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-   -   Return of the Brain Eating Worms. AKA Ceeroque and Creyje, the Biological Trio. ALIAS (http://www.lucasforums.com/showthread.php?t=141496)

Ray Jones 12-16-2004 04:32 AM

Return of the Brain Eating Worms. AKA Ceeroque and Creyje, the Biological Trio. ALIAS
 
Yes, indeed, as you might have noticed from this threads title, it's riddle time, i know, again, but.. yes, it's been too long and you're all just come over like ment..uuh..brai..tse..think-thingy
pygmies. So come nearer, join in and solve this meaning-of-the-world changing riddle. Those who, although impossible, already know the answer, will post it in spoilers or be punished, i know mods, super-mods, admins and super-admins, you know?? Obey.

So, here it is:

There are 12 balls..







..and one chick, what a mess. http://drmccoy.xtrem3.de/santasmilies/dozey_santa.gif




*banned, thread over, GO HOME!*








---



Now seriously. There are 12 cubes, and make it gamepads if you like.
However, it should be spheres, because that's not so easy as using cubes or gamepads.
11 are complete equal in color, weight etc. 1 looks equal but has a different weight.
You have a balance and 3 measurings.

How do you find the different cube/gamepad/MAAANWHATEVER!!, and is it heavier or lighter?


[edit]

What the? "It" shortened the title. CRAP! Now i look like a dumbnut.

DEAR MOD!

Take the ALIAS out, now! Stop my thread's title from looking STRANGE.

please? http://drmccoy.xtrem3.de/santasmilies/puppy_santa.gif

DrMcCoy 12-16-2004 04:49 AM

Re: Return of the Brain Eating Worms. AKA Ceeroque and Creyje, the Biological Trio. A
 
Quote:

Originally posted by RayJones
Now i look like a dumbnut.
...could be cause you are a dumbnut... http://drmccoy.xtrem3.de/santasmilies/tongue_santa.gif

Ray Jones 12-16-2004 04:52 AM

But i do not look like one!

DrMcCoy 12-16-2004 05:12 AM

Quote:

Originally posted by RayJones
Now i look like a dumbnut.
Quote:

Originally posted by RayJones
But i do not look like one!
. . .

Ray Jones 12-16-2004 05:25 AM

Quote:

Originally posted by RayJones
But i do not look like one!
Read: But i usually do not look like one!


*is Das Mole'd*

Fealiks 12-16-2004 05:36 AM

solved: is it, you knock it on your temple consecutively, and whichever one knocks you out fastest is heaviest. can I have my prize now?

Alien426 12-16-2004 05:36 AM

spoiler:

1. You divide the balls into 3 groups of 4 balls each
2. weigh group 1 and group 2
group 1 heavier than group 2:
-> proceed with group 2
group 2 heavier than group 1:
-> proceed with group 1
group 2 and group 1 weigh the same:
-> proceed with group 3
3. divide the group into 2 sub-groups of 2 balls each
4. weigh sub-group a and b
5. the rest should be easy enough for everyone to figure out

Ray Jones 12-16-2004 06:04 AM

That doesn't work exactly.

Because you must find out if it's heavier or lighter. And
spoiler:
you'll need the second measurement to find out if the cube is in it, although if it is lighter, then of course it's easy to find out with the third measurement. But if you chose the "wrong" group, you'll know the cube is heavier, but have only one measurement left to find out which one of four it is. Impossible. In case 3 things look good from there on, but not using the method you described. Because you don't even know if the cube would be heavier or not.


Failed! Try again! Or i'll have to assume you're brainless.


Fealiks: another post like that and i'll have to do .. nothing. I have no moderator powers. Sadly enough. But be informed, this thread is for using your brain, only. And for making clear that i usually don't look like a dumbnut. On the other hand.. you have only 3 tries, not 11.

Fealiks 12-16-2004 06:20 AM

well technically, i will be usin my brain.
spoiler:
would it work if...*drumrolls*you weighed each ball and the heaviest is the heaviest? of couse it would!!!!



...oh, wait, you only get 3 tries...

Ray Jones 12-16-2004 07:08 AM

Come on, people. Have a try. And don't be afraid to look as stupid and smallbrained as alien does at the moment. Look at the thread title. It's looking too dumb, until someone will take the "ALIAS" out.
Yeah, and Thrik didn't change it, assumed he read this thread, he's wether too lazy, or just likes it if i look like a dumbnut. Grr..reat job, Mister.

*hears a maniacally evil laughter, then Thrik's voice from the far: "What a dumbnut!"



