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On April 10 2011 07:38 Talanthalos wrote:Show nested quote +On April 10 2011 07:27 shadowy wrote:On April 10 2011 07:03 LastPrime wrote:On April 10 2011 06:32 tomnov wrote:10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer how do they split the money and how many pirates die? Does the vote include the person making the split suggestion? Answer: + Show Spoiler + There are 2 pirates left (8 die). The ninth pirate will get all the gold.
Basically to satisfy all conditions it will go like this:
10= 9x11 + 1x1 9= 8x12 + 1x4 8= 7x14 + 1x3 7= 6x16 + 1x4 6= 5x20 + 1x0 5= 4x25 + 1x0 4= 3x33 + 1x1 3= 2x25 + 1x0 2= 1x100 + 1x0
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
+ Show Spoiler +sry for really getting on your nerves today, but this is also wrong. just to give you 1 example of what you havent thought of: when 3 pirates are left, the 1. pirate will be able to get 99gold and give the last pirate 1 gold. the last pirate obv. knows that he cant get any gold from declining this offer (as you stated) so he will take the 1 gold.
ofc. this isnt the finals answer cause this way of thinking will continue to the higher first levels also.
ill will try an figure it out now
You are correct. Don't worry about nerves - as I said earlier today it's just riddles. However before I read your post, I already saw a flaw in my logic based on terr13 posts.
I posted a new answer, which I will repost here:
+ Show Spoiler +Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all. ------- When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted! ------- When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails. -------- When there are 4 people: 4th will propose: 1 - 0, 2-0, 3 - 0, 4-100 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will stay alive and will try to make the most of the money. He will vote yes 2:2 split accepted. --------- When there are 5 people: 5th will propose - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will try to kill more and he know he can stay alive. He will vote No. 5th will vote yes, to stay alive. 2:3 - split denied. ---------- When there are 6 people: 6th will propose 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will can safely vote No, because he can safely kill more pirates. 4th will try to kill more - what he gets doesn't matter. He will vote No. 5th will vote yes, in oder to stay alive. 6th will try yo stay alive. He will vote Yes" 2:4- split denied. At this point, there is no point to keep going onward. The final answer is there will be 4 pirates left, and 4th one (counted backwards) will get all the money.
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Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
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When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
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Agreed.
[quote]
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails. [\quote]
Here is where your logic fails to hold. 3rd person's offer is not irrelevant. If he offers 1 to 1st person, the first person will accept, because it is better than getting 0 from 2nd person's offer.
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[QUOTE]On April 10 2011 08:10 LastPrime wrote: [quote]Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
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When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
------- [/quote]
Agreed.
[quote]
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails. [\quote]
Here is where your logic fails to hold. 3rd person's offer is not irrelevant. If he offers 1 to 1st person, the first person will accept, because it is better than getting 0 from 2nd person's offer. [/QUOTE]
You are absolutely correct. Let me work this again and see where I arrive. Will post in edit.
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On April 10 2011 08:06 stiknork wrote:Show nested quote +On April 10 2011 07:59 LastPrime wrote: A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route. + Show Spoiler +http://en.wikipedia.org/wiki/Hamiltonian_path
Yes now prove there has to be at least another Hamiltonian path given the conditions.
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1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
+ Show Spoiler +Fill in the 5 liter, then spill it into the 3 liter. When the 3 liter is full, you have 2 liters left in the big one. Do it twice, and you got 4 liters.
2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
+ Show Spoiler +Take 1 from the first bag, 2 from the second bag, 3 from the third, etc... Multiply the number of grams missing (the 55 (1+2+3+...+10=55) coins should weigh 55 grams if they were all real) by 10 and you have the number of the bag with fake coins.
2.b. Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
+ Show Spoiler +Do the same thing with 1/2/4/8 coins from bags #1/2/3/4.
