O.k. Riddle time

wprager wrote:If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
kill 100 rats in 50 minutes?

1 cat per rat per minute. It would take 2 cats to kill 100 rats in 50 minutes.

wprager wrote:Yeah, most of the hard ones I ended up "looking up". Just too impatient. But there was one puzzle that I would put in the fairly hard category, which I did without help, and I felt quite good about it. I'm not sure it's in the archive, so here goes:

A camp councilor is on a hike with 8 campers. They get terribly lost until they come to a 4-way intersection of two roads, with a sign post. The sign, which had fallen to the ground, says "Camp 1 hour walk this way". It's getting late, and his only choice is to send scouting parties in each of the four directions with instructions to walk for an hour and come right back to report their findings. Then, he makes a decision based on the findings and leads the troupe to safety before nightfall.

Problem is, one of the campers is a liar. Worse, he doesn't lie all the time, very inconsistent about that. He can obviously trust himself, but he'd need to send three campers down each of the other three remaining paths (if the liar lies the other two in his scouting party would outvote him).

What does he do?

I don't understand this. If the liar is going to get outvoted in his own scouting party, then what is the problem exactly?

Assuming the councilor takes a path, and then he sends the remaining campers on the other three paths, then how does the liar even get his voice heard?

EDIT: I think I understand the question now. Instead of sending parties down all 4 of the paths, he could split up the campers into 3 groups. The councilor and two campers down road 1, 3 campers down road 2, and 3 campers down road 3. If they come back in an hour and none of them have found the campsite, that means road 4 is the only way left to get back. And since there would be more than 2 people per group, the liar would get outvoted.

Have I understood this question right?

PKC wrote:

Tukker wrote:I have two US coins totaling 30 cents. One is not a nickel. What are the coins?

Haha, too easy Tukk, one's a quarter the other is a nickel.

damn, thats either one you get right away or never get

Tukker wrote:

PKC wrote:

Tukker wrote:I have two US coins totaling 30 cents. One is not a nickel. What are the coins?

Haha, too easy Tukk, one's a quarter the other is a nickel.

damn, thats either one you get right away or never get

Good point. Its easy to look for weaknesses in the statement when you know its a riddle. In other words, you're looking for the trick because its always inside the riddle.

Amnesia021 wrote:

Maybe i'm missing something, but didn't you answer you're own question?

There are 9 people total and 4 paths.

Councilor takes path 1
3 of the remaining 8 take path 2
3 of the remaining 5 take path 3
and the last 2 take path 4

In fact, you could probably skip the fourth path and just send 4 on each of the remaining 2...

Umm, maybe I mis-remembered something. Would it work better with only seven campers? Or do we need five paths? I'm really sorry, I know the solution I'm just struggling to come up with the problem.

OK, let's try this. There are only five campers, four paths and one councelor.

wprager wrote:

OK, let's try this. There are only five campers, four paths and one councelor.

Path one is for the Councelor
Path two you send 2 of 5 down
Path three you send 2 of the 3 down
Path four you send one down

Since only one can be a liar, if any one in paths 2 or 3 disagree, you know the liar is in that goup.
Path 1, the counselor doesn't find it
Path 2, the two agree that the camp isn't there
Path 3, the two disagree that the camp isn't there
Path 4, the one says it isn't there.

Or, they all are in agreement that the camp isn't in 1, 2, or 3, but the path 4 guy also says that there is no camp, therefore he would be lying...?

Amnesia021 wrote:
Path one is for the Councelor
Path two you send 2 of 5 down
Path three you send 2 of the 3 down
Path four you send one down

Since only one can be a liar, if any one in paths 2 or 3 disagree, you know the liar is in that goup.
Path 1, the counselor doesn't find it
Path 2, the two agree that the camp isn't there
Path 3, the two disagree that the camp isn't there
Path 4, the one says it isn't there.

Or, they all are in agreement that the camp isn't in 1, 2, or 3, but the path 4 guy also says that there is no camp, therefore he would be lying...?

