This is topic A Riddle in forum Books, Films, Food and Culture at Hatrack River Forum.


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Posted by Tante Shvester (Member # 8202) on :
 
A sundial is the timepiece with the fewest moving parts. What is the timepiece with the most moving parts?
 

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Posted by Strider (Member # 1807) on :
 
define timepiece.
 

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Posted by Elmer's Glue (Member # 9313) on :
 
Hourglass.
 

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Posted by Tante Shvester (Member # 8202) on :
 
A device used to measure the passage of time, I guess. Does this suffice?
 

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Posted by EmpSquared (Member # 10890) on :
 
The Earth?
 

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Posted by Strider (Member # 1807) on :
 
damn, i was going to go with that one.

the universe?
 

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Posted by Lyrhawn (Member # 7039) on :
 
A tree.

(I think hourglass is the best answer so far)
 

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Posted by Elmer's Glue (Member # 9313) on :
 
This watch.
 

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Posted by Tante Shvester (Member # 8202) on :
 
Elmer's got it!

Good job.


Edit:
Clarification --
The correct answer is hourglass.
 

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Posted by Elmer's Glue (Member # 9313) on :
 

 

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Posted by Tante Shvester (Member # 8202) on :
 
Do you have any good riddles for me?
 

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Posted by Tatiana (Member # 6776) on :
 
That's a good riddle. I think one of those water clocks might have even more moving parts, if you think of the water as a very fine sand of molecules.
 

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Posted by Humean316 (Member # 8175) on :
 
I have one:

Three grainsilos have the following capacity:

A 8000 kilos
B 5000 kilos
C 3000 kilos

A is full, B and C are empty.

Can you without weighing put 4000 kilos in silo A and 4000 kilos in silo B?
 

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Posted by EmpSquared (Member # 10890) on :
 
Put 3000 in C
Transfer that 3000 to B
Fill C up with the rest from A so that:

A=2000
B=3000
C=3000

Fill up B to the top with C, so the remainder of C is 1000.

A=2000
B=5000
C=1000

Dump C into A.

A=3000
B=5000
C=0

Fill up C with B.

A=3000
B=2000
C=3000

Fill up A with C.

A=6000
B=2000
C=0

Fill up C with B.

A=6000
B=0
C=2000

Fill up B with A.

A=1000
B=5000
C=2000

Fill up C with B.

A=1000
B=4000
C=3000

Fill up A with C.

A=4000
B=4000
C=0

Voila. Was that too many steps?
 

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Posted by Humean316 (Member # 8175) on :
 
Well, when you dump C into A and get 3,5,0 consecutively, you could have done that first by dumping A into B, but essentially you are correct. The website where I got the riddle solved it another way, so I'll make that a challenge to someone else to solve it another way.
 

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Posted by TomDavidson (Member # 124) on :
 
A: 8000
B: 0
C: 0

A: 5000
B: 0
C: 3000

A: 5000
B: 3000
C: 0

A: 2000
B: 3000
C: 3000

A: 2000
B: 5000
C: 1000

A: 7000
B: 0
C: 1000

A: 7000
B: 1000
C: 0

A: 4000
B: 1000
C: 3000

A: 4000
B: 4000
C: 0

A: 0
B: 4000
C: 4000
 

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Posted by Javert (Member # 3076) on :
 

quote:

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Originally posted by TomDavidson:

A: 0
B: 4000
C: 4000

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Impressive.

How'd you get 4000 in the silo that could only hold 3000?
 

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Posted by scifibum (Member # 7625) on :
 

quote:

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Originally posted by Humean316:
I have one:

Three grainsilos have the following capacity:

A 8000 kilos
B 5000 kilos
C 3000 kilos

A is full, B and C are empty.

Can you without weighing put 4000 kilos in silo A and 4000 kilos in silo B?

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A 8000
B 0
C 0

A 3000
B 5000 <--count the grains as you add them, total is N
C 0

Move N/5 back into A

No weighing.
 

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Posted by Strider (Member # 1807) on :
 
I'll do my own alternate method now!

