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  #276  
Old 2015-03-22, 19:38
Dino-Fly's Avatar
Dino-Fly Dino-Fly is offline
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Quote:
Originally Posted by marcosmapf View Post
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.
I'm fairly certain jumping off or on top of the seesaw makes it count as an extra weighing.

That said, if we assume that those actions are allowed, I would say that the time the person jumping in the air before they land on the seesaw again is too quick for the seesaw to move sufficiently up or down to extract useful data from it.

If we say that the seesaw moves instantaneously when a person jumps on it, wouldn't it be quicker then to make it 6 vs 6, and ask two people on either side to jump up at the same time? You'll be able to narrow it down to 2 potential people to be deviants, and then make them individually jump at the same time as one of the other people who we already know have the regular weight who are on their opposite side. When the seesaw balances when one of those two people jump you'll have the solution. So using your method, and allowing for a few things that I think are against the rules, you could solve the problem in 1 "weighing."
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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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  #277  
Old 2015-03-22, 22:30
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marcosmapf marcosmapf is offline
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Quote:
Originally Posted by Dino-Fly View Post
I'm fairly certain jumping off or on top of the seesaw makes it count as an extra weighing.

That said, if we assume that those actions are allowed, I would say that the time the person jumping in the air before they land on the seesaw again is too quick for the seesaw to move sufficiently up or down to extract useful data from it.

If we say that the seesaw moves instantaneously when a person jumps on it, wouldn't it be quicker then to make it 6 vs 6, and ask two people on either side to jump up at the same time? You'll be able to narrow it down to 2 potential people to be deviants, and then make them individually jump at the same time as one of the other people who we already know have the regular weight who are on their opposite side. When the seesaw balances when one of those two people jump you'll have the solution. So using your method, and allowing for a few things that I think are against the rules, you could solve the problem in 1 "weighing."
The thing is, I wanted to use it as little as possible... I also think it is somewhat cheating, but I imagine that normal means won't get you too far in this puzzle. Finding out who has the different weight out of 8 people with only 2 weighing just looks impossible. It wouldn't be an issue if we knew that he weights more or less, but we don't have that information...
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  #278  
Old 2015-03-22, 23:04
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Polaris Polaris is offline
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Do you happen to have a solution for 6 with 2 ? Or 3 with 1 ?

So frustrating, I have a solution for 11 person in 3 use x)
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  #279  
Old 2015-03-23, 00:43
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marcosmapf marcosmapf is offline
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Quote:
Originally Posted by Polaris View Post
Do you happen to have a solution for 6 with 2 ? Or 3 with 1 ?

So frustrating, I have a solution for 11 person in 3 use x)
This is exactly my issue. I think I have ended up with the same results as you, since I also lack a solution for 6 with 2 and 3 with 1, and I also have a solution for 11 with 3
That's why I suggested the jump solution, it would solve all of the issues while not exactly going against any rules.
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  #280  
Old 2015-03-31, 16:51
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Been thinking about this. I'm not 100% there, but here's my working.

Spoiler:
1. Divide 12 into 3 groups of 4. Weigh 4v4.
2. If 4v4 is even, the deviant is in the third group.
3. Divide the group of 4 into two groups of 2.
4. Weigh 1v1.
5. If 1v1 is even, the deviant is in the remaining 2.
6. You weigh 1 of the remaining 2 against someone who is known not to be deviant. If it's even, the deviant is the remaining person. If not, you've still found your deviant.

5. If 1v1 is uneven, the deviant is in that group.
6. Weigh 1 of the 2 against a known non-deviant and you'll find the deviant.

2. If 4v4 is uneven, the third group is discounted.
3. Divide 8 into 3, 3 and 2.
4. Weigh 3v3. If 3v3 is even, the deviant is in the third group.
5. Weigh 1 of the remaining 2 against a known non-deviant to find the deviant.

5. If 3v3 is uneven... no, I don't know. Can't get further than that.
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  #281  
Old 2015-03-31, 20:44
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We must be missing something...

