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Originally posted by Resnick ya i know, but they said the answer is 15...which i dont see how it can be

I thought it was 14. The answer's been explained: drop the first egg at floor 14, then 27, 39, 50, 60, 69, 77, 84, 90, 95, 99. Then once it breaks, just count up from the last floor where it didn't break (or from 96 if it never broke). The max you'll ever have to do is 14.

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Originally posted by DigiNut I thought it was 14. The answer's been explained: drop the first egg at floor 14, then 27, 39, 50, 60, 69, 77, 84, 90, 95, 99. Then once it breaks, just count up from the last floor where it didn't break (or from 96 if it never broke). The max you'll ever have to do is 14.

nah too complicated, drop 1 at 50, if it breaks do it at 25...and so on...do it for both

edit* check my last post's edie..lol sorry too lazy to repost

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Originally posted by Resnick ok how is the first building one not log n?? first of all, is N unique to both eggs or each egg has its own different height? cuz it doesnt really say, if it is not unique, then its logn of each... taking worse case, log[base2] n = 8 ..*2 = 16 which is the right answer

First, the problem says that each breaks from the height N (i.e., they are the same). See DigiNut's reply for a solution requiring only 14 drops. Also, log2(100) = 7. This solution would work if you had 7 eggs, but you don't. Try actually devising a solution using binary search and you'll see why it fails (both of your eggs will be broken before you learn anything).

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the genie one, u can just do anything and assume at inf it will work out???? i dunno

This is a tempting (but wrong) assumption. It is true that a random solution will work approaching 100% probability, but this is not a "guaranteed" solution. As an analogy, imagine the real number line. If I ask you to pick a point on it at random, you will with 100% probability choose a non-integer (or, for that matter, irrational) value. This does not preclude you from picking an integer though. Similarly, 100% of the integers contain the digit "3", but this doesn't mean that they all do. It's an interesting and widely-misunderstood point in mathematics...

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and the coin one depends on the coins you choose.... it asks for the most expensive u can buy with exact change?? so if u pick coin A = 10^10 $ then thats more expensive than a lower denomination edit* ok i read the question wrong, but it still depends on the coins, for instance, if u pick 2,3, then most expensive is 1$ cuz anything after that u have exact change for, now if u pick 10$ and 9$ u can buy 11$ and u wont have exact change, but 11>A>b , but in first case 1$

Yes, the value is dependent on A and B; I'm looking for a value in terms of A and B. (You may tell me, for example, that the value of the most expensive item you can't buy is A*B-3. But that's not right :-))

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Originally posted by Resnick nah too complicated, drop 1 at 50, if it breaks do it at 25...and so on...do it for both edit* check my last post's edie..lol sorry too lazy to repost

You only have two eggs though. Suppose your first one breaks at 50, and then your next one breaks at 25. Now what? You're out of eggs... game over. That's the point of the problem.

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Originally posted by Flyboy217 First, the problem says that each breaks from the height N (i.e., they are the same). See DigiNut's reply for a solution requiring only 14 drops. Also, log2(100) = 7. This solution would work if you had 7 eggs, but you don't. Try actually devising a solution using binary search and you'll see why it fails (both of your eggs will be broken before you learn anything).

oh i though u had infinite eggs...k

Re: Re: Re: Smart?

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Originally posted by DigiNut I'm pretty sure that's what it means. Those are both subsets, with the same sum, except that the set you picked has far *more* than two subsets with that some. You only need to prove that there's at least two.

This is correct. As a point of admission, I will say that "same subset sums" is completely nonstandard lingo. I'm sorry, that wasn't clear.

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Originally posted by Flyboy217 This is a tempting (but wrong) assumption. It is true that a random solution will work approaching 100% probability, but this is not a "guaranteed" solution. As an analogy, imagine the real number line. If I ask you to pick a point on it at random, you will with 100% probability choose a non-integer (or, for that matter, irrational) value. This does not preclude you from picking an integer though. Similarly, 100% of the integers contain the digit "3", but this doesn't mean that they all do. It's an interesting and widely-misunderstood point in mathematics...

wtf? why would it be 100% prob that i pick a non-integer? why cant i pick 3 or 4or 5?