3. (accessible) You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
+ Show Spoiler +Flip them both. When 7 is empty, flip the 7. 7 minutes have gone by. When 11 is empty, 11 minutes have gone by, and 4 minutes since you flipped the 7. Flip the 7 again, and wait for it to be empty. 15 minutes have gone by.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
+ Show Spoiler +It's easy to find that it's around 1h05. To find the exact time, lets call x the angle that the hands do with the 12. Hour and Minutes are on the x. "60+x minutes" has to be equal to "x hours". "x hours" is x*12/360 hours, or x*12/6 minutes, or 2x minutes (for example, 90° = 180 minutes = 3 hours). "x minutes" is x*60/360 minutes or x/6 minutes (for example 90° = 15 minutes). So we need to have 60+x/6 = 2x, which gives 11x/6 = 60 or x = 360/11 degrees. Both hands are on 360/11 degrees, which means it's 1:05:05
5. (my favorite :p) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Hard one!
6. What's your eyes color? (hard problem)
+ Show Spoiler +All 100 guys with blue eyes will leave on the 100th day. Let's assume there's only one guy with blue eyes. He'll look around him and see no one has blue eyes, but he knows someone does, so he knows it has to be him, so he leaves on the first day. Now, if there was 2 guys with blue eyes, each of them would think "This dude is the only one I can see who has blue eyes. If I don't have blue eyes, he doesn't see anyone with blue eyes, so he'll leave. Let's see if he's still here tomorrow..." The next day, the other dude's still here, so it means I too have blue eyes. The other guy thinks the exact same thing, and we both leave on the 2nd day. [...] Now, with 100 guys with blue eyes, each one of them will think "I see 99 guys with blue eyes. If I don't have blue eyes, they'll all leave on the 99th day. Let's see if that happens. On the 100th day, the other 99 guys are still here, which means I too have blue eyes. The other 99 guys think the same thing, we all figure out we have blue eyes, and all leave on the 100th day.
7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
+ Show Spoiler +Split them in 3 piles of 3. Weigh one of the piles against the other. 2 possibilities : - it's balanced : the lighter coin is in the 3rd pile - it's not balanced : the lighter coin is on the lighter side Either way, we now know in which pile of 3 the lighter coin is. Weigh one coin of this pile against another of this pile. - it's balanced : the 3rd coin is lighter. - it's not balanced : we found the lighter coin.
8. (Day9 wants you to do this one) On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes? Note: the string possibly burns unevenly.
+ Show Spoiler +Light one on one end, and the other on both hands. When the one you lighted on both end is done burning, 30 minutes have gone by. Light the other hand of the first one, when it's done burning, 15 more minutes have gone by, total 45 minutes.
9. (if you know what prime numbers are) When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
+ Show Spoiler +Lets call A our prime number greater than 32. R < 30. Lets suppose R is not a prime number.
10. On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
+ Show Spoiler +The first vizir makes 2 steps forward. The second vizir comes close to the other. Now the third vizir has 2 options : - if the first 2 have different colors, he goes between them - if the first 2 have the same color hats, he goes to either end of this new line. The 4th vizir does the same : - if all 3 hats he sees in the new line are the same color, he goes to one end of the line - if there are hats of both colors, he goes between the red and blue that are next to one another. In the end, if they all do that, all reds will be on one side, all blues on the other. They will all announce their color starting from the most left, then the most right, then the second-most left, then the second most-right. Only one will be wrong (or maybe they'll all be right, if they're lucky).
11. Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
Too complicated for 1:10 am, can't be bothered :D
12. (Monty Hall?) On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
+ Show Spoiler +Yes you should switch. If you picked the door with the car first, switching means losing and not switching means winning. If you picked the door with the goat first, switching means winning and not switching means losing. But the second situation happens twice more often, which means you double your chance of winning by switching (2/3 vs 1/3).
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
+ Show Spoiler +Of course, it's exactly like writing in binary.
14. On April 10 2011 02:14 ghrur wrote: Question: What is the maximum number of times 10 lines can intersect in the same plane? paraphrased: What is the maximum number of intersection points between 10 distinct lines on a plane?
+ Show Spoiler +45? Each line can cross every other once, and I don't think there's a reason why some intersection points should be common, so for each line there is 9 intersection points. But you must not count twice the point of intersection of line A and B and line B and A, so half of 10*9. It seems too easy, maybe there's a catch.
