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This one is a classic, but can you do it?
You have 2 identical eggs, and a 100 floor building. You wish to find out the highest floor number in which the egg can be dropped without breaking. Describe a strategy which you can precisely find that floor number in the fewest drops.
Q&A:
Q: But, what do you mean "fewest?" A: I mean at the worst case. For instance, your strategy might be try fl1, then fl2, then 3 4 5... Under this strategy, the worst that can happen is the egg never breaks, i.e. it can be dropped from over 100 floors, which will take you all 100 drops to find out. So, this strategy takes 100 drops.
Q: Does the egg crack if dropped repeatidly? A: No, egg is good as new as long as it don't break yet.
Again, put answers in spoilers per usual. Collaborate, have fun!
--evan
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+ Show Spoiler + I would assume that since you don't want to run out of eggs you would start at floor 2. If it doesn't break, go to floor 4. When you finally get a break, you go down a floor. If that egg doesn't break you are at the floor that is highest without breaking. Otherwise it's the floor underneath you.
Of course, this takes 50 drops, so it's probably suboptimal.
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+ Show Spoiler + 1. Drop egg on Floor 3. If it doesn't break, jump to 4. 2. If it breaks, drop on floor 2. If no break - done. 3. If it broke on floor 2, done... it's floor 1. 4. Repeat step #1 on floor X + 2 where X is the floor from step #1.
**edit** I guess it is possible that the egg breaks even from floor 1, so this wouldn't work
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United States24345 Posts
+ Show Spoiler +I'd do the first drop on 33 (1/3). If it breaks, go up from 2. If it doesn't break, go to 55 (1/3). If it breaks, go up from 34. If it doesn't break, go up to 70. If it breaks, go up from 56. If it doesn't break, go up to 80. Etc. No idea if it's ideal but I think it's decent XD
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Straight outta Johto18973 Posts
+ Show Spoiler +Let us define a process A that averages the value of the highest known failure level, and lowest known success value. In the case of non-integer values, round up. We initially start with the assumption an egg dropped from floor 0 will not crack.
Start at the highest level, 100. Drop Egg. If success, end test. If failure, run process A. Return 50 and repeat test at 50.
If failure or success at 50, run Process A again. So if 50 was a success, you go to 75. If it was a failure you go to 25. Repeat this continuously until Process A returns the same value as the floor you are currently on. This is your maximum floor.
This should only require around 8 tries maximum. Too lazy to calculate the expected value and variance of the number of tries for a scenario where each floor has a equal probability of being the successful level. Also, if you consider the probability of success as the floor level increases to be a decreasing function (as common sense would dictate), other methodologies would be more efficient at solving this problem as the expected value would be lower. However, this methodology returns the lowest number of maximum attempts necessary.
If you don't want the assumption an egg at level 0 does not crack, then run at test there first. If it cracks, then no floor exists where the egg will not crack. This brings the maximum number of tests to run to 9, but not a big deal. Still less than 10.
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United States24345 Posts
On November 13 2011 12:50 MoonBear wrote:+ Show Spoiler +Let us define a process A that averages the value of the highest known failure level, and lowest known success value. In the case of non-integer values, round up. We initially start with the assumption an egg dropped from floor 0 will not crack.
Start at the highest level, 100. Drop Egg. If success, end test. If failure, run process A. Return 50 and repeat test at 50.
If failure or success at 50, run Process A again. So if 50 was a success, you go to 75. If it was a failure you go to 25. Repeat this continuously until Process A returns the same value as the floor you are currently on. This is your maximum floor.
This should only require around 8 tries maximum. Too lazy to calculate the expected value and variance of the number of tries for a scenario where each floor has a equal probability of being the successful level. You will need more than two eggs for this method.
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On November 13 2011 12:50 MoonBear wrote:+ Show Spoiler +Let us define a process A that averages the value of the highest known failure level, and lowest known success value. In the case of non-integer values, round up. We initially start with the assumption an egg dropped from floor 0 will not crack.
Start at the highest level, 100. Drop Egg. If success, end test. If failure, run process A. Return 50 and repeat test at 50.
If failure or success at 50, run Process A again. So if 50 was a success, you go to 75. If it was a failure you go to 25. Repeat this continuously until Process A returns the same value as the floor you are currently on. This is your maximum floor.
This should only require around 8 tries maximum. Too lazy to calculate the expected value and variance of the number of tries for a scenario where each floor has a equal probability of being the successful level.
If you don't want the assumption an egg at level 0 does not crack, then run at test there first. If it cracks, then no floor exists where the egg will not crack. This brings the maximum number of tests to run to 9, but not a big deal. Still less than 10.
+ Show Spoiler +If you drop at 50 and it breaks and then at 25 and it breaks then you start dropping scrambled eggs.
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Straight outta Johto18973 Posts
Oh, my bad. Didn't read the question properly, lol. Need sleep. Assumed infinite eggs.
Then yah, Smurphy/Hasu's method is probably optimal.
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On November 13 2011 12:49 micronesia wrote:+ Show Spoiler +I'd do the first drop on 33 (1/3). If it breaks, go up from 2. If it doesn't break, go to 55 (1/3). If it breaks, go up from 34. If it doesn't break, go up to 70. If it breaks, go up from 56. If it doesn't break, go up to 80. Etc. No idea if it's ideal but I think it's decent XD
+ Show Spoiler +I think this is the right idea, but there has to be some way to pick what the correct ratio is to minimize the number of drops needed (I'm assuming you picked 1/3 relatively arbitrarily)
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On November 13 2011 12:55 MoonBear wrote: Oh, my bad. Didn't read the question properly, lol. Need sleep. Assumed infinite eggs.
Then yah, Smurphy/Hasu's method is probably optimal.
Mine's the same as the one above me and mine is less effective than the one below me.
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+ Show Spoiler +Think about it this way: let h be the highest floor you know it won't break at. Let b be the floor for which the first egg breaks. You will need to then start at h+1 and work up to (b-1) with the second egg.
So, you want to minimize the # of drops of the first egg + b-1-(h+1). Assume you employ a strategy where you pick an increment i and drop the first egg at floors I, 2i, 3i, etc. When the first egg breaks, you will then need I-1 drops (at most) for the second egg. The first egg would have to be dropped 100/I times. So, we wish to minimize 100/I + (I-1). A quick check for reasonable values (you could use calculus as well) will show 10 is the ideal increment. Drop the first egg at floors 10,20,30. First egg will be dropped at most 10 times, and second egg dropped at most 9. Notice that if the increment is 7,8,9,11,12,13, the total number of drops necessary in worst case is also 19 (I think).
Sorry, wrote this on my phone so hard to double check this math. I think 19 is the best you can do
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+ Show Spoiler +you drop from 10 if it doesnt break you move up 10 (10-100max 10 drops) then once that egg breaks say at like 20 then you drop from 11 and increment by 1 (11-19, max 9) so you're max number of drops is 19
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Ok, stolen from Micronesia's but I think this is better: + Show Spoiler + Drop from 9th floor. If breaks, start at floor 1, if not go to floor 18. If it breaks, restart at floor 10, if not go to floor 27. Then the worst case scenario is it takes you getting to the 99th floor on testing which is 18 drops, as opposed to the potential 34 from Micronesia.
Also, going to the 10th provides a worst case of 19, and dropping from the 8th is a worst case of 19 as well, so I'm pretty sure the 9 floor steps is optimal. But there may be another method that's more optimal than this.
EDIT** Un-even dropping intervals may be more optimal but I'm not sure how to calculate this (i.e., we start at 20, then move up to say like 37, then 53, etc.).
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On November 13 2011 13:15 hasuprotoss wrote:Ok, stolen from Micronesia's but I think this is better: + Show Spoiler + Drop from 9th floor. If breaks, start at floor 1, if not go to floor 18. If it breaks, restart at floor 10, if not go to floor 27. Then the worst case scenario is it takes you getting to the 99th floor on testing which is 18 drops, as opposed to the potential 34 from Micronesia.
Also, going to the 10th provides a worst case of 19, and dropping from the 8th is a worst case of 19 as well, so I'm pretty sure the 9 floor steps is optimal. But there may be another method that's more optimal than this.
too late
edit: for both of us...
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On November 13 2011 13:11 JDub wrote:+ Show Spoiler +Think about it this way: let h be the highest floor you know it won't break at. Let b be the floor for which the first egg breaks. You will need to then start at h+1 and work up to (b-1) with the second egg.
So, you want to minimize the # of drops of the first egg + b-1-(h+1). Assume you employ a strategy where you pick an increment i and drop the first egg at floors I, 2i, 3i, etc. When the first egg breaks, you will then need I-1 drops (at most) for the second egg. The first egg would have to be dropped 100/I times. So, we wish to minimize 100/I + (I-1). A quick check for reasonable values (you could use calculus as well) will show 10 is the ideal increment. Drop the first egg at floors 10,20,30. First egg will be dropped at most 10 times, and second egg dropped at most 9. Notice that if the increment is 7,8,9,11,12,13, the total number of drops necessary in worst case is also 19 (I think).
Sorry, wrote this on my phone so hard to double check this math. I think 19 is the best you can do
Don't increments of 8, 9, 10, 11, 12, and 13 have the same worst case scenario of 19? I haven't found any better than this ^~^
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United States7481 Posts
+ Show Spoiler + drop your first egg at the following floors, in order. if it breaks at 1 go down to the previous one and start incrementing by 1 15 29 42 54 65 75 84 92 100 your max drops are 16. i don't think this is optimal... i tried starting at floor 14 but then you peter out at 95. i'm probably doing math wrong somewhere. e: flamewheel is better at addition than i
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FREEAGLELAND26780 Posts
Oh hmm I think I remember reading about something like this a year or so ago...
