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Riddle:
You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn).
Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
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More Riddle:
You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a daughter?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a daughter. If they do have a daughter, I'll give you $100."
Which family should you pick, or does it not matter?
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On April 20 2012 14:49 Aelfric wrote:More Riddle: Show nested quote +You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a girl?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter?
+ Show Spoiler +Family on the right because they just had a child and the mother is a girl. Family on the left could be a family of all men with gay parents.
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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 + Your story has holes.
Where do they arrive? You didn't state at the prison.
Where can they meet and plan a strategy? Therefore it can be anywhere.
You didn't specify if the switches are fastened to a wall/switchboard. Therefore I can move them.
A very simple answer is that they all go into the switch room on planning day because the warden didn't say that you couldn't. Then they say "We have all visited the switch room."
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On April 20 2012 11:51 Whitewing wrote:Show nested quote +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.
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On April 20 2012 15:00 LAN-f34r wrote:Show nested quote +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.
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On April 20 2012 14:42 Aelfric wrote:
Riddle:
You have two lengths of rope. Each rope has the property that if you light it on fire at one end, it will take exactly 60 minutes to burn to the other end. Note that the ropes will not burn at a consistent speed the entire time (for example, it's possible that the first 90% of a rope will burn in 1 minute, and the last 10% will take the additional 59 minutes to burn).
Given these two ropes and a matchbook, can you find a way to measure out exactly 45 minutes?
+ Show Spoiler +you light both ends of 1 rope and 1 end of the other rope at the same time. when the rope you light at both ends burns out, you light the other end of the other rope. the total time it takes to burn will be 45 mins.
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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?
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On April 20 2012 14:56 zJayy962 wrote:Show nested quote +On April 20 2012 14:49 Aelfric wrote:More Riddle: You and a friend are standing in front of two houses. In each house lives a family with two children. "The family on the left has a boy who loves history, but their other child prefers math," your friend tells you. "The family on the right has a 7-year old boy, and they just had a new baby," he explains. "Does either family have a girl?" you ask. "I'm not sure," your friend says. "But pick the family that you think is more likely to have a girl. If they do have a girl, I'll give you $100."
Which family should you pick, or does it not matter? + Show Spoiler +Family on the right because they just had a child and the mother is a girl. Family on the left could be a family of all men with gay parents.
+ Show Spoiler +The riddle is talking about children not the parents, it has nothing to do with being gay or anything.
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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 +Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
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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
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On April 20 2012 15:10 zJayy962 wrote:Show nested quote +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 +Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
+ Show Spoiler +No because group 1 can have 2 of the 3 fastest horses and only 1 moves on.
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On April 20 2012 15:10 zJayy962 wrote:Show nested quote +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 +Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses.
+ Show Spoiler +No, it is a bit harder than that. You only find the fastest horse with that methot. Not the rest 2 fastest ones. Try to figure out why.
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On April 20 2012 15:13 Aelfric wrote:Show nested quote +On April 20 2012 15:10 zJayy962 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 +Is it 6? First take the horses into 5 different races and then race the 5 winners against each other in 1 race. Then the top 3 in your last race will be your 3 fastest horses. + Show Spoiler +No, it is a bit harder than that. You only find the fastest horse with that methot. Not the rest 2 fastest ones. Try to figure out why.
+ Show Spoiler +I see. The 2nd and 3rd place could be in the original race with the fastest horse. So from my method from before you split the horses into 5 different groups and race them all (thats 5 races) then you race the fastest one from each group (that will give you #1 spot). Now take the 2nd and 3rd from the 6th race and gather the horses from the race you got the fastest horse and have them race (thats 7). Answer is 7 races.
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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 +
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.
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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?
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On April 20 2012 15:13 sausageslayer wrote:Show nested quote +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.
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On April 20 2012 15:36 Mr.F. wrote:Show nested quote +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
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[QUOTE]On April 20 2012 11:30 whatwhatanut wrote:
[QUOTE]On April 20 2012 10:51 drew-chan wrote:
[QUOTE]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?
[/QUOTE]
+ 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.
[QUOTE]
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 +
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....
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On April 20 2012 15:31 iTzSnypah wrote:Show nested quote +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.
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EPT![[image loading]](http://images.wikia.com/wowwiki/images/a/a4/Warden_artwork.jpg)