(just ideas, [spoiler] for hardcore solvers )
Spoiler:
However I divide it up, be it 4-4-4, 3-3-4-2, I always come out one step too short in only one case... I even have two subdivision for your 8, Mia, still the same...
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  #282  
Old 2015-04-01, 13:04
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Another hint:

Spoiler:
I worked out my solution based off of a mathematical magic I learnt years ago. Here's a video of Matt Parker explaining it:


You won't be able to solve it working with just that trick, but it helped me to start thinking in the right direction.
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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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  #283  
Old 2015-04-01, 17:26
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I gave up and read both of your hints, the first one didn't teach me much, only that Spoiler:
4-4-4 is how the first split-up must be

And I didn't get the second one... Do I need to go to that website he advertises/find out how that magic trick worked ?
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  #284  
Old 2015-04-01, 18:20
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Quote:
Originally Posted by Polaris View Post
I gave up and read both of your hints, the first one didn't teach me much, only that Spoiler:
4-4-4 is how the first split-up must be

And I didn't get the second one... Do I need to go to that website he advertises/find out how that magic trick worked ?
Whoops sorry, I was in a rush this morning and posted the wrong video. Here's the right one (I think, I'm still a bit rushed):

Spoiler:
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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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  #285  
Old 2015-04-15, 16:31
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Here's one that came up in the news here yesterday, as an exam question in Singapore:

Albert and Bernard want to know when Cheryl's Birthday is. She gives them 10 possible dates:
May 15, May 16, May 19
June 17, June 18
July 14, July 16
August 14, August 15, August 17.

Cheryl tells Albert the month of her birthday and Bernard the day.

Albert says "I do not know when Cheryl's birthday is, but I also know that Bernard does not know."
Bernard says "I didn't know Cheryl's birthday before, but I do now.
Albert says "Then I now know when Cheryl's birthday is."

So when is Cheryl's birthday?
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Last edited by Mia; 2015-04-16 at 09:58.
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  #286  
Old 2015-04-16, 03:06
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Spoiler:
Bernard would know instantly if the day was the 17th,18th, or 19th as they each only show up once.
So for Albert to know for sure Bernard does not know, he must know it isnt in June or May at all.

However, if Bernard knowing that it isnt in June/May gives him the answer then he must have a number that, when those months are removed only shows up once;
15th.
So, ergo, 15th of August.
At least I think so. Good puzzle.
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  #287  
Old 2015-04-16, 10:06
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Whoops, that should be August 17 not August 16. Edited. Sorry!
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  #288  
Old 2015-04-16, 15:16
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Spoiler:
July 16th
The beginning is the same as DF's, then Bernard knows, so his number is either 15, 16 or 17, then Albert knows, so the month must be July.


Not bad but quite easy for me. I guess it just suits the way I think.

P.S. : Still thinking about the deviant one
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  #289  
Old 2015-06-02, 01:00
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Still thinking, I have come up with a solution, but I'm missing two non-deviant individual to carry it out
(And I haven't looked at the second hint yet.)
Spoiler:
So, first you divide them up as follows : 4-4-4
I will focus only on the part that is problematic for me : if the deviant is among the 4-4 you test first.
So first use : 4-4
Take 1 from the heavier side and 1 from the lighter side, put them down. Put the remaining 3 from the heavier side to the lighter.
Place 6 non-deviant on the empty plate (we've got only 4 ).
-They are balanced : I've isolated 2 among which 1 is the deviant -> Can be determied with 1 use.
-The lighter side stayed lighter : the deviant is among the 3 that were on the lighter side, and I know the deviant is lighter -> Can be determined with 1 use.
-The lighter side got heavier : the deviant is among the 3 that were on the heavier side, and I know the deviant is heavier -> Can be determined with 1 use.
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  #290  
Old 2015-06-04, 01:19
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I HAS IT !!! (I think)

Spoiler:

Mini tasks first (in each tasks you have determined that the deviant is among that mini group, thus having the rest that are non-deviant) :
-2 with one use : test one of the twos with another individual. If equal, it's the non tested, if not equal it's the tested who is deviant.
-3 with one use, when you know if the deviant is heavier/lighter : test two, if they are equal, it's the third one, if they are not, it's the heavier/lighter.

-4 with two uses : take 2, test them with 2 non-deviant ones. If equal, the deviant is among the other 2, if unequal, the deviant is among these 2.

Now then, the goal is to get one mini-group of 2 or 3 at the end of the second use.
-1st use : 4-4.
If equal, deviant is in the other four, run previous mini-task.
If unequal (I'll call them this way : heavier 4 := h4, lighter 4 := l4)
-Take down h3 (3 from the heavier side) ; switch h1 (the 1 that's left on the heavier side) and l1 (1 from the lighter side) ; put 3 non-deviant ones on the side that held the heaviers.
-2nd use : the 5-5 you just created (non-deviant 3 and l1 - l3 and h1)
If equal, run mini task on h3.
If unequal :
If heavier side stayed heavier, run mini task on l3.
If heavier side became lighter, run mini task on l1 and h1.

