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On April 20 2012 15:40 Aelfric wrote:Show nested quote +On April 20 2012 15:31 iTzSnypah wrote:On April 20 2012 15:06 Aelfric wrote:Wow you guys are fast, then lets try something harder: You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest? + Show Spoiler +
solution: 12 Races.
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out.
you have all the horses that placed 2nd race against eachother. 4/5 are out.
you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
+ Show Spoiler +While i can say i liked the beginning of your solution, the answer you give is wrong.
I fixed it after realising something. EDIT nvm i failed
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On April 20 2012 15:37 Taekwon wrote:Show nested quote +On April 20 2012 15:13 sausageslayer wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? + Show Spoiler +Boy B, he just said A and D didn´t do it. but no one said he didn´t do it + Show Spoiler +lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking.
+ Show Spoiler + Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example:
I live in the US and own a computer. That is a true statement.
I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one:
I am a man who attends college. That is true.
I am a woman who attends college. A lie, even though i do attend college.
The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
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On April 20 2012 15:39 sausageslayer wrote:Show nested quote +On April 20 2012 15:36 Mr.F. wrote:On April 20 2012 15:06 Aelfric wrote:
Wow you guys are fast, then lets try something harder:
You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest? + Show Spoiler +i think the answer is 11. you have 5 races initially, and since you only want the top 3 fastest horses, you can say that the slowest 2 horses from each group couldn't possibly be the "top 3" so you can count them out. you are left with 15 horses. you then have 3 more races between the remaining horses. again repeat the process of negating the slowest 2 from each group. after this you are left with 9 horses. have 1 race with 5 of the horses and negate the bottom two again. after that you are only left with 7 horses. have another race with 5 of them, and negate the bottom two. there are now 5 horses left. have one more race and pick the top 3.
does this make sense?
+ Show Spoiler +Was thinking something like this aswell, as far as i can see this is the least amount of races
+ Show Spoiler +Well, let me tell you this. The sollution you give might work. But you can find top 3 with less than 11 races. So the answer is lesser.
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On April 20 2012 15:40 Aelfric wrote:Show nested quote +On April 20 2012 15:31 iTzSnypah wrote:On April 20 2012 15:06 Aelfric wrote:Wow you guys are fast, then lets try something harder: You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest? + Show Spoiler +
solution: 12 Races.
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out.
you have all the horses that placed 2nd race against eachother. 4/5 are out.
you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
+ Show Spoiler +While i can say i liked the beginning of your solution, the answer you give is wrong.
+ Show Spoiler +
iTzSnypah's answer inspires me
Answer is 7 races (surprisingly few IMO)
#1) 5 Races - Divide into 5 group , 5 horse each
the bottom 2 of each race are eliminated, 15/ 25 Left
#2)The winner of each group go into a second round
the winner of the race is the top horse
the 1st runner of this race stays (1st runner run of his group in #1 stays, 2nd runner up of his group in #2 eliminated as no chance to be in top 3)
the 2nd runner of this race up stays (all other from his group in #1 got eliminated )
4th / 5th guy and all others belong to their group in #1 eliminated
6 lefts, with the top 1 determined
#3) the remaining five with uncertain seating run the last one
<--this sounds a lot like dual tournament format for some part
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On April 20 2012 15:02 solidbebe wrote:Show nested quote +On April 20 2012 15:00 LAN-f34r wrote:On April 20 2012 11:51 Whitewing wrote:On April 20 2012 11:30 whatwhatanut wrote:On April 20 2012 10:51 drew-chan wrote:On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? + Show Spoiler +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory. + Show Spoiler +i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years. + Show Spoiler + Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small. + Show Spoiler +The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him. + Show Spoiler +Then he just flips it down and starts counting from there.
+ Show Spoiler +No he can't. He doesn't know if the switch is up because someone was in before him, or if its because he is the first one in there and it started up.