+ Show Spoiler [My thought process] +Guess and check since I'm too dumb to prove stuff!
I'm a fan of square roots. Go with every 10 floors. Start from floor 10, and if the egg breaks check linearly up from floor 1 with egg 2. Barring a break on floor 10, go up by 10 at a time. So 10, 20, 30, 40... If egg 1 breaks, go down 9 floors from the multiple of ten you're on, and check upward with egg 2.
Maximum number of steps is 10 + 9 = 19 if the breaking floor is 99.
I dont't this is optimal. Let's see...
15 floors at a time?
15 30 45 60 75 90 100 If egg breaks on 89 then that's 6 + 14 = 20.
Hmm somewhere in between that. What about 13?
13 26 39 52 65 78 91 100 Breaking at floor 90... 7 + 12 = 19
14 28 42 56 70 84 98 100 Breaking at floor 97 means 6 + 13 = 19 as well
I still don't think this is right. Hmm... maybe a decreasing jump size?
So 10 floors, 9 floors, 8 floors, 7 floors...
Trying with jump size initial equal to 10 is too many 'jumps'.
Maybe 15?
15 29 42 54 65 75 84 92 99 100 Breaking at floor 15 means 1 + 14 = 15 Breaking at floor 29 means 2 + 13 = 15 ... Breaking at floor 99 means 9 + 6 = 15 Breaking at floor 100 means 10
So perhaps initial jump size is max number of tries needed?
Try jump size initial equal to 14 then
14 27 39 50 60 69 77 84 90 95 99 100 @floor 14: 1 + 13 = 14 @floor 27: 2 + 12 = 14 ... @floor 99: 11+3 = 14 @floor 100: 12
Hmm so perhaps it is better to have jump size decrease to close to one when we near floor 100
13 25 36 46 55 63 70 76 81 85 88 90 91 So initial jump size of 13 fails
Seems like 14 looks nice. + Show Spoiler [My Solution] +We will be jumping floors. The first jump is from ground level to floor 14, so a jump size of 14 floors. Decrease jump size by 1 floor after each jump. Therefore, from 14, jump by 13 floors to 27. From there, jump by 12 to 39... So forth and so on. When the egg breaks, start from the floor higher than the last pre-jump floor (so if egg 1 breaks on floor 39, start dropping egg 2 from floor 28). It should take a maximum number of 14 tries.
Using the decreasing jump size method, we find it should only take as many tries as the initial jump size, but only if we can reach 100 before jump size becomes zero. An initial jump size of 13 fails since you only reach 91 floors, and a jump size of 15 takes 15 tries maximum. I'm a noob and can't do proofs though.
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+ Show Spoiler + We know that the sum of the first n integers is [n(n+1)]/2, setting this equal to 100 yields us n^2 + n - 200 = 0. Solving this for n gives us 14.65. So we take 15 because we need to round up. 14 is optimal actually, I'm noob.
Then we start at floor 15. If it breaks go to floor 1 and work up. If it doesn't go to 29 (15 + 14) and test. This worst case is 15, which is the best I've seen (hopefully nobody beat me to it!)
Flamewheel > me
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United States1719 Posts
On November 13 2011 13:21 flamewheel wrote:Oh hmm I think I remember reading about something like this a year or so ago... + Show Spoiler [My thought process] +Guess and check since I'm too dumb to prove stuff!
I'm a fan of square roots. Go with every 10 floors. Start from floor 10, and if the egg breaks check linearly up from floor 1 with egg 2. Barring a break on floor 10, go up by 10 at a time. So 10, 20, 30, 40... If egg 1 breaks, go down 9 floors from the multiple of ten you're on, and check upward with egg 2.
Maximum number of steps is 10 + 9 = 19 if the breaking floor is 99.
I dont't this is optimal. Let's see...
15 floors at a time?
15 30 45 60 75 90 100 If egg breaks on 89 then that's 6 + 14 = 20.
Hmm somewhere in between that. What about 13?
13 26 39 52 65 78 91 100 Breaking at floor 90... 7 + 12 = 19
14 28 42 56 70 84 98 100 Breaking at floor 97 means 6 + 13 = 19 as well
I still don't think this is right. Hmm... maybe a decreasing jump size?
So 10 floors, 9 floors, 8 floors, 7 floors...
Trying with jump size initial equal to 10 is too many 'jumps'.
Maybe 15?
15 29 42 54 65 75 84 92 99 100 Breaking at floor 15 means 1 + 14 = 15 Breaking at floor 29 means 2 + 13 = 15 ... Breaking at floor 99 means 9 + 6 = 15 Breaking at floor 100 means 10
So perhaps initial jump size is max number of tries needed?
Try jump size initial equal to 14 then
14 27 39 50 60 69 77 84 90 95 99 100 @floor 14: 1 + 13 = 14 @floor 27: 2 + 12 = 14 ... @floor 99: 11+3 = 14 @floor 100: 12
Hmm so perhaps it is better to have jump size decrease to close to one when we near floor 100
13 25 36 46 55 63 70 76 81 85 88 90 91 So initial jump size of 13 fails
Seems like 14 looks nice. + Show Spoiler [My Solution] +We will be jumping floors. The first jump is from ground level to floor 14, so a jump size of 14 floors. Decrease jump size by 1 floor after each jump. Therefore, from 14, jump by 13 floors to 27. From there, jump by 12 to 39... So forth and so on. When the egg breaks, start from the floor higher than the last pre-jump floor (so if egg 1 breaks on floor 39, start dropping egg 2 from floor 28). It should take a maximum number of 14 tries.
Using the decreasing jump size method, we find it should only take as many tries as the initial jump size, but only if we can reach 100 before jump size becomes zero. An initial jump size of 13 fails since you only reach 91 floors, and a jump size of 15 takes 15 tries maximum. I'm a noob and can't do proofs though. + Show Spoiler +you are right. the answer is 14. the point of the question is to normalize the worst case scenario. the way u would formalize the process of finding 14 is using the n(n+1)/2 = 100, which is the equation for the sum of numbers from 1 to n. Since 200 ~= n^2, choose n = 14 which produces n^2 = 196. I didn't do it that way the first time i solved it though; only figured it out after looking at the question after having solved it lol.
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On November 13 2011 13:24 darunia wrote: maths is hard i agree, the egg would obviously crack from either the first or second floor, there's your answer
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We've had this question like 4 times on this site already... and I was so pumped...
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Hint: (Answer but no solution) + Show Spoiler + The # of tries in a worse case scenario would be 14. If your solution takes more than 14 tries, you should probably revise it
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United States1719 Posts
On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go.
edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!'
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14,27,39,50,60,69,77,84,90,95,99,100 worst case 14
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On November 13 2011 13:32 rotinegg wrote:Show nested quote +On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go.
Hire an exterminator?
Came after your edit
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Nevermind. I understand what I did wrong.
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On November 13 2011 13:32 rotinegg wrote:Show nested quote +On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + simple solution: go 500 miles drop off 500 go back go to 500 and pick up 500 then got to B (+500) repeat this 8 times and waste 4000 apples (24 trips total)
you can optimize this I'm sure
...do you only have 4000 apples?
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On November 13 2011 13:32 rotinegg wrote:Show nested quote +On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!'
+ Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
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On November 13 2011 14:17 hacklebeast wrote:Show nested quote +On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Do you realize how bad for the environment that is? Your are trading apples for polar bears at that point
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where are number puzzles 1-19?
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Drop from the 14th floor, if breaks then go up from 1 up to 13. if doesn't break, drop from the 14+13 = 27 floor.
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On November 13 2011 14:36 darunia wrote:Show nested quote +On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Do you realize how bad for the environment that is? Your are trading apples for polar bears at that point
oh sure it really isn't necessary to go back untill you hit one of the key points + Show Spoiler + but where's the fun in that? And it's not that bad, i think it's a total of 4666 miles, so less that 2.5 round rips.
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United States1719 Posts
On November 13 2011 14:17 hacklebeast wrote:Show nested quote +On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad...
You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles?
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+ Show Spoiler +The question reads "in which," not "from." Therefore, the 100th floor is the highest floor number in which the egg can be dropped without breaking.
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On November 13 2011 15:05 rotinegg wrote:Show nested quote +On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles?
+ Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out
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On November 13 2011 15:05 rotinegg wrote:Show nested quote +On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? Drink all 13 bottles of wine and show up mysteriously dead to the party.
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On November 13 2011 15:05 rotinegg wrote:Show nested quote +On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles?
alright let me try + Show Spoiler + lets number rats : 1 2 3 4 lets label bottles: ABCDEFGHIJKLM
so u follow this, ( you give the rats (number) a drink from each of the bottles) 1-AEFGHI 2-BEHIJK 3-CFHIJL 4-DGIKL
then wait and take results. The labeled bottle is poisoned if: none of the rats died: M then (to the left bottle poisoned, to the right rats which died) A:1 B:2 C:3 D:4 E:1,2 F:1,3 G:1,4 H:1,2,3 I:1,2,3,4 J:2,3 K:2,4 L:3,4
edit:aww to slow t.t
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United States1719 Posts
On November 13 2011 15:16 hacklebeast wrote:Show nested quote +On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet
Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble
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On November 13 2011 15:34 rotinegg wrote:Show nested quote +On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble + Show Spoiler + 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer
forgot spoiler sorry
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United States1719 Posts
On November 13 2011 15:35 Obelisco wrote:Show nested quote +On November 13 2011 15:34 rotinegg wrote:On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer Part 1 is right, part 2 is wrong
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On November 13 2011 15:42 rotinegg wrote:Show nested quote +On November 13 2011 15:35 Obelisco wrote:On November 13 2011 15:34 rotinegg wrote:On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer Part 1 is right, part 2 is wrong
+ Show Spoiler + Put all 100 marbles into 1 jar.
