Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
]0,c[ isn't dense in [0,1] since it doesn't include elements >c, so I don't think that will do. I think the answer is some kind of exotic set, since it seems pretty counter intuitive to me that the set has to be dense in [0,1] and yet have a lebesgue measure <1.
Perhaps it involves the cantor set, or some strange set including only irrational numbers? I'll give this some more thought...
ITT we do someone's graduate real analysis homework
Cmuben wrote:
Randil wrote:
Perhaps it involves the cantor set, or some strange set including only irrational numbers? I'll give this some more thought...
cantor set is a good hint
I originally thought about a process involving translations of portions of the Cantor set to "fill up" the holes that had been removed, but then the measure would still be 0
Another idea may be to start with a similar infinite process in which, at each step, the middle third of the hole left behind is filled in again with a closed interval, so the second step of the construction would be
[0,1/9]U[2/9,1/3]U[4/9,5/9]U[2/3,7/9]U[8/9,1]
and the measures of the sets at each step, starting from step 0, would be 1, 2/3, 5/9, 14/27, ..., and would go to 1/2 in the limit, and so would its complement within [0,1], which IIRC would be an open set dense in [0,1].
More generally, consider this type of process with a more general factor r (above was the special case r=1/3); at each step n the measure of the set is An and the measure of its complement is 1-An, and then An+1=(1-r)An+(1-An)r=r+(1-2r)An, with A0=1.
Then A0=1, A1=1-r, A2=1-2r+2r^2, and more generally...
An=r+(1-2r)An-1=r+(1-2r)(r+(1-2r)An-2)=...=r+(1-2r)r+(1-2r)^2*r+(1-2r)^3*r+...+(1-2r)^(n-1)*r+(1-2r)^n=sum((1-2r)^j,j,0,n-1)r+(1-2r)^n=(1-(1-2r)^n)r/(2r)+(1-2r)^n=(1+(1-2r)^n)/2.
...so I guess this one doesn't work either, because regardless of the factor chosen (of course 0<r<1>infinity, and because as I said, 0<r<1, this means -2<-2r<0, so 1<3-2r<3, so 1/3<3-2r<1>infinity is 1/(2r^2-5r+4), and because 0<r<1>infinity of r/(1-(1-2r)(1-r)^(m-1)), and indeed, as r->1 (if m is finite) this expression approaches 1, and as r->0 this expression approaches 1/(1+m), and approaching the case of a more Cantor-like set, as m->infinity regardless of r, it approaches 0.
Then the complement of the set so constructed is an open dense set with area between 0 and m/(1+m), so pick a value of m, an integer greater than 1, so that 1/(1+m)<1-c, and then pick r, a real number between 0 and 1, so that 1/(1+m)<1-r<1-c and so that An=1-c (I realize now that maybe I should triply-index A by n, m, and r but whatev).
Since we seem to be short on questions about questions, here is another one:
You are a prince that is about to be married to another country. The country has three princesses. You are allowed to chose which princes to marry (but not to marry more or less than one).
One of the princesses always lies, one always tells the truth and one can answer whatever she feels like.
You have heard a rumor that the king of the other country already has given his favorite princess away to someone. You conclude that in order to keep your head, you should avoid choosing that princes. The rumor also says that the favorite princes is the one that can chose as she pleases.
How many yes/no questions do you need to ask the princesses in order to keep your head?
And a bonus one that some of you already know the answer to:
You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
One based on a failed assumption of mine:
You have glass tiles made of Unobtainuim. As they are hit, they absorb 50 % of the impact. How many such tiles do you need to stop a jumbojet? Assume full physical realism (Earth and reasonable flight height), aside from the Unobtainium.
And here is one I made up today:
You are an alien corespondent. The aliens have a similar numerical system to us, but they uses other symbols. You have gotten access to a standard 4 function calculator. Find the algorithm that determinates what weights each symbol has.
Bonus complication 1:
Make the algorithm do no assumptions about the number of symbols.
Bonus complication 2:
Some symbols are duplicates.
