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Comments:
<0> what makes a linear pair different then supplementar angles?
<1> Pair of what ?
<2> of angles, I imagine
<3> MrPockets: a linear pair by definition share a side, suppl angles don't have to
<1> What is a linear pair of angles ?
<1> I found out.
<4> 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
<5> i hope the answer is "magic"
<4> HiLander: my answer?
<1> The answer is magic.
<1> The question you asked.
<5> the humor has been beaten out of the joke.
<6> what is strict weak ordering?
<7> Does "a *-autonomous category" mean * in the Linear Algebra sense?
<1> monochrom : All the elements must be comparable with all the others.
<1> http://www.sgi.com/tech/stl/StrictWeakOrdering.html
<7> Really, anybody familiar with the concept of "*-autonomous categories" ? I can't infer from the context.
<6> I don't think binary search trees etc. have really been defined or studied using partial orders.
<4> Olathe: that isn't true.
<1> What isn't ?
<8> monochrom: a strict weak ordering is C++'
<8> monochrom: a strict weak ordering is C++'s name for a total ordering on equivalence cl***es.
<4> Olathe: the part about all objs have to be comparable.
<1> But they do, since f(x, y) and f(y, x) gives their relative positions.
<1> Because transitivity is there, you can't have loops.
<1> So, what those two say is correct.
<4> evilgeek: i thot strict weak ordering was http://en.wikipedia.org/wiki/Strict_weak_ordering
<6> The SGI page and the wikipedia page agree with each other.
<8> 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.
<6> And evilgeek's statement is an informal way of saying both, if you are an insider.
<4> really. strict weak ordering seems closer to partial ordering than total ordering.
<4> i don't understand the part about comparability
<1> Show a nontotal, partial order of some elements that fits.
<8> Olathe: the integers where a < b is false for all a and b.
<1> Hmmm...a better phrasing.
<1> Show a nontotal, partial order of some equivalence cl***es that fits.
<5> 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?
<4> HiLander: correct.
<8> HiLander: yes.
<5> ok. then this definition makes sense to me, regardless as to whatever it's used for
<6> No, I don't think we intend to put incomparable elements into the same cl***.
<5> no, i'm just saying that elements in different cl***es have an order induced on them by their cl***es
<8> monochrom: we do. they are incomparable, and incomparability is an equivalence relation.
<5> 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
<5> i'm just saying that if they're incomparable, they have to be in the same cl***
<7> evilgeek: "being different" is not an equivalence relation, how is "incomparable" one?
<8> a and b are incomparable if a<b and b<a are both false.
<7> That only leaves a == b if we are subjected to trichotomy.
<1> Yay trichotomy !
<6> 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.)
<9> ok, what's this channel all about?
<5> guns
<5> that's why it's #guns
<4> lmao
<3> sometimes we talk about llamas too
<3> but, only llamas with guns
<1> Llamas rock.
<9> oh kay... I've fallen into the twilight zone.
<6> 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?
<9> llamas, guns, and rock-and-roll.
<6> Perhaps even better, you are to store a subset of them in a binary search tree.
<3> no, No. not rock and roll. just guns and llamas
<9> sorry, got carried away. wow, actual math! excellent.
<6> Now someone hands you such a binary search tree and ask, "is b in the tree"? What do you do?
<4> monochrom: no clue what you mean.
<1> Wouldn't it be best to store equivalence cl***es in the BST ?
<6> no clue how to make you know what I mean.
<1> So, you'd just have one element, then you'd search that.
<6> 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.
<7> monochrom - What does the datatype of the Tree you are thinking of look like in Haskell?
<6> 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.
<4> monochrom: thanks much clearer.
<6> The former, stronger question corresponds to defining a,b,c,d to be "incomparable and non-equivalent". The latter, "incomparable and equivalent".
<6> Olathe's question begs the question: how do you store a subset of an equivalence cl***?
<1> You can have a reference to some set.
<1> So, you have a tree of elements that are sets.
<6> How do you store a set?
<6> As a list? As a binary search tree? As a hash table?
<1> You can use pointers in C, for instance.
<1> Any way you want.
<4> a specialization of a bst
<1> A hash table would probably be best, if the equivalence cl***es would be large.
<1> If you're just trying to determine whether you have an element, a tree probably isn't the best option.
<6> Sigh. It's very hard to explain things on freenode #math.
<2> heh
<5> ha
<10> on the vertical line test to see if a relationship is a function, does this mean that a circle is not a function?
<11> If your question is "is a circle the graph of a function f:R->R", then the answer is no
<10> thanks
<12> To each value in the domain of the function, there can be no more that one "related" point in the image ... intuitively.
<10> thanks
<9> the definition of a function is that each unique value of x produces one and only one y.
<9> 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.
<9> if it p***es through 2 points, it fails the test and is therefore not a function.
<10> I see the "concepts of domain and range apply to functions since all functions are relations but not all relations are functions"
<10> schuams Inrtermediate algebra
<13> 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.
<13> If someone knew of a site or something, that could help me a lot.
<14> http://wikipedia.org/wiki/Laplace_expansion
<15> http://www.maths.surrey.ac.uk/interactivemaths/emmaspages/option1.html
<1> Do Gauss-Jordan elimination to get a triangular matrix, then multiply the entries in the diagonal.
<13> The Gu***-Jordan isn't Programming Friendly(at least for me,altho exiting.), Thanks people.
<16> how about row reduction? each row operation makes a simple transform to the determinant
<17> xerox: The categoryt of vector sapces with *-autonomous with V* being the dual space as normal
<16> _llll_, was that the answer to a questtion someone asked several hours ago?
<17> vageuly
<16> lol
<17> the wonders of the awaylog and /hilight
<7> _llll_: what does it mean to be *-autonomous?
<7> Is * something like the unit... something?
<18> dual
<7> I'm being very precise... :)
<18> yep
<17> * is a dual
<7> Sorry.
<7> OK, good.
<18> wow i got something right
<17> *:C-->C is a functor
<7> And what does it feel like to be dual-autonomous?
<18> gimme a cookie
<18> i dont want your pat
<16> sorry that answer did'n;t merit a cookie
<18> neither does your spelling :)
<16> everyone knows that * designates dual
<18> [19:11] <7> Is * something like the unit... something?
<19> i'm not "everyone"
<18> everyone - 1
<17> hmm, i dont know too much about these *-thingies, i used to know something but i cant remember it
<16> the dual of a space is the set of (linear) maps on that space
<17> it;'s kind of like flipping things upside down in some sense i know longer remember
<16> canonical example is inversion through a circle in the plane
<7> _llll_ - Thank you for the informations.
<7> Thanks GuestBlah too :)
<16> which maps points to lines and lines to points
<18> hahahaha you're welcome
<16> wait that example contradicted that definition
<16> but they are both duals..
<17> http://www.math.ucr.edu/home/baez/week227.html
<7> mnvl: do you have that explained somewhere?
<16> in my head
<16> vaguely
<17> there is some algebraic cl***ification theorem, maybe some kindof morita equivalence result, but i cant remember it
<20> 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
<11> all quadratic equations are factorizable
<3> er, yeah.
<3> if you can't factor them, you can't solve with quadratic, why do you ask?
<21> Unless you're talking difficulty, as in devise something where you can't work it out in your head.
<20> Whiteknight: I suppose
<11> that's relative
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