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  #251  
Old 2015-03-09, 16:19
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Oooooooooooh I get it now. Thanks Polly.


Mia, she will take the seat of her friend?
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Polaris: "And what is a guitar doing in the middle of an asteroïd anyway?"
sgk: Think of it this way: it's like a message in a bottle. In our world, we put a message inside a bottle to protect it while it travels through the oceans to reach some other island. In other worlds, they put a message inside an asteroid to protect it while it travels through space to reach some other planet. In this case it is a gift, a guitar, rather than just a message.
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  #252  
Old 2015-03-09, 16:27
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Quote:
Originally Posted by SpaceGuitarist View Post
)


Mia, she will take the seat of her friend?
It's not hard, right??? She was utterly stumped.

Sorry. Thought I'd throw in an easy, lighthearted one there.
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  #253  
Old 2015-03-09, 20:55
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I was watching Brooklyn Nine-Nine, and there was a nice riddle in it. It took me about a half an hour to get a solution that solved every possible situation. Here's the riddle:

There are 12 people on the island. 11 have the exact same weight and 1 either weighs more or weighs less than the rest. There is a see-saw on the island, but you can only use it 3 times. Find the person with the different weight.
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #254  
Old 2015-03-09, 22:14
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1. Put 6 people on the seesaw (3 vs 3) to determine which group of 6 has the deviant one. If equal, the other 6 has the deviant.
2. From the deviant group of 6 put two vs two on the seesaw, if equal then put one of those guys vs one of the two left = solved.
3. If two vs two is unequal...


Hmm shit....
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  #255  
Old 2015-03-10, 00:42
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I played way too much Layton, you can do nothing against me

But it is indeed a nice one, took me a while to figure out the first time I stumbled upon it.
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  #256  
Old 2015-03-11, 19:58
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Quote:
Originally Posted by Dino-Fly View Post
I was watching Brooklyn Nine-Nine, and there was a nice riddle in it. It took me about a half an hour to get a solution that solved every possible situation. Here's the riddle:

There are 12 people on the island. 11 have the exact same weight and 1 either weighs more or weighs less than the rest. There is a see-saw on the island, but you can only use it 3 times. Find the person with the different weight.
Hmm. I'm really bad at puzzles but I'll give this a try:
  1. Weight 5 vs 5
    • If equal, then the deviant is on the remaining 2. Weight them once.
      • Now remove randomly one of them and replace with someone else and weight again. If the result is equal, then the deviant is the one that was removed. If the result is different, then the deviant is the one that remained.
        .
    • If unequal, then the deviant is on either group.
      • Now... hmm... shit.
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Polaris: "And what is a guitar doing in the middle of an asteroïd anyway?"
sgk: Think of it this way: it's like a message in a bottle. In our world, we put a message inside a bottle to protect it while it travels through the oceans to reach some other island. In other worlds, they put a message inside an asteroid to protect it while it travels through space to reach some other planet. In this case it is a gift, a guitar, rather than just a message.

Last edited by SpaceGuitarist; 2015-03-11 at 20:29.
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  #257  
Old 2015-03-11, 21:44
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That could be hilarious, cook the carrots, add sauce, do this, do that now.......hmmmm ...... shit.
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  #258  
Old 2015-03-11, 22:41
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Quote:
Originally Posted by Jesse View Post
That could be hilarious, cook the carrots, add sauce, do this, do that now.......hmmmm ...... shit.
Tbh this sometimes happens when I forget I don't have a particular ingredient I need. I live 5 mins from a 24 hour supermarket, though, and the dish is usually salvagable!
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  #259  
Old 2015-03-12, 00:29
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For me the problem is that I walk out the kitchen and get distracted by something on the internet. By the time I get back I find out my once promising pot of stew has become a flaming portal to hell.

