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On April 14 2011 09:51 JeeJee wrote:Show nested quote +On April 13 2011 14:18 MusicalPulse wrote:
Friend asked me this one, kinda like the other 3 person truth/lie one.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm.. here we go i think + Show Spoiler +
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
if da=yes
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B)
JAJA -> C is random god
DAJA -> A is random god
JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case)
"if i were to ask you, are you the lying god, would you say da?
if da=yes
trueGod: ja
falseGod: da
if da=no
trueGod: ja
falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
I don't see how this works.
+ Show Spoiler +If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
Onto a puzzle of my own!
You are an engineer at a light bulb company in charge of quality control and are given two identical light bulbs and a building N stories tall to test them with. You are to determine the exact floor upon which the bulbs will break (call it K, if you drop the bulbs from floor K or any floor greater than K, they break, but if you drop them from a floor less than K, they will remain intact). A broken bulb cannot be retested but an intact bulb can be. Since you are an engineer, you want to accomplish this task as fast as possible. What is the method that will guarantee a minimum maximum amount of tests needed to determine K? (Say the building has 3 floors and you drop the bulb first on floor 3, then 1, then 2. the number of tries is 3 because if K is equal to 3 or 2, it will take 3 tests to determine K. The fact that it would take two tests to determine K if K = 1 is irrelevant)
I hope this is stated clear enough so that no one needs to ask questions
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I've discovered the thread yesterday and i've spent most of today working on the xkcd one and i have the solution.
It's not easy, but damn it's a brilliant problem.
And the solution is so brilliantly symmetric and triangular. :D
+ Show Spoiler +Imagine you are one blue eyed person.
If on the island there was only one blue eyed person, then on the first night you would leave because you would see nobody else with blue eyes.
If on the island there were 2 ppl with blue eyes they would see each other, so you would not leave on the first night, because the other one could be the only blue eyed person on the island. But then you would leave on the second night, because the only possible reason the other blue eyed person didn't leave would be if he saw a blue eyed person other than himself, but you only see him with blue eyes, so you must be the other blue eyed person.
Now it gets tricky.
If on the island there were 3 ppl with blue eyes then you would see 2 blue eyed people and you would imagine that each of them saw only 1. By the same reasoning as above they would leave on the second day, but they do not. Why? That's because they must see someone else with blue eyes, and that lucky guy must be you.
And so on if there were 4 ppl. You would see only 3 and they would leave on the third day, because they would follow the reasoning above. But they do not leave on the third day, there must be an extra blue eyed person, and that would be you.
Now it gets very tricky (but also unnecessary).
Imagine you are one of the 100 blue eyed people. You would see 99 people who are blue eyed, and you would think that each of those blue eyed people thinks that there are 98 blue eyed people.
Now imagine you are one of the 99 imaginary blue eyed people. You would see 98 blue eyed people, and you would think that each of those people saw 97 blue eyed people.
Now again imagine you are one of those imaginary 98 blue eyed guys, you would see 97 blue eyed people and you would think that each of those ppl saw 96 blue eyed people.
And so on.
Those imaginary people by the logic i showed before would leave after as many days as there are blue eyed people on the island.
So you wait untill the 99th day and you would think that the 99 blue eyed people you see would all leave because those people imagined that there were 98 but those 98 didn't leave on the 98th day.
And then you are ecstatic to find out that they don't leave! That must mean that there are 100 blue eyed people on the island, and you are one of them.
Now imagine you are a brown eyed person. And now it gets sad.
You are at the 100th day are you are very nervous, because you see 100 blue eyed people and you know that if those suckers don't leave tonight, then, oh baby baby, you have a ferry ticket off the god awful silent island. Miami beach here i come! Right?
And you know what? Those lucky bastards on that night pack their bags and wave you good bye. No wait, they do fucking not! They don't even communicate to you.
And to think that the only thing you need to leave the island is that someone speaks up and says:
"I can see someone who has brown eyes."
And yet, you see so many of them.
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About the Monty hall problem.
(Monty Hall?)
On April 10 2011 02:00 Tunks wrote:
How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though.
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Assume an extreme alternative: 1000 pigeons on a fence, 999 males and one female.
I gotta try and find the female pigeon. I pick one with a chance of 1/1000, now the host says "let me remove some males" and he removes 998 male pigeons from the fence. Now its clear that I will have a better chance switching my choice (999/1000)
Now imagine the same experiment, but right after I picked my first pigeon, lightning strikes on the fence, and 998 pigeons drops dead from the electrocution, only 2 remains, my first pick and some other pigeon. Now the host and his helpers discovers that all the dead pigeons are in fact males. If I assume the lightning was completely independent from my first pigeon-pick, and all the dead pigeons are males, my original pigeon has now changed probability to ½, and I will not gain anything from a switch (but wont lose anything either)
I know the difference between these two scenarios has to do with conditional probability, but I´m not a mathematician so I cant explain exactly how my pigeon went from 1/1000 to ½ probability. Interesting isn't it?
