OT - Interesting Mind Teaser

[loveinvain]

Definitely does not take 3 pages.

I solved this a few years back. I came up with a tricky solution and thinking about it years later came up with a more simple, almost elegant solution.

Great puzzle, it took me a while.

[loveinvain]

Just wanted to add that if you solved this in a mere couple of innings then you are a much better man than I (though I guess I knew this already).

Me personally, a friend at work gave me this to chew on, I spent an hour or 2 at work mulling it over when I had a chance, thought about it on the way home from work, another 45 minutes or so, thought about it when falling asleep at night, another half hour or so, let my subconcsious work on it while sleeping and then on the way to work the next day it finally came to me.

I enjoy thinking about stuff like this. If you (or anyone else) has more logic problems like this I would love to hear them. So far this particular one is my favorite.

I will also add that I have presented this problem to several people, so far noone has solved it so it's not all that easy though I suspect several were capable of doing so but they have lives to live unlike me I suppose. I really should get a life. Sigh.

[TeckChavez]

Here is a fun one.

RiverIQGame
and then click the BLUE CIRCLE to start the game.

Here are the rules -
Everybody has to cross the river.
Only two persons on the raft at a time.
The father cannot stay with any of the daughters without their mother's presence.
The mother cannot stay with any of the sons without their father's presence.
The thief (striped shirt) cannot stay with any family member if the policeman is not there.
Only the father, the mother and the policeman know how to operate the raft.

To start, click on the big blue circle on the right.

To move the people, click on them.

To move the raft, click on the pole on the opposite side of the river.

~Teck

[luthar]

Is it too easy?!

Step 1: Separate in 2 groups of 6. Use scale.. Take the group that weights more.

Step 2: Separate remaining 6 in 2 groups of 3. Use Scale, take the group that weights more

Step 3: Take 2 of the 3 left. Use scale. If they are equal, it means the last stone you didn't weight is the one. If the stones aren't the same, you found out witch one it is!

Edit: Oups.. I guess it doesn't work since we don't know if it's more or less [img]/images/graemlins/wink.gif[/img]... But the solution is easy tought

[freemoney]

yeah but when you weigh the 2 out of 3 left on last turn you will find they are the same or that one is more than the other, if you get the same you dont know if the one you did not weigh is more or less than the other two and if you get that they weigh different you dont know which one weighs different then the one you didnt weigh.

[Hornacek]

Pretty damn easy...

1) Break 12 into three groups of four.

1234 5678 90ab

b) Weight group 1 vs group 2
1234 vs 5678

c) if same, weight group 90ab has the bad one.

ci) 34 vs 90 -> 34 is normal, find if 90 is off. if it weighs the same, ab has abnormal one, and just weigh 1 vs a.

cii) if it weighs different, 90 has the bad one, weight 1 vs 9

d) if different, culprit is within these two groups. 90ab has all normal balls. note which side is heavier.

di) 9012 vs 3456. If they weigh the same, then culprit is either 7 or 8.

dii) if 9012 weighs more, and 1234 was heavier, then 1 or 2 is heavier than the rest.

diii) if 9012 weighs more, and 5678 was heavier, then 3 or 4 was the heavier culprit.

and so on. its a little tricky, but you can get it a lot easier if you actually try it out with 12 balls.

[loveinvain]

Wrong. For example:

dii) if 9012 weighs more, and 1234 was heavier, then 1 or 2 is heavier than the rest. .. OR 5678 contains one that is lighter.

[Tickner]

That was really fun!

Thanks.

[durron597]

Fun puzzle! Answer below

<font color="white">
1) two groups, 1234 and 5678, with 90AB off the scale

2) If they balance, compare 123 with 90A
2a) If those balance, then
2ai) B is the fake coin, compare it to a good coin to determine heavier or lighter
2b) If the don't balance
2bi) then we know that if 123 is heavier then the coin is light and in 90A, and if it's lighter then the coin is heavy (and in 90A)
2bii) Weigh 9 vs. 0. If the coin was determined to be heavy in 2b heavy then the heavier coin is the fake, otherwise the light one is
2biii) Otherwise A is the fake with heaviness determined in step 2b

3) If the left side went down, compare 1678 with 5OAB
3a) If they balance
3ai) then we know that the bad coin is in the group 234 and is light. So weight 2 vs 3 and the lighter is the fake, if they balance then 4 is the fake (and is light)
3b) If the left side goes down again, then we know for sure that either 1 is heavy or 5 is light (and 9 is of course real). So weigh 1 vs. 9, if 1 is heavy then that's the fake, if they balance then 5 is the fake and it's light.
3c) If the right side now goes down, then we know that the fake moved from one side to the other (which means it's in group 678) and is light.
3ci) Use the procedure from 2bii to determine which the fake coin is

4) If the right side went down, we can compare 1678 with 5OAB here also
4a) The solution is symetrical, just replace all "light" with "heavy" and vice versa.
</font>

[Gramps]

[ QUOTE ]
Wrong. For example:

dii) if 9012 weighs more, and 1234 was heavier, then 1 or 2 is heavier than the rest. .. OR 5678 contains one that is lighter.

[/ QUOTE ]

Yeah, it's the heavier or lighter part that is the tricky part.