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Topic: Perfect Cylinder and unlimited H2O problem (Read 6251 times) |
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ChuckHL
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Perfect Cylinder and unlimited H2O problem
« on: Oct 24th, 2006, 1:02pm » |
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This was a question from microsoft asked at the UofA on the interviews on October.
Assume you have a perfect cylinder. If you want, assume its transparent (but its not fragil at all) so that you can see through it easily. This perfect cylinder is placed vertical like this drawing:
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The bottom part of the cylinder is closed and there is a 2OZ mark close to that end of the cylinder. The other end is open and there is a 10OZ mark close to that end (see picture above).
Hint: The cylinder can be moved however you want to move it.
You have unlimited water to keep filling and tossing the water away. You do not have a second container, so the water you toss out of the cylinder is gone. You do not any other instrument except a pen for marking (the pen cannot be used as a ruler). You cannot ballpark. You cannot use your cloths to help you measure. How do you get all of the other marks for 1,3,5,6,7,8, and 9 OZ.
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towr
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #1 on: Oct 24th, 2006, 1:33pm » |
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Can we ignore the water creeps up the sides? Because that would really make it difficult..
hint: Tilt the cylinder.
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Sameer
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #2 on: Oct 24th, 2006, 1:45pm » |
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this would really require a precise tilting and holding equipment
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towr
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #3 on: Oct 24th, 2006, 2:00pm » |
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As a followup, in how many different order can you find your marks. (and let's include 4, which was probably only left out by mistake)
You don't even need the 10 mark, to be honest. And given just the 2, you can get all the others in order.
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ChuckHL
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #4 on: Oct 24th, 2006, 2:27pm » |
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Sorry for forgetting to put the 4. Indeed 4 is also required. I also agree that you do not really need 10 to get the other numbers. However, I added it as that was given by the interviewer.
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towr
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #5 on: Oct 24th, 2006, 2:37pm » |
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It might be interesting to look what pairs of numbers would work if neither is a power of two (in that case you only need that number)
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Ajax
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #6 on: Oct 26th, 2006, 2:54pm » |
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I believe, correct me if I'm wrong, that if you have the 2 marking, you don't need the 4.
if you put water until the 2 mark and by inclining the cylinder you manage to have water just covering the bottom, the point where it is just touching it's the 4 mark, imagine like a diagonal
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| « Last Edit: Oct 26th, 2006, 3:14pm by Ajax » |
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mmm
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Ajax
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #7 on: Oct 26th, 2006, 3:18pm » |
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some questions:
can you flip the container without spilling the water? I mean, can you secure it somehow?
some other thing. Are you sure that 10 is not really required, provided that the cylinder does not end at 10 oz but a bit higher?
Just reply with yes or no, I don't want hints
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| « Last Edit: Oct 26th, 2006, 3:30pm by Ajax » |
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mmm
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Icarus
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #8 on: Oct 26th, 2006, 4:05pm » |
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Yes, they are sure. (It took me a while to figure out the trick for getting beyond your highest current mark, too.)
Nice to hear from you again!
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| « Last Edit: Oct 26th, 2006, 4:05pm by Icarus » |
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Ajax
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Thanks.
I've been missing for so long due to military obligations (one year of compulsory service  )
So, do you agree that mark 4 can be calculated?
And an image of what I meant...
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| « Last Edit: Oct 28th, 2006, 9:59am by Ajax » |
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towr
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #10 on: Oct 29th, 2006, 7:06am » |
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on Oct 28th, 2006, 9:17am, Ajax wrote:| So, do you agree that mark 4 can be calculated? |
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The thing with the 4-mark, is that it accidentally wasn't asked for in the riddle; not that it has to be given to find all the other marks (In case you thought that's what I meant).
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Whiskey Tango Foxtrot
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #11 on: Oct 29th, 2006, 7:19am » |
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Aha! Now I understand. This is a great problem. Were you asked this in an interview?
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TimMann
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Re: Perfect Cylinder and unlimited H2O problem
« Reply #12 on: Oct 29th, 2006, 3:45pm » |
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on Oct 24th, 2006, 2:37pm, towr wrote:It might be interesting to look what pairs of numbers would work if neither is a power of two (in that case you only need that number)
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Nice. I think this answers the question:
Given marks a and b with a < b, using the tilting trick once, we can directly form new marks a/2, b/2, 2a, 2b, a + (b - a)/2 = a/2 + b/2, and a + 2(b - a) = 2b - a. (Did I miss any?)
For all of these, if a and b have an odd common factor c, the new mark is either half-integral or still has c as a factor. So, if we don't allow non-integral marks, clearly the two initial marks must not have an odd common factor.
What if we do allow non-integral marks? This doesn't help, as all the marks we can generate are rational numbers that can be written as n/d with n being divisible by c and d being a power of two, so we can never generate the 1 mark.
OK, so what if the two initial marks are positive integers with no odd common factors? Then we can always eventually form the 1 mark using the following algorithm (simplifications welcome).
1) While a is even, replace a by a/2.
2) If a = 1, stop.
3) While b is even, replace b by b/2.
4) If b = 1, stop.
5) Note that a != b because the original a and b had no odd common factor, and that a and b are both odd by steps 1 to 4.
6) Replace a by min(a, b) and replace b by a/2 + b/2. Note that these two values still must have no odd common factor.
7) Go to step 1.
This algorithm must terminate because every step that changes a or b reduces a + b, but never makes either value less than 1. So eventually one or the other must reach 1.
Of course 1 is a power of two and we can form all other postive integral marks easily once we have it.
So, having no odd common factor is necessary and sufficiient for two initial marks to generate all positive integral marks.
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| « Last Edit: Oct 29th, 2006, 3:49pm by TimMann » |
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