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Originally posted by Flyboy217 1) There is a 100-story building. You have two identical eggs, each of which will break if dropped from some integer story N (or greater). You are allowed to drop each egg from any height you wish, until it breaks. By the time both eggs are broken, you must determine N. What is the fewest number of drops needed in the worst case? [10 min.]

log[base 2] of 100, think that is near 7...

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2) There is a balance in front of you. I will first ask you to choose a set of integer weights. You will then be given a block of unknown integer weight from 1 to 100, and asked to balance it using only weights from your set. You will be allowed to place your weights on one or both pans at your discretion, until it balances. What is the fewest number of weights you need in your set to guarantee that you can balance the given weight? [15 min.]

um log[base2] of 100 again ? i think that is about 7 weights. 1,2,4,8,16,32,64

oops.. typo it's 16 of course

time to go back to my problem at work ... nice ones though!

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Last edited by drizzt81 on Oct-10-2003 at 06:02

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Originally posted by biznology why are these all math questions? is math the only difficult academy?

not really, but the idea is that math is simple enough that you can 'theoretically' solve all problems by just THINKING about them. None of these involve any really complicated mathematics. They are all basic logic/ reasoning type questions.

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Ok i didnt go through every single number but im sure if u pick the weights 1 2 5 10 20 50 100 u can guarentee a balance

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Originally posted by drizzt81 not really, but the idea is that math is simple enough that you can 'theoretically' solve all problems by just THINKING about them. None of these involve any really complicated mathematics. They are all basic logic/ reasoning type questions.

but thats still not it...these are 'difficult' questions but they can be answered. not by everyone, necessarily, but still answered. what about everything else without an answer? sorry, im a philosophical asshole|

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Originally posted by drizzt81 log[base 2] of 100, think that is near 7... um log[base2] of 100 again ? i think that is about 7 weights. 1,2,4,8,16,32,64 oops.. typo it's 16 of course time to go back to my problem at work ... nice ones though!

Wrong and wrong . Thanks to modern-day CS, many people love thinking along the lines of log2. Get off your log2 horses and start THINKING, people! That's what's nice about these problems... they make you think in a way you're most likely not used to thinking.

C'mon, someone solve at least one!

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Originally posted by Flyboy217 18 is not optimal for #1.

I suppose you could get it down to 17, if you used increments of 12. Once you hit floor 84 and it doesn't break, instead of going to 96, you go to 92. That way you only have to drop the next one 6 times.

This doesn't lower the worst case to 12, though, since this is no longer the worst case. The worst case will be when it does break at 84 (after 7 drops) and you need 10 to determine what's in between.

There is no possible way it can be lower unless there's a trick in the question. Here's the proof:

Assuming you drop the first egg at a regular interval n until it breaks, and I am sure that this is required, then the total number of drops required is going to be Floor[100 / n] + (n - 2). You can't make this lower than 18 for any value of n, unless you use the modification mentioned above, which nets you 17.

Flyboy, I'm bothered how you don't respond to any of my reasoning or proofs to tell me how the answers are wrong - I'm starting to wonder if you actually thought these questions through yourself or just dug them up somewhere along with the answers?

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Originally posted by biznology but thats still not it...these are 'difficult' questions but they can be answered. not by everyone, necessarily, but still answered. what about everything else without an answer? sorry, im a philosophical asshole|

what biz said.

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Originally posted by DigiNut I suppose you could get it down to 17, if you used increments of 12. Once you hit floor 84 and it doesn't break, instead of going to 96, you go to 92. That way you only have to drop the next one 6 times. This doesn't lower the worst case to 12, though, since this is no longer the worst case. The worst case will be when it does break at 84 (after 7 drops) and you need 10 to determine what's in between. There is no possible way it can be lower unless there's a trick in the question. Here's the proof: Assuming you drop the first egg at a regular interval n until it breaks, and I am sure that this is required, then the total number of drops required is going to be Floor[100 / n] + (n - 2). You can't make this lower than 18 for any value of n, unless you use the modification mentioned above, which nets you 17. Flyboy, I'm bothered how you don't respond to any of my reasoning or proofs to tell me how the answers are wrong - I'm starting to wonder if you actually thought these questions through yourself or just dug them up somewhere along with the answers?

I apologize for not responding sooner. In general, I prefer to simply say "wrong" so that the person may figure out the solution entirely by himself. In this case, your proof was a good idea, but it is flawed. While I would normally just leave it at that, I'll go one step further and tell you that you make a faulty assumption in your first sentence. Any more and I'd be telling you the solution.

I've got plenty more where these came from, if anyone is interested. Yes, I'm a big math nerd :-). Served me damn well in school though ;-)

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Originally posted by Flyboy217 Wrong and wrong . Thanks to modern-day CS, many people love thinking along the lines of log2. Get off your log2 horses and start THINKING, people! That's what's nice about these problems... they make you think in a way you're most likely not used to thinking. C'mon, someone solve at least one!

ok ok, i agree I am stupid, but seeing how i have been up since noon yesterday (irish time, it's now 4pm) i guess we all make mistakes

I have done it out.. logically and not the idiot blind, it's 3am and i don't want to program anymore way..:

i think it is do-able with 5 weights:
1,3,7,17,24 How about that?

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Originally posted by drizzt81 ok ok, i agree I am stupid, but seeing how i have been up since noon yesterday (irish time, it's now 4pm) i guess we all make mistakes I have done it out.. logically and not the idiot blind, it's 3am and i don't want to program anymore way..: i think it is do-able with 5 weights: 1,3,7,17,24 How about that?

How do you plan on balancing my 100 weight with those? They sum up to only 52 :-)

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Originally posted by Flyboy217 How do you plan on balancing my 100 weight with those? They sum up to only 52 :-)

oops today isn;t my day, i was thinking that i had to DETERMINE the weight of yours.. not balance it..

with a sum of 52 i will be able to tell you how much your weight is :/

still not bad, if that had been teh question

try 1,3,7,17,31,79

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Last edited by drizzt81 on Oct-10-2003 at 16:22

quote:

Originally posted by Flyboy217 1) There is a 100-story building. You have two identical eggs, each of which will break if dropped from some integer story N (or greater). You are allowed to drop each egg from any height you wish, until it breaks. By the time both eggs are broken, you must determine N. What is the fewest number of drops needed in the worst case? [10 min.]

What's the "worst case"?