Bonus complication 3:
Some weights lack a symbol.
Bonus complication 4:
Some weights are fractions.
And since we where talking dies:
You are a manufacturer of biased dies. Your client has asked for a set of biased dies for a Yahtzee style game using an alien numerical system as above. But you don't know the system more than the symbols used.
Investigate if there is a solution where you are only allowed to make one set of dies, yet can grantee a cheating result. You are allowed to provide the client with an exact algorithm for how to play the game with these biased dies.
The result is considered cheating if you get a better score if you know how the dies are loaded than if you had played optimally assuming that they where fair.
look, if I was about to spend my entire life with a princess that's too stupid to answer anything but yes/no-questions and has a high chance of being a dishonest brat, I'd just shag the most beautiful one, then saddle my horse and ride off to a distant country.
The real question would be: would you prefer the honest one, or would you rather go with the one that's bound to tell you every day how smart and beautiful you are, and how happy she is to be married to you? :p
(of course, 4 questions are enough to determine all three princesses. There's probably a solution with less, but given the smarts of those princesses, the kind of questions needed might make their heads explode.)
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Solution to the alien symbol problem:
I assume that the aliens' numerical system also includes negative numbers. This solution is not the fastest one, but it works.
For simplicity's sake, let's call the symbols A,B,...,J (10 symbols for the digits 0,1,...,9).
To find what symbol stands for 0, calculate A-B, and if the result is A, we know that B is 0. If not, continue with A-C and so on, up to A-J, until you find the symbol so that the result is A. If you don't find it, you know that A is 0.
In order find the digits 1 to 9, first remove the zero symbol. Let's assume that J was the zero symbol. To find what A is, calculate the 8 values A-B, A-C,...,A-I. Find how many of these results were negative, and call this number n. Now A=9-n. Do this step for all remaining symbols, and the problem is solved.
This algorithm will work in any basis, not only basis 10. Duplicates can be found if the resulting difference between two symbols is 0, which the algorithm will find when calculating A-B, A-C, etc.
I can do that faster:
calculate A-A (1 operation)
Now we know which symbol is 0
calculate B/B (1 operation)
where B is any symbol that's not the zero, now we know the 1
Calculate 1+1 = 2, 2+1 = 3, ... (7 operations)
Now we have all numbers from 0 to 8. The remaining one is number 9.
-> 9 symbols determined in 9 operations.
Bonus 1: This works with any amount of symbols. If we don't know in advance how many symbols there are (even though we could count the buttons on the calculator), we need 2 more calculations in step 3, adding up until we get a 2-digit result.
Bonus 2: if there are leftover symbols (duplicates), each's weight can be determined by adding 1, thus 1 calculation per symbol. If we pick a 0-duplicate in step 2, we just note it as 0 and pick another.
This relies on the calculator to always choose a single representation for each digits in its results. If the calculator chooses the symbols in its results randomly, there's no algorithm to determine all weights that's guaranteed to terminate.
Bonus 3: uhm, so if there's no symbol for 7, what would the calculator say when I type 6+1? If the result is empty, I'll just try 6+2 next until I find a symbol. Doesn't work if symbols for 0 or 1 are lacking, or if a larger range of symbols is missing.
Bonus 4: First determine all integers as above, then determine any remaining fractional symbol. Enumerate |N x |N, try each fraction and see if it produces a missing symbol. Stop when all symbols are found. Guaranteed to terminate, someday, unless the calculator's batteries run out or you die of old age.
>You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
Ask "is the pope catholic?".
Or, since you are a rude warrior, you can just push them aside.
EDIT:
I am interested in how hard that alien correspondent thing would be if the symbols for +-*/ are also different.
I have an idea how to do it, but it takes a lot of tries..
I am interested in how hard that alien correspondent thing would be if the symbols for +-*/ are also different.
First, figure out which symbols are digits and which are operators. If it's not obvious from the layout, just type something and see what the calculator does.
