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On March 18 2009 16:58 motbob wrote:Show nested quote +On March 18 2009 03:57 Geo.Rion wrote:
Riddle:
You are on the top of a mountain, which is 1000m high. On halfway down you have some sort of rock where you can stop. You got a rope which is 750m long, and a knife. How do you get down from the mountain alive, without using any other stuff. You can seize your rope only at halfway/top of the mountain. + Show Spoiler +Cut the rope into 2 pieces, one 500m and one 250m.
Make a noose at the end of the 250m piece
Slip the 500m piece through. You now have a rope that is 500m.
Slide down to the rock and stop. Pull the rope through the noose
Slide down the the bottom
fixed
+ Show Spoiler +Cut the rope into 2 pieces, one 500m and one 250m.
Make a noose at the end of the 250m piece
Slip neck through the noose.
Slide down to the rock and stop. Pull the rope tight
Jump down to the bottom
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On March 17 2009 01:18 fanatacist wrote:Show nested quote +On March 17 2009 00:51 gondolin wrote:+ Show Spoiler +On March 17 2009 00:32 fanatacist wrote: Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p) What the FUCK are you talking about??
As i said i was just being a smart-ass. The game in riddles is to find a way not to follow the rules, so I could have said "hey if the monk can teleport/phase shift" the riddle is false, and this is exactly what i did except i only changed the mathematical model.
(Warning: there have been 4 years since i took a logical course, so what i say there may be somewhat inexact/false).
As you may know, when you have a true (recursively enumerable) formal system sufficiently strong to modelize arithmetic (let's take Peano), it is incomplete, namely it can't prove the theorem:
G: the theorem (G) is non provable
So even if G is true (in the naturals), there are models of the system where G is false (for instance Peano+Not Consis(Peano) which is consistent by Godel theorem). This model will be like N, except there will be non standard integers greater than all natural numbers (and in fact the godel number of the proof of G will be a non standard integer, since G is true so non provable in the integers).
This may be easier to conceptualize when you work with ZF, there will be models with AC or non AC, the continuum hypothesis, etc, and you can choose your model.
In each of these models, the TVI is true, so to "cheat" i have to relax one of the axiom of classical logic which is the Law of excluded middle. This is called intuitive logic because you can't use Reductio ad absurdum (you don't have Not (Not P)) => P).
Models of classical logic are given by boolean algebras, models of intuitive logic are given by heyting algebras, which are more general. One ("universal") example of heyting algebra are the open sets of a topological space X. If P and Q are two open sets of X, P and Q is given by P meet Q, P or Q by P join Q, and Not(P) by the interior of X\P (True is given by X and False by the empty set of course). This algebra is boolean (classic) iff every open set is regular. So if you prove something in intuitive calculus, this give you statements about open sets of a topological space, and convercely if a statement is true in every topological space, it can be deduced by intuitive logic. (so for instance you can prove that even if Not Not P != P in intuitive logic, you always have Not Not Not P = Not P).
Like in classical logic, you can construct the natural numbers, then the integers, then the rationnals, the the real (but you have to use Dedekinds cut rather than cauchy sequences). What is nice is that if you do this construction on the topos of the topological set X, then |R is the sheaf of continuous functions X -> R (the real R this time... i will denote |R the intuitive model of R in my topos).
Here the TVI may fail, let's say you take X=[0,1], and construct a continuous function F:[0,1]*R->R such that F(0,.) =-1, F(1,.)=1 and there is no continuous function c:U->R such that F(c(t),t)=0 on U (U open).
You then have a function f:|R->|R that takes a section (V,s) and sends it to F(s(t),t). Then f is continuous, f(-1)=-1, f(1)=1, but f(x) ne 0 pour tout x dans [-1,1]. (or maybe you just have Not(it exist x in [-1,1] such that f(x)=0, I think this formulation is weaker in intuitive logic, intuitive logic is just so weird, i always get confused so i just learn of cool results you can get with it :p)/
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Was that really necessary?
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Dude, I am SO glad I passed my math exam last week and will never have to deal with that crap ever again.