But then you get to the part where you "pick a random marble from that jar" and if it's the empty jar your head explodes because you're a 1st generation android.
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United States1719 Posts
On November 13 2011 15:55 Warble wrote:Show nested quote +On November 13 2011 15:42 rotinegg wrote:On November 13 2011 15:35 Obelisco wrote:On November 13 2011 15:34 rotinegg wrote:On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer Part 1 is right, part 2 is wrong + Show Spoiler + Put all 100 marbles into 1 jar.
But then you get to the part where you "pick a random marble from that jar" and if it's the empty jar your head explodes because you're a 1st generation android.
if you can't pick a marble then so be it. answer's correct.
you got 2 2-gallon buckets. you pour 1 gallon of water in bucket one, and 1 gallon of JD in bucket two. Then, you pour x ounces (not more than half a gallon) of water from the bucket one to bucket two. Then you slosh bucket two around some and pour x ounces (same amount as last time) of that mixture into bucket one, so they end up having 1 gallon each again. Your fraternity brother bets you his girlfriend for yours, that there is more water in bucket one than there is JD in bucket two. Do you take the bet? Why?
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On November 13 2011 16:15 rotinegg wrote:Show nested quote +On November 13 2011 15:55 Warble wrote:On November 13 2011 15:42 rotinegg wrote:On November 13 2011 15:35 Obelisco wrote:On November 13 2011 15:34 rotinegg wrote:On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote:On November 13 2011 13:29 Kiarip wrote: We've had this question like 4 times on this site already... and I was so pumped... got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go. edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer Part 1 is right, part 2 is wrong + Show Spoiler + Put all 100 marbles into 1 jar.
But then you get to the part where you "pick a random marble from that jar" and if it's the empty jar your head explodes because you're a 1st generation android.
if you can't pick a marble then so be it. answer's correct. you got 2 2-gallon buckets. you pour 1 gallon of water in bucket one, and 1 gallon of JD in bucket two. Then, you pour x ounces (not more than half a gallon) of water from the bucket one to bucket two. Then you slosh bucket two around some and pour x ounces (same amount as last time) of that mixture into bucket one, so they end up having 1 gallon each again. Your fraternity brother bets you his girlfriend for yours, that there is more water in bucket one than there is JD in bucket two. Do you take the bet? Why?
The amount of water in bucket 1 = the amount of JD in bucket 2, since there is 1 gallon each of water and JD in total, and 1 gallon of mixture in each bucket at the end (assuming both have equal densities). So if you take the bet, you would win. But there is not enough information given in the question to determine if you should take the bet, since the quality of his girlfriend is unknown.
edit: oh wait, actually whiskey is less dense so x ounces from the second bucket is of greater volume than x ounces from the first bucket; there would be more water in bucket one than JD in bucket 2 by volume. Don't take the bet?
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+ Show Spoiler +Let's try the following general strategy:
We start from a height called h(1). If the egg breaks we try all floors from 1 to h(1)-1. If it doesn't break we chose a different floor, h(2)>h(1). If it broke try everything from h(1)+1 to h(2)-1. The same way if the egg broke at h(i) we try every floor from h(i-1) to h(i)-1.
So if the first egg broke at h(k) our worst case scenario is the second egg breaking at h(k)-1, which would lead to k+[h(k)-h(k-1)]-1 steps. Since the problem asks us to find the best worst case scenario let's just chose the function h such that the expression k+h(k)-h(k-1) doesn't depend on k. That's just a series where the difference between subsequent elements decreases by 1. E.g 5, 9, 12, 14, 15
We know that the last floor number in the sequence h(l) is 100 and h(0)<0. Solving for n(n-1)/2)>100 n>14. So if the egg never breaks we need 14 tries, starting with floors 9, 22, 34 etc. The choice of the first floor could be anything between 9 and 14, it doesn't change the final tally.
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Gah, Rotin, you had me doing complicated maths before I took a second look at your question.
+ Show Spoiler + Both buckets must have the same amount of both substances in the end, assuming a V/V basis and ignoring real-world effects.
This is because w1 + w2 = 1 = j1 + j2 ...(1)
Where w = water and j = JD.
We are told that in the end w1 + j1 = 1 = w2 + j2 ...(2)
His claim is that w1 > j2. However, we can show that any case where w1 != j2 cannot occur.
If w1 > j2, then from (1) we know that j1 > w2. But if j1 > w2, then we get a contradiction at (2) because w1 + j1 > w2 + j2 even though they're supposed to be equal.
Another way to solve it is to plug (2) into (1) and get:
w1 + j1 = 1 = w1 + w2 j1 = w2
And likewise we can prove that j2 = w1.
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On November 13 2011 16:30 Pangpootata wrote:Show nested quote +On November 13 2011 16:15 rotinegg wrote:On November 13 2011 15:55 Warble wrote:On November 13 2011 15:42 rotinegg wrote:On November 13 2011 15:35 Obelisco wrote:On November 13 2011 15:34 rotinegg wrote:On November 13 2011 15:16 hacklebeast wrote:On November 13 2011 15:05 rotinegg wrote:On November 13 2011 14:17 hacklebeast wrote:On November 13 2011 13:32 rotinegg wrote: [quote] got city A full of apples. You want to transport 4000 apples to city B, which is 1000 miles away. U got a truck that can carry 1000 apples at any given time, but it also has a squirrel infestation and 1 apple gets eaten per mile driven. What's your strategy to transport the most apples from city A to B? You can drop off apples along the way and they will be there for you to come pick up later, untouched. Go.
edit: don't even expect a response if you write something like 'kill the squirrel!' or 'YOU CAN'T SAVE ANY OF THE APPLES!!!@! LOLZ!Q!!!' + Show Spoiler + Move 1000 apples 1 mile, repeat 4 time so 3996 apples are one mile in. Move 1000 apples 1 mile, repeat 4 time so 3992 apples are two miles in. contine this prosses after the first 250 miles you will have lost 1000 apples, so each mile only loses you 3 additional apples. after 583 (250 + the 333 to lose another thousand) you lose only two, and the rest of the way every mile will only lose you 2 apples. 4000-250*4-333*3-417*2=1164 apples remaining.
correct? I had seen the op question before, but not this one.
edit, fixed
Not bad... You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles? + Show Spoiler +
rat 1 2 3 4 wine 1 X X X X wine 2 O X X X wine 3 X O X X wine 4 X X O X wine 5 X X X O wine 6 O O X X wine 7 O X O X wine 8 O X X O wine 9 X O O X wine 10 X O X O wine 11 X X O O wine 12 O O O X wine 13 O O X O
I could test 3 more wines with this method
aww tl messed up my spacing, but I bet you can figure it out sweet Part 1) You are given 50 black marbles and 50 white marbles, and two empty, identical jars. You are told that once you distribute the jars, they will be shaken up while you are blindfolded, and you will pick a jar at random and pick a random marble from that jar. Distribute the marbles so that you have the greatest chance of picking a black marble. Part 2) Redistribute the marbles so that you the least chance of picking a white marble 1)50 white and 49 black in one jar, 1 black marble alone in the other, equals almost 75 %.. like 74 something part two is same answer Part 1 is right, part 2 is wrong + Show Spoiler + Put all 100 marbles into 1 jar.
But then you get to the part where you "pick a random marble from that jar" and if it's the empty jar your head explodes because you're a 1st generation android.
if you can't pick a marble then so be it. answer's correct. you got 2 2-gallon buckets. you pour 1 gallon of water in bucket one, and 1 gallon of JD in bucket two. Then, you pour x ounces (not more than half a gallon) of water from the bucket one to bucket two. Then you slosh bucket two around some and pour x ounces (same amount as last time) of that mixture into bucket one, so they end up having 1 gallon each again. Your fraternity brother bets you his girlfriend for yours, that there is more water in bucket one than there is JD in bucket two. Do you take the bet? Why? The amount of water in bucket 1 = the amount of JD in bucket 2, since there is 1 gallon each of water and JD in total, and 1 gallon of mixture in each bucket at the end (assuming both have equal densities). So if you take the bet, you would win. But there is not enough information given in the question to determine if you should take the bet, since the quality of his girlfriend is unknown. edit: oh wait, actually whiskey is less dense so x ounces from the second bucket is of greater volume than x ounces from the first bucket; there would be more water in bucket one than JD in bucket 2 by volume. Don't take the bet?
I am not sure I would chance drinking any amount of JaeDong. That just seems weird.
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United States1719 Posts
On November 13 2011 16:56 Warble wrote:Gah, Rotin, you had me doing complicated maths before I took a second look at your question. + Show Spoiler + Both buckets must have the same amount of both substances in the end, assuming a V/V basis and ignoring real-world effects.
This is because w1 + w2 = 1 = j1 + j2 ...(1)
Where w = water and j = JD.
We are told that in the end w1 + j1 = 1 = w2 + j2 ...(2)
His claim is that w1 > j2. However, we can show that any case where w1 != j2 cannot occur.
If w1 > j2, then from (1) we know that j1 > w2. But if j1 > w2, then we get a contradiction at (2) because w1 + j1 > w2 + j2 even though they're supposed to be equal.
Another way to solve it is to plug (2) into (1) and get:
w1 + j1 = 1 = w1 + w2 j1 = w2
And likewise we can prove that j2 = w1.
nice job, it's a simple question that trips most people up.
You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black?
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Okay, I have one.
You're probably all familiar with the pirates gold puzzle, but I'm going to add a twist to it.