Then calculate
ABCDE ? ABCDE
for all 4 operators. The result with 7 or more digits was done through multiplication, the one with 4 to 6 digits through addition.
You've got two other results, one of them resulted in 1 and the other in 0. Calculate 0+1.
With 5 operations, you now know all 4 operators and the digits 0 and 1, adding just 3 operations to the previous case.
And a bonus one that some of you already know the answer to:
You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
Does "fastest" mean "the least amount of questions" (which would be one), or is it some kind of trick question (which the "you are a rude warrior" leads me to suspect)?
>You are a rude warrior that needs to pass two men. One always tells the truth and the other always lies. They wont let you pass until you have figured out who is the lying one. What is the fastest way to do so?
Ask "is the pope catholic?".
Or, since you are a rude warrior, you can just push them aside.
The riddle is there are 2 doors , one door leads to where you need to go, the other door leads to instant death or something. there are 2 beings, one which always tells the truth the other always lies. you get to ask one of the beings ONE question. what do you ask to always know which door to go through? The riddle and answer can be found in the movie "the labyrinth". I would ask "if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?" lol tautologies
here is a good riddle: "a census taker goes to a house, and a lady answers the door. He asks her how many children she has, she says she has 3. He then asks, how old they are. She says the product of their ages is 36. He says, well that's not enough information. She responds, saying "the sum of their ages is equal to the number of the house next door." The census taker looks at the house, thinks for a minute, and says that is sill not enough information. The woman responds, "the oldest plays piano". The census take says thank you and leaves. How old are her children?
The answer to my previous math question is the set E is a modified version of the union of the removed intervals of this: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set
instead of summing (1/4 + 1/8 + 1/16 +...), we need the sum to be (c/2 + c/4 + c/8 + c/16 +...) which is easy to do.
Joined: 11/9/2007
Posts: 375
Location: Varberg, Sweden
Cmuben wrote:
here is a good riddle: "a census taker goes to a house, and a lady answers the door. He asks her how many children she has, she says she has 3. He then asks, how old they are. She says the product of their ages is 36. He says, well that's not enough information. She responds, saying "the sum of their ages is equal to the number of the house next door." The census taker looks at the house, thinks for a minute, and says that is sill not enough information. The woman responds, "the oldest plays piano". The census take says thank you and leaves. How old are her children?
One quick guess is 9, 2 and 2 since he can't decide the ages by the first two clues which means that the house number has more than one series of 3 numbers whose product is 36. Doing some fast calculations in the head I think 13 is the only house number where this is the case with the series 6, 6, 1 and 9, 2, 2. And since one of them is the oldest (which I take for granted means being at least one years older than the others) the answer should be the second of those.
Also wouldn't it be enough to just give out the first and last clue to solve this, since I think there only exist one series of numbers that meets the first case and having a tie for the highest number?
I would ask "if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?"
That doesn't work.
Let A represent whether he would like to have sex with you, and let B represent whether he is a liar. Then let C be the answer to "Would you like to have sex with me" and D be the answer to "if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?" This has the following solutions:
A=true, B=false, C=true, D=true
A=true, B=false, C=true, D=false
A=true, B=true, C=false, D=true
A=true, B=true, C=false, D=false
Therefore the question has no answer if he doesn't want to have sex with you, and is undecided if he does.
One quick guess is 9, 2 and 2 since he can't decide the ages by the first two clues which means that the house number has more than one series of 3 numbers whose product is 36. Doing some fast calculations in the head I think 13 is the only house number where this is the case with the series 6, 6, 1 and 9, 2, 2. And since one of them is the oldest (which I take for granted means being at least one years older than the others) the answer should be the second of those.
Also wouldn't it be enough to just give out the first and last clue to solve this, since I think there only exist one series of numbers that meets the first case and having a tie for the highest number?