A hint for the riddle: while it is true that you can't tell which side the deviant person is on when the seasaw is unbalanced, you do however know who went up and who went down.
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #260  
Old 2015-03-12, 02:01
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Oooh, your hint made me understand that I misunderstood your riddle
I thought we knew the deviant was lighter '
The solution is not so different, but forced me to think again. Yep, definetely a nice one.
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  #261  
Old 2015-03-12, 13:06
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Quote:
Originally Posted by Polaris View Post
I thought we knew the deviant was lighter '
Oh if that was the case it would be so much easier... we'd balance 6 vs 6, check who goes up, then go 3 vs 3, again who goes up and then 1 vs 1, done.

But this "either weights less OR more" aaaargh, adds so many variables.
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Polaris: "And what is a guitar doing in the middle of an asteroïd anyway?"
sgk: Think of it this way: it's like a message in a bottle. In our world, we put a message inside a bottle to protect it while it travels through the oceans to reach some other island. In other worlds, they put a message inside an asteroid to protect it while it travels through space to reach some other planet. In this case it is a gift, a guitar, rather than just a message.
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  #262  
Old 2015-03-12, 17:29
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Try a slightly easier one first then, 9 peeps, 1 heavier, only 2 uses of the seasaw.
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  #263  
Old 2015-03-12, 20:31
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Quote:
Originally Posted by Polaris View Post
Try a slightly easier one first then, 9 peeps, 1 heavier, only 2 uses of the seasaw.
...are you sure that that is solvable in all permutations? I'm struggling to find a guaranteed solution. Which troubles me, since I posed the riddle this is supposed to help solve. [emoji14]
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Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #264  
Old 2015-03-12, 21:10
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Quote:
Originally Posted by Polaris View Post
Try a slightly easier one first then, 9 peeps, 1 heavier, only 2 uses of the seasaw.
Oh, that one is easy. I can do it with 1 use of the seasaw...
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  #265  
Old 2015-03-12, 21:29
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They all look the same

Quote:
Originally Posted by Dino-Fly View Post
...are you sure that that is solvable in all permutations? I'm struggling to find a guaranteed solution. Which troubles me, since I posed the riddle this is supposed to help solve. [emoji14]
Mmm... Yes I'm sure. Could it be that I have a solution to your riddle which is different to yours ? I find the way of solving these two being similar, yours requiring one more idea.

My solution to yours :
Spoiler:
Divide them into three groups of four.
First use of the seasaw : determine in which group of four the deviant is.
Divide those four in two groups of two.
Second use of the seasaw : determine the group of two in which the deviant is.
Third use of the seasaw : try one of the two guy with any other.
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  #266  
Old 2015-03-12, 23:07
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Quote:
Originally Posted by Polaris View Post
My solution to yours :
Spoiler:
Divide them into three groups of four.
First use of the seasaw : determine in which group of four the deviant is.
Divide those four in two groups of two.
Second use of the seasaw : determine the group of two in which the deviant is.
Third use of the seasaw : try one of the two guy with any other.
Your solution is faulty. How would you find out which group the deviant is in in the first use of the seasaw? You can only do that if the seasaw balances with a group of four on either side (therefore the group that wasn't weighed has the deviant in). If the seesaw is unbalanced the deviant could be on either side, depending on whether the deviant is heavier or lighter. Thus your solution only works 83% of the time.
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #267  
Old 2015-03-13, 00:21
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Damn, you're right... Give me two minutes

edit : I hereby solemny recognise that I was being cocky and will need more time to think about this
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Last edited by Polaris; 2015-03-13 at 01:20.
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  #268  
Old 2015-03-13, 01:31
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Quote:
Originally Posted by Polaris View Post
I hereby solemny recognise that I was being cocky and will need more time to think about this
Welcome to the "hmm... shit" club!
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Polaris: "And what is a guitar doing in the middle of an asteroïd anyway?"
sgk: Think of it this way: it's like a message in a bottle. In our world, we put a message inside a bottle to protect it while it travels through the oceans to reach some other island. In other worlds, they put a message inside an asteroid to protect it while it travels through space to reach some other planet. In this case it is a gift, a guitar, rather than just a message.
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  #269  
Old 2015-03-16, 12:09
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Here is a hint:

Spoiler:
NOTE: I have arbitrarily decided that the person who is deviant is person 'd' and that 'd' is heavier than everyone else. This is entirely arbitrary, the same system for finding the solution applies if the person is lighter or a different person is the deviant.