15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
I'll think about it tomorrow :D Too late and too hard
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On April 10 2011 07:11 rexob wrote: the second post about blue eyes is gonna keep me up all night :S
If you just want to know the solution, here it is: + Show Spoiler +http://xkcd.com/solution.html
Very interesting and much more simplistic than I thought it would be.
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+ Show Spoiler + 1 person /w blue --> understandable 2 people /w blue---> understandable >3 people /w blue ---> ????
The 1 person example; no other blue eyes, obviously. The 2 people example; the 2nd person knows the first person can see a 2nd set of blue eyes. easy peazy. The 3 people example; the 3rd person knows the other two can see each others eyes. Why would anybody leave?
+ Show Spoiler + If there's you, B(lue)1, B(lue)2, B(lue)3. Now there is 2 cases, either you have blue eyes or you don't. If you don't have blue eyes the only people with blue eyes that B2 will see is B1 and B3. B2 knows that if he didn't have blue eyes B1 and B3 would have left on the second night. If B1 and B3 doesn't leave on the second night B2 knows that he must have blue eyes aswell. And if B1, B2 and B3 doesn't leave on the third night then you know that you must be blue eyed. Using this logic you can figure it out.
The key is not thinking that they see eachother, instead take it in steps thinking, "If I don't have blue eyes, B1 only sees 2 people with blue eyes" and then you put yourself in the perspective of B1 with the assumption that you do not have blue eyes which would lead to him thinking "If I don't have blue eyes, B2 only sees 1 person with blue eyes which would make him leave on the second night"
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On April 10 2011 06:19 Frigo wrote:Show nested quote +10. On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)? + Show Spoiler +All of them see the hats of those later in the line, the first one in line sees them all, except his. He communicates the parity of the sum of the binary representation of hats (0 or 1) by saying the respective color of the parity (it doesn't matter which color is 0 and which is 1). The guy next in the line can deduce his hat color based on this parity and what he sees. The others can deduce their hat color from the parity, and what others have said before him. Only the first in the line might die, the others surely live. Unless one of them fucks up
Very nice. The solution to this one eluded me.
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Ive got a couple love those things
On April 10 2011 00:05 stepover12 wrote: 7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
I'll add more when I can.
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
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On April 10 2011 05:46 cmpcmp wrote:Solution to the 3 princesses / most eligible bachelor puzzle Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking. Solution + Show Spoiler +You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger. Explanation + Show Spoiler +There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
Wow, beautiful!
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On April 10 2011 08:43 Darkren wrote: 4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
+ Show Spoiler + Number1 and Number2 forward - 2min Number2 back - 4min Number3 and Number4 forward - 14min Number1 back - 15min Number1 and Number2 forward - 17min
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10.) Easier than you think I had this in a leadership class. + Show Spoiler + You will not be able to see the hat on your head. Is the statement you should focus on, if you are smart enough you'd just take the hat off your head so you can see it. These riddles usually go with something so obvious that you feel a little wetarded when you have to solve it.
Reminds me on "How fast can everyone touch the ball trick?" Our class of about 30 students all touched it in about 1 seconds flat after we figured it out, first we were passing the ball to each other got that down to about 20 something seconds, then we tried getting closer, got it down more, then we tried just handing it off in a circle got it down faster. Then we all put a finger on the ball and let go at the same time.
It just shows there is more than one solution, but which is best? In other words a lot of the solutions weren't wrong, and a lot weren't right, the ones where you have to sacrifice someone is just silly =P Nobody needed to die! just take the hat off and put it on your hand and say the color
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I've edited the new ones to the 1st post <3
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On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
I always forget the correct solution to this particular puzzle and instead I try to escort the 10-minute one with the 1-minute one, which leads to an incorrect solution -_-
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
+ Show Spoiler +The 2$ in the boy's pocket was paid by the men, it is already in the 3*9$ they paid. The missing 3$ was given back to the men, 3*9$+3$=30$. Deceptive puzzle, but can't be misunderstood when read online.