To make it fair, I will specify the entire problem for those who may not have seen it before.
Here is the classic version:
- We have 10 pirates who have 100 gold coins in booty to distribute between them.
- The pirates are democratic. All pirates get 1 vote, and in the case of a tie, the "Ayes" have it. (That means if a vote is 50/50, the proposal wins.)
- The pirates have a strict hierarchy. They each know where they stand, and there is only one pirate of each rank.
- Only the most senior pirate can make a proposal.
- If a pirate's proposal is rejected, he is killed.
- Pirates love being alive.
- Pirates love gold the most. Any gold is better than everything else in the world. Of course, they can't enjoy gold if they're dead, so the above rule holds.
- Pirates enjoy killing other pirates and there is no love lost between them. Given an option to kill another pirate or to do nothing, they will choose to kill another pirate. They don't enjoy this as much as gold or their own life, however.
- All pirates are rational and know that all other pirates are rational. This means they will always make the best decisions given these rules, and they know that other pirates know all this.
My version is this:
The same as above, but now there are potentially infinite pirates.
What happens to each pirate for each starting number of pirates?
And the advanced version for the serious players:
+ Show Spoiler + I have actually not given you enough information to completely solve the puzzle. What bit of information did I intentionally leave out? What is the solution for all variations of this extra information?
Note: I may have unintentionally left some information out, but I've tried not to.
Note 2: It is possible to solve this puzzle without this extra piece of information (since I assumed not everyone would try the advanced version) - it's just that there will be multiple valid solutions. Hence the advanced version is essentially to identify what this missing information is and find all solutions. It is best to solve the simpler version first.
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United States1719 Posts
On November 13 2011 17:09 Warble wrote:Okay, I have one. You're probably all familiar with the pirates gold puzzle, but I'm going to add a twist to it. To make it fair, I will specify the entire problem for those who may not have seen it before. Here is the classic version: - We have 10 pirates who have 100 gold coins in booty to distribute between them.
- The pirates are democratic. All pirates get 1 vote, and in the case of a tie, the "Ayes" have it. (That means if a vote is 50/50, the proposal wins.)
- The pirates have a strict hierarchy. They each know where they stand.
- Only the most senior pirate can make a proposal.
- If a pirate's proposal is rejected, he is killed.
- Pirates love being alive.
- Pirates love gold the most. Any gold is better than everything else in the world. Of course, they can't enjoy gold if they're dead, so the above rule holds.
- Pirates enjoy killing other pirates and there is no love lost between them. Given an option to kill another pirate or to do nothing, they will choose to kill another pirate. They don't enjoy this as much as gold, however.
- All pirates are rational and know that all other pirates are rational. This means they will always make the best decisions given these rules, and they know that other pirates know this and know these rules.
My version is this: The same as above, but now there are potentially infinite pirates. What happens to each pirate for each starting number of pirates? And the advanced version for the serious players: + Show Spoiler + I have actually not given you enough information to solve the puzzle. What bit of information did I intentionally leave out? What is the solution for all variations of this extra information?
Note: I may have unintentionally left some information out, but I've tried not to.
+ Show Spoiler +Did this one before... Start from the case of 2 pirates and work your way up: - For 2 pirates, senior takes 100 gold, junior can't do shit.
- For 3 pirates, senior gives 1 gold to the lowest rank, and takes 99 gold for himself. Lowest agrees because if not, then senior is killed and we end up with the 2 pirate version proposal in which the lower rank gets no gold.
- For 4 pirates, senior gives 1 gold to 2nd lowest ranked pirate, and takes 99 gold for himself. Second lowest agrees because if not, senior is killed and we end up with the 3 pirate version proposal in which the second lowest gets no gold.
- For 5 pirates, senior needs 2 partners to corroborate, so he gives 1 gold to lowest and 3rd lowest rank, and takes 98 gold for himself. They will accept because if not, they will revert to scenario with 4 pirates in which they get no gold.
- For 6 pirates, senior needs 2 partners so he gives the 2nd lowest and 4th lowest 1 gold each.
This continues on and can be generalized for infinite number of pirates as: - if there are an even number of pirates -> the senior pirate gives 1 gold to every pirate who is an even number'th rank from the bottom, and the rest to himself.
- if there are an odd number of pirates -> the senior pirate gives 1 gold to every pirate who is an odd number'th rank from the bottom, and the rest to himself.
I'm not sure what you left out; I'm guessing it's the number of actual gold coins to distribute, but I'm not sure why you would... this will bug me tonight... anyway, assuming the thing u left out is indeed the number of coins, i'll call that quantity x, and the number of pirates y. General solution is: Senior partner gets n-(y-2+y%2)/2 coins and the rest of the pirates who fit the rules stated above get 1 coin each.
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On November 13 2011 17:04 rotinegg wrote: You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black?
Is this a trick question? It looks very straightforward.
+ Show Spoiler + This only trick I can think of is that the cube must be hollow.
Then to get the number of cubes, we take the surface area, subtract 1 for each edge, and add 1 for each corner.
n = 2 * (95*85 + 95*2000 + 85*2000) - 4 * (95 + 85 + 2000) + 4 = 727,434
The surface area is simply the first term in that equation, so we just take that over 6n (the number of faces) and get:
P = 16.9%.
Or roughly 1/6 since the 2nd and 3rd terms are tiny compared to the 1st.
EDIT: I just realised there are 8 corners, not 4. Oh well.
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There's only 100 gold coins in both versions so your generalisation won't work as n gets larger.
In particular, your answer should be able to identify what happens when we start with any number of pirates, whether it be 100 pirates, 1,000 pirates, 10,000 pirates...
The only information that changes between my version and the classic version I outlined is the number of pirates. However, in specifying my version, I complicate the puzzle enough that it actually requires another piece of information that has not been given in either version. However, this bit of information isn't crucial since, as I hinted, I doubt many people will realise I'd left it out and will still end up finding one solution (but they wouldn't have been aware there were other solutions).
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On November 13 2011 17:04 rotinegg wrote:Show nested quote +On November 13 2011 16:56 Warble wrote:Gah, Rotin, you had me doing complicated maths before I took a second look at your question. + Show Spoiler + Both buckets must have the same amount of both substances in the end, assuming a V/V basis and ignoring real-world effects.
This is because w1 + w2 = 1 = j1 + j2 ...(1)
Where w = water and j = JD.
We are told that in the end w1 + j1 = 1 = w2 + j2 ...(2)
His claim is that w1 > j2. However, we can show that any case where w1 != j2 cannot occur.
If w1 > j2, then from (1) we know that j1 > w2. But if j1 > w2, then we get a contradiction at (2) because w1 + j1 > w2 + j2 even though they're supposed to be equal.
Another way to solve it is to plug (2) into (1) and get:
w1 + j1 = 1 = w1 + w2 j1 = w2
And likewise we can prove that j2 = w1.
nice job, it's a simple question that trips most people up. You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black?
+ Show Spoiler +16150000 cubes in total
To get the number of cubes that could have been painted, we take the surface area of the cube before it got broken. 16150 = 95 * 85 * 2 (there are two faces with this size) 340000 = 2000 * 85 * 2 380000 = 2000 * 95 * 2 736150 = total
Each face has two instances of overlap, while the 8 corners of the cube have 3 instances of overlap. 727422=736150 - 4(95) - 4(85) - 4(2000) - 8, this should give the amount of cubes with a black face. I think subtracting 8 there was probably an error.
Now we need to determine the amount of cubes that have 3, 2, and 1 possible painted side. 8 cubes have 3 sides painted by definition of a cube.
If my math up top is correct, you could subtract that number from the total painted cubes and arrive with the correct figure. I don't think it is, so we can: 8696=4(95-2) + 4(85-2) + 4(2000-2). This should give us the amount of non-corners. To get the rest I'll use my probably flawed math from up top. 718718=727422 - 8 - 8696
# of sides : # of cubes 3: 8 2: 8696 1: 718718
Now onto probabilities # of sides : Probability of rolling a black face 3: 50% 2: 33% 1: 17%
If I didn't already fail I'm really going to fail now. Total probability = probability of picking a cube with black * probability of it being rolled and coming up black. x= total cubes = 16150000 answer = .5(8/x) + .33(8696/x) + .17(718718/x) answer = .0000002 + .00018 + .007 (didn't round up because it is .007 :D) answer = 0.72%
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United States1719 Posts
On November 13 2011 17:29 Warble wrote:Show nested quote +On November 13 2011 17:04 rotinegg wrote: You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black? Is this a trick question? It looks very straightforward. + Show Spoiler + This only trick I can think of is that the cube must be hollow.
Then to get the number of cubes, we take the surface area, subtract 1 for each edge, and add 1 for each corner.
n = 2 * (95*85 + 95*2000 + 85*2000) - 4 * (95 + 85 + 2000) + 4 = 727,434
The surface area is simply the first term in that equation, so we just take that over 6n (the number of faces) and get:
P = 16.9%.
Or roughly 1/6 since the 2nd and 3rd terms are tiny compared to the 1st.
Your logic is correct. I haven't done the actual calculations myself, though. I'm a college student preparing to work in the financial sector and my buddies and I give each other brainteasers. One of the popular (and simpler) ones is: you have an nxnxn cube, and you paint the outside; how many 1x1x1 cubes are painted? When you get into that mode of thinking, it's easy for you to try to solve any cube problem with the same approach of detaching the surface and figuring out the rest... So you see alot of people break the cubes up into special cases for flat surfaces with 1 side painted, edges with two sides painted and corners with 3 sides painted, and end up doing hideous mathematical calculations... including myself :p good for you though, glad to see smart ppl on TL
edit: The post above did exactly what I first tried to do haha
You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar?
