x-y plane) |
| 19:13:00 | <tomsmeding> | but that's cheating again |
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| 19:13:30 | <dolio> | Yeah, I'm not sure how to see geometrically that the projection is still orthogonal. |
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| 19:14:28 | <tomsmeding> | it is relevant that B-A lies in the x-y plane, because without that restriction, it's easy to think of two orthogonal vectors in R^3 that are not orthogonal after projection to the x-y plane |
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| 19:15:34 | <tomsmeding> | ah: B-A lies in the x-y plane, so rotate and scale the whole system so that B-A is (1,0); then the (rotated) cross product is still orthogonal, so it's in the y-z plane |
| 19:16:06 | <tomsmeding> | it's not parallel to the z-axis because that would mean that OAB is parallel to the x-y plane, which it isn't, so it has a nonzero y component |
| 19:16:23 | <tomsmeding> | and angles in the projection are preserved by rotations around the z-axis and scalings |
| 19:16:36 | <tomsmeding> | that feels like an argument |
| 19:17:02 | <tomsmeding> | convoluted, though |
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| 19:18:49 | <tomsmeding> | "so that B-A is (1,0)" -> "so that B-A is (1,0,0)" |
| 19:19:13 | <tomsmeding> | I guess the scaling is superfluous, but it helps intuition |
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| 23:41:10 | <Lears> | Oh. I should have figured this out a lot sooner, but it looks like I forgot to /anti/-commute one of cross products (z k x (b1 i + b2 j))---that's what I get for doing it in my head. Distributivity indeed gives us a projection orthogonal to a-b. |
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All times are in UTC on 2024-10-09.