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Solution to pirates + Show Spoiler +After a grueling process of keeping track of every possible offer each pirate member can have for every split, I've finally come to a conclusion. So working backwards, we start with the case with 2 pirates left. The one doing the splitting will be labeled as pirate#2, with the one lower in rank being pirate#1 (so the rest will be labeled in accordance to their ranks with the captain being pirate#10. As it has been discussed before, #2 has the power to split everything to himself, since his vote is enough as the majority. so:
Split1: #2-100, #1-0
From here a lot of reasoning will be cumulative and dependent upon the possible offers of previous splits. So now when #3 is doing the splitting, he can easily persuade #1 to agree with him with one coin, since it's better than getting 0 with #2 splitting. #2 will be inclined to always disagree until it is his turn to split, but his vote does not matter. #3 and #1 are enough as the majority. So:
Split2: #3-99, #2-0, #1-1
So #4 has to try and persuade at least 1 person to his side yet again like with #3. #2 and #3 are going to disagree no matter what since they stand to benefit much greater when it is their turn. So similarly with #3, he can just offer #1 a bit more than what #3 will have to offer.
Split3: #4-98, #3-0, #2-0, #1-2
Now things change up. #5 is in a bad position as #1 isn't enough for majority. In fact, #4, #3, and #2 will benefit tremendously from their splits only, so they'll never agree. so #5 is destined to die.
Split4: #5 will die
Things change greatly again. #6 is in a much better position than #5, since he can offer #5 0 as #5 will still have his life, and #1 a bit more than what #3 would have offered for majority!
Split5: #6-97, #5-0, #4-0, #3-0, #2-0, #1-3
Now like #5, #7 is also screwed since he can not ever get #6, #4, #3, or #2 to agree. He'll never reach majority and die.
Split6: #7-dies.
Do you see the pattern now? The odd ranks greater than 3 can not pull majority no matter what. They will die when they try to split. The even ranks can abuse this, offer minimum to those destined to die and #1 a bit more than what they would get so far. So moving along, #5 accepts 1 as it's now the best offer he can get after just his life in the previous offer. #7's new best offer is simply his life. This pattern will repeat itself until it's the captain doing the splitting.
Split7: #8-95, #7-0, #6-0, #5-1, #4-0, #3-0, #2-0, #1-4
Split8: #9 will die
Split9: #10-92, #9-0, #8-0, #7-1, #6-0, #5-2, #4-0, #3-0, #2-0, #1-5
The end result is one cunning captain with most of the treasure for himself, and the majority formed by #9, #7, #5, and #1 sucking it up and agreeing since there's absolutely no better possible outcome for them unless they agree.
One important possible flaw: I'm assuming the pirates will only accept the offer if it's GREATER than any possible future offer, not equal to. If the pirates are willing to agree even if future offers are the exact same, then the captain would just end up with all 100 and still get his majority to agree.
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On April 10 2011 08:46 Husnan wrote:Show nested quote +On April 10 2011 05:46 cmpcmp wrote:Solution to the 3 princesses / most eligible bachelor puzzle Yes, there is a logical, reasonable, and not stupid solution if that's what ur thinking. Solution + Show Spoiler +You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger. Explanation + Show Spoiler +There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her. Wow, beautiful!
That logic doesnt work because the younger sister could lie and tell u the middle one is the youngest. U could also ask the older sister if the middle one is younger and than her and she will answer u yes.
+ Show Spoiler +U could ask do u think lying to someone is ok? If u pick the younger one she will tell u no(lie) If u pick the older one she will tell u no If u pick the middle one she will answer yes because she lies sometimes and tell the truth other times
U then know if she answers yes that she is the middle sister and if she answers no u marrie her
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nm, re-thought my solution to one, will have to reconsider, heh
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hmm in the pirate puzzle do we assume that the governing 2 rules are money and life? or money and kills? because for example: if there are 3 pirates left let's call them A B and C with A making the call the split like this should work A9 B:0 C:1. my reason for this is that A wants to keep as much of it as possible to himself and since he needs another vote to survive he gives the pirate with no chances of gaining money if he dies a piece of the loot in order to get the needed votes, i dont think it will ever come to only 2 remaining pirates.
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[QUOTE]On April 10 2011 09:41 Frigo wrote: [QUOTE]On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint? [/quote] You sure this is possible?
[quote] [/QUOTE]
Yes it's possible ill give a hint.
+ Show Spoiler +U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it
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