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On April 14 2011 15:06 xxpack09 wrote:Show nested quote +On April 14 2011 09:51 JeeJee wrote:On April 13 2011 14:18 MusicalPulse wrote:
Friend asked me this one, kinda like the other 3 person truth/lie one.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm.. here we go i think + Show Spoiler +
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
if da=yes
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B)
JAJA -> C is random god
DAJA -> A is random god
JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case)
"if i were to ask you, are you the lying god, would you say da?
if da=yes
trueGod: ja
falseGod: da
if da=no
trueGod: ja
falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
I don't see how this works. + Show Spoiler +If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
+ Show Spoiler +
he's not randomly deciding to say yes or no
he's randomly deciding to lie/tell the truth as i understand it
big difference.
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If A is made of consecutive integers then the sum/difference can be written as Hankle matrices, where |A+A| = |A-A|.
A+A
1 2 3 4
1 [2 3 4 5]
2 [3 4 5 6]
3 [4 5 6 7]
4 [5 6 7 8]
A-A
1 2 3 4
4 [3 2 1 0]
3 [2 1 0 -1]
2 [1 0 -1 -2]
1 [0 -1 -2 -3]
Things get interesting when the integers are non-consecutive.
This is equivalent to removing rows/columns from the matrices.
Since the column and rows removed from the A+A matrix are the same that will always leave you with a symmetric matrix. The A-A will possibly be symmetric and in those cases the number of elements will be the same. But in all cases where A-A is not symmetric it will have more distinct elements than A+A.
A+A
1 # 3 4
1 [2 # 4 5] [2 4 5]
# [# # # #] = [4 6 7]
3 [4 # 6 7] [5 7 8]
4 [5 # 7 8]
A-A
1 # 3 4
4 [3 # 1 0] [3 1 0]
3 [2 # 0 -1] = [2 0 -1]
# [# # # #] [0 -2 -3]
1 [0 # -2 -3]
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On April 15 2011 14:16 JeeJee wrote:Show nested quote +On April 14 2011 15:06 xxpack09 wrote:On April 14 2011 09:51 JeeJee wrote:On April 13 2011 14:18 MusicalPulse wrote:
Friend asked me this one, kinda like the other 3 person truth/lie one.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm.. here we go i think + Show Spoiler +
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
if da=yes
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B)
JAJA -> C is random god
DAJA -> A is random god
JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case)
"if i were to ask you, are you the lying god, would you say da?
if da=yes
trueGod: ja
falseGod: da
if da=no
trueGod: ja
falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
I don't see how this works. + Show Spoiler +If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no. + Show Spoiler +
he's not randomly deciding to say yes or no
he's randomly deciding to lie/tell the truth as i understand it
big difference.
+ Show Spoiler +But think about it this way
Say he decides to tell the truth.
He doesn't know whether he would say da or ja since it's completely up to random chance. Just because he has randomly decided to tell the truth when being asked "if i were to ask you, are you the lying god, would you say da?" doesn't mean he would tell the truth when being asked "are you the lying god." It seems to me like this second randomization breaks the problem making him unable to give an adequate answer
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On April 19 2011 11:50 xxpack09 wrote:Show nested quote +On April 15 2011 14:16 JeeJee wrote:On April 14 2011 15:06 xxpack09 wrote:On April 14 2011 09:51 JeeJee wrote:On April 13 2011 14:18 MusicalPulse wrote:
Friend asked me this one, kinda like the other 3 person truth/lie one.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which. edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm.. here we go i think + Show Spoiler +
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
if da=yes
trueGod: ja
falseGod: ja
randomGodDecidingToLie:da
randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B)
JAJA -> C is random god
DAJA -> A is random god
JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case)
"if i were to ask you, are you the lying god, would you say da?
if da=yes
trueGod: ja
falseGod: da
if da=no
trueGod: ja
falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
I don't see how this works. + Show Spoiler +If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no. + Show Spoiler +
he's not randomly deciding to say yes or no
he's randomly deciding to lie/tell the truth as i understand it
big difference.
+ Show Spoiler +But think about it this way
Say he decides to tell the truth.