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On April 20 2012 15:51 Taulon wrote:Show nested quote +On April 20 2012 15:37 Taekwon wrote:On April 20 2012 15:13 sausageslayer wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? + Show Spoiler +Boy B, he just said A and D didn´t do it. but no one said he didn´t do it + Show Spoiler +lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking. + Show Spoiler + Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example:
I live in the US and own a computer. That is a true statement.
I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one:
I am a man who attends college. That is true.
I am a woman who attends college. A lie, even though i do attend college.
The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie.
+ Show Spoiler +Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
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On April 20 2012 16:20 Taekwon wrote:Show nested quote +On April 20 2012 15:51 Taulon wrote:On April 20 2012 15:37 Taekwon wrote:On April 20 2012 15:13 sausageslayer wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? + Show Spoiler +Boy B, he just said A and D didn´t do it. but no one said he didn´t do it + Show Spoiler +lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking. + Show Spoiler + Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example:
I live in the US and own a computer. That is a true statement.
I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one:
I am a man who attends college. That is true.
I am a woman who attends college. A lie, even though i do attend college.
The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie. + Show Spoiler +Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation.
+ Show Spoiler +It is still in 2 parts, even though it is worded differently.
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On April 20 2012 15:40 CarelessPride wrote:Show nested quote +On April 20 2012 15:00 LAN-f34r wrote:On April 20 2012 11:51 Whitewing wrote:On April 20 2012 11:30 whatwhatanut wrote:On April 20 2012 10:51 drew-chan wrote:On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? + Show Spoiler +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory. + Show Spoiler +i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years. + Show Spoiler + Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small. + Show Spoiler +The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him. + Show Spoiler +
1 person is the counter. the1st time the counter is in the room he moves switch no1 to off does not count anything but notes the position of the switches. Anyone else who comes in the room moves switch no1 to on if the switch is off. if its on then just moves switch no2. Everytime the counter goes in the room. He adds +1 if switch no1 is moved or adds 2 if both switches have changed position since he was last here. the minimum time it would take would be 11 visits or the maximum of 66. 66 being he visits 3 times ina row in between every unique visit..
btw i really want to know if there's an more efficient way to do this....
+ Show Spoiler +
I don't think that works. In fact I can't see how this riddle is suppose to be solved with a counter. I think it needs to be more elaborate then that, and I don't have a clue as to how it's suppose to be solved. I would go ahead and just wait for a really really long time as someone else suggested and just bet on it
Lets say the counter comes in and sees
ON ON-so he turns on 1
#1 (the counter) leaves the switches as OFF ON
#2 changes it to ON ON
#3 changes it to ON OFF
#1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
#2 changes it to ON ON
#3 changes it to ON OFF
#1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
The problem here is that theoretically #2 and #3, followed by #1, in an endless cycle. 1,2,3,1,2,3,1,2,3, and eventually #1 will reach any agreed upon number, (1000) and say "we've all been to the switch room" and be wrong, since in fact only 3 out of the 23 people have been there.
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Here is an easy one:
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?
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On April 20 2012 16:40 phiinix wrote:Show nested quote +On April 20 2012 15:40 CarelessPride wrote:On April 20 2012 15:00 LAN-f34r wrote:On April 20 2012 11:51 Whitewing wrote:On April 20 2012 11:30 whatwhatanut wrote:On April 20 2012 10:51 drew-chan wrote:On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? + Show Spoiler +
Someone is elected as the counter. Everytime the counter enters the room, he sees the position of the switches. If it is odd, as in 1 on and the other off, he counts it as 1, and sets it to even, both on or off. Other prisoners will set the switch to odd if even, and not touch it if odd. Prisoners who have already set it to odd once will not do it again. When the counter counts 22 he knows that all prisoners have set it to odd once.
Also if the switches are known at the beginning you can just tell everyone to only change switches twice and count up till 44 or something.
i'm pretty darn sure you're horribly wrong. The prisoners being required to switch a switch ruins yoru theory. + Show Spoiler +i think the only way to do it is nvr can switch 1 unless it is up so that after time when all are in the down position you can assume from the duration of time that they are all down i would say ~5 years. + Show Spoiler + Solution is more or less the same, simply have one of the two switches be the one the counter uses, and have people who aren't going to modify it flip the other switch. Say there's one on the left and one on the right, have the counter only count when the left switch is up, and have him flip it down when he goes in. Everyone else only flips the right switch, unless the left switch is down, then they flip it up. They only ever flip the left switch up twice, never more than that, even if it is down.