And just to let you know, i don't know the solution myself.

Fealiks 12-16-2004 07:11 AM


























































V__V

Ray Jones 12-16-2004 07:13 AM

I won't warn you again. I'll tell my mom! I'm really serious with this thread.

Fealiks 12-16-2004 07:20 AM

teeheehee!

okay.

what if you grabbed two balls at a time, and felt how heavy each one was?

;p

DrMcCoy 12-16-2004 07:35 AM

*ignores the thread*











































...

Fealiks 12-16-2004 07:59 AM

*claps*

Ray Jones 12-16-2004 08:51 AM

ignore list + 2

DrMcCoy 12-16-2004 08:54 AM

Quote:

Originally posted by --------
***************
...

Fealiks 12-16-2004 09:19 AM

Ray, I was joking!


I'm sowwy.http://www.punjablinks.com/1sr/sorry/4.jpg

Alien426 12-16-2004 09:45 PM

OK, this time I actually thought it through.
spoiler:
You divide into 4 groups.
if(groupA != groupB)
 {
  if(groupA != groupC)
   {
    X = groupA
   }
  else
   {
    X = groupB
   }
 }
else
 {
  if(groupA != groupC)
   {
    X = groupC
   }
  else
   {
    // you are ****ed, since you can't tell
    // if the resulting group is heavier or
    // lighter than the rest
    X = groupD
   }
 }

You note whether the X-group is heavier or lighter than the other groups it's measured against. Then weigh two balls of the X-group (a and b).
If they weigh the same, the result is ball c. Otherwise the one that has the same traits that the X-group had (i.e. heavier or lighter) is the result.

Ray Jones 12-17-2004 12:09 AM

This was also my first attempt, but i soon had to discover that there is one case were it also fails..


Hmm.. i am pretty convinced i have the solution. Shall i post it, so this thread is ready to spam it to death, or anyone want to have another try so we have at least three serious attempts?

DrMcCoy 12-17-2004 12:12 AM

...i'm for spamming...

NOW!

...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...
...spam...

*dies*

Ray Jones 12-17-2004 12:20 AM

*looks at corpse-o-mccoy(TM), complete armed with grenades, knives, a photograph of martha and two m16's, loaded and ready to kill*

http://drmccoy.xtrem3.de/santasmilies/dozey_santa.gif

it wasn't me.

DrMcCoy 12-17-2004 12:28 AM

*doesn't move/notice/blink/live*

Ray Jones 12-17-2004 12:42 AM

*kicks lifeless body*

Alien426 12-17-2004 01:23 AM

Quote:

Originally posted by RayJones
but i soon had to discover that there is one case were it also fails..
Take a look at the X = groupD branch of my attempt. It's not perfect... yet. You can describe your solution, but please use spoiler tags. I think I'll write a PHP script to solve this puzzle.

*picks up the M-16s from McCoy's very dead body*

Now I have a machine gun. Ho ho ho. http://drmccoy.xtrem3.de/santasmilies/cool_santa.gif

Ray Jones 12-17-2004 01:43 AM

Hmm. I'll post it.. later.

Hintus numeros unos:

spoiler:
Psssst. Take 3 groups 4 balls!

DrMcCoy 12-17-2004 03:21 AM

Quote:

Originally posted by Alien426
*picks up the M-16s from McCoy's very dead body*
*tries to say "THAT'S MY M-16, DAMNIT!"*
*but is dead, so nothing is coming out of his mouth*

...

Ray Jones 12-17-2004 03:27 AM

*kicks dead body, really hard and in the nuts*

Alien426 12-17-2004 03:29 AM

*keeps reading Stiff to get ideas of what to do with McCoy's cadaver*

DrMcCoy 12-17-2004 03:34 AM

*tries to whisper: "you could resurrect me..."*
*but is still dead*

Ray Jones 12-17-2004 03:45 AM

Maybe we should burn his corpse, huh?

DrMcCoy 12-17-2004 03:49 AM

*wants to scream: "NO!"*
*but is, yeah you guessed it, dead*

...

Ray Jones 12-17-2004 05:27 AM

Ok, you've heard what Thrik said. I need this thread to solve the riddle. And after that it will just fade away.

So BEHAVE.

Wait. I've already got the solution. OK. I will post it soon, this evening (CET), i think.

And after that it will just fade away.

Guess along, folks!!

Skinkie 12-17-2004 06:13 AM

I got it, you pick up allthe spheres one at a time and select the lightest one. This way you accomplish it with zero, I repeat, ZERO measurments. Give me a prize.