I really buggered it up and I'm sorry. Essentially, you've got the required approach.

Now, let me re-state the problem in its correct form.

There is 1 councillor, 8 campers and 4 paths, as I originally stated. However there are 2 unreliable campers (in other words they may or may not lie).

Now the problem gets a lot harder, doesn't it? There is still a solution, though.

wprager wrote:I really buggered it up and I'm sorry. Essentially, you've got the required approach.

Now, let me re-state the problem in its correct form.

There is 1 councillor, 8 campers and 4 paths, as I originally stated. However there are 2 unreliable campers (in other words they may or may not lie).

Now the problem gets a lot harder, doesn't it? There is still a solution, though.

I still think my solution applies to this problem.

Councilor takes path 1.
4 campers take path 2.
4 campers take path 3.

Since the two liars would get outvoted in their own groups, when they return in an hour each from their path, the councilor can make a decision of which way to go. If no one has found the campsite in the first three paths, then it must be the 4th path that leads them back.

PKC wrote:

wprager wrote:I really buggered it up and I'm sorry. Essentially, you've got the required approach.

Now, let me re-state the problem in its correct form.

There is 1 councillor, 8 campers and 4 paths, as I originally stated. However there are 2 unreliable campers (in other words they may or may not lie).

Now the problem gets a lot harder, doesn't it? There is still a solution, though.

I still think my solution applies to this problem.

Councilor takes path 1.
4 campers take path 2.
4 campers take path 3.

Since the two liars would get outvoted in their own groups, when they return in an hour each from their path, the councilor can make a decision of which way to go. If no one has found the campsite in the first three paths, then it must be the 4th path that leads them back.

if the two liars are on the same path, then you will have 2 vs 2. and since no one went on the 4th path, then you still won't be sure.

In the case of two liars, you would need to send people to all 4 paths.

coucillor takes path 1
3 of the 8 take path 2
3 of the 5 take path 3
and the last 2 take path 4

If the 2 liars are on 1 path then that path becomes the process of elimination, if the other 3 paths agree, then you know the two liars are on the one path, if the other 3 paths are not the campground, then the last path is.

If the 2 liars on on seperate paths, but in the groups of 3, then you will notice that two groups have dissenting opinions (thus allowing you two know that there is a liar in each group). The other 2 members in each group would outvote the liar in the group.

It works the same way if the one liar is in a group of 3 and the other liar is in the group of two. You would have two dissenting groups. You can't be sure who is lying on the path with two people but you can be sure of the other 3 outcomes, thus eliminating the need to be sure of the 4th.

The real questions is, is it possible that the councillor could be one of the liars? :pirat:

Yeah but there's also the possibility that the two liars, if they happened to be in the same group, would lie to contradict each other. In which case, the majority of that group, in my scenario, would still be right.

Plus, I don't even understand your solution, because I'm fairly certain that you contradict yourself. The councilor doesn't know who the two liars are, so how do you know which path they are going on. If you send the two liars in your scenario on path 4, then they might all come back in an hour and still be nowhere.

I'm almost positive my solution is correct. Guess we just have to wait for wprager.

PKC wrote:Yeah but there's also the possibility that the two liars, if they happened to be in the same group, would lie to contradict each other. In which case, the majority of that group, in my scenario, would still be right.

Plus, I don't even understand your solution, because I'm fairly certain that you contradict yourself. The councilor doesn't know who the two liars are, so how do you know which path they are going on. If you send the two liars in your scenario on path 4, then they might all come back in an hour and still be nowhere.

I'm almost positive my solution is correct. Guess we just have to wait for wprager.

If the two liars are contradicting themselves, then one of them isn't lying, and therefore isn't a liar...

In my solution, if everyone comes back from their path and everyone says that the camp isn't down their path, then you know the two liars are on path 4 and that path 4 is the correct way to go. The councillor is not the liar, paths 2 and 3 have 3 people, so if all three agree, then you know that there are no liars in that group, leaving path 4 as the group of liars, which means the opposite of what they say is true.