Personally, I prefer the method of "wait until I've gone 50 times." Assuming the selection is truly random and that the probability of choosing any given prisoner is approximately the same as choosing any other prisoner, the odds of having been chosen 50 times before every other person has gone at least once is absurdly small. + Show Spoiler +The problem is that the first time the counter goes in, he doesn't know if the switch is up because it started that way or because someone else entered before him. + Show Spoiler +
1 person is the counter. the1st time the counter is in the room he moves switch no1 to off does not count anything but notes the position of the switches. Anyone else who comes in the room moves switch no1 to on if the switch is off. if its on then just moves switch no2. Everytime the counter goes in the room. He adds +1 if switch no1 is moved or adds 2 if both switches have changed position since he was last here. the minimum time it would take would be 11 visits or the maximum of 66. 66 being he visits 3 times ina row in between every unique visit..
btw i really want to know if there's an more efficient way to do this.... + Show Spoiler +
I don't think that works. In fact I can't see how this riddle is suppose to be solved with a counter. I think it needs to be more elaborate then that, and I don't have a clue as to how it's suppose to be solved. I would go ahead and just wait for a really really long time as someone else suggested and just bet on it
Lets say the counter comes in and sees
ON ON-so he turns on 1
#1 (the counter) leaves the switches as OFF ON
#2 changes it to ON ON
#3 changes it to ON OFF
#1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
#2 changes it to ON ON
#3 changes it to ON OFF
#1 (the counter) sees that the switches are both different from when he was there, so he adds +2 and changes the switches back to OFF ON
The problem here is that theoretically #2 and #3, followed by #1, in an endless cycle. 1,2,3,1,2,3,1,2,3, and eventually #1 will reach any agreed upon number, (1000) and say "we've all been to the switch room" and be wrong, since in fact only 3 out of the 23 people have been there.
+ Show Spoiler +I got stuck for a while with this...i could get the count to 22 but couldn't get the last one due to not knowing the starting positions of the switches.
The theory i used was pretty much the same as anyone elses, if the left switch was up, the prisoner would flick it down. they could only do it once, and had to be the first time they visited with the switch being up. IF they visited and it was down then they just flick the right switch. The counter always flicks the left switch up. The only circumstance in which the counter would see the switch was UP when he got there, was if he was the first prisoner, in which case he could easily count the 23 prisoners. Seeing as you dont know whether it started up or down, you have to assume it started down, and cant guarantee that the 23rd prisoner has flicked it down( the very first to enter the room, but #23 for your count).
BUT if each prisoner flicks the left switch down TWICE and the count goes to 45 he can guarantee that all 23 prisoners have been atleast once.
He would get a count of 44 for 22 prisoners(including himself, and the 45th would be the gaurantee that the last person had entered the room too. You can't say a count of 46 because again you wouldn't know if that very first time the counter entered and saw the switch as down whether it was due to a prisoner or starting position.
I hope that makes sense to someone else @@
so simply my solution is:
If the counter enters the first time and sees the switch UP, he counts to 23(including himself twice) from his next visit.
If the counter enters the first time and sees the switch DOWN, he counts to 45(including himself twice) from his next visit.
The strategy being:
The first 2 times that each prinsoner enters and sees the left switch UP, they flick it down.
Every time after that, or if they visit and have not flicked it down twice, but the switch is already down, they flick the right switch instead.
Only the counter flicks the left switch up, and only flicks it if it is down, otherwise flicks the right switch.
The counter adds +1 for every visit after the first, in which he finds the left switch down, and flicks it up.
