Ray Jones Posted December 16, 2004 Posted December 16, 2004 Yes, indeed, as you might have noticed from this threads title, it's riddle time, i know, again, but.. yes, it's been too long and you're all just come over like ment..uuh..brai..tse..think-thingy pygmies. So come nearer, join in and solve this meaning-of-the-world changing riddle. Those who, although impossible, already know the answer, will post it in spoilers or be punished, i know mods, super-mods, admins and super-admins, you know?? Obey. So, here it is: There are 12 balls.. ..and one chick, what a mess. *banned, thread over, GO HOME!* --- Now seriously. There are 12 cubes, and make it gamepads if you like. However, it should be spheres, because that's not so easy as using cubes or gamepads. 11 are complete equal in color, weight etc. 1 looks equal but has a different weight. You have a balance and 3 measurings. How do you find the different cube/gamepad/MAAANWHATEVER!!, and is it heavier or lighter? [edit] What the? "It" shortened the title. CRAP! Now i look like a dumbnut. DEAR MOD! Take the ALIAS out, now! Stop my thread's title from looking STRANGE. please?
DrMcCoy Posted December 16, 2004 Posted December 16, 2004 Originally posted by RayJones Now i look like a dumbnut. ...could be cause you are a dumbnut...
DrMcCoy Posted December 16, 2004 Posted December 16, 2004 Originally posted by RayJones Now i look like a dumbnut. Originally posted by RayJones But i do not look like one! . . .
Ray Jones Posted December 16, 2004 Author Posted December 16, 2004 Originally posted by RayJones But i do not look like one! Read: But i usually do not look like one! *is Das Mole'd*
Fealiks Posted December 16, 2004 Posted December 16, 2004 solved: is it, you knock it on your temple consecutively, and whichever one knocks you out fastest is heaviest. can I have my prize now?
Alien426 Posted December 16, 2004 Posted December 16, 2004 1. You divide the balls into 3 groups of 4 balls each 2. weigh group 1 and group 2 group 1 heavier than group 2: -> proceed with group 2 group 2 heavier than group 1: -> proceed with group 1 group 2 and group 1 weigh the same: -> proceed with group 3 3. divide the group into 2 sub-groups of 2 balls each 4. weigh sub-group a and b 5. the rest should be easy enough for everyone to figure out
Ray Jones Posted December 16, 2004 Author Posted December 16, 2004 That doesn't work exactly. Because you must find out if it's heavier or lighter. And you'll need the second measurement to find out if the cube is in it, although if it is lighter, then of course it's easy to find out with the third measurement. But if you chose the "wrong" group, you'll know the cube is heavier, but have only one measurement left to find out which one of four it is. Impossible. In case 3 things look good from there on, but not using the method you described. Because you don't even know if the cube would be heavier or not. Failed! Try again! Or i'll have to assume you're brainless. Fealiks: another post like that and i'll have to do .. nothing. I have no moderator powers. Sadly enough. But be informed, this thread is for using your brain, only. And for making clear that i usually don't look like a dumbnut. On the other hand.. you have only 3 tries, not 11.
Fealiks Posted December 16, 2004 Posted December 16, 2004 well technically, i will be usin my brain. would it work if...*drumrolls*you weighed each ball and the heaviest is the heaviest? of couse it would!!!! ...oh, wait, you only get 3 tries...
Ray Jones Posted December 16, 2004 Author Posted December 16, 2004 Come on, people. Have a try. And don't be afraid to look as stupid and smallbrained as alien does at the moment. Look at the thread title. It's looking too dumb, until someone will take the "ALIAS" out. Yeah, and Thrik didn't change it, assumed he read this thread, he's wether too lazy, or just likes it if i look like a dumbnut. Grr..reat job, Mister. *hears a maniacally evil laughter, then Thrik's voice from the far: "What a dumbnut!" And just to let you know, i don't know the solution myself.
Ray Jones Posted December 16, 2004 Author Posted December 16, 2004 I won't warn you again. I'll tell my mom! I'm really serious with this thread.
Fealiks Posted December 16, 2004 Posted December 16, 2004 teeheehee! okay. what if you grabbed two balls at a time, and felt how heavy each one was?
DrMcCoy Posted December 16, 2004 Posted December 16, 2004 Originally posted by -------- *************** ...
Alien426 Posted December 17, 2004 Posted December 17, 2004 OK, this time I actually thought it through. You divide into 4 groups. if(groupA != groupB) { if(groupA != groupC) { X = groupA } else { X = groupB } } else { if(groupA != groupC) { X = groupC } else { // you are fucked, since you can't tell // if the resulting group is heavier or // lighter than the rest X = groupD } } You note whether the X-group is heavier or lighter than the other groups it's measured against. Then weigh two balls of the X-group (a and b). If they weigh the same, the result is ball c. Otherwise the one that has the same traits that the X-group had (i.e. heavier or lighter) is the result.
Ray Jones Posted December 17, 2004 Author Posted December 17, 2004 This was also my first attempt, but i soon had to discover that there is one case were it also fails.. Hmm.. i am pretty convinced i have the solution. Shall i post it, so this thread is ready to spam it to death, or anyone want to have another try so we have at least three serious attempts?
DrMcCoy Posted December 17, 2004 Posted December 17, 2004 ...i'm for spamming... NOW! ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... ...spam... *dies*
Ray Jones Posted December 17, 2004 Author Posted December 17, 2004 *looks at corpse-o-mccoy, complete armed with grenades, knives, a photograph of martha and two m16's, loaded and ready to kill* it wasn't me.
Alien426 Posted December 17, 2004 Posted December 17, 2004 Originally posted by RayJones but i soon had to discover that there is one case were it also fails.. Take a look at the X = groupD branch of my attempt. It's not perfect... yet. You can describe your solution, but please use spoiler tags. I think I'll write a PHP script to solve this puzzle. *picks up the M-16s from McCoy's very dead body* Now I have a machine gun. Ho ho ho.
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