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Comments:
<MrPockets> what makes a linear pair different then supplementar angles?
<Olathe> Pair of what ?
<cyby> of angles, I imagine
<Flannel> MrPockets: a linear pair by definition share a side, suppl angles don't have to
<Olathe> What is a linear pair of angles ?
<Olathe> I found out.
<roods> can anyone tell me why ***ocitive constainers in C++ use strict weak ordering instead of partial ordering. eg: used in the implementation of a red-black tree
<HiLander> i hope the answer is "magic"
<roods> HiLander: my answer?
<Olathe> The answer is magic.
<Olathe> The question you asked.
<HiLander> the humor has been beaten out of the joke.
<monochrom> what is strict weak ordering?
<xerox> Does "a *-autonomous category" mean * in the Linear Algebra sense?
<Olathe> monochrom : All the elements must be comparable with all the others.
<Olathe> http://www.sgi.com/tech/stl/StrictWeakOrdering.html
<xerox> Really, anybody familiar with the concept of "*-autonomous categories" ? I can't infer from the context.
<monochrom> I don't think binary search trees etc. have really been defined or studied using partial orders.
<roods> Olathe: that isn't true.
<Olathe> What isn't ?
<evilgeek> monochrom: a strict weak ordering is C++'
<evilgeek> monochrom: a strict weak ordering is C++'s name for a total ordering on equivalence cl***es.
<roods> Olathe: the part about all objs have to be comparable.
<Olathe> But they do, since f(x, y) and f(y, x) gives their relative positions.
<Olathe> Because transitivity is there, you can't have loops.
<Olathe> So, what those two say is correct.
<roods> evilgeek: i thot strict weak ordering was http://en.wikipedia.org/wiki/Strict_weak_ordering
<monochrom> The SGI page and the wikipedia page agree with each other.
<evilgeek> roods: there is no point in thinking about them as anything other than total orderings on equivalence cl***es. there isn't even any point in thinking of them as anything other than just total orderings.
<monochrom> And evilgeek's statement is an informal way of saying both, if you are an insider.
<roods> really. strict weak ordering seems closer to partial ordering than total ordering.
<roods> i don't understand the part about comparability
<Olathe> Show a nontotal, partial order of some elements that fits.
<evilgeek> Olathe: the integers where a < b is false for all a and b.
<Olathe> Hmmm...a better phrasing.
<Olathe> Show a nontotal, partial order of some equivalence cl***es that fits.
<HiLander> a total order on equivalence cl***es would imply a partial order on elements, right? the only possibly incomparable elements would lie in the same cl***, right?
<roods> HiLander: correct.
<evilgeek> HiLander: yes.
<HiLander> ok. then this definition makes sense to me, regardless as to whatever it's used for
<monochrom> No, I don't think we intend to put incomparable elements into the same cl***.
<HiLander> no, i'm just saying that elements in different cl***es have an order induced on them by their cl***es
<evilgeek> monochrom: we do. they are incomparable, and incomparability is an equivalence relation.
<HiLander> whether or not elements in the same equivalence cl*** (part of the partition) are comparable is not part of my argument, even if it is of the idea
<HiLander> i'm just saying that if they're incomparable, they have to be in the same cl***
<xerox> evilgeek: "being different" is not an equivalence relation, how is "incomparable" one?
<evilgeek> a and b are incomparable if a<b and b<a are both false.
<xerox> That only leaves a == b if we are subjected to trichotomy.
<Olathe> Yay trichotomy !
<monochrom> Ok, I now have a story that is based on the math and, where math leaves us freedom, on what we use it for. (This doesn't mean I fix what we use it for; rather, it means once you decide what you use it for, you see the consequence.)
<BitWitty> ok, what's this channel all about?
<HiLander> guns
<HiLander> that's why it's #guns
<roods> lmao
<Flannel> sometimes we talk about llamas too
<Flannel> but, only llamas with guns
<Olathe> Llamas rock.
<BitWitty> oh kay... I've fallen into the twilight zone.
<monochrom> Suppose there are four distinct elements a,b,c,d. They satisfy not(x<y) forall x,y. You are to store them in a binary search tree. What do the valid binary search trees look like?
<BitWitty> llamas, guns, and rock-and-roll.
<monochrom> Perhaps even better, you are to store a subset of them in a binary search tree.
<Flannel> no, No. not rock and roll. just guns and llamas
<BitWitty> sorry, got carried away. wow, actual math! excellent.
<monochrom> Now someone hands you such a binary search tree and ask, "is b in the tree"? What do you do?
<roods> monochrom: no clue what you mean.
<Olathe> Wouldn't it be best to store equivalence cl***es in the BST ?
<monochrom> no clue how to make you know what I mean.
<Olathe> So, you'd just have one element, then you'd search that.
<monochrom> Since the elements satisfy not(x<y) forall x,y, the tree you receive is really in arbitrary shape. If you insist on looking for b and nothing else, you degenerate to exhaustive search.
<xerox> monochrom - What does the datatype of the Tree you are thinking of look like in Haskell?