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On April 20 2012 16:09 Phantom_Sky wrote:Show nested quote +On April 20 2012 15:40 Aelfric wrote:On April 20 2012 15:31 iTzSnypah wrote:On April 20 2012 15:06 Aelfric wrote:Wow you guys are fast, then lets try something harder: You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest? + Show Spoiler +
solution: 12 Races.
you have 5 5horse races. horses that place 4/5 are out. you write down what places the remaining horses place.
15 horses left in contention.
you have all the horses that placed 1st race against eachother. 4/5 are out.
you have all the horses that placed 2nd race against eachother. 4/5 are out.
you have all the horses that placed 3rd race against eachother. 4/5 are out.
you write down what places the remaining horses place.
The horse that placed 1st both races is your fastest horse.
8 horses left in contention.
you have 2 races of 4 horses each. 4th is out
6 horses left in contention.
you race 5 of your last 6 horses. 1 horse sits out. 4/5 are out.
4 horses left.
you race them. 1/2 are your 2nd/3rd fastest horse in that order.
+ Show Spoiler +While i can say i liked the beginning of your solution, the answer you give is wrong. + Show Spoiler +
iTzSnypah's answer inspires me
Answer is 7 races (surprisingly few IMO)
#1) 5 Races - Divide into 5 group , 5 horse each
the bottom 2 of each race are eliminated, 15/ 25 Left
#2)The winner of each group go into a second round
the winner of the race is the top horse
the 1st runner of this race stays (1st runner run of his group in #1 stays, 2nd runner up of his group in #2 eliminated as no chance to be in top 3)
the 2nd runner of this race up stays (all other from his group in #1 got eliminated )
4th / 5th guy and all others belong to their group in #1 eliminated
6 lefts, with the top 1 determined
#3) the remaining five with uncertain seating run the last one
<--this sounds a lot like dual tournament format for some part
horse race
+ Show Spoiler +
5 races with groups of 5.
Race the winners of each group against each other. You got the fastest horse. Keep 2nd and 3rd (Lets call them #1, #2)
Race the 2nd place finishers from the group stages with each other. Keep 1st and 2nd (#3, #4)
Race the 3rd place from the group stages with each other. Keep the winner here (#5)
Race #1-#5. Winner here is the second fastest. Runner up is third fastest.
So thats... 9?
actually im wrong lol. I just couldnt understand your wording.
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On April 20 2012 16:30 LAN-f34r wrote:Show nested quote +On April 20 2012 16:20 Taekwon wrote:On April 20 2012 15:51 Taulon wrote:On April 20 2012 15:37 Taekwon wrote:On April 20 2012 15:13 sausageslayer wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? + Show Spoiler +Boy B, he just said A and D didn´t do it. but no one said he didn´t do it + Show Spoiler +lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking. + Show Spoiler + Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example:
I live in the US and own a computer. That is a true statement.
I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one:
I am a man who attends college. That is true.
I am a woman who attends college. A lie, even though i do attend college.
The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie. + Show Spoiler +Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation. + Show Spoiler +It is still in 2 parts, even though it is worded differently.
+ Show Spoiler + Taekwon have you taken a sentential logic class before? Just curious. Because I still think Taulon is correct in his logic.
A=I will punch you in the face
B=I will take your wallet
Either A or B
What makes this sentence true?
1. A (I punch you in the face)
(or)
2. B (I take your wallet)
What makes this sentence false?
1. A&B (I both punch you in the face and take your wallet) This makes my bolded statement false because I said I will do either A OR B, not A AND B)
(or)
2. Not A&B (I don't punch you in the face, and I also don't take your wallet) This makes the bolded statement false because I said Either A Or B. I will pick one.
-In other words, I can do one or the other to be true. If I do both or neither, it's false. This is because I used the word "or" and not "and"
Now, lets say instead:
Neither A nor B
What makes this sentence true?
1. Not A and Not B (I don't go to the store, and don't play sc2)
What makes this sentence false?
1. A and Not B (I go to the store and don't play sc2)
2. Not A and B (I go to the store and play sc2)
3. Not A&B (I don't go to the store and play sc2)
Long story short: In sentential logic, the phrase A nor B translates to Not A and Not B, or not (A or B)
For reference:
A: "I didn't eat it!"