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Originally posted by Flyboy217 Yes, the value is dependent on A and B; I'm looking for a value in terms of A and B. (You may tell me, for example, that the value of the most expensive item you can't buy is A*B-3. But that's not right :-))

but re-read my post, i just proved that u cant find a general solution in terms of A and B , so either someothings wrong with the questino, or with my proof

I think I figured out the answer to number 2, it's 5 weights.

If you have a weight, there are 3 things you can do with it: put it on the left, put it on the right, or put it nowhere. This gives you 3^n (where n is the number of weights) possible options AT MOST. You also have to account for the fact that doing nothing with all weights gets you zero, which won't balance anything, and that for each setup there's also a mirror image setup. So the maximum number of different total weights you can get from n weights is (3^n - 1)/2. Obviously to get the maximum, you have to pick the proper weights...

For 4 weights, this gets 40; for 5, it gets you 121, which is enough.

Flyboy, you didn't say we had to actually identify the weights, and I'm not sure how to go about doing that... I'm pretty sure my answer above is correct, though.

Edit: actually, I think it would make sense to pick those weights as powers of 3, i.e. 1, 3, 9, 27, and 81... would that do it?

k flyboy i think i know what your saying about the irration number everytime, but thats the limit as u pick infinite random numbers...its not an actualy value, which is what im saying is the solution to the genie one...

but its different case, all im saying is at the end it will work at, and in fact it will

but ur saying 'the limit' or irrational / ration will go to infinite, so ur doesnt actually reach 100% irrational, where as mine does

forget it tho, tell us the answer

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Originally posted by Resnick wtf? why would it be 100% prob that i pick a non-integer? why cant i pick 3 or 4or 5?

The point is that you *can* pick 3 or 4 or 5, but that if you're choosing truly randomly, then you will pick one of those with exactly 0% probability. I can direct you to some neat sites on this topic.

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but re-read my post, i just proved that u cant find a general solution in terms of A and B , so either someothings wrong with the questino, or with my proof

Not sure what your proof is meant to show. There are plenty of functions for which f(A,B) is less than A and B in some cases, while being greater than A and B in others. f(A,B) = A*B is one such example (although not for the natural numbers).

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Originally posted by DigiNut I think I figured out the answer to number 2, it's 5 weights. If you have a weight, there are 3 things you can do with it: put it on the left, put it on the right, or put it nowhere. This gives you 3^n (where n is the number of weights) possible options AT MOST. You also have to account for the fact that doing nothing with all weights gets you zero, which won't balance anything, and that for each setup there's also a mirror image setup. So the maximum number of different total weights you can get from n weights is (3^n - 1)/2. Obviously to get the maximum, you have to pick the proper weights... For 4 weights, this gets 40; for 5, it gets you 121, which is enough. Flyboy, you didn't say we had to actually identify the weights, and I'm not sure how to go about doing that... I'm pretty sure my answer above is correct, though.

Good work! Your solution set will be weights that are powers of 3.

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Originally posted by DigiNut I think I figured out the answer to number 2, it's 5 weights. If you have a weight, there are 3 things you can do with it: put it on the left, put it on the right, or put it nowhere. This gives you 3^n (where n is the number of weights) possible options AT MOST. You also have to account for the fact that doing nothing with all weights gets you zero, which won't balance anything, and that for each setup there's also a mirror image setup. So the maximum number of different total weights you can get from n weights is (3^n - 1)/2. Obviously to get the maximum, you have to pick the proper weights... For 4 weights, this gets 40; for 5, it gets you 121, which is enough. Flyboy, you didn't say we had to actually identify the weights, and I'm not sure how to go about doing that... I'm pretty sure my answer above is correct, though. Edit: actually, I think it would make sense to pick those weights as powers of 3, i.e. 1, 3, 9, 27, and 81... would that do it?