He doesn't know whether he would say da or ja since it's completely up to random chance. Just because he has randomly decided to tell the truth when being asked "if i were to ask you, are you the lying god, would you say da?" doesn't mean he would tell the truth when being asked "are you the lying god." It seems to me like this second randomization breaks the problem making him unable to give an adequate answer
kind of a bump since i forgot about this but
+ Show Spoiler +
huh?
i don't think you get it
it doesn't matter if he's lying or not, he'll end up giving the same answer due to the way the question is structured
i guess the main thing is that you're thinking he can choose to both lie AND tell the truth in order to answer this one question, but the original question, as given, explicitly forbids that. it's random as to which one he's doing, but he is doing one of them.
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It's about Problem #4 (when do the clock hands (hour and minute hand) meet again after 12pm?)
Edit: Oh, wait, I just totally failed and somewhat mistook minute and second hand, zomfg x___x
No surprise I mistook so many right answers for wrong ones with these incredible flaw in my idea.
Solution as I intended, with right scales:
-> They won't meet again before 13:00 (1:00) for obvious reasons
-> will meet before 14:00 (2:00)
Defining "1" as one time around the clock:
For hour hand:
1/12 (starting point) + 1/720m (change per minute)
For minute hand:
0 (starting point) + 1/60m (change per minute)
12+1/720x = 1/60m
-> m = 60/11 = 5+5/11 (approximately 5 and a half)
-> Hands will meet again at approx. 13:05:30
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(editing fail, delete please)
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Here is the classic solution to the hour minute hand problem (I hope this is the one you were referring to):
After one revolution, the minute hand has hit the hour hand 0 times. After two revolutions, it has hit the hour hand 1 time.
So after 12 revolutions of the minute hand, it has hit the hour hand exactly 11 times. Because of even spacing, we can deduce that it takes 12/11th of an hour for the minute hand to hit the hour hand. So the next time we have an intersection is about 1 o clock + 1/11th of an hour (almost 6 minutes)
Of course, this is just a prettier restatement of the algebraic solution:
In one hour, the hour hand moves 30 degrees while the minute hand moves 360 degrees. Let x be the number of hours so we should write:
360+30x = 360x or x = 360/330 = 12/11 of an hour.
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Hello gents!
I'm having some trouble solving a problem and I hope someone of you fine people can shed some light over it. It probably require that you have studied some mathematics.
Determine a positive integer and a negative integer such as both integers by division with 153 gives the rest/remainder 7 and such as both integers by division with 177 gives the rest/remainder 4.
I'm sorry if it doesn't make any sense but I've translated the text from 4 different language and I finally understand the meaning of, lost in translation.
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On April 15 2011 14:14 noob4ever wrote: but I´m not a mathematician so I cant explain exactly how my pigeon went from 1/1000 to ½ probability. Interesting isn't it?
It didn't.
That situation is identical to selective removal of pigeons.
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OK one of my favorite problems (hopefully not posted here already--it's somewhat of a classic).
Find all partitions of the positive integers into 2 sets S and T such for any positive integer n, there are exactly as many ways of expressing it as the sum of 2 distinct integers from S as there are from T.
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Ok, ever since I've found this thread I've been trying to remember some puzzles, here are a few...
An odd number of soldiers are stationed in a field, in such a way that all the pairwise distances between all soldiers are distinct. Each soldier is instructed to watch the soldier closest to himself.
Prove that at least one soldier is not being watched.
Solution:
+ Show Spoiler + (Proof by contradiction): Consider the two soldiers at the shortest distance from each other. They are watching each other; if anyone else is watching one of them, then we have a soldier being watched twice (and therefore another would not be watched). Otherwise, we could remove the pair from the field without affecting the watching patterns of the others. Continuing in this manner, we would be left with one soldier left in the field, not watching anyone, the contradiction.
Mike and Jane go to a dinner party with four other couples; each person there shakes hands with everyone he or she doesn't know. Later, Mike does a survey and discovers that every one of the nine other attendees shook hands with a different number of people.
How many people did Jane shake hands with?
Solution:
+ Show Spoiler +The key here is that Jane is the partner of the one person NOT polled. Since each person shook hands with at most 8 people, the answers Mike receives are one each from the set [0, 1,..., 8]. Consider a couple named A and B, who scored 8 and 0, respectively. They must have been partners, since otherwise the opportunity to shake each other's hands would've ruined their scores. Consider C and D, as well, with scores 1 and 7. They too must be a couple, since C shook A's hand, and D must've missed B; if C and D weren't couples, their scores would be different. Similarily, the scores 2-6 and 3-5 must be couples, leaving Mike and Jane shaking hands with only the high scorers. Therefore, both Mike and Jane shook hands with 4 people.
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EPT