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I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water - the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position, even while he is on land. (The original monster only moved when Day9 was in the water. This monster has no such restriction. It moves all the time - but can only move in water.)
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy?
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On November 13 2011 17:32 Warble wrote: There's only 100 gold coins in both versions so your generalisation won't work as n gets larger.
In particular, your answer should be able to identify what happens when we start with any number of pirates, whether it be 100 pirates, 1,000 pirates, 10,000 pirates...
The only information that changes between my version and the classic version I outlined is the number of pirates. However, in specifying my version, I complicate the puzzle enough that it actually requires another piece of information that has not been given in either version. However, this bit of information isn't crucial since, as I hinted, I doubt many people will realise I'd left it out and will still end up finding one solution (but they wouldn't have been aware there were other solutions).
If it is limited to 100 coins even as the amount of pirates increases the solution is somewhat simple. The most senior pirates get killed until the proposal can reach a 50/50 with the most senior getting 2 gold and the other (half of pirates - 1) get 1 gold. For 10 pirates and 100 coins I don't see why his generalization doesn't work, and nothing crucial seems to be missing. Spill the beans already though =p. My guess would be that there was no ranking established, so maybe two pirates could share the same rank, and would have to work together when making proposals. This would also force the most senior pirate (or group) to appease groups instead of giving 1 gold to every other pirate.
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On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water - the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position, even while he is on land.
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy? + Show Spoiler + If we're talking amnesia, the monster doesn't swim unless Day9 is in the water. The way to prolong his life is to go into the water on the opposite side of the island of the gate. Lure the monster there, then run straight across the island and to the gate. The water monster will have to run around half the circumference - a tangent line towards Day9 if he still hasn't reached the gate.
Even if the monster knows where Day9 is, the fastest path to him until he reaches the center of the island is still the radius line. This will keep the monster as far from the gate as possible until Day reaches the center of the island and moves off the center towards the gate. Only then will it start to circle around.
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Each pirate stands alone in his rank.
The unspecified information isn't crucial for solving the problem - you can still solve it. It's just that it means there are multiple valid solutions. (The information would restrict the number of solutions.)
This monster moves even when Day9 is on land.
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I read that after I posted, but my solution still works, maybe. If we do some math, (pi)*r = the arc that the monster must run. So the monster has to move 3.141r while Day has to move 1r. Damnit The monster can move 4r in the time Day moves 1r. If the monster 1 shots him, there is no solution because he can't get off the island with the monster able to traverse half the circle before Day can traverse the radius. If the monster is rational and makes the best decision for the situation, which would be to stay on the same "radius line" (if that makes any sense) as Day. Basically if you drew a line from the center of the circle to Day and extended it out to the edge of the circle, that is where the monster would be.
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It's almost like I rigged it against Day9.
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Yea, I don't really want to do rotinegg's problem I could understand beefing up the cube one with bigger numbers because the amount of calculations basically remains the same. Get total cubes, get number with a side painted, add up probabilities. Each cat kind of adds new calculations unless there is some overarching generalization, which probably exists.It is kind of like your teacher assigning addition problems that have you adding up 10 numbers instead of 2. It doesn't require any more knowledge, the problem just takes longer.
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On November 13 2011 17:29 Warble wrote:Show nested quote +On November 13 2011 17:04 rotinegg wrote: You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black? Is this a trick question? It looks very straightforward. + Show Spoiler + This only trick I can think of is that the cube must be hollow.
Then to get the number of cubes, we take the surface area, subtract 1 for each edge, and add 1 for each corner.
n = 2 * (95*85 + 95*2000 + 85*2000) - 4 * (95 + 85 + 2000) + 4 = 727,434
The surface area is simply the first term in that equation, so we just take that over 6n (the number of faces) and get:
P = 16.9%.
Or roughly 1/6 since the 2nd and 3rd terms are tiny compared to the 1st.
EDIT: I just realised there are 8 corners, not 4. Oh well.
+ Show Spoiler +I think it's a "trick" question because you can overthink by looking at edges, corners and whatnot. There are 6x85x95x2000 sides and 2x(85x95+85x2000+95x2000) sides are painted black. There's no need to treat edges and corners differently: you want to double and triple count them because they have 2 and 3 sides painted black respectively.
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United States1719 Posts
On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water, the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is a tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position.
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy? + Show Spoiler [watermonster vs Day9] +Day9 would start to run in a spiral until he reaches the inner circle with radius length r/4. He would time make the spiral pattern in a way so that when he reaches r/4 away from the center, the monster would be on the other side of the island, lagging a full 180degrees behind. That is when Day9 would make a run for it straight to the coast (no more spiraling), and hopefully find a gate in the water. He can definitely make it to the coast, but not much more.
I think this is the best strategy because the max distance you can get the water monster to swim around is half of the island's circumference, or pi*r. At the circle with radius length r/4, you can keep up with the monster in terms of angular velocity (2*pi*r / 4s = (2*pi*r / 4) / s), so you can definitely outrun the monster inside that circle. Since Day9 is so smart, I will leave it up to him to come up with the spiral formula that will land him at the r/4 point with exactly a half circle lead on the monster. The distance Day9 covers from that point to the point on the coast along the same radial line is 3r/4, which will take 3r/4s time, while the monster must swim pi*r, which will take pi*r/4s, about .14r/4s more than it takes Day9 to reach the coast. If Day9 doesn't see a gate nearby, he must repeat the same process for every degree, minute and second of the island's entire 360 degrees.
Not sure if this is the solution you were looking for... something feels amiss
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On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar?
+ Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
Yes, there's more that Day9 can do to outsmart the monster. You know its behaviour...use it...
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On November 13 2011 18:21 rotinegg wrote:Show nested quote +On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water, the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is a tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position.
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy? + Show Spoiler [watermonster vs Day9] +Day9 would start to run in a spiral until he reaches the inner circle with radius length r/4. He would time make the spiral pattern in a way so that when he reaches r/4 away from the center, the monster would be on the other side of the island, lagging a full 180degrees behind. That is when Day9 would make a run for it straight to the coast (no more spiraling), and hopefully find a gate in the water. He can definitely make it to the coast, but not much more.
I think this is the best strategy because the max distance you can get the water monster to swim around is half of the island's circumference, or pi*r. At the circle with radius length r/4, you can keep up with the monster in terms of angular velocity (2*pi*r / 4s = (2*pi*r / 4) / s), so you can definitely outrun the monster inside that circle. Since Day9 is so smart, I will leave it up to him to come up with the spiral formula that will land him at the r/4 point with exactly a half circle lead on the monster. The distance Day9 covers from that point to the point on the coast along the same radial line is 3r/4, which will take 3r/4s time, while the monster must swim pi*r, which will take pi*r/4s, about .14r/4s more than it takes Day9 to reach the coast. If Day9 doesn't see a gate nearby, he must repeat the same process for every degree, minute and second of the island's entire 360 degrees.
Not sure if this is the solution you were looking for... something feels amiss
I don't think that solution works. Unless I'm not understanding something, the monster can't lag 180 degrees behind unless Day runs through the center of the circle. I think I see where you're trying to go with this, but it doesn't quite work because the monster would simply change directions if Day did the spiraling thing.
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United States1719 Posts
On November 13 2011 18:33 Frozenhelfire wrote:Show nested quote +On November 13 2011 18:21 rotinegg wrote:On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water, the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is a tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position.
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy? + Show Spoiler [watermonster vs Day9] +Day9 would start to run in a spiral until he reaches the inner circle with radius length r/4. He would time make the spiral pattern in a way so that when he reaches r/4 away from the center, the monster would be on the other side of the island, lagging a full 180degrees behind. That is when Day9 would make a run for it straight to the coast (no more spiraling), and hopefully find a gate in the water. He can definitely make it to the coast, but not much more.
I think this is the best strategy because the max distance you can get the water monster to swim around is half of the island's circumference, or pi*r. At the circle with radius length r/4, you can keep up with the monster in terms of angular velocity (2*pi*r / 4s = (2*pi*r / 4) / s), so you can definitely outrun the monster inside that circle. Since Day9 is so smart, I will leave it up to him to come up with the spiral formula that will land him at the r/4 point with exactly a half circle lead on the monster. The distance Day9 covers from that point to the point on the coast along the same radial line is 3r/4, which will take 3r/4s time, while the monster must swim pi*r, which will take pi*r/4s, about .14r/4s more than it takes Day9 to reach the coast. If Day9 doesn't see a gate nearby, he must repeat the same process for every degree, minute and second of the island's entire 360 degrees.
Not sure if this is the solution you were looking for... something feels amiss I don't think that solution works. Unless I'm not understanding something, the monster can't lag 180 degrees behind unless Day runs through the center of the circle. I think I see where you're trying to go with this, but it doesn't quite work because the monster would simply change directions if Day did the spiraling thing. Would the monster know what Day9 is doing? I was simply going off the assumption that the water monster only knows how to reduce the distance between Day9 and itself at any given point and nothing more... In which case it wouldn't turn around when it is less than half a lap behind Day9, because turning around would mean there would be more ground to cover.
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United States1719 Posts
On November 13 2011 18:32 Warble wrote:Show nested quote +On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off
I'm still thinking about the pirate booty one... grr + Show Spoiler +So my generalization still holds for pirates <= 202, because the seniors would still want to live even if it means getting no gold. Now, I'm not sure how I would go about it afterwards... When n=203, senior needs 101 votes to not be killed, but only has 100 coins... T_T
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+ Show Spoiler +I think the first egg need to go up the floor in increments of 3. This is the best solution I can think of.