DONE !
The first hint certainly helped in that I took the first step for granted, but I think I would've found out nonetheless (with more time of course). Haven't looked at the second hint though
Are you still around Dino-Fly ? I'd like to know if I got it right

(The process is much more complicated than I expected, everyone feel free to ask questions if I wasn't clear/made a mistake somewhere/etc...)
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  #291  
Old 2015-06-05, 02:11
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You're lucky I popped in for a change!

Anyway, I didn't completely follow what you wrote, but it looks fundamentally correct. So yup, I think you got it right!

As for a brief treatise on my solution(Polaris, only if you're completely certain of your solution take a look at the spoiler, I don't want you being angry with me for giving you a solution without you solving it yourself correctly. ):

Spoiler:
Note that I am going to rig this case study that the seesaw is unbalanced when weighing. If the seesaw had to balance, the deviant is in the other group of 4, and the solution is easy. So an always unbalanced seesaw provides the most difficult problem.

Let's call the set of people A, B, C, D, E, F, G, H, I, J, K, L. Let's arbitrarily choose that person F is heavier than everyone else.

1. Divide the people into 3 groups of four, so 1:[A, B, C, D], 2:[E, F, G, H] and 3:[I, J, K, L].

2. Compare 2 group, let's use 1 and 2. Side 1 will go up, side 2 will go down. We have learnt the following data from this:
a) Deviant is not in 3.
b) A, B, C or D potentially has someone lighter in the group.
c) E, F, G or H potentially has someone heavier in the group.

Points b) and c) are very important. I will mark the people that could be lighter based on a) with an '*', thus: A*, B*, C*, D*. And potentially heavier people with an '=', thus: E=, F=, G=, H=. I'll also mark people who are definitely not the deviant by not making them bold.

3. What I now do is jumble the 3 groups up, making three new cominbations, with each side with a roughly equal number of people who have a '*', '@' or '=', so let's say:
1: A*, E=,
I, B*;
2: F=,
J, C*, G=;
3:
K, D*, H=, L.

Now we compare the two groups leaving out the one we made with the most potential deviants (thus either 1, or 2). I'll leave out 1 in this case as leaving out 2 is an easy solution.

Since F is the heavy person, group 2 will go down and group 3 will go up. This is where points a) and b) are important. C can't be a light deviant because it went down now, and H can't be a heavy deviant because it just went up. Also, A, E, and B are now all confirmed non-deviants.

So this is how things stand now: A, B, C, D*, E, F=, G=, H, I, J, K, L

Again, we repeat the same process of making a new mixed up collection of 3 groups of 4, with an evenly balanced number of potential deviants in each. So: 1: A, B, C, D*; 2: E, F=, H, I; 3: G=, J, K, L

We'll now weigh the two potential heavy deviants against each other (a potential heavy against the potential light deviant wouldn't yield any new information). So we weigh group 2 vs group 3.

Group 2 will go down, and group 3 will go up, confirming that F is the deviant. Not only that we also know that he is the deviant by virtue of being heavier than the rest.



You can fiddle around with the potential scenarios, but you'll find this method will always yield the right result.
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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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  #292  
Old 2015-06-05, 12:00
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Polaris Polaris is offline
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Question,
Spoiler:
why do you need to put a non-deviant in each sub-group ? You can use the seesaw only with 4 people on a side ?
Nice solution btw, didn't think about that subprogram (2heavier+1lighter with one go).
The basic idea of mine is more or less that if when you exchange 2 person the balance switches aswell, then the deviant is among those two, the rest is finding a way to set-up the right configuration.


Very nice puzzle by the way, do you have any other ones ? Else I won't have anything to think about before I fall asleep now
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  #293  
Old 2015-06-05, 12:43
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Dino-Fly Dino-Fly is offline
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Spoiler:
It's not strictly needed, but I applied primum non nocere. Having groups of 4 was the status quo, and since it would still work I didn't change it. But it would have been more efficient if I had to make it 3 vs. 3 using just potential deviants in the second weighing I suppose.


Sadly no, don't know any other puzzles! Personally I find doing probability calculations an effective method to fall asleep, though. Eg. in billiards you have 16 numbered balls. How many possible combinations can the first three balls go into the holes, but the order doesn't matter. So, [7, 13, 1] is the same as [1, 7, 13] and [13, 1, 17] etc.

Obviously if you had lessons in probability in school this is relatively easy to calculate, but usually you fall asleep halfway through doing the mental calculations!
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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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  #294  
Old 2015-06-05, 22:22
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Never thought of actually calculating, I myself usually prefer to think or reflect
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