Idea?

[Valeanx Madpath]

Instead of asking for a "destroy all on corpse" button, sure to lead to TRAGEDY at some point, ask for a "garbage can" tradeskill container that eats incorrect (and all combines are incorrect in a "garbage can" of course) combines. Tada. (Gosh, that was hard to code.)

I use collapsed sewing kits. Summon a sewing kit, fill it with crap, destroy it, summon another. I know people who also use the no rent trophy bags for this.

[VxM]

[Maevenniia]

I use collapsed sewing kits. Summon a sewing kit, fill it with crap, destroy it, summon another. I know people who also use the no rent trophy bags for this.

[VxM]

I do exactly that, it's a nice convenient little trick =)

[Chenier]

I do exactly that, it's a nice convenient little trick =)

Me too.

I also, when either farming or when I was still skilling up (especially pottery stuff that you just going to throw away and not fire), turn off the drop confirmation box. Makes getting rid of crap really fast.

I'm too much of a compulsive bag organizer, so I'll loot up stacks and then toss them; in a crowd of mobs, I'll loot everything and then drop/destroy.

I was more annoyed when farming in Hollowshade - lore, no drop crap SUCKS.

[Elyssanda]

my best friend when farming?
/hidecorpses looted

then I don;'t have to bother with the junk I don't want.

[Itek]

**Itek smacks himself in the forehead**

Sorry guys, I'd been up for 3 hours (2:30am to 5:30am) studying "Discrete math" for one of my two finals Monday when I wrote that. Forgot about trophy/collapseable bags.

And I've never turned "hide corpses" on and very rarely AE stuff except in PoHate and SSra basement where there usually isn't a problem with "too much junk" on piles of corpses.

Here's a question from my math final which will explain why I was slipping gears...

"Use mathmatical induction to prove that [{2^(2*N)}-1]/3 results in an integer for all positive integers N"

Yeah, and I blew that one on the final. I forgot, somehow, that ((4*Q)-1) is equal to 3*Q + Q-1 where Q-1 is proven to be divisible by 3. Hope I got partial credit for proving Q-1/3 when N, and ((4*Q)-1) when N+1

[Maevenniia]

<snip>
Here's a question from my math final which will explain why I was slipping gears...

"Use mathmatical induction to prove that [{2^(2*N)}-1]/3 results in an integer for all positive integers N"

Yeah, and I blew that one on the final. I forgot, somehow, that ((4*Q)-1) is equal to 3*Q + Q-1 where Q-1 is proven to be divisible by 3. Hope I got partial credit for proving Q-1/3 when N, and ((4*Q)-1) when N+1

*head explodes from math this early in the morning*

[Dunthor Warsmith]

I might be missing something, apologies if it's been removed and I'm behind the times... but up 'til about 2 weeks ago "/hidecorpse looted" did exactly that. There have been a couple patches since my last farming run... I really hope this isn't gone...

Oooh, that should be quite handy! I only remembered {group,all,none} options ... "looted" sounds like exactly what I want. Thanks

"Use mathmatical induction to prove that [{2^(2*N)}-1]/3 results in an integer for all positive integers N"

Induction proofs always feel like they're cheating

One thing I've been wondering for a while, is it more formal to prove N given N-1, or prove N+1 given N? Teachers, professors, and texts tend to strongly prefer one over the other, but vary as to which it is. Just wondering if that ever came up.

[mewkus]

Ya know, if you tell people you're taking "Discrete math", you're missing the point of it. You need to be a little more discrete about it.
Hehe, an old joke from when I took it. Err, wait a minute, I didn't take discrete math. No, really, I didn't.

[Eleena Transient]

[{2^(2*N)}-1]/3

N = 0: 2^(2*1)-1 / 3
(2^2 - 1)/3
(4-1)/3
3/3
1 = integer


Assume N works

[{2^(2*[N+1])}-1]/3
[{2^(2N+2)}-1]/3
[{2^(2N)*2^2)}-1]/3
[{2^(2N)*4)}-1]/3

Set X = 2^2N

(4x-1)/3
4x/3 - 1/3
(3x+x)/3 - 1/3
3x/3 + x/3 - 1/x
x + x/3 - 1/3
x + (x-1)/3

replace x

2^2n + [{2^(2*N)}-1]/3
4^n + [{2^(2*N)}-1]/3
4 to an integer is an integer
[{2^(2*N)}-1]/3 is an integer (by assumption)
sum of 2 integers is an integer
thus, N+1 works

That look right?

[Chenier]

OMG STOP WITH THE MATH!!!

I got D+ in Math and I'm proud of that!

[Dunthor Warsmith]

[{2^(2*N)}-1]/3

N = 0: 2^(2*1)-1 / 3
(2^2 - 1)/3
(4-1)/3
3/3
1 = integer

Since the question is to prove for positive integers, wouldn't 1 be a more appropriate starting case? True, this is actually a broader proof contain the concept desired, but I think this would be a point of technicality.

[Itek]

Eleena-

Yep. But I stopped at (2^2N)*(2^2)-1 and "brain farted" and stared at the page for the last 15 minutes of the test going "but... but.... but...."

Dunthor-

You must use the method

1) prove a base case, usually N=0 or N=1 or -alternatively- if the proposition is true for N=>4 (for example) N=4

2) assume (inductive hypothesis) that the proposition also is true for some arbitrary (and not explicit) N

3) write out the equivalent statement using N+1 and show (using the inductive hypothesis statement) that the proposition is also true for N+1

Reason: You have to prove the statement for ALL N and showing N and N-1 only shows it's true from the base case TO the number N, where proving it for N+1 shows it's true for the base case and ALL other N which are 1 or more larger than the base case. (If it's true for 0 it's true for 0+1, 1+1, 2+1, etc.)

Here's one from my CompSci class that I missed. See if you can catch my error.

Given:

public class Node
{
Object data;
Node left, right;
}

Write a method of counting leaves in the resulting binary tree formation from a given Node node. You may use recursion.

My code

public static int totalLeafNodes (Node node)
{
int count = 0;

if (left == null && right == null) return 1;

if (left != null) count = count + totalLeafNodes(left);

if (right != null) count = count + totalLeafNodes(right);

return count;
}



Finals are done, got nearly 3 weeks off, unless you count a total review of Calc One for my 8 week summer course Calc Two. (yes, yes, I'm a glutton for pain)

[Itek]

Since the question is to prove for positive integers, wouldn't 1 be a more appropriate starting case? True, this is actually a broader proof contain the concept desired, but I think this would be a point of technicality.

I think they meant N=0 to be the base case, because the math shows they did use, as required for N -is-a-member-of-positive-integers-, 1 when they did 2^(2*1)

(for N=0 it's not true as ((2^(2*0))-1) is 0 which is not divisible by 3)

[KyrosKrane]

See if you can catch my error.

You forgot to count the current node if left or right is not null. Initialize count = 1 to fix it. =)

Edit: Oh, and you're also assuming left and right are in scope as sub-elements of node. Depending on the language, you should have something like node.left and node.right, or node->left and node->right.

[Eleena Transient]

typo in my post, i set N=1 for base case but listed N=0, oops