Ray Jones 12-17-2004 01:10 PM

OK, folks, here it is, my solution for the riddle and i found out on my own. And i cannot remember when i used the terms "equal" and "different balls" with so countless numberings this often the last time. ^_______^;;;;;;

I divided it into three spoilers so that, if maybe somewhere out there, someone, lonely and curious, someone like me, who just needs a hint to do the rest alone.. uh.. can do so. I mean, you should really try it. It's F.U.N.!




First weighting:
spoiler:
Divide the balls into 3 groups A, B and C 4 balls and weight groups A and B.

Possible results:

a) The balls of group A and B are equal: the sought-after ball is in goup C.

b) The balls of group A and B are different: the sought-after ball is in wether in group A or B.



Second weighting:
spoiler:
case a) Take 3 balls of group C and replace them with 3 from group B on the balance.

Possible results:

a1) All balls on the balance are equal, again. The ball we're looking for is the left over one: now we already now which ball it is, but not wether if it's heavier or lighter.

a2) The balance shows a difference, so the sought-after ball is 1 of the 3 replaced balls: we don't know which ball exactly it is, but we know if it's heavier or lighter.


case b) Take 3 balls of group A from the balance, replace them with 3 balls from group B and fill group B up with 3 balls from group C (we know they're equal because the "right" ball is already on the balance).

Possible results:

b1) All balls are now equal. The ball we're looking for is 1 of the 3 from group A: as in result a2) we don't know which ball exactly it is, but we know if it's heavier or lighter.

b2) The weights are still different and the balance shows the same as in the first weighting: the 3 balls from group A we took out are equal to those from group B which replaced them and those again are equal the "new" ones from group C. The ball we're looking for must be 1 of the 2 from group A or B which haven't been exchanged. So we know it's 1 out of 2 and don't know if it would be heavier or lighter, so we have to keep in mind which one of the 2 balls is heavier (and which one lighter).

b3) The weights are still different, but have "changed sides": the ball we're looking for is 1 of the 3 from group B that have been moved from one side of the balance to the other. Again we don't know which ball exactly it is, but we know if it's heavier or lighter.



Third weighting:
spoiler:
Case a2), b1), b3) Put 2 of the 3 balls on the balance.

Possible results:

I) Both balls are equal: the ball we're looking for is the third one. And according to the previous results we kow if it's heavier or lighter.

II) The balls are different: 1 of the balls on the balance is the sought-after and according to the previous results we kow if it's the heavier or lighter one.


Case a1) Take the already known to be different ball and weight it against an "equal one".

Result III): now we know if it's heavier or lighter.


Case b2) Take heavier one of the 2 balls and weight it against an "equal ball".

Possible results:

IV) The balls are equal: the left over ball is it. And it's lighter.

V) The balls weight different: it is the heavier ball. And it's err.. heavier.



You may now excuse me, I'll have a Martini or three to compensate my brain's rotation.

Ray Jones 12-20-2004 02:20 AM

Since i figured out that I like threads where I tend to talk to myself, here is yet another one. I expect the same participating like that i am used to.

Here it is.

There are 10 packs of coins 10 coins. 9 of the packs consist of genuine coins, the coins of other one are bogus. We know the genuine coins weight 10g each and the false coins weight 11g each. Plus we've got a letter balacne and one try.

How to find the bogus pack-o-coins?

Go Ray. XD

Alien426 12-20-2004 03:01 AM

That's easy.

spoiler:
You number the packs of cocains from 1 to 10 and take the corresponding number of coins from the pack.

1 coin from pack 1
2 coins from pack 2
3 coins from pack 3...

Then you weigh the whole heap and subtract 550 grams from the result. The difference is equal to the number of the pack with the counterfeit tickets for the Number Nine coins

Alien426 12-20-2004 03:08 AM

By the way, the previous puzzle wasn't solved. Hate to burst your bubble, but you started with "3 groups A, B and C 4 balls" and the last weighing was with "2 of the 3 balls"...

Ray Jones 12-20-2004 03:41 AM

:confused:

spoiler:
When you have 3 balls and know that 1 of them is heavier (or in the other case lighter) and you put 2 on the balance, it's either the 1 of those on the balance or the other one. And you know if it's heavier or lighter.


[edit]

Ahh. You mean, FOUR and then THREE. Look at step 2.

Alien426 12-20-2004 04:53 AM

Try to crack this die hard puzzle:

You have a five gallon bottle and a three gallon bottle. You must put exactly four gallons of water in the bigger bottle.


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