The Big Programming Thread - Page 880

On May 08 2017 14:39 Manit0u wrote:
Was reading a bit on the Ruby object model and it really cracked me up:

the superclass of an eigenclass of a class is the eigenclass of the class’s superclass

Some crazy shit right there.

> class << Strin­g
.. puts self.­superclass­.__id__
.. end
=> "11488"
> class << Strin­g.supercla­ss
.. puts self.­__id__
.. end
=> "1184611846"

On May 09 2017 05:39 AKnopf wrote:


Haha, gotta love that! But where did you read that?

The superclass of any eigenclass is always class (that's why it's a class in the first place). And the eigenclass of anything can never be class*. But that's what any classes superclass is. So that definitely is wrong. Funny sentence though :D

* Also there is no eigenclass that is shared with anything (it is "eigen" to its object). But your sentence would mean that the class "C" and its superclass "S" share the same eigenclass "e". This cannot be true, though.

I guess the sentence could be true if by "is" the author means "I *is* an instance if class C or any of its subclasses". But even that would be a stretch since then an eigenclass would have to be the superclass to its object, which it is technically not.




Sorry if I'm wrong here. Did I miss anything? Also sorry for being triggered. Ruby's object model is just too much fun :D




Edit: Just to be sure, I just tried it on tryruby.org

> class << Strin­g
.. puts self.­superclass­.__id__
.. end
=> "11488"
> class << Strin­g.supercla­ss
.. puts self.­__id__
.. end
=> "1184611846"

Not the same

class Object; def eigenclass; class << self; self; end; end; end

E.eigenclass
# => #<Class:E>
E.eigenclass.superclass
# => #<Class:C>
C.eigenclass.superclass
# => #<Class:Object>
Object.eigenclass.superclass
# => #<Class:BasicObject>
BasicObject.eigenclass.superclass
# => Class
Class.superclass
# => Module
Module.superclass
# => Object

On May 09 2017 18:25 Manit0u wrote:


https://www.sitepoint.com/understanding-object-model/

class Object; def eigenclass; class << self; self; end; end; end

E.eigenclass
# => #<Class:E>
E.eigenclass.superclass
# => #<Class:C>
C.eigenclass.superclass
# => #<Class:Object>
Object.eigenclass.superclass
# => #<Class:BasicObject>
BasicObject.eigenclass.superclass
# => Class
Class.superclass
# => Module
Module.superclass
# => Object

Edit:

Also, super useful thingie:
+ Show Spoiler +

On May 13 2017 10:22 travis wrote:
doing a practice final for linear algebra. True or false

If T is a linear transformation from R^n to R^n which is one-to-one, then for
every y in R^n there exists x such that T(x) = y.

I said false, the solution page says this is true.
but isn't this the definition of onto, not one-to-one ?

On May 13 2017 12:26 ZigguratOfUr wrote:


Yeah, the solution is wrong.

On May 13 2017 18:41 CoughingHydra wrote:

Actually, no! The solution is right - if you have a linear transformation from R^n to R^n (or more generally from a linear space V to a linear space W with dim V = dim W < infinity), then being one to one is equivalent to being onto. This follows from the Rank-nullity theorem. This is very similar to having a mapping from an n-element set to an n-element set - it is onto if and only if it is one to one.

On May 13 2017 19:56 Hanh wrote:
CoughingHydra is correct.

On May 13 2017 19:06 Acrofales wrote:

This doesn't sound right, because they're not finite sets.

My counterexample was wrong for R^n. It'd hold for Z^n. Need to come up with one that works with real numbers.

On May 13 2017 20:20 Prillan wrote:

Yup, a map being linear is a pretty strong property.

It's a nicely put question since it tests students on: injectivity (one-to-one), surjectivity (onto), rank, nullity and the rank-nullity theorem.


EDIT:


Z is not a field, so you wouldn't even have a vector space to begin with.