Amnesia021 wrote:

if the two liars are on the same path, then you will have 2 vs 2. and since no one went on the 4th path, then you still won't be sure.

In the case of two liars, you would need to send people to all 4 paths.

coucillor takes path 1
3 of the 8 take path 2
3 of the 5 take path 3
and the last 2 take path 4

If the 2 liars are on 1 path then that path becomes the process of elimination, if the other 3 paths agree, then you know the two liars are on the one path, if the other 3 paths are not the campground, then the last path is.

If the 2 liars on on seperate paths, but in the groups of 3, then you will notice that two groups have dissenting opinions (thus allowing you two know that there is a liar in each group). The other 2 members in each group would outvote the liar in the group.

It works the same way if the one liar is in a group of 3 and the other liar is in the group of two. You would have two dissenting groups. You can't be sure who is lying on the path with two people but you can be sure of the other 3 outcomes, thus eliminating the need to be sure of the 4th.

The real questions is, is it possible that the councillor could be one of the liars? :pirat:

First, the councillor is not a liar

Your solution is correct, and I'm pretty sure so is your explanation. Sorry, I'm in a hurry and cannot take enough time to fully explore it. Here is the answer with fairly concise explanations of the decision "algorithm". I'll spoilerize it:

Spoiler:

My solution still works though. Look:

1. Councilor is obviously not a liar. His report of his findings down path 1 are indisputable.
2. Since he sent 4 people each down paths 2 and 3:
a. If there is a dispute in path 2 there are two options:
i. One of the liars is in the group, in which case he would be outvoted 3-to-1
ii. Both of the liars are in this group, in which case path 3's group is telling the truth about their findings and path 2 can be discredited because both liars are in that group.

That leads me to the conclusion that path 4 is the path to take assuming none of the other groups found that their path led back to the campsite.

Bah, this riddle sucks. The councilor should tell everyone to go Diddle themselves for being a bunch of distrustful Diddle. Like who gets lost and decides to lie to the rest of the group about where their path leads. What's the point? Their still lost too.

You know what a better solution is? Send everyone down all 4 paths, come back in an hour and wait for the liars to head down the right path and just follow them. Like why is everyone power tripping, especially this piece of Dung councilor who's trying to run the show and everything. Diddle him, who's he to say who goes where and for how long? He's the liar. And the dumbass for getting lost and not having enough decency to have an idea of the layout of the campsite they are at. Like who goes for a long Donkey walk but doesn't remember how to get back?

Diddle it!

PKC wrote:My solution still works though. Look:

1. Councilor is obviously not a liar. His report of his findings down path 1 are indisputable.
2. Since he sent 4 people each down paths 2 and 3:
a. If there is a dispute in path 2 there are two options:
i. One of the liars is in the group, in which case he would be outvoted 3-to-1
ii. Both of the liars are in this group, in which case path 3's group is telling the truth about their findings and path 2 can be discredited because both liars are in that group.

That leads me to the conclusion that path 4 is the path to take assuming none of the other groups found that their path led back to the campsite.

Bah, this riddle sucks. The councilor should tell everyone to go Diddle themselves for being a bunch of distrustful Diddle. Like who gets lost and decides to lie to the rest of the group about where their path leads. What's the point? Their still lost too.

You know what a better solution is? Send everyone down all 4 paths, come back in an hour and wait for the liars to head down the right path and just follow them. Like why is everyone power tripping, especially this piece of Dung councilor who's trying to run the show and everything. Diddle him, who's he to say who goes where and for how long? He's the liar. And the dumbass for getting lost and not having enough decency to have an idea of the layout of the campsite they are at. Like who goes for a long Donkey walk but doesn't remember how to get back?

Diddle it!

