Interview, what to expect?

[chrisname]

helios wrote:

Are you saying that there is some x such that ¬((A && B) => C)?

No. I proved C by contradiction by assuming A and B. If x is a manhole cover and manhole covers are round, x must be round. Now I realise that there was no need to prove it by contradiction, because the way I just said is simpler.

Boy, lately you've taken to use logic to say useless things, haven't you?

No, I was posting a joke response to an interview question (since in essence I'd be saying "manhole covers are round because manhole covers are round"). I don't know what you mean by "lately" anyway, since that implies that I've been doing it often.

[ResidentBiscuit]

To prove a valid argument, dont you negate the conclusion and look for an instance where that holds true? If there isn't one, then it's valid sice you never get taken from true premises to a false conclusion.

[Catfish3]

Ugh.

[helios]

1. If A and B and not C, contradiction.
2. A and B.
3. Therefore C.

This is not a proof by contradiction. Statement 1 is equivalent to ¬(A && B && ¬C).
¬(A && B && ¬C)
¬(A && B) || C
A && B => C (definition of =>)
Thus the argument is just modus ponens.

A proof by contradiction would involve assuming that there exists a manhole cover which is not round and then arriving at a contradiction:
Suppose there exists a manhole cover which is not round and call it X. Since all manhole covers are round and X is not round, X is not a manhole cover. But by definition, X is a manhole cover, therefore X does not exist.
There's little point in doing this because you're just one step away from the definition of manhole cover to begin with.

I don't know what you mean by "lately" anyway, since that implies that I've been doing it often.

http://cplusplus.com/forum/lounge/87798/

[chrisname]

That was one time, and I was clearly trolling (or maybe I'm a better troll than I thought). It was a joke on something Turing said about a lookup table that contained the answers to every possible question. I think it was mentioned in AI - A Modern Approach. It's akin to if I had posted a program that sends the questions to a person and then sends the answers back to the operator, and called it a "human-aided Turing test passer".

[helios]

That was one time

And this is another. Therefore, lately.

and I was clearly trolling

Obviously, but you still said it.

[xerzi]

The only way to ensure a 'square' cover can't fall into a hole is to make it larger than the hole.

And the same doesn't apply to a circle ? Pretty sure if the cover is smaller then the hole regardless of the shape it will still fall through. In either case it probably has a groove that the cover sits on.

[iseeplusplus]

The side of a square is shorter than its diagonal. A square cover, over a square hole, would need to have sides longer than the diagonal of the hole.

[Script Coder]

idk what is going on here, but my answer:
"circles cover the most space per unit perimeter"
probably an incorrect one, but nonetheless my answer.

[cire]

cire wrote:

The only way to ensure a 'square' cover can't fall into a hole is to make it larger than the hole.

xerzi wrote:

And the same doesn't apply to a circle?

Deliberate obtuseness is not a virtue.

If you would prefer an answer that didn't assume one knew the basic way a manhole cover works:

The only way to ensure a 'square' cover can't fall into a same-shaped hole is to make it considerably larger than the hole. The same isn't true of a round cover and a round hole, which require only a minute increase in size to eliminate the chance of the cover falling into the hole.

[xerzi]

@cire

I'm being obtuse ?

Read this:

The only way to ensure a 'square' cover can't fall into a same-shaped hole is to make it considerably larger than the hole.

I can think of a few ways to make a square cover not fall through, you can round out the corners of the hole and the cover won't fall through. How's that making it considerably larger than the hole ?

[ne555]

The square cover case:

(rotations don't translate the center of mass)
_ rotate \pi/2 over the x axis
_ rotate \pi/4 over the z axis
_ drop the cap and kill the man in the hole

In order for the cover to not fall, its side should be bigger than the diagonal of the hole.

[helios]

you can round out the corners of the hole and the cover won't fall through.

This will only make a difference once the cover is almost exactly a circle, depending on the size of the groove on the edge. At lower levels of roundedness, the "diagonals" (maximum distance between opposing arcs) are still longer than the "sides" (minimum distance between opposing straight segments), so it's still possible for the cover to fit into the hole.

[cire]

I can think of a few ways to make a square cover not fall through, you can round out the corners of the hole and the cover won't fall through.

If a given size square cover is capable of falling through a square hole, rounding the corners of the hole far enough to prevent the square cover from falling through only increases the ratio of cover-to-hole surface area, meaning that the relative size of the cover increases.

How's that making it considerably larger than the hole ?

How is it not?

[Edit: And I'm completely sidestepping the issue of same-shapedness you violate with your "solution."]

[devonrevenge]

how did it go? did they say anything about manhole covers??

[rapidcoder]

We do have round, oval and rectangular manholes here in Poland. I haven't seen triangular, though. The answer to that question is probably: tradition. Traditionally the telecom covers are rectangular, water/gas system covers are oval, while central heating covers are round. :D