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On November 13 2011 18:41 rotinegg wrote:Show nested quote +On November 13 2011 18:32 Warble wrote:On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off
The distance between cat 49 and cat 50 is 1 m and doesn't change until cat 50 bumps into cat 51, after which 49 and 50 approach each other at 2 m/s, which means 0.5 s to close the gap.
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United States1719 Posts
On November 13 2011 18:44 Warble wrote:Show nested quote +On November 13 2011 18:41 rotinegg wrote:On November 13 2011 18:32 Warble wrote:On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off The distance between cat 49 and cat 50 is 1 m and doesn't change until cat 50 bumps into cat 51, after which 49 and 50 approach each other at 2 m/s, which means 0.5 s to close the gap. Ooh, you're right, my brain farted for a second... there's still a mistake somewhere because the correct answer is off.. hmmm.. also there is a very simple way of solving this, which made me go nuts after I heard it, because by that point my brain was full of cats walking back and forth, bouncing off of eachother and falling off poles.. haha
edit: your mistake was at 99/2 = 45.5m and no idea where you got the 22.5s from... I get 24.25s...
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I guess it depends on how rational we assume the monster to be. I actually misunderstood what you were getting at, but I think i see it now. He basically starts at some offset between 0 degrees and 180 degrees, and runs in a semi-circle with the end-point being a tangent line from the inner circle to the gate. He does this at a distance in which he can move around the circle faster than the monster, so that the tangent line distance is less than the radius. This would require him to be in a distance such that the circumference of the circle he runs in is more than 4x smaller than the island circumference. Thinking about it practically, it would have to be small(er) enough to get enough of a lead. I'm not sure if the math could work out for Day to get off the island.
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United States1719 Posts
On November 13 2011 18:51 Frozenhelfire wrote: I guess it depends on how rational we assume the monster to be. I actually misunderstood what you were getting at, but I think i see it now. He basically starts at some offset between 0 degrees and 180 degrees, and runs in a semi-circle with the end-point being a tangent line from the inner circle to the gate. He does this at a distance in which he can move around the circle faster than the monster, so that the tangent line distance is less than the radius. This would require him to be in a distance such that the circumference of the circle he runs in is more than 4x smaller than the island circumference. Thinking about it practically, it would have to be small(er) enough to get enough of a lead. I'm not sure if the math could work out for Day to get off the island. The math works out so he can barely reach the coast before the monster does, (pi-3)r/4s faster to be precise, but since there is no information on how far off the gate is from the coast, I couldn't proceed any further from that.
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Yea, he could also do one of those dealies, and make a fillet between the gray and red lines to increase the distance he can cover. That might end up being the same as the spiral thing, but this is the general idea you were getting at, right?
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United States1719 Posts
Yup, looks about right :p traveling in a straight line from the center to the circle with 1/4 circumference like the one you drew won't work, though, because by the time u reach the circle with 1/4 the circumference, the monster will be closer than 180 degrees behind. The only way to guarantee a half lap is to keep changing the spot the monster needs to be at to be closest to Day9 at a faster rate than the monster can catch up, hence i chose a spiral
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I guess my solution would be to open developers console, apply noclip or spawn a whole bunch of dead flesh to distract the monster and then get to the gate.
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On November 13 2011 12:31 evanthebouncy! wrote: This one is a classic, but can you do it?
You have 2 identical eggs, and a 100 floor building. You wish to find out the highest floor number in which the egg can be dropped without breaking. Describe a strategy which you can precisely find that floor number in the fewest drops.
Q&A:
Q: But, what do you mean "fewest?" A: I mean at the worst case. For instance, your strategy might be try fl1, then fl2, then 3 4 5... Under this strategy, the worst that can happen is the egg never breaks, i.e. it can be dropped from over 100 floors, which will take you all 100 drops to find out. So, this strategy takes 100 drops.
Q: Does the egg crack if dropped repeatidly? A: No, egg is good as new as long as it don't break yet.
Again, put answers in spoilers per usual. Collaborate, have fun!
--evan
+ Show Spoiler + The first thing I notice is that we need decreasing intervals between drops.
So if we start on 10, then we want both break and non-break scenarios to have similar remaining tries.
Break means we try 1->9, worst case scenario of 9 subsequent tries (10 total).
So 10 is only an optimum start if non-break also has a worst case of 9 subsequent tries.
And we know this is no longer possible because this means we then pick 19, then 27, and it's obvious we're not going to get very far.
Looking at the high end, we know that our last try is 100 to check if the eggs ever break. And we want to pick n such that we minimise the number of iterations before reaching 100.
We can solve this by working backwards.
We must end on 100.
2nd last is 99. It can't be 98 because if 100 breaks then we have possibilities [99,100] and no tries remaining.
3rd last is 97. That way, if 99 breaks, we can try 98 with our final try.
4th last is 94. This is because if 97 breaks, we have 2 tries left, for 95 and 96.
From here it's clear that the size of the spread increases by 1 for each try we go back.
This gives the sequence:
9, 22, 34, 45, 55. 64, 72, 79, 85, 90, 94, 97, 99, 100.
This involves a total of 14 tries in the worst case.
So the minimum number of tries is 14. We can begin on any floor [9,14]. My solution covers the lowest floors possible for each attempt in the primary sequence. The highest floors possible can be found by starting on 14 and increasing in increments of 13, 12, 11, etc. and ending on floor 100.
As for Day9:
+ Show Spoiler + So you've worked out that he can safely make it to the water. But he's still got more tricks up his sleeves. What else can he do to increase the time he stays alive after reaching the water?
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Here's an idea for Day9:
+ Show Spoiler +If you don't assume he runs on the line connecting the gate and the center of the island, and instead runs on a curve from the r/4 circle, he might get closer to the gate before the monster reaches him. Of course, exactly how much of a lead this would give him would diminish based on how far the gate is, but would still be >0 as long as the gate is not infinitely far away. I can't think of how to model this though, I think it would involve some sort of curve.
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On November 13 2011 18:48 rotinegg wrote:Show nested quote +On November 13 2011 18:44 Warble wrote:On November 13 2011 18:41 rotinegg wrote:On November 13 2011 18:32 Warble wrote:On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off The distance between cat 49 and cat 50 is 1 m and doesn't change until cat 50 bumps into cat 51, after which 49 and 50 approach each other at 2 m/s, which means 0.5 s to close the gap. Ooh, you're right, my brain farted for a second... there's still a mistake somewhere because the correct answer is off.. hmmm.. also there is a very simple way of solving this, which made me go nuts after I heard it, because by that point my brain was full of cats walking back and forth, bouncing off of eachother and falling off poles.. haha edit: your mistake was at 99/2 = 45.5m and no idea where you got the 22.5s from... I get 24.25s...
I'm guessing this is the simple solution you're talking about:
+ Show Spoiler +Don't consider collisions as bouncing cats. Instead, every collision has the same effect as having the two cats pass through one another. Thus, the time will be the maximum of the times it would take each individual cat to fall off if no other cats were present. We have a 99m pole and a cat on the left facing right moving at 1m/s, so this is 99s.
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United States1719 Posts
On November 13 2011 21:06 sidr wrote:Show nested quote +On November 13 2011 18:48 rotinegg wrote:On November 13 2011 18:44 Warble wrote:On November 13 2011 18:41 rotinegg wrote:On November 13 2011 18:32 Warble wrote:On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off The distance between cat 49 and cat 50 is 1 m and doesn't change until cat 50 bumps into cat 51, after which 49 and 50 approach each other at 2 m/s, which means 0.5 s to close the gap. Ooh, you're right, my brain farted for a second... there's still a mistake somewhere because the correct answer is off.. hmmm.. also there is a very simple way of solving this, which made me go nuts after I heard it, because by that point my brain was full of cats walking back and forth, bouncing off of eachother and falling off poles.. haha edit: your mistake was at 99/2 = 45.5m and no idea where you got the 22.5s from... I get 24.25s... I'm guessing this is the simple solution you're talking about: + Show Spoiler +Don't consider collisions as bouncing cats. Instead, every collision has the same effect as having the two cats pass through one another. Thus, the time will be the maximum of the times it would take each individual cat to fall off if no other cats were present. We have a 99m pole and a cat on the left facing right moving at 1m/s, so this is 99s. Bingo.
you have 20 prisoners on death row. The day before the execution, the executioner decides to have a bit of mercy and lets them know exactly what is going to go down. He says:
- They will be lined up in one row, so that they stand behind one another.
- The executioner will come up behind each of them and randomly put on a red or blue hat.
- You will not be able to see your own hat, or any of those behind you, but you will be able to see those in front of you.
- The executioner will start from the very back, and ask you what your hat color is. You are permitted to only say either red or blue, and if correct, you will live. If not, your brains will be blown out.
- You will be given one night to come up with a strategy as a group to save the most men.
- You may not communicate with each other during the execution day; if you so much as blink or breathe in a peculiar pattern, all 20 of you will be shot immediately.
All the prisoners are very astute and logical, and place no importance on their own life over others'. How would you save the most people, and how many people would you save on average using said strategy?
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Here's a naive solution to the prisoners' problem:
+ Show Spoiler +The first ten will, in order, announce that they are wearing the color of the hats of the first ten men. Then the last ten men will know their own colors.
Alternatively, the odd-numbered prisoners will announce the color of the person in front of them, saving the even-numbered prisoners. Either way, you're sacrificing one for one.
Assuming uniform distribution, this will save 7.5 prisoners on average.
.... I hope guessing your own color is done out loud and is not considered extra communication. Without communication and knowing the distribution, I don't think this has any real solution.