:lol!:


censorship

PKC wrote:My solution still works though. Look:

1. Councilor is obviously not a liar. His report of his findings down path 1 are indisputable.
2. Since he sent 4 people each down paths 2 and 3:
a. If there is a dispute in path 2 there are two options:
i. One of the liars is in the group, in which case he would be outvoted 3-to-1
ii. Both of the liars are in this group, in which case path 3's group is telling the truth about their findings and path 2 can be discredited because both liars are in that group.

That leads me to the conclusion that path 4 is the path to take assuming none of the other groups found that their path led back to the campsite.

Bah, this riddle sucks. The councilor should tell everyone to go Diddle themselves for being a bunch of distrustful Diddle. Like who gets lost and decides to lie to the rest of the group about where their path leads. What's the point? Their still lost too.

You know what a better solution is? Send everyone down all 4 paths, come back in an hour and wait for the liars to head down the right path and just follow them. Like why is everyone power tripping, especially this piece of Dung councilor who's trying to run the show and everything. Diddle him, who's he to say who goes where and for how long? He's the liar. And the dumbass for getting lost and not having enough decency to have an idea of the layout of the campsite they are at. Like who goes for a long Donkey walk but doesn't remember how to get back?

Diddle it!

OK, let's analyze your solution.

The councilor comes back wihtout having found the camp. The first group of campers comes back with a split vote: two of them say camp's that way but the other two say nuh-huh. At this point you know that both liars are in this group and are both, in fact, lying. However you still have no idea about their path -- does it lead to the camp or not? So if the last group comes back and says they did not find the camp either you only know for sure about two paths -- not enough info to draw a conclusion.

By the way, the riddle is a nice analogy for error correction scheme that handles up to two bit errors.

wprager wrote:

PKC wrote:My solution still works though. Look:

1. Councilor is obviously not a liar. His report of his findings down path 1 are indisputable.
2. Since he sent 4 people each down paths 2 and 3:
a. If there is a dispute in path 2 there are two options:
i. One of the liars is in the group, in which case he would be outvoted 3-to-1
ii. Both of the liars are in this group, in which case path 3's group is telling the truth about their findings and path 2 can be discredited because both liars are in that group.

That leads me to the conclusion that path 4 is the path to take assuming none of the other groups found that their path led back to the campsite.

Bah, this riddle sucks. The councilor should tell everyone to go Diddle themselves for being a bunch of distrustful Diddle. Like who gets lost and decides to lie to the rest of the group about where their path leads. What's the point? Their still lost too.

You know what a better solution is? Send everyone down all 4 paths, come back in an hour and wait for the liars to head down the right path and just follow them. Like why is everyone power tripping, especially this piece of Dung councilor who's trying to run the show and everything. Diddle him, who's he to say who goes where and for how long? He's the liar. And the dumbass for getting lost and not having enough decency to have an idea of the layout of the campsite they are at. Like who goes for a long Donkey walk but doesn't remember how to get back?

Diddle it!

OK, let's analyze your solution.

The councilor comes back wihtout having found the camp. The first group of campers comes back with a split vote: two of them say camp's that way but the other two say nuh-huh. At this point you know that both liars are in this group and are both, in fact, lying. However you still have no idea about their path -- does it lead to the camp or not? So if the last group comes back and says they did not find the camp either you only know for sure about two paths -- not enough info to draw a conclusion.

By the way, the riddle is a nice analogy for error correction scheme that handles up to two bit errors.

tell em you're going down each path yourself, and will kill each of the lying sons of Wing Dang Doodle with a rusty hunting knife if they don't fess up now

Tukker wrote:
tell em you're going down each path yourself, and will kill each of the lying sons of Wing Dang Doodle with a rusty hunting knife if they don't fess up now

That would work, too. You might have to look for another summer job next year, though.

wprager wrote:

Tukker wrote:
tell em you're going down each path yourself, and will kill each of the lying sons of Wing Dang Doodle with a rusty hunting knife if they don't fess up now

That would work, too. You might have to look for another summer job next year, though.

their already lost, just tell the parents they ran off on there own :D