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It's funny how when you put an unknown riddle up nobody gets it, but these famous common ones get answers quite fast... You guys need to google a lot less dudes (generalizing, of course)
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On March 19 2009 11:08 Z-BosoN wrote:
It's funny how when you put an unknown riddle up nobody gets it, but these famous common ones get answers quite fast... You guys need to google a lot less dudes (generalizing, of course)
Common = usually easy
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On March 19 2009 11:08 Z-BosoN wrote:
It's funny how when you put an unknown riddle up nobody gets it, but these famous common ones get answers quite fast... You guys need to google a lot less dudes (generalizing, of course)
Do that mean my second attempt is wrong too? Or wasn't you referring to your fan riddle? Now I have to start thinking of a third solution 
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On March 17 2009 22:06 opsayo wrote:
[spoiler]Your solution has one "leader" to tally up counts.
This usually takes around 10,000+ days, or ~28 years. What's interesting is that we can optimize even more. You chose the leader to be the person who enters the room on the second day. If we choose the leader to be the first person to enter the room twice. That is, if the light is still on, then only unique prisoners have entered on everyday. This means that if I enter the room on the 40th day (and it's my second time) and the light is still on, then I know for a fact that 39 unique prisoners have entered the room already. This does a somewhat significant slimming on the solution time (~9000 days instead of ~10,000).
Now with this track of mind, we have been using only one leader to count. But...
What happens if you use more than one? How much does this reduce the average time?
Lastly, what do you need to add or change to the prisoner's planning in order to make it work with the new plan?
[spoiler]
The first prsoner who enters the room twice does not know that he is the first who have done it.
Does the solution with more than 1 leader really exist, i dont understand how several leaders can communicate with each other or with supreme leader without screwing up all procedure.
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On March 20 2009 01:28 15vs1 wrote:Show nested quote +On March 17 2009 22:06 opsayo wrote:+ Show Spoiler +Your solution has one "leader" to tally up counts.
This usually takes around 10,000+ days, or ~28 years. What's interesting is that we can optimize even more. You chose the leader to be the person who enters the room on the second day. If we choose the leader to be the first person to enter the room twice. That is, if the light is still on, then only unique prisoners have entered on everyday. This means that if I enter the room on the 40th day (and it's my second time) and the light is still on, then I know for a fact that 39 unique prisoners have entered the room already. This does a somewhat significant slimming on the solution time (~9000 days instead of ~10,000).
Now with this track of mind, we have been using only one leader to count. But...
What happens if you use more than one? How much does this reduce the average time?
Lastly, what do you need to add or change to the prisoner's planning in order to make it work with the new plan?
The first prsoner who enters the room twice does not know that he is the first who have done it. Does the solution with more than 1 leader really exist, i dont understand how several leaders can communicate with each other or with supreme leader without screwing up all procedure.
This riddle is really hard to explain! It's more a programming problem than a riddle but I still think it's awesome.
You are right the first prisoner to return twice would not know he is first.
That is why you would set aside the first 100 days for this purpose.
Everyone leaves the light off until someone turns up twice. That person then turns the light on and assumes the leader role.
When it gets to day 100 the person on that day turns the light off and you start with the system described before by TanGeng:
+ Show Spoiler +
The first time an ordinary prisoner enters the room and finds the light bulb turned off, the prisoner turns the light bulb on. If the light bulb is on, the prisoner does nothing. If the prisoner enters the room and finds the light bulb turned off a second time, the prisoner does nothing.
Whenever the Leader enters the room, the Leader turns the light bulb off if it is on. If the Leader does this, the Leader adds 1 to the tally. When the tally reaches 99, the leader declares that all prisoners have been inside the room.
Except this time the leader will already know lots of people.. and those people will also know they have been counted so they won't turn the light on again.
Here's an example of how to use multiple people. Let's say you just wanted to use the leader and one other counter. I'm going to call them Leader and Counter.
You have to set this up before it begins.
First you give the Counter a set number of people to count... say 10.
Then you would set aside days for the Counter to check back in:
Say days 900-1000 and every 800-900 after that. "CheckBack days"
The Counter then acts like the Leader. He turns off light bulbs and adds 1 to his tally whenever he does this. When he reaches 9 he stops turning bulbs off.