B: "Neither A nor D ate it."
C: "I swear on my momma's grave I didn't eat it!"
D: "C is telling the truth!"
Lets take B's sentence. Neither A nor D ate it. This means that A did not eat it, AND D did not eat it. They both have to be in conjunction, or in other words, you can't have one without the other.
Lets pretend we KNOW B is lying. We can't 100% deduce that both A and D ate the cake. If A ate the cake, B is lying. If D ate the cake, B is lying. If both ate the cake, B is lying.
Imo the riddle is worded a bit poorly if Not (Not A nor D) is suppose to translate to A&D
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On April 20 2012 15:06 Aelfric wrote:Wow you guys are fast, then lets try something harder: Show nested quote +You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
+ Show Spoiler +
7 races. Break the group into 5 sets called A-E, and number determine their ranking in that set.
Race1-5:A,B,C,D,E.
Race6: A1,B1,C1,D1,E1.
Say that ABC are the fastest in Race 6. We know A1 is the fastest horse.
Race 7: A2,A3,B1,B2,C1
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On April 20 2012 17:33 phiinix wrote:Show nested quote +On April 20 2012 16:30 LAN-f34r wrote:On April 20 2012 16:20 Taekwon wrote:On April 20 2012 15:51 Taulon wrote:On April 20 2012 15:37 Taekwon wrote:On April 20 2012 15:13 sausageslayer wrote:On April 20 2012 03:36 Taekwon wrote:
A boy comes around and yells at four of his friends: "HEY! Who ate my cake?! I was saving it for later!"
Girl A said: "I didn't eat it!"
Boy B said: "Neither A nor D ate it."
Boy C said: "I swear on my momma's grave I didn't eat it!"
Girl D said: "C is telling the truth!"
One of these children is lying. Who is it? + Show Spoiler +Boy B, he just said A and D didn´t do it. but no one said he didn´t do it + Show Spoiler +lol, since this is the third time B has been selected as the answer (including the same logic), I'll clarify something in regards to the matter. It's a simple logical oversight: if B is the liar, that would also mean that there are two cake eaters, which as indicated in the original riddle, is not the case. Remember to focus on what the riddle is asking. + Show Spoiler + Not true, A person can be lying without every single thing in their sentence being untrue. If half the sentence is false, it is still a lie. For example:
I live in the US and own a computer. That is a true statement.
I live in the US, and do not own a computer. That is a lie, even though part of the statement is true.
If you don't like that example, try this one:
I am a man who attends college. That is true.
I am a woman who attends college. A lie, even though i do attend college.
The same applies to this scenario. B's statement is "A didn't eat it" and "D didn't eat it". For him to be lying, only one of those has to be false. A didn't eat the cake, that must be true. However, D did eat the cake. since part of his statement was false the statement was a lie. + Show Spoiler +Why uh...thank you for the rather untoward presentation of lies but what you're disputing has honestly has no relevancy to the riddle at hand. At point blank, you're examples are also a facetious misrepresentation of girl B's statement. She sad neither A nor D ate the cake as one statement. If I were to take examples to your, I suppose flippant, level of abstract, it would be: Neither China nor the US nuked Russia. She made only a single claim that involves both parties, there is no "part" to it. If one of the two nuked Russia, that would have to mean the other nation did too. Or else the girl is lying; ergo there are more than two nukers. Juxtapose that with the cake, and you presto. Hence, your lecture didn't..establish much of anything.
P.S. No offense to Russians, nobody was nuked in the crafting of this explanation. + Show Spoiler +It is still in 2 parts, even though it is worded differently. + Show Spoiler + Taekwon have you taken a sentential logic class before? Just curious. Because I still think Taulon is correct in his logic.
A=I will punch you in the face
B=I will take your wallet
Either A or B
What makes this sentence true?
1. A (I punch you in the face)
(or)
2. B (I take your wallet)
What makes this sentence false?
