The Big Programming Thread - Page 879

On May 04 2017 06:34 Shield wrote:
OOP question.

Is there a case when it's normal for base class to return an instance of its derived classes? Kind of like factory - you pass a parameter and you get an instance of a specific derived class. I find it a bit weird, but I'm wondering if it's just me.

On May 04 2017 06:34 Shield wrote:
OOP question.

Is there a case when it's normal for base class to return an instance of its derived classes? Kind of like factory - you pass a parameter and you get an instance of a specific derived class. I find it a bit weird, but I'm wondering if it's just me.

On May 04 2017 06:24 travis wrote:
I am a bit confused about a couple questions I am being asked about countability

the first one asks about the (empty set )* N
the second one asks about (empty set) * R

It asks if these (separate question for each) are:

1.) empty
2.) finite (and thus countable)
3.) infinite and countable
4.) uncountable

I feel like they are both 1 and 2
but we are only allowed to pick one answer
Where is my thinking wrong?


edit: my thinking was right, got a quick reply from professor, their wording was incorrect

On May 04 2017 19:39 Acrofales wrote:

Definitely not normal. Possible? Only if you know what subclasses exist. It just seems like a bizar way of doing it. It doesn't sound like coherent functionality for the superclass, and it sounds like something that should either be in each subclass, or in a factory.

On May 05 2017 03:49 spinesheath wrote:

Doesn't seem weird to me, for certain setups. Obviously it wouldn't return subclasses that it shouldn't know about, that would be very strange indeed. But I find it perfectly reasonable to have a public base class that represents some sort of object used in calculations and a couple of internal subclasses that implement different behavior based on the contents of this object. The class hierarchy is sealed off by not providing any public constructors.

Like say some sort of currency class that also transparently deals with undefined values (null object pattern). You wouldn't want to have a CurrencyFactory for that, you just put a factory method in the currency class.

On May 05 2017 07:23 travis wrote:
*beckons at acrofales* (or anyone adventurous)

I have some hard questions about cardinality

"let D be the set of all functions with domain N and co-domain Q"

"Does D have the same cardinality as R" ?

I don't even know where to begin with this question


Another question: If F is a finite set, do Q and Q-F have the same cardinality? (I think yes for this one, it would seem to me that it is still bijective.

Final one: let B be the set of numbers of the form ax + b√2 where are a and b are in Q.
Do B and Z have the same cardinality? I think this one is also yes, I think they are trying to get fancy with the √2, but it's just a linear function.

First, I assume that you are using the following notation.
N = natural numbers, Q = rational numbers, Z = integers, R = real numbers.
What does Q-F mean? The set of functions from Q to F or Q minus F?

Also, I'll use
P(S) = the power set of S, i.e. the set of all subsets of S,
|S| = the cardinality of the set S,
S_1 -> S_2 = the set of all functions from S_1 to S_2
AxB = cartesian product of A and B, i.e. the set of all pairs (a,b)
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Next, let us recall some facts I assume you know.
1) for each set S, |P(S)| > |S|
2) for each set S, |P(S)| = cardinality of the set of all functions from S to {0,1},
i.e. in our notation |P(S)| = |S -> {0,1}|
3) |N| = |Z| = |Q| = countable infinity
4) |P(N)| = (|P(Z)| = |P(Q)|) = |R| = continuum infinity

Recall also that one can consider a function f:A->B as a set of pairs (x,f(x)),
but that means that each f is an element of AxB, which in turn implies that
for the set of all function A->B it is true that
5) A->B is a subset of P(AxB), and in particular
6) |A->B| <= |P(AxB)|

Also I assume you know that |NxN| = |N|, etc.

Now we are ready.
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1. question. Yes, D has the same cardinality as the set R.
Proof.
|D| = |N->Q| >= |N->{0,1}| = |P(N)| = continuum = |R|
Therefore |D| >= |R|. On the other hand,
|D| = |N->Q| <= |P(NxQ)| = |P(NxN)| = |P(N)| = continuum = |R|,
and hence |D| <= |R|.
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2. question. I don't know what you mean by Q-F
If it means Q minus F, then yes, they have the same cardinality.
If it means the set of functions Q->F, then it has continuum cardinality,
so it is not the same as Q.
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3. question.
Yes, they have since |Z| = |Q| <= |B| <= |QxQ| = |NxN| = |N| = |Z|.
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