From then on if the Counter enters the room during any CheckBack day he turns the bulb on. If the Leader enters the room during those days and sees the light on, he knows that the Counter has found 10 people, so he can add a 10 to his tally, all in one go. But of course, he must remember not to add 10 to his tally during the next CheckBack.
Now with computer software you can determine the optimal number of Counters, the optimal number for them to count to and the correct spacing and number of the CheckBack days.
Most likely you would want to wait til close to the end before having check back days so that you give the Counters plenty of time to count people.
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This one made me cry laughing:
the hardest interview puzzle question ever :
A hundred prisoners are each locked in a room with three pirates, one of whom will walk the plank in the morning. Each prisoner has 10 bottles of wine, one of which has been poisoned; and each pirate has 12 coins, one of which is counterfeit and weighs either more or less than a genuine coin. In the room is a single switch, which the prisoner may either leave as it is, or flip. Before being led into the rooms, the prisoners are all made to wear either a red hat or a blue hat; they can see all the other prisoners' hats, but not their own. Meanwhile, a six-digit prime number of monkeys multiply until their digits reverse, then all have to get across a river using a canoe that can hold at most two monkeys at a time. But half the monkeys always lie and the other half always tell the truth. Given that the Nth prisoner knows that one of the monkeys doesn't know that a pirate doesn't know the product of two numbers between 1 and 100 without knowing that the N+1th prisoner has flipped the switch in his room or not after having determined which bottle of wine was poisoned and what color his hat is, what is the solution to this puzzle?
got from http://wiki.xkcd.com/irc/Puzzles
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On March 20 2009 01:28 15vs1 wrote:
The first prsoner who enters the room twice does not know that he is the first who have done it.
Does the solution with more than 1 leader really exist, i dont understand how several leaders can communicate with each other or with supreme leader without screwing up all procedure.
Wrong! + Show Spoiler +If it is your first time in the room, leave the light on! If it is your second time in the room and the light is still on, turn it off and count yourself as the new leader, and tally the amount of days that have passed.
This is commonly referred to as a "snowball round." Here is an in-depth explanation (this is the first introduction to having stages in data collection):
+ Show Spoiler +A common atypical stage, called a snowball round can be applied to several schemes. The snowball round occurs at the beginning of the scheme and (for schemes with leaders) passes out a single badge. The prisoner on day 1 leaves his simple token by turning on the light. Later visitors leave their token by leaving the light on. The first prisoner to visit who is unable to leave a token (either because he has been in the room before and no longer has a token to leave, or because he is the last visitor of the stage, or because special rules of the scheme prevent him from doing so) turns off the light and picks up all the tokens left in the room. This number is determined by the day number. If he visits on day k, then he picks up k-1 tokens. In addition, he is also given a badge (usually, the crown). No one is allowed to turn on the light during the snowball round except the prisoner on day 1, so once the light is turned off, the rest of the round is inactive. The effect of the snowball round is transfer a large number of tokens to a single prisoner in just a few days. Unfortunately, it is only worthwhile when a large majority of the prisoners have tokens to leave, which is why it is only done at the start of the scheme.
+ Show Spoiler +In general doing things this way - using stages to create a leader rather than relying on a pre-defined leader can slim down solutions significantly.
For example if we were two use four counters that each collect 25 tokens (light switch counts) each. During the token collection stage they all have the ability to turn off the light switch.
During the counter collection stage, the first person in the counter room leaves his counter (higher level token worth 25 single tokens), and removes his count. The very next person in the room who finds the light on during this sequence turns off the light, becomes the master token counter, and adds the other players tokens to his total. This is more optimal than relying on one specific person to tally the tokens (since any master counter entering the room can do it).
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[QUOTE]On March 17 2009 16:15 semioldguy wrote:
#2 You come to a fork in the road. You know that one direction leads to certain death while the other will provide safe passage. At this fork you know there is a pair of identical twins except that one of them will always tell the truth and the other will always tell a lie. You can only ask one twin one question before choosing which direction to go. What do you ask to ensure safe passage?