A. 8
B. 0
C. 0

A. 3
B. 5
C. 0

A. 3
B. 2
C. 3

A. 6
B. 2
C. 0

A. 6
B. 0
C. 2

A. 1
B. 5
C. 2

A. 1
B. 4
C. 3

A. 4
B. 4
C. 0
 

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Posted by Paul Goldner (Member # 1910) on :
 
My alternative solution.

Chop A in half, slip a sheet of steel between the halves, flip the top half over, call it B.
 

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Posted by Strider (Member # 1807) on :
 
I think I win for least amount of steps, and least amount of insanity.
 

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Posted by scifibum (Member # 7625) on :
 
Can I win for most amount of insanity?
 

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Posted by Humean316 (Member # 8175) on :
 

quote:

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Originally posted by scifibum:
Can I win for most amount of insanity?

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quote:

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Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

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quote:

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Originally posted by andi330:

quote:

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Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

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That is totally from The Hobbit. It's a human.

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quote:

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Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

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Sarcasmometer reading: 9.3 on the 10-scale

( )
 

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Posted by Flaming Toad on a Stick (Member # 9302) on :
 
Rabbit:
Bombs numbered 1-10
1-3vs4-6(M1)
If even, 1-3vs7-9(M2). If even, 10 is your dud. If (M2) not even, you now know that dud is in 7-9 group, and whether it's lighter or heavier. 7vs9 (M3), if even, 8 is dud. If uneven, the dud will be apparent based on (M2), either 7 or 9.

If (M1) uneven, 1-3vs7-9(M2). If even, dud is in 4-6 group. If uneven, dud is in 1-3 group. In either case, the dud's discrepancy will be known.

In short:
1-3vs4-6(M1)
1-3vs7-9(M2)
E=even, U=uneven
EE, 10.
EU, 7,8 or 9, discrepancy known.
UE, 4, 5 or 6, discrepancy known.
UU, 1, 2 or 3, discrepancy known.
(M3), if necessary, is any two bombs from the group containing the dud. The dud will be obvious in either case.
 

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Posted by The Rabbit (Member # 671) on :
 
Flaming Toad: You have a problem.

(M3) Won't work as you have it written. If you select two of the possibly bad bombs and they come out even -- then its the third bomb. But if they come out uneven, all you know is that one of those two is bad. You need one more detail.
 

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Posted by scholarette (Member # 11540) on :
 
But from the first two weighings, you know if the bomb is too heavy or too light.
 

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Posted by Threads (Member # 10863) on :
 
I think Flaming Toad's method provides all the information we need. Here's another way of looking at it.

[1,2,3] vs. [4,5,6]
[4,5,6] vs. [7,8,9]

If the groups are all equal then 10 is the dud.

If they aren't then one of them either weighs less than or more than the other two and the dud therefore weighs less than or more than the non-duds respectively (without loss of generality we have A=B and (B<C or B>C), if B<C then the dud weighs more than non-duds and if B>C then the dud weighs less). Toad's M3 works in both cases you listed. If the two bombs picked weigh the same then the dud is the third. If they don't then choose the lighter one if the dud is lighter and the greater one if the dud is greater.
 

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Posted by kmbboots (Member # 8576) on :
 

quote:

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Originally posted by andi330:

quote:

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Originally posted by Noemon:
Oh, oh, I've got a great riddle. It's really obscure, too!

What walks on four legs in the morning, two legs at noon, and three legs in the evening?

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That is totally from The Hobbit. It's a human.

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quote:

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Originally posted by scholarette:
But from the first two weighings, you know if the bomb is too heavy or too light.

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That's what I was looking for.


Now the harder puzzle. Same as above but you have 12 bombs.
 

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Posted by vonk (Member # 9027) on :
 
Re: the original riddle, I was going to guess the Solar System. Does that not have more parts than an hourglass, or is it not an accurate enough timepiece?
 

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Posted by anti_maven (Member # 9789) on :
 
To continue the theme:

"What have I got in my pocket?"


Sadly only 2.70€. No rings of power this time. Must keep trying...
 