<monochrom> However, if the question is weakened to: looking for b, but a,c,d are also ok, basically any x satisfying not(x<b) and not(b<x) is ok, then you can still use the usual logarithmic search algorithm. You can even just return the element at the root, whichever it is.
<roods> monochrom: thanks much clearer.
<monochrom> The former, stronger question corresponds to defining a,b,c,d to be "incomparable and non-equivalent". The latter, "incomparable and equivalent".
<monochrom> Olathe's question begs the question: how do you store a subset of an equivalence cl***?
<Olathe> You can have a reference to some set.
<Olathe> So, you have a tree of elements that are sets.
<monochrom> How do you store a set?
<monochrom> As a list? As a binary search tree? As a hash table?
<Olathe> You can use pointers in C, for instance.
<Olathe> Any way you want.
<roods> a specialization of a bst
<Olathe> A hash table would probably be best, if the equivalence cl***es would be large.
<Olathe> If you're just trying to determine whether you have an element, a tree probably isn't the best option.
<monochrom> Sigh. It's very hard to explain things on freenode #math.
<cyby> heh
<HiLander> ha
<toko123> on the vertical line test to see if a relationship is a function, does this mean that a circle is not a function?
<Sturgeon> If your question is "is a circle the graph of a function f:R->R", then the answer is no
<toko123> thanks
<cathper> To each value in the domain of the function, there can be no more that one "related" point in the image ... intuitively.
<toko123> thanks
<BitWitty> the definition of a function is that each unique value of x produces one and only one y.
<BitWitty> the vertical line test is merely testing to see if for each value of x the line p***es through, there is only one y ***ociated with it.
<BitWitty> if it p***es through 2 points, it fails the test and is therefore not a function.
<toko123> I see the "concepts of domain and range apply to functions since all functions are relations but not all relations are functions"
<toko123> schuams Inrtermediate algebra
<Afat> Hello, I know how to get the determinant of a matrix 3x3, but I don't know how to get the determinant from a 4x4 matrix.
<Afat> If someone knew of a site or something, that could help me a lot.
<Catfive> http://wikipedia.org/wiki/Laplace_expansion
<asasd> http://www.maths.surrey.ac.uk/interactivemaths/emmaspages/option1.html
<Olathe> Do Gauss-Jordan elimination to get a triangular matrix, then multiply the entries in the diagonal.
<Afat> The Gu***-Jordan isn't Programming Friendly(at least for me,altho exiting.), Thanks people.
<mnvl> how about row reduction? each row operation makes a simple transform to the determinant
<_llll_> xerox: The categoryt of vector sapces with *-autonomous with V* being the dual space as normal
<mnvl> _llll_, was that the answer to a questtion someone asked several hours ago?
<_llll_> vageuly
<mnvl> lol
<_llll_> the wonders of the awaylog and /hilight
<xerox> _llll_: what does it mean to be *-autonomous?
<xerox> Is * something like the unit... something?
<GuestBlah> dual
<xerox> I'm being very precise... :)
<GuestBlah> yep
<_llll_> * is a dual
<xerox> Sorry.
<xerox> OK, good.
<GuestBlah> wow i got something right
<_llll_> *:C-->C is a functor
<xerox> And what does it feel like to be dual-autonomous?
<GuestBlah> gimme a cookie
<GuestBlah> i dont want your pat
<mnvl> sorry that answer did'n;t merit a cookie
<GuestBlah> neither does your spelling :)
<mnvl> everyone knows that * designates dual
<GuestBlah> [19:11] <xerox> Is * something like the unit... something?
<Kahdloc> i'm not "everyone"
<GuestBlah> everyone - 1
<_llll_> hmm, i dont know too much about these *-thingies, i used to know something but i cant remember it
<mnvl> the dual of a space is the set of (linear) maps on that space
<_llll_> it;'s kind of like flipping things upside down in some sense i know longer remember
<mnvl> canonical example is inversion through a circle in the plane
<xerox> _llll_ - Thank you for the informations.
<xerox> Thanks GuestBlah too :)
<mnvl> which maps points to lines and lines to points
<GuestBlah> hahahaha you're welcome
<mnvl> wait that example contradicted that definition
<mnvl> but they are both duals..
<_llll_> http://www.math.ucr.edu/home/baez/week227.html
<xerox> mnvl: do you have that explained somewhere?
<mnvl> in my head
<mnvl> vaguely
<_llll_> there is some algebraic cl***ification theorem, maybe some kindof morita equivalence result, but i cant remember it
<quell> is there a methodical way of constructing a quadratic equation that can only be solved by using the quadratic formula, i.e., it is not factorable
<Sturgeon> all quadratic equations are factorizable
<Flannel> er, yeah.
<Flannel> if you can't factor them, you can't solve with quadratic, why do you ask?
<Whiteknight> Unless you're talking difficulty, as in devise something where you can't work it out in your head.
<quell> Whiteknight: I suppose
<Sturgeon> that's relative
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