

Cute riddle: Four kids holding one toy balloon - cool-RR
http://blog.ram.rachum.com/post/34561608024/cute-riddle-four-kids-holding-one-toy-balloon

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sown
I used to find these questions amusing and even stimulating but after a decade
of getting refused jobs because I couldn't answer questions such as this or
some question about pirates to the satisfaction of an interviewer has made me
slightly embittered.

And remember, no matter what, there's only one correct answer: _the one the
interviewer wants to hear_. Very rarely do I see a new and correct answer to
these puzzle questions provoke a positive response.

I hit the back button and moved on with my life.

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RiderOfGiraffes
Problem with puzzles like this - if you see the answer quickly, what do you
do? Any conversation fills up with uninformed, idle speculation, then someone
kills it with the answer, then it's kind of all over.

What would be interesting would be to have a site that solved the problem of
puzzle sharing and cooperation. Has anyone done that?

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chris_wot
You know, that would make an awesome riddle... How do you pose a riddle and
only accept answers that won't stop the discussion about the riddle...

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cliveholloway
It's called "NoScript". I have yet to see the solution on the site :D

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donutdan4114
[SPOILER] If you're a visual person, the solution is expertly drawn out here:
<http://i.imgur.com/Jpn1k.png>

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nitid_name
Of course, this assumes there is no friction or that friction is negligible
compared to the upward pull of the balloon...

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jules
Spoiler.

Apply this recursively: <https://dl.dropbox.com/u/388822/balloon.png>

Works for any number of children.

~~~
MojoJolo
Does this work for odd number of children?

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jules
Sure: <https://dl.dropbox.com/u/388822/oddballoon.png>

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bartkappenburg
Assumptions:

\- we have balloons with no rings (just a knot ;-))

\- we can't add any rings (cheating!)

\- we can make rings out of strings

\--------------------------------------------------

Solution: 1 string that makes a ring under the balloon, 1 that goes through
that ring and has two rings at each end. 1 string through 1 ring for kid #1
and #2, 1 string through the other ring for kid #3 and #4. 4 strings in total
( general function f(N,M) => f(4,1) = 4 )

For 1 balloon an three kids you need 3 strings (1 as a ring under the balloon,
1 string through the ring. One end is held by kid #1, other end is a ring. 1
string through that ring and the ends are hold by kid #2 and #3) f(3,1) = 3.
Likewise: f(2,1) = 2 f(1,1) = 1 <= trivial

Now for M balloons with M > 0\. For 1 balloon you can have 1 string that makes
a ring, for 2 balloons 1 string as well (interconnected). For 3 balloons you
need 2 strings from 1 to 2 and from 2 to 3. Ergo: for M balloons you need M-1
strings.

Consider the fact that having 2 balloons or 100 balloons, you only need to
attach a string to one of these interconnecting strings to float all the
balloons if you release this string.

Hence: M balloons and 2 kids needs the same amount of strings for for example
1 balloon and 2 kids plus the interconnecting strings.

From here it follows that for M > 1 (and N>0) the general function is: f(N,M)
= (N-1) + (M-1) (in words: you need M-1 interconnecting strings and one string
less than the number of kids)

3D plot:
[http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e98...](http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ecs5d2lubni)

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kang
One of the possible ways is like
<http://erikdemaine.org/papers/PictureHanging_FUN2012/>

"We show how to hang a picture by wrapping rope around n nails, making a
polynomial number of twists, such that the picture falls whenever any k out of
the n nails get removed, and the picture remains hanging when fewer than k
nails get removed"

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paulgb
The two-nails problem is my favourite nerd-snipe. It can be explained to
anyone without a math background, it can be solved on paper, it isn't a
"trick" problem, and it isn't immediately easy.

The problem: hang a painting with one unbroken piece of rope going from one
corner of the painting to another around two nails such that if either nail is
removed the painting will fall.

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MindTwister
The balloon has a loop underneath, you then let 2 kids (A and B), hold a
thread between them, take a second thread, feed it through the loop, once
around the thread between A and B, back through the loop and hands the ends to
kid C and D. If any of the kids let go, the thread will unwind and the balloon
is free.

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hdra
I don't really get it. Why would the AB thread go off if any one of the CD let
go?

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regularfry
Because the CD thread is the only thing stopping the AB thread from pulling
back through the loop.

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nadam
1: holded

0: not holded

binary AND: ==O--

binary OR: ==>\--

So the generalized riddle for any monotone boolean function can be easily
solved because all boolean monotone functions can be built up of binary ORs
and ANDs.

(Edit: accidentally mixed up OR and AND, fixed now.)

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danmaz74
Possible solution (I can't see the comments in the original post; are those
facebook comments?). The balloon is attached to a ring, and a thread (A)
passes through that ring. This thread has a ring at each end, and these rings
are smaller than the ring attached to the balloon.

Finally, a thread passes through each ring of thread A; let's call them
threads B1 and B2. Each kid holds one end of threads B1 and B2.

If for example one of the ends of thread B1 is lost by its kid, B1 will slip
out of the ring on thread A, and thread A will subsequently slip out of the
ring on the balloon. The ballon flies away :)

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chengsun
This looks similar to the picture-hanging puzzle, where a picture is hung
around a number of nails, and the removal of any one of them will cause the
picture to fall.

~~~
amatus
I believe it is exactly the picture-hanging puzzle, upside-down.

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lucvh
Tie a knot in the thread to the balloon, then trap this knot in the middle of
a gap between two twisted threads, the ends of these held by the children.

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bornon5
I think this is related to Borromean ring system - three inseparable rings,
yet no two are directly linked to each other. Cut any one ring, and the other
two fall apart.

<http://en.wikipedia.org/wiki/Borromean_rings>

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shill
A small, square cloth or net covers the balloon. Each kid holds a string
attached to a corner of the net. If just one kid lets go of a corner, the
balloon rolls out of the net and into the sky.

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xuhu
The balloon will go in unstable equillibrium in this case.

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brador
Chain link system.

Each string acts as one chain link. Kid technically could hold both ends of
their string, but you could tie it if kid holding one string was an absolute
condition.

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alexkus
Feel free to remove/censor/hide my posted solution if you want to keep it
going for a while longer. Mind you, there are almost certainly other solutions
to this puzzle.

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MojoJolo
I find the riddle and the answers amazing! But the answers I saw were only for
even number of children. Is there a good solution for odd numbers?

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dutchbrit
Who says it's the way the thread is arranged? Maybe the helium would lift the
other kids up if one decides to let go.

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hdra
The question did ask how the thread is arranged..

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fein
Well, he did write "threads"

> "How are the threads arranged?"

So the whole 4 separate threads idea still holds if we assume the balloon will
lift the kids off the ground. That was a bit of a huge hole in the riddle. It
should have been noted that the children were immovable anchors. Feynman would
have had a field day with this.

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jsn
Easy! Make the balloon large enough to lift N - 1 kids!