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Posted by The Rabbit (Member # 671) on :
 
Will no one attempt the 12 bomb problem?
 

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Posted by scifibum (Member # 7625) on :
 
I'm thinking about it, but having trouble so far.
 

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Posted by The Pixiest (Member # 1863) on :
 
A riddle?

http://tinyurl.com/6q9bct
 

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Posted by The Rabbit (Member # 671) on :
 

quote:

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Originally posted by Tante Shvester:
Elmer's got it!

Good job.


Edit:
Clarification --
The correct answer is hourglass.

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quote:

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Originally posted by The Rabbit:

quote:

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Originally posted by scholarette:
But from the first two weighings, you know if the bomb is too heavy or too light.

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That's what I was looking for.


Now the harder puzzle. Same as above but you have 12 bombs.

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I did mention that. I used the word "discrepancy".

12 bombs:1-12
M1: 1-4 vs 5-8.
If even, dud is in 9-12 group.
If so, 1-8 are regular. The second weighing will use one of the bombs 1-8 *say 1*.
M2: 1 and 9 vs 10 and 11.
If even, 12 is dud. You can use M3 to determine if it's lighter or heavier then any other bomb.
If uneven, M3 will be 10 vs 11. If 1+9 was heavier:
10=11, 9 is heavy dud.
10>11, 11 is light dud.
10<11, 10 is light dud.
If 10+11 was heavier,
10=11, 9 is light dud.
10>11, 10 is heavy dud.
10<11, 11 is heavy dud.
That's the easy part...

If M1 is uneven, dud is in 1-8. 9-12 are normal.
For M2, we use one of those bombs *say 9*. This part took me the greater part of an hour.

M2 1+2+5 vs 3+6+9.

From M1, 1-4>5-8, result A. 1-4<5-8, result B.
from M2, 1,2,5=3+6+9 Result C, 1,2,5>3+6+9 Result D, 1,2,5<3+6+9 Result E.
Cases:
AC, M3 is 7vs8.
7=8, 4 is heavy dud. 7>8, 8 is light dud. 7<8, 7 is light dud.
AD, M3 is 1vs2
1=2, 6 is light dud. 1>2, 1 is heavy dud. 1<2, 2 is heavy dud.
AE, M3 is 3vs12
3=12, 5 is light dud. 3>12, 3 is heavy dud.
BC, M3 is 7vs8
7=8, 4 is light dud. 7>8, 7 is heavy dud. 7<8, 8 is heavy dud.
BD, M3 is 1vs2
1=2, 6 is heavy dud. 1>2, 2 is light dud. 1<2, 1 is light dud.
BE, M3 is 3vs12
3=12, 5 is heavy dud. 3<12, 3 is light dud.

In case AE, 3<12 is not possible given the parameters of the question. In case BE, 3>12 is similarly not possible.

There goes an hour and a half. Thanks, Rabbit.
 

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Posted by scifibum (Member # 7625) on :
 
Nice work. I had the damnedest time figuring out how to do M2 in order to limit the possible results sufficiently.
 

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Posted by Flaming Toad on a Stick (Member # 9302) on :
 
Like I said, the greater part of an hour. Gah!
 

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Posted by Nighthawk (Member # 4176) on :
 
I got asked something similar to the "10 bomb" and "12 bomb" riddles in a recent job interview.

They used simple weights instead of high explosives, though.
 

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Posted by scifibum (Member # 7625) on :
 
"I got asked something similar to the "10 bomb" and "12 bomb" riddles in a recent job interview."

I also got presented with a brain teaser in a recent job interview.

I think it'd be much better to ask people to find a suitable algorithm for solving a problem than to come up with a narrowly defined solution to a deliberately tricky problem. These things are great for making people think hard and exercise their brains, but don't map very well to real world problems, in my opinion.

(Imagine: you're tasked with designing efficient manufacturing processes. As a precondition of any solution, are you going to decide that you can only weigh items within a batch 3 times? Or you're writing a program: should we say that you get 15 lines of code for your function, or just let you write one that works?)