how would u measure 5 tho?

edit * nevermind, i guess that is right then

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Originally posted by Resnick k flyboy i think i know what your saying about the irration number everytime, but thats the limit as u pick infinite random numbers...its not an actualy value, which is what im saying is the solution to the genie one... but its different case, all im saying is at the end it will work at, and in fact it will but ur saying 'the limit' or irrational / ration will go to infinite, so ur doesnt actually reach 100% irrational, where as mine does forget it tho, tell us the answer

Not really sure what you mean. The actual value of the probability of choosing an irrational number is 1. Not just close to 1, but 1. You will, with 100% probability, pick an irrational number. But you still may pick a rational one. (This sounds strange, but read on).

Similarly, the probability that your random solution will work is *exactly* 100%. This is not the same as saying that it will work though, because with 0% probability it won't . But just as you may pick an integer value, you may be given an unsolved case.

This is highly counterintuitive, and I fear I won't convince you of its validity by myself. I'm stepping out right now, but I'll dig up some references online to show you why this is so.

I won't spoil the fun by answering yet though

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Originally posted by Flyboy217 Not really sure what you mean. The actual value of the probability of choosing an irrational number is 1. Not just close to 1, but 1. You will, with 100% probability, pick an irrational number. But you still may pick a rational one. (This sounds strange, but read on). Similarly, the probability that your random solution will work is *exactly* 100%. This is not the same as saying that it will work though, because with 0% probability it won't . But just as you may pick an integer value, you may be given an unsolved case. This is highly counterintuitive, and I fear I won't convince you of its validity by myself. I'm stepping out right now, but I'll dig up some references online to show you why this is so. I won't spoil the fun by answering yet though

so what ur saying is, if i ask a billion people a billion times each, that every single time, every single one of them will pick an irrational number? thats bullshit

For #3, could you just tell me whether or not the solution has anything to do with Gaussian distribution? It's the only area of statistics I know of relating an infinite array of random numbers to a constant distribution.

I believe you could prove 4(a) using a similar format as (2). I'll have to think about it for a few minutes...

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Originally posted by Resnick so what ur saying is, if i ask a billion people a billion times each, that every single time, every single one of them will pick an irrational number? thats bullshit

What you're speaking of is not actually random. Humans can't generate random numbers. Asking your friend to pick a number between 1 and 10 is considered void in almost any statistical experiment.

Ok, so for 4(a), if you pick 10 numbers, then there are 2^n possible subsets including the null set. This gets you 1024 different sets, and you can only take values up to 1000, so at least 24 of those cannot be unique.

Edit: crap, that doesn't work, it's the individual elements that only go up to 1000, not the sum.

I'm sure my answer is close, I just haven't figured out how to adapt it.

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Originally posted by DigiNut What you're speaking of is not actually random. Humans can't generate random numbers. Asking your friend to pick a number between 1 and 10 is considered void in almost any statistical experiment.

wow i cant believe im having this discussion, first of all, what ur talking about isnt math, its philosophy

once u bring that into the picture, NONE of these problems have solutions (its like saying for the building one, what if i picked n=37 without dropping either egg...that using 0 droppings) and infact that would be the right answer


so is this a math problem or a waste of our time?

If the main set has two nondistinct values, are two subsets using the "other" nondistinct value considered distinct?

i.e. if the main set is {2, 2, 3} then can {2, 3} and {2, 3} be considered distinct?

I'm assuming the answer has to be yes, because otherwise you could just pick your main set as {2, 2, 2, 2, 2, 2, 2, 2, 2, 2} which would essentially disprove your statement... but I'm asking anyway.

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Originally posted by Resnick wow i cant believe im having this discussion, first of all, what ur talking about isnt math, its philosophy once u bring that into the picture, NONE of these problems have solutions (its like saying for the building one, what if i picked n=37 without dropping either egg...that using 0 droppings) and infact that would be the right answer so is this a math problem or a waste of our time?