"if i asked you which road to go down, which road would you tell me to down?" (the brother who lies would negate his first lie and tell you the right door)
and "what road would your brother tell me to go down" (go to the opposite door they say the brother would tell you to go through)
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On March 20 2009 16:15 NoNameLoser wrote:This one made me cry laughing: the hardest interview puzzle question ever :A hundred prisoners are each locked in a room with three pirates, one of whom will walk the plank in the morning. Each prisoner has 10 bottles of wine, one of which has been poisoned; and each pirate has 12 coins, one of which is counterfeit and weighs either more or less than a genuine coin. In the room is a single switch, which the prisoner may either leave as it is, or flip. Before being led into the rooms, the prisoners are all made to wear either a red hat or a blue hat; they can see all the other prisoners' hats, but not their own. Meanwhile, a six-digit prime number of monkeys multiply until their digits reverse, then all have to get across a river using a canoe that can hold at most two monkeys at a time. But half the monkeys always lie and the other half always tell the truth. Given that the Nth prisoner knows that one of the monkeys doesn't know that a pirate doesn't know the product of two numbers between 1 and 100 without knowing that the N+1th prisoner has flipped the switch in his room or not after having determined which bottle of wine was poisoned and what color his hat is, what is the solution to this puzzle? got from http://wiki.xkcd.com/irc/Puzzles
God, it took me till the part with the hats to realize that this is a joke. Fail
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Here are a few riddles from one of the games I enjoyed in my youth. None of them are super difficult but they seemed difficult at that age.
A precious gift, this,
Yet it has no end or beginning,
And in the middle, nothing.
+ Show Spoiler +
Answers its caller without being
asked. Responds within seconds, and
speaks all languages with equal ease.
+ Show Spoiler +
Black when bought.
Red when used.
Grey when thrown away.
+ Show Spoiler +
Bloodless and boneless
it travels about.
Yet it never leaves home.
+ Show Spoiler +
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There is a flying boat; the 4 wheels exploded; how many bananas are left?
Answer
+ Show Spoiler +
None!
The elephant ate them all
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On March 24 2009 02:12 Durak wrote:A precious gift, this, Yet it has no end or beginning, And in the middle, nothing. + Show Spoiler +
+ Show Spoiler +reminds me of chuck and larry 
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On March 17 2009 10:46 terr13 wrote:
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion?
somebody please tell me the point/reason/wtfdoesitmean "equally at random" ???
On March 27 2009 17:18 terr13 wrote:
Equally at random means that every day, the chance for any individual to enter the room is 1/100, meaning that they are independent of each other.
As opposed to just "at random" which means.... exactly the same thing? :S
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Equally at random means that every day, the chance for any individual to enter the room is 1/100, meaning that they are independent of each other.
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Durak! You played Betrayal At Krondor also. I loved that 
terr13's riddle is awesome btw. I like the solution.
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On March 27 2009 17:02 Reason wrote:Show nested quote +On March 17 2009 10:46 terr13 wrote:
There are 100 prisoners in solitary cells. There's a central living room with one light bulb; this bulb is initially off. No prisoner can see the light bulb from his or her own cell. Everyday, the warden picks a prisoner equally at random, and that prisoner visits the living room. While there, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting that all 100 prisoners have been to the living room by now. If this assertion is false, all 100 prisoners are shot. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world could always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity. The prisoners are allowed to get together one night in the courtyard, to discuss a plan. What plan should they agree on, so that eventually, someone will make a correct assertion? somebody please tell me the point/reason/wtfdoesitmean "equally at random" ??? Show nested quote +On March 27 2009 17:18 terr13 wrote:
Equally at random means that every day, the chance for any individual to enter the room is 1/100, meaning that they are independent of each other. As opposed to just "at random" which means.... exactly the same thing? :S
English is a language where people put a lot of unnecessary words. Deal with it.
It could mean that every prisoner, including ones that have already gone, would be equally up for choosing. Equally kind of prevents dumbasses from asking that question ("can people go twice lawl?").
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EPT![[image loading]](http://i43.tinypic.com/2cckuh5.png)