1. A&B (I both punch you in the face and take your wallet) This makes my bolded statement false because I said I will do either A OR B, not A AND B)
(or)
2. Not A&B (I don't punch you in the face, and I also don't take your wallet) This makes the bolded statement false because I said Either A Or B. I will pick one.
-In other words, I can do one or the other to be true. If I do both or neither, it's false. This is because I used the word "or" and not "and"
Now, lets say instead:
Neither A nor B
What makes this sentence true?
1. Not A and Not B (I don't go to the store, and don't play sc2)
What makes this sentence false?
1. A and Not B (I go to the store and don't play sc2)
2. Not A and B (I go to the store and play sc2)
3. Not A&B (I don't go to the store and play sc2)
Long story short: In sentential logic, the phrase A nor B translates to Not A and Not B, or not (A or B)
For reference:
A: "I didn't eat it!"
B: "Neither A nor D ate it."
C: "I swear on my momma's grave I didn't eat it!"
D: "C is telling the truth!"
Lets take B's sentence. Neither A nor D ate it. This means that A did not eat it, AND D did not eat it. They both have to be in conjunction, or in other words, you can't have one without the other.
Lets pretend we KNOW B is lying. We can't 100% deduce that both A and D ate the cake. If A ate the cake, B is lying. If D ate the cake, B is lying. If both ate the cake, B is lying.
Imo the riddle is worded a bit poorly if Not (Not A nor D) is suppose to translate to A&D
+ Show Spoiler + Its Girl C. Because her mother isn't dead because you didn't supply background on any of the kids.
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Hmmm the warden problem is difficult. I wrote a really long solution, then before I posted it I realized there was a hole in my theory and scrapped it 
The biggest problem is the lack of communication between the prisoners, since they have no way to relay information (unless there's something I'm missing)
EDIT: I missed the part where they were able to collude with each other at the start, before they go into individual cells. So the solution I thought up was correct after all haha.
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On April 20 2012 15:06 Aelfric wrote:Wow you guys are fast, then lets try something harder: Show nested quote +You have 25 horses. When they race, each horse runs at a different, constant pace. A horse will always run at the same pace no matter how many times it races.
You want to figure out which are your 3 fastest horses. You are allowed to race at most 5 horses against each other at a time. You don't have a stopwatch so all you can learn from each race is which order the horses finish in.
What is the least number of races you can conduct to figure out which 3 horses are fastest?
Just realized A very IMPORTANT detail.
+ Show Spoiler +
9 Races total
5 races of 5 horses. 4/5 out
15 left
Winners of first race race. winner is fastest horse. 4/5 out.
2nd Place of first race race. 3/4/5 out. (if a horse got second place in the first race he CAN'T be fastest horse so only 2 go on)
3rd Place of first race race. 2/3/4/5 out. (if a horse got 3rd place in the first race he can ONLY at best be 3rd fastest horse 1 goes 1)
5 left
Race 5. 1/2 are 2nd/3rd fastest.
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On April 20 2012 07:06 Excludos wrote:Show nested quote +On April 20 2012 06:51 TheAngryZergling wrote:On April 20 2012 02:33 XiGua wrote:I am betting that nobody will solve this without cheating. I present to you, THE WARDEN The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell. "No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead. "Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you" What is the strategy they come up with so that they can be free? + Show Spoiler +
there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.
when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.
+ Show Spoiler +Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.
I address that point explicitly in my original solution in the following sentence:
+ Show Spoiler +
his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
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On April 20 2012 17:59 iTzSnypah wrote:ABOUT the Warden Puzzle: The OP did not include 1 Crucial statement. + Show Spoiler [THIS IS THE ANSWER] +http://www.braingle.com/brainteasers/teaser.php?op=2&id=18012&comm=0
please provide the crucial statement, not only the answer which I would rather not see.
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On April 20 2012 18:13 TheAngryZergling wrote:please provide the crucial statement, not only the answer which I would rather not see.
Switches are down position to begin with.
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EPT![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