You need the second part, because without the second part you don't know that the sum of the ages is not unique. if she said the product is 36 and the oldest played piano, then (1,1,36), (1,2,18), (1,3,12), (1,4,9),(2,2,9), (2,3,6), and (3,3,4) could all be possibilities. the second clue implies that the number on the house next door is 13, which implies the answer is (2,2,9)
"if i asked you to have sex with me, would the answer to that question be the same as the answer to this question?", no matter if you say yes or no, you imply that the answer to the "if i asked you to have sex with me" is yes. you cant find out if who is the liar or which door to go through, but you at least get to have sex with them before you risk your life playing their stupid game, which is what life is all about anyways.
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Here's one I've been thinking about for quite some time now, but haven't been able to solve:
Suppose you have n dies, with k sides each. You roll all of them, and remove the dice with the lowest result. You add all the other results, and call this sum X. What is P(X=k)?
The condition n<k isnt quite right, the case n-1=k still works, corresponding to the minimum working value of X, so the restriction needs to be n-1<= k. k can still be made arbitrarily high, since a working dice combination can be done using the general formula
(1,1,1, ... ,1, k-n+2)
This gives X=k, as long as k>=n-1 as claimed.
The answer to my previous math question is the set E is a modified version of the union of the removed intervals of this: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set
instead of summing (1/4 + 1/8 + 1/16 +...), we need the sum to be (c/2 + c/4 + c/8 + c/16 +...) which is easy to do.
Does my solution also work?
Here it is in more concise form:
Let a sequence of subsets of [0,1] Cn(m,r), where n is in W, m is in N, and 0<r<1, be defined recursively by setting C0(m,r)=[0,1] and constructing, for 0<n, Cn(m,r) by removing from each closed sub-interval of Cn-1(m,r) the centered open interval of r times its length, and additionally, if m|n, by adding into each open interval between two closed intervals in the set the centered closed interval of r times its length. Let C be the set that Cn(m,r) approaches in the limit as n approaches infinity, and let A=[0,1]\C; then it can be proven that r can be chosen so that the measure of A lies between 0 and m/(1+m), so m merely needs to be chosen large enough so that c<m/(1+m), with the warning that m must be finite, for otherwise the measure of A would be 1/2 regardless of r.
the post in which I typed out some of the work involved got a bit mangled because I forgot to check "Disable HTML in this post" :-(
Joined: 4/20/2005
Posts: 2161
Location: Norrköping, Sweden
Derakon wrote:
You should specify that n<k; otherwise you could have 100 d4s and a probability of zero that X=k.
I'm terribly sorry, this was my mistake.. the probability I was looking for wasn't P(X=k), but instead P(X=m), for any integer n-1<=m<=k*(n-1). Basically, I want to know the distribution of X. Sorry for the confusion.
Suppose that you are inside a room, and your task is to determine whether you are on the surface of Earth or inside some kind of space station where artificial gravity is achieved by rotation. You are provided with any tools you want (as long as they don't help you communicate or observe outside the room). The perceived gravity matches that of Earth, so that alone cannot be done to determine the answer.
Ok, that's a relatively interesting problem in itself, but not the actual problem I'm posing, as the solution is rather trivial: You can use, for example, a pendulum or a rotating wheel which can also rotate on a perpendicular axis in order to determine whether the gravity is achieved by rotating a space station or not. (To people who haven't studied lots of physics it might not be self-evident why this is so, though.)
But a pendulum or rotating wheel will also show rotation on Earth too: After all, the Earth is rotating once every 24 hours, which the pendulum/wheel is going to show. The difference is, of course, that the space station is going to rotate significantly faster (probably in the order of minutes instead of 24 hours), which is what gives it away.
So the problem I'm presenting consists of two parts:
1) Assume that there's no physical limit (imposed by material strength, etc) to the size of the space station. To make the problem more difficult, let's build a space station which makes one full rotation in 24 hours, like Earth, and produces the same perceived gravity like that. Question: What would the radius of the space station have to be in order to achieve this (iow. it makes a full rotation in 24 hours and the perceived gravity at the outer edge is the same as on the surface of the Earth)?
2) In this situation, is there any way of determining whether you are in such a spacestation or on Earth?