Perhaps I'm just bitter that I failed my test.

Out of curiosity, Nighthawk, what was the job?
 

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Posted by Nighthawk (Member # 4176) on :
 
Senior software development position at a major pharmaceutical company in Miami.

Honestly, for a position such as this they don't care of you get the answer right or wrong; they care about how you come to the conclusion. First they asked what was the minimum amount of weighings and I told them "three" almost instantly, but they wanted to know how I came to that conclusion in vivid detail. It's meant to display deductive reasoning, not memorization.

On a related note, I was asked by Yahn Bernier at Valve Software something similar to a riddle, but designed to analyze algorithmic ability: you have a deck of cards and five cards drawn from it. Systematically and programmatically (not through code, but through design), how would you determine whether you have a straight or not. This puzzle involves defining the deck, the cards and the method to scan for the yes/no result, and those methods could be either really optimized or horribly impractical. For example, I came up with 23 different solutions, perhaps 20 of which I would never dream of implementing.

I've heard of numerous companies, especially programming shops, doing similar things. Microsoft was infamous at one point for asking questions like that in their upper tier job interviews. They would ask questions like, put simply, "how would you test an elevator?" and expect you to write or speak a virtual thesis of a response.
 

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Posted by The Rabbit (Member # 671) on :
 
Nice solution Flaming Toad.

When I solved this I started with the hypothesis that you have to have narrowed it down to three bombs in the first two weighings and you have to know whether those bombs are heavy or light. Then the problem comes down to how you eliminate 9 bombs in the first two weighings.

Prior to the final weighing you always end up with 9 good bombs, 2 potentially heavy bombs and 1 potentially light bomb (or vice versa). I solved this by putting the one of the light bombs and one of the heavy bombs on the right side of the scale and two good bombs on the left side. If the right side rises, the light bomb is bad. If is falls, the heavy bomb is bad. I'm not sure if that solution is more elegant or just more complicated than putting the two light or two heavy bombs on opposite sides of the scale.

My current problem involves determining the maximum starting number that you can solve with only 3 weighings. I know it can be done with 13 easily (at least its easy after you've done the solution for 12).

For 13 bombs you start of by dividing the bombs into three groups of 4, 4 and 5. The two groups of four go on the scale if they are unequal, you proceed as in the 12 bomb problem. i.e you put one heavy bomb and 2 light bombs aside. You put two heavy bombs and one light bomb on one side of the scale and one heavy, one light and one good on the other side of the scale.

If the first 2 sets of 4 bombs have equal weight. You put 2 of the remaining five on one side of the scale and one of the remaining five plus 1 good bomb on the right side. If they are unequal you have 2 heavy and 1 light bomb and you solve it in M3. If they are equal, you weigh one of the remaining 2 bombs against one of the good bombs in M3.

So now I need to figure out if it can be done with more than 13 bombs.
 

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Posted by Flaming Toad on a Stick (Member # 9302) on :
 
With 13, it's possible to determine which one's the dud, but not always possible to determine whether it's lighter or heavier. The only way I've seen to solve the 14 bomb problem is by adding another known good bomb to the equation, leaving 15 bombs, with one dud and one known good one.
 

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Posted by Seatarsprayan (Member # 7634) on :
 

quote:

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quote:

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What walks on four legs in the morning, two legs at noon, and three legs in the evening?

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That is totally from The Hobbit. It's a human.

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The answer is indeed a human, but it's not from The Hobbit. It's the riddle the Sphinx asked Oedipus.

The Hobbit riddle was: "No-legs lay on one-leg, two-legs sat near on three-legs, four-legs got some."

The answer was "Fish on a little table, man at table sitting on a stool, the cat has the bones."
 

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Posted by Dobbie (Member # 3881) on :
 
What word am I thinking of, and it's not "kitty"?
 

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Posted by Elmer's Glue (Member # 9313) on :
 
Kitty.
 

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Posted by Dobbie (Member # 3881) on :
 
Get out of my head.
 

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Posted by Elmer's Glue (Member # 9313) on :
 
But it's nice and roomy in here.

 

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