9 ms·
Cute riddle: Four kids holding one toy balloon
- jules 14y agoSpoiler. Apply this recursively: https://dl.dropbox.com/u/388822/balloon.png https://dl.dropbox.com/u/388822/balloon.png Works for any number of children.
- MojoJolo 14y agoDoes this work for odd number of children?
- andrewflnr 14y agoThe odd kid out can hold both ends of a string.
- AndrewHampton 14y agoYes, just tie the last child's string to the last loop.
- jules 14y agoSure: https://dl.dropbox.com/u/388822/oddballoon.png https://dl.dropbox.com/u/388822/oddballoon.png
- sown 14y agoI 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.
- RiderOfGiraffes 14y agoProblem 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?
- indiecore 14y agoThen it's your turn to pose a riddle.
- chris_wot 14y agoYou 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...
- cliveholloway 14y agoIt's called "NoScript". I have yet to see the solution on the site :D
- indiecore 14y agoThat riddle is NP-hard
- romain_dardour 14y agoCrowdsourced "spoiler" button that hides the answer to everyone unless they choose to reveal it ?
- lumberjack 14y agoI think you are supposed to simply share it. It's the privilege you get for being the first to find the answer. At least that how it works in real life. However on the Internet I guess that mentality doesn't bind well. Hackers are not content with simply knowing the answer. They want of find it themselves. That's an awesome trait in a community.
- alexkus 14y agoFeel 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.
- chengsun 14y agoThis 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 14y agoI believe it is exactly the picture-hanging puzzle, upside-down.
- MindTwister 14y agoThe 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.
- regularfry 14y agoAh, you beat me to it - that's exactly the arrangement I posted on the site.
- hdra 14y agoI don't really get it. Why would the AB thread go off if any one of the CD let go?
- regularfry 14y agoBecause the CD thread is the only thing stopping the AB thread from pulling back through the loop.
- lucvh 14y agoTie 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.
- nadam 14y ago1: 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.)
- brador 14y agoChain 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.
- dutchbrit 14y agoWho says it's the way the thread is arranged? Maybe the helium would lift the other kids up if one decides to let go.
- hdra 14y agoThe question did ask how the thread is arranged..
- dutchbrit 14y agoVery true :). Some interesting answers here so far!
- fein 14y agoWell, 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.
- deleted 14y ago[deleted]
- kang 14y agoOne of the possible ways is like http://erikdemaine.org/papers/PictureHanging_FUN2012/ 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"
- paulgb 14y agoThe 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.
- donutdan4114 14y ago[SPOILER] If you're a visual person, the solution is expertly drawn out here: http://i.imgur.com/Jpn1k.png http://i.imgur.com/Jpn1k.png
- nitid_name 14y agoOf course, this assumes there is no friction or that friction is negligible compared to the upward pull of the balloon...
- ericmason 14y agoOr maybe the threads are crossed over the top of the balloon.
- bornon5 14y agoI 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 http://en.wikipedia.org/wiki/Borromean_rings
- danmaz74 14y agoPossible 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 :)
- shill 14y agoA 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.
- xuhu 14y agoThe balloon will go in unstable equillibrium in this case.
- MojoJolo 14y agoI 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?
- jsn 14y agoEasy! Make the balloon large enough to lift N - 1 kids!
- bartkappenburg 14y agoAssumptions: - 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=d41d8cd98f00b204e9800998ecf8427ecs5d2lubni http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e98...