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Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now
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NOTE: a very cheap solution (and i think beyond the intentions of the question) - lets say for the front 19 people there are 19^2 possible combinations. Well the first guy can simply "count" up to the combination number in seconds, wait that amount of time, and then make his answer. He could halve the time required by making his answer either red/blue to signal two different combo sets Then every person in front having memorized that set would be able to easily win.
Thus from now on, i am assuming each prisoner must answer instantly, however each prisoner will be allowed to know what happened to the guys before them (shot or alive) -----
Continuing from the previous solution if i'm on the right track
ASSUMING the prisoners get to hear whether the guy behind them is shot or not
in a 4 person game, first guy can use a red blue to signal one of 2 possible combos, RRR/BBB (all same) or 2 of one colour and 1 of the other colour Given an all same, the next part is easy, and all 3 are saved
Given a 2:1 combo, the next guy will see one of two things - either RR/BB (both same) or RB/BR In an RR/BB situation, this is ideal as he knows exactly what his colour is Then the no3 and no4 will know they are holding the same colour hats, and will have an easy time answering
in a RB/BR situation, no2 will be in a bit of a pickle. Now, if prisoner 1 was allowed to use the "wait 1 second then answer or wait 2 seconds and answer" - he would be able to announce both the combination, and what colour no2 is wearing (wait 1 = red, wait 2 = blue) If you play this waiting game (on a reasonable time frame) then no2 will always be saved, and no3 will know the combo (2:1) and two of the colours, allowing him to save himself and the guy in front
Repeating this style, you get 15 saved, and 5 prisoners on a 50/50, resulting in 17.5 saved (bad answer) ----
However, this hinges on the "wait 1 second and answer or wait 2 seconds answer"
If you must answer instantly, as i presumed before, then you have a few combinations. No3 and no4 are always saved as they know what happens to the guy before him. No1 is always 50/50, no matter what The possible combos for no2 are RRR RRB RBR BRR BBR BRB RBB BBB
Thus, at least 2/8 times he is saved. For the remaining 6 combinations, he is saved in an additional 2 of them BRR and RBB
Leaving 4 tricky ones, RBR, BBR, RRB, BRB In these situations, he has a 50/50 chance to survive
Thus, out of 8 combos, he survives 6 of them on average, and has a 75% chance of survival. Thus for 4 players, we have 50%, 75%, 100% 100%, giving us 3.25 saved on average, giving us 16.25 saved overall. This is a worse solution than the one i posted above, leaving me to believe that iterative solutions in general, will fail
Thus, i think the solution i posted above is the best
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+ Show Spoiler +Super Easy.... let x be whether the egg breaks (true or false), and y be the floor number... Step 1. Start at y=10, test x, if x=true continue to step 3, if x=false continue to step 2 Step 2, add 10 to y, test x, if x=false repeat step 2, if x=true continue to step 3 Step 3. subtract 9 from y, test x. if x=true continue to step 5, if x=false continue to step 4 Step 4. Add 1 to y, test x, if x=true continue to step 5, if x=false repeat step 4 Step 5. we conclude our answer to be what y was equal to when x=true minus 1. Doing this method will give you a maximum value for fewest drops of 19, and a minimum of 1.
DJ WILMA OUT!
I may have misunderstood the question, but the average outcome of drops is equal to 10 drops, if thats what your looking for I'm pretty sure mines the best
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On November 14 2011 00:19 DJWilma wrote:+ Show Spoiler +Super Easy.... let x be whether the egg breaks (true or false), and y be the floor number... Step 1. Start at y=10, test x, if x=true continue to step 3, if x=false continue to step 2 Step 2, add 10 to y, test x, if x=false repeat step 2, if x=true continue to step 3 Step 3. subtract 9 from y, test x. if x=true continue to step 5, if x=false continue to step 4 Step 4. Add 1 to y, test x, if x=true continue to step 5, if x=false repeat step 4 Step 5. we conclude our answer to be what y was equal to when x=true minus 1. Doing this method will give you a maximum value for fewest drops of 19, and a minimum of 1.
DJ WILMA OUT!
I may have misunderstood the question, but the average outcome of drops is equal to 10 drops, if thats what your looking for I'm pretty sure mines the best
weve had solutions where the maximum is 14 and that's the best
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United States1719 Posts
On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case
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I hope that rule doesn't count voice inflections :d Prisoners:
The last prisoner says the color of the hat the person in front of him is wearing. Odd numbers inflect the color of their hat as a question i.e. blue? if the person in front of them has the same color hat. Even numbers inflect if the person has a different hat.
As for Day
+ Show Spoiler +Once he reaches the water he can rip off a finger and throw it at the monster. The monster will stop to eat it while Day runs. :D I guess he could do that multiple times if needed.
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United States1719 Posts
On November 14 2011 03:36 Frozenhelfire wrote: I hope that rule doesn't count voice inflections :d Prisoners:
The last prisoner says the color of the hat the person in front of him is wearing. Odd numbers inflect the color of their hat as a question i.e. blue? if the person in front of them has the same color hat. Even numbers inflect if the person has a different hat.
voice inflections also count as cheating
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I guess I'm stumped here. Are there ten hats of each color or could all the prisoners end up with the same hat?
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On November 14 2011 03:12 rotinegg wrote:Show nested quote +On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case
here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution
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On November 14 2011 06:04 JeeJee wrote:Show nested quote +On November 14 2011 03:12 rotinegg wrote:On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution
Good solution. I thought it would be something simple. I forgot that everyone can see what is in front of them instead of just the last prisoner.
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I'm fairly certain if you drop an egg from the first floor it will break because you can drop eggs from like 4 feet off the group and they break...
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On November 14 2011 07:21 Frozenhelfire wrote:Show nested quote +On November 14 2011 06:04 JeeJee wrote:On November 14 2011 03:12 rotinegg wrote:On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution Good solution. I thought it would be something simple. I forgot that everyone can see what is in front of them instead of just the last prisoner. A nice variant is that everyone can see everyone' else's colors but their own. But instead of two colors, you can work a solution for n colors, and in fact if I remember well the "paper" solutions with "infinite" colors/prisonners, at least in some special cases.
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United States1719 Posts
On November 14 2011 06:04 JeeJee wrote:Show nested quote +On November 14 2011 03:12 rotinegg wrote:On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution Nice job!
I don't remember any more off the top of my head at the moment haha
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On November 13 2011 22:05 rotinegg wrote:Show nested quote +On November 13 2011 21:06 sidr wrote:On November 13 2011 18:48 rotinegg wrote:On November 13 2011 18:44 Warble wrote:On November 13 2011 18:41 rotinegg wrote:On November 13 2011 18:32 Warble wrote:On November 13 2011 17:46 rotinegg wrote: You have a 99 meter pole, and it is one-dimensional. There are 100 cats that occupy the pole, all evenly spaced so that there is a cat every meter. These ain't no normal cats: they have no width or length; just height. They also walk at a constant speed of 1m/s, and have infinite acceleration so they can start walking or turn in the opposite direction instantaneously. Now, 50 cats on the left side of the bar start out facing right, and the other 50 on the right side start out facing left. When you blow a whistle, they will all start walking in the direction they are facing, but turn around and start walking the other way once they hit another cat. If they reach the end of the bar, they will just fall off. Assuming this, how long does it take for all the cats to fall off the bar? + Show Spoiler + So we want to find out how long it takes for the middle cats to fall off. Time and distance are equivalent in this puzzle.
Let's consider the 50th cat.
It walks for 0.5 s right and bumps. Turns and walks 0.5 s left and bumps into cat 49. This means cat 50 ends up back where it started after 1 s and cat 49 is now facing left.
We don't really need to follow the analysis for cat 49 because now it's obvious that the "bump wave" travels left at 1 m/s starting from the middle after 0.5 s.
Now let's consider cat 1. It's travelling right at 1 m/s. Combine this with the wave and we see that they're approaching each other at 2 m/s. So cat 1 is bumped after 22.5 s + 0.5 s. After which cat 1 is doomed.
What about cat 2? There is a new "bump wave" every 1 s, ...
Due to symmetry, we can treat this puzzle as a 45.5 metre pole with a wall on one end. So the analysis for this side applies to the other side.
Actually, forget all that. This puzzle is simple. We can already see that each bump wave will knock off 1 cat. There are 50 cats, so cat 50 will carry the 50th bump wave all the way to the end. Since each bump wave lasts 1 s, and begins after 0.5 s, the 50th bump wave occurs at 50.5 s, whereupon cat 50 has 99/2 = 45.5 m to walk.
So the last pair of cats fall off at 96 seconds.
The 50th cat would walk 0.5s right and bump, correct, but it would walk less than 0.5s to the left because cat 49 also walked during that time, no? hehe answer is off The distance between cat 49 and cat 50 is 1 m and doesn't change until cat 50 bumps into cat 51, after which 49 and 50 approach each other at 2 m/s, which means 0.5 s to close the gap. Ooh, you're right, my brain farted for a second... there's still a mistake somewhere because the correct answer is off.. hmmm.. also there is a very simple way of solving this, which made me go nuts after I heard it, because by that point my brain was full of cats walking back and forth, bouncing off of eachother and falling off poles.. haha edit: your mistake was at 99/2 = 45.5m and no idea where you got the 22.5s from... I get 24.25s... I'm guessing this is the simple solution you're talking about: + Show Spoiler +Don't consider collisions as bouncing cats. Instead, every collision has the same effect as having the two cats pass through one another. Thus, the time will be the maximum of the times it would take each individual cat to fall off if no other cats were present. We have a 99m pole and a cat on the left facing right moving at 1m/s, so this is 99s. Bingo. you have 20 prisoners on death row. The day before the execution, the executioner decides to have a bit of mercy and lets them know exactly what is going to go down. He says: - They will be lined up in one row, so that they stand behind one another.
- The executioner will come up behind each of them and randomly put on a red or blue hat.
- You will not be able to see your own hat, or any of those behind you, but you will be able to see those in front of you.
- The executioner will start from the very back, and ask you what your hat color is. You are permitted to only say either red or blue, and if correct, you will live. If not, your brains will be blown out.
- You will be given one night to come up with a strategy as a group to save the most men.
- You may not communicate with each other during the execution day; if you so much as blink or breathe in a peculiar pattern, all 20 of you will be shot immediately.
All the prisoners are very astute and logical, and place no importance on their own life over others'. How would you save the most people, and how many people would you save on average using said strategy?
+ Show Spoiler + you can save 19 at least. Last guy shout the parity of the blue hat in front (encoded as blue/red). And the ppl in front knows everything, they can subtract the ppl they see in front to find his own color, and once he declares his own color, again, everyone in front of him knows the parity exactly.
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On November 14 2011 08:39 corumjhaelen wrote:Show nested quote +On November 14 2011 07:21 Frozenhelfire wrote:On November 14 2011 06:04 JeeJee wrote:On November 14 2011 03:12 rotinegg wrote:On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution Good solution. I thought it would be something simple. I forgot that everyone can see what is in front of them instead of just the last prisoner. A nice variant is that everyone can see everyone' else's colors but their own. But instead of two colors, you can work a solution for n colors, and in fact if I remember well the "paper" solutions with "infinite" colors/prisonners, at least in some special cases.
yeah i think that was puzzle number 14 or so.
I really should link every puzzles together haha
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On November 14 2011 09:13 evanthebouncy! wrote:Show nested quote +On November 14 2011 08:39 corumjhaelen wrote:On November 14 2011 07:21 Frozenhelfire wrote:On November 14 2011 06:04 JeeJee wrote:On November 14 2011 03:12 rotinegg wrote:On November 13 2011 23:05 BrTarolg wrote: Working iteratively,
with 1 person, its 50/50 with 2 people, the first guy is always 50/50, but the second guy can be saved by the first with 3 people, the first guy can announce a "same/different" signal, and save the following two (i.e, red = following two have same colour, blue = following two have different colour)
With this strategy, you have a 50/50 chance for 1/3 prisoners, and 100% chance for everyone else
With 18 players, you save 12 of them and 50/50 for 6 of them, and with the last two you save 1 and 50/50 the other meaning you save 13 and 50/50 7 of them giving you an average of 16.5 saved
A pretty decent solution, thinking of a better one now you can save an expected 19.5 prisoners with 19 as the worst case here's what I'm thinking now, I think this solves it + Show Spoiler + yeah i wrote something long here but I deleted it anyway last prisoner simply announces RED if the number of red hats in front of him is odd, or BLUE if the number of red hats is even. knowing this info, everyone else can make the correct guess about their hat color, assuming everyone can hear everyone else's guess. i only really thought of this because you gave away the expected number of prisoners that live
e: shit maybe not e2: nope nevermind it works, i'm crazy. re-added tldr version of solution Good solution. I thought it would be something simple. I forgot that everyone can see what is in front of them instead of just the last prisoner. A nice variant is that everyone can see everyone' else's colors but their own. But instead of two colors, you can work a solution for n colors, and in fact if I remember well the "paper" solutions with "infinite" colors/prisonners, at least in some special cases. yeah i think that was puzzle number 14 or so. I really should link every puzzles together haha Way too hard to follow, sorry lol.
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On November 14 2011 00:41 BrTarolg wrote:Show nested quote +On November 14 2011 00:19 DJWilma wrote:+ Show Spoiler +Super Easy.... let x be whether the egg breaks (true or false), and y be the floor number... Step 1. Start at y=10, test x, if x=true continue to step 3, if x=false continue to step 2 Step 2, add 10 to y, test x, if x=false repeat step 2, if x=true continue to step 3 Step 3. subtract 9 from y, test x. if x=true continue to step 5, if x=false continue to step 4 Step 4. Add 1 to y, test x, if x=true continue to step 5, if x=false repeat step 4 Step 5. we conclude our answer to be what y was equal to when x=true minus 1. Doing this method will give you a maximum value for fewest drops of 19, and a minimum of 1.
DJ WILMA OUT!
I may have misunderstood the question, but the average outcome of drops is equal to 10 drops, if thats what your looking for I'm pretty sure mines the best
weve had solutions where the maximum is 14 and that's the best
i think it can be achieved in 9 steps?
so like, doing a drop on 1st and 100th floor, and then subsequently cut the range into half (e.g 50th floor).. then if it breaks, do the upper half (75th floor), if it doesnt, do the lower half (e.g 25th), and repeat the process..
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On November 14 2011 10:31 kaleidoscope wrote:Show nested quote +On November 14 2011 00:41 BrTarolg wrote:On November 14 2011 00:19 DJWilma wrote:+ Show Spoiler +Super Easy.... let x be whether the egg breaks (true or false), and y be the floor number... Step 1. Start at y=10, test x, if x=true continue to step 3, if x=false continue to step 2 Step 2, add 10 to y, test x, if x=false repeat step 2, if x=true continue to step 3 Step 3. subtract 9 from y, test x. if x=true continue to step 5, if x=false continue to step 4 Step 4. Add 1 to y, test x, if x=true continue to step 5, if x=false repeat step 4 Step 5. we conclude our answer to be what y was equal to when x=true minus 1. Doing this method will give you a maximum value for fewest drops of 19, and a minimum of 1.
DJ WILMA OUT!
I may have misunderstood the question, but the average outcome of drops is equal to 10 drops, if thats what your looking for I'm pretty sure mines the best
weve had solutions where the maximum is 14 and that's the best i think it can be achieved in 9 steps? so like, doing a drop on 1st and 100th floor, and then subsequently cut the range into half (e.g 50th floor).. then if it breaks, do the upper half (75th floor), if it doesnt, do the lower half (e.g 25th), and repeat the process..
unfortunately you don't have infinite eggs. if you did, then yes that would be the best approach. e: random puzzle/teaser I recall reading in a book. It's kinda famous but if you haven't heard it before, it's a good illustration of how you approach problems. Reminds me of the ants on a stick problem that was posted earlier in this thread.
You have two houses side by side, with windows facing each other. Owners decide to hang a rope between the windows to dry clothes on. Luckily for the owners, the windows are exactly level with each other. They have a 20 foot rope.
So they tack the rope to each window, such that it hangs down, forming a hump due to gravity. Note that a hanging rope does NOT form a parabola. It forms a catenary shape: http://en.wikipedia.org/wiki/Catenary Anyway, the vertical distance of this hump is ten feet. How far apart are the houses?
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Sanya12364 Posts
On November 14 2011 03:45 rotinegg wrote:Show nested quote +On November 14 2011 03:36 Frozenhelfire wrote: I hope that rule doesn't count voice inflections :d Prisoners:
The last prisoner says the color of the hat the person in front of him is wearing. Odd numbers inflect the color of their hat as a question i.e. blue? if the person in front of them has the same color hat. Even numbers inflect if the person has a different hat.
voice inflections also count as cheating + Show Spoiler + Use red/blue parity - even or odd number of red/blue. Only the last one is at risk of being killed for announcing the initial parity of the 19 ahead of him. The everyone else announces based on parity changes or non-change which will be the color of the hat that he is wearing.
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Sanya12364 Posts
On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water - the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position, even while he is on land. (The original monster only moved when Day9 was in the water. This monster has no such restriction. It moves all the time - but can only move in water.)
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy?
Initial strategy: (touch the water) + Show Spoiler + The monster follows Day9's movements inside the circle of safety and will try to be at the edge of the circle closest to Day9. The monsters covers the arc of the circle at 4s/r. If Day9 can cover arc faster, then he can leave the monster behind on the circle. Day9 can do this by running a smaller circle at less than radius r/4 (covering arc at faster than s/(r/4)). Via this method, Day9 can never beat the monster by more than pi arcs or the monster turns around.
Naively, we can say that Day9 can touch the water safely. Day9 is 3/4r away from the edge of the island. The monster is pi*r from that point and can only cover 3r in the time that Day9 needs to reach that point and get in the water. Here Day9 is making a break perpendicularly from the r/4 circle.
Optimally, Day9 should make his break on a tangent to the r/4 circle in a direct line for the gate. This causes the monster to swim more of arcs of the circle in pursuit (the monster is assumed to be pursuing the long way) while Day9 makes progress in getting off the island.
The math gets complicated from there and there's some pursuit diff. eqs. I'll add more later.
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not true. Binary Search can easily break 2 eggs np np.
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Sanya12364 Posts
On November 13 2011 17:50 Warble wrote: Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water - the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position, even while he is on land. (The original monster only moved when Day9 was in the water. This monster has no such restriction. It moves all the time - but can only move in water.)
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy?
Intermediate Strat: (still on the island) + Show Spoiler +First of all, let's start by agreeing that Day9 is really fucking smart, and all we're trying to figure out is What Would Day9 Do. WWDD? That's right. He's already miles ahead of us by running full speed as soon as he starts the strategy because he knows that it doesn't make sense to run any slower. Plus he's boss because of his ungodly stamina. Look at those legs go. He's worked through the strategy in part 1: http://www.teamliquid.net/blogs/viewblog.php?topic_id=285063¤tpage=6#102 to get ahead of the monster by running around in circles inside the 1/4r circle and if he messed up he could always come back and try again. Day9's thought ahead, but how did he know when to make a dash for the gate? First he busted out the polar coordinates and simplified the problem. Radius of the island becomes 1. His own speed becomes C. The speed of the monster becