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Smart? (pg. 6)
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| Flyboy217 |
| quote: | 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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| Resnick |
| quote: | 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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| Flyboy217 |
| quote: | 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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| Resnick |
| quote: | 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 bull |
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| DigiNut |
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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| DigiNut |
| quote: | 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 bull |
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. |
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| DigiNut |
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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| Resnick |
| quote: | 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? |
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| DigiNut |
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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| DigiNut |
| quote: | 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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| Resnick |
| quote: | 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 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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| DigiNut |
| quote: | 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 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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