What on earth are you talking about?

The egg problem was specified as WORST CASE, i.e. the HIGHEST possible number of drops, assuming it's the LAST one you "guess".

I'm not talking about philosophy at all. Any elementary study of statistics will tell you that human guessing cannot be used to create a set of random numbers.

I'm still frustrated with the wording of some of these questions, but it sounds like you're just sour grapes over not being able to figure them out.

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Originally posted by DigiNut What on earth are you talking about? The egg problem was specified as WORST CASE, i.e. the HIGHEST possible number of drops, assuming it's the LAST one you "guess". I'm not talking about philosophy at all. Any elementary study of statistics will tell you that human guessing cannot be used to create a set of random numbers. I'm still frustrated with the wording of some of these questions, but it sounds like you're just sour grapes over not being able to figure them out.

yes and in the egg problem the worst case is 0, BECAUSE the number n doesnt change depending on which floor u drop it on...but fuck that, thats philosophy and has nothing do to with math

ok, so what is random? and if u say computers can produce a random number then i want ur proof on that

or pick something else that is completely random

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Originally posted by Resnick yes and in the egg problem the worst case is 0, BECAUSE the number n doesnt change depending on which floor u drop it on...but fuck that, thats philosophy and has nothing do to with math ok, so what is random? and if u say computers can produce a random number then i want ur proof on that or pick something else that is completely random

Ugh, I'll let Flyboy argue with you over the egg thing, because that question's been answered and you seem to neither understand what "worst case" means, nor the nature of random numbers.

No, a computer does not generate random numbers, in fact, if you read any textbook, it will refer to them as "pseudo-random" numbers because it's based on an algorithm.

Very little in this world is actually 100% random, independent of outside sources. But lots of things can be modelled that way, like the movement/position of subatomic particles. Human physiology is random in the sense of being chaotic, but if you ask someone to pick a number between 1 and 10, they'll amost invariably pick an integer value, which just goes to show that the value is NOT random, the probability of picking an integer is heavily weighted.

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Originally posted by DigiNut Ugh, I'll let Flyboy argue with you over the egg thing, because that question's been answered and you seem to neither understand what "worst case" means, nor the nature of random numbers. No, a computer does not generate random numbers, in fact, if you read any textbook, it will refer to them as "pseudo-random" numbers because it's based on an algorithm. Very little in this world is actually 100% random, independent of outside sources. But lots of things can be modelled that way, like the movement/position of subatomic particles.

very little in this world is 100% random? name ONE case thats 100% random!!

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Originally posted by Resnick very little in this world is 100% random? name ONE case thats 100% random!!

Argh... aren't you studying engineering?

You're going to have a lotta trouble if you have so much trouble grasping the concept of a random number. I'm not going to get into a whole discussion of random number theory and stochastic processes, you can read that in a textbook.

Human guesses are not useful as random numbers. Period.

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Originally posted by DigiNut Argh... aren't you studying engineering? You're going to have a lotta trouble if you have so much trouble grasping the concept of a random number. I'm not going to get into a whole discussion of random number theory and stochastic processes, you can read that in a textbook. Human guesses are not useful as random numbers. Period.

hmm it seems that ur not getting my point either, let me make it nice and simple...probabilityes are basically x/y+x so like x = heads and y = tails.. in a basic flip of a coin, the probability of heads coming up is x / x+y so its 1/1+1 = 1/2 ...thats a really simple concept

so now if u take a range, say [1,10] and pick a random number from there, then x = rational #'s, y = irrational numbers, no i know for a fact that there are rational numbers between 1 and 10, so lets just say there is one..even tho there are infinite but forget that.

so 1/1+y is the probability, which means no matter what y is, the fraction CANT be zero (mind u y cant be infinite cuz thats not a number)

and of course this is a really shitty proof cuz both x and y are infinity and u cant really evaluate it, BUT its the probability is not zero

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