First step:

Divide the 12 into 3 groups of 4. I'll represent the people with letters 'a' through to 'l', thus one group has a, b, c, d and so on.

Weigh two groups against each other at random (let's say [a, b, c, d] vs [e, f, g, h]), note that if we randomly selected groups [e, f, g, h] vs [i, j, k, l], you can apply Polaris' solution.

From this first weighing we will now know:
a, b, c, d could be heavier
e, f, g, h could be lighter
i, j, l, k are all the same weight


***

Hopefully that should get the cogs working for someone.
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}

Last edited by Dino-Fly; 2015-03-16 at 18:45.
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  #270  
Old 2015-03-16, 17:38
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Gah
Use spoilers ! I want to solve this
(so yeah, didn't read your post)
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  #271  
Old 2015-03-16, 18:45
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Whoops, I figured everyone had forgotten about it, sorry! Spoilers added!
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #272  
Old 2015-03-16, 18:50
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Oh, that was only a hint ?

I feel like I'm close btw, I can solve a lot of underridles (4 with 2 go, 3 with one when having a 4th non-deviant person), but I can't seem to find the good combination I need. I feel like a hint now would just be giving me the solution straight-out.

Either that or I'm still missing an idea (maybe something weghing the same person twice in two different groups ?)...
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  #273  
Old 2015-03-16, 18:56
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In the hint I have written down who you weigh in the first weighing and exactly what relevant information you gain from it. You can decide if that will give too much away!
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Spoiler:
Code:
#include "stdafx.h"
#include <iostream>
 
using namespace std;
 
int main()
{
  cout<<"Regards,\nDino-Fly\n";
 cin.get();
}
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  #274  
Old 2015-03-22, 10:50
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This hint gets seriously tempting as I still haven't found the solution >.<
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  #275  
Old 2015-03-22, 17:08
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Can someone jump from the seesaw without counting as an extra use of the seesaw?

If so, I have found a solution. You might call it cheating, but I call it creativity.

**Spoiler** people jump when using the seesaw. They never leave it, only stay mid-air for a second

I just can't make it more clear then it already is, I'm sorry if its too confusing. It makes perfect sense on my mind, but I can't translate it into english correctly...

Spoiler:


*Divide the total group in 3 groups of 4 people: Group 1 (A B C D), Group 2 (E F G H), Group 3 (I J K L)
*Get Group 1 and Group 2 on the seesaw, oposing to each other



----> The solution splits into two answers at this point <----




*****Group 1 and Group 2 weight the same******

Group 3 (I J K L) has, therefore, the "culprit"

*Split group 3 in two groups of two members (I J) and (K L)
*Match (I J) on the seesaw; if they have different weights, match I or J with anyone from any other group and you'll get your answer; otherwise, match K and L with anyone from the other groups and you'll get the answer.





*****Group 1 and 2 have different weights*****

*Split Group 2 into two groups (E F G) and (H) (Lets call it "Group H")
*Get Group 1 on one side of the see-saw (A B C D) and Group 2 on the other (E F G);

The logic is the following: There are 7 people on the seesaw. The side with four people will, obviously, be on the ground. If everyone but the "culprit" weights the same, then the seesaw should go up and as soon as someone jumps.


*****Group 1 and 2 have different weights when someone jumps (One of the sides sink)*****


----> Solution splits into two paths <----

*Ask one of the members from Group to jump.

*If someone from the 1st group jumps and the seesaw stays even, he/she is the culprit. Match him with someone else to know if he weights more or less

*else, match someone two people from the 1st group and ask them to jump together; ask someone from the 2nd group to jump aswell. The one who moves the seesaw is the culprit.

*****Group 1 and 2 are always even when someone jumps*****

*Culprit is Member H from Group 2.





Phew, I hope everyone understands.
It isn't a cool solution, but it works.

Last edited by marcosmapf; 2015-03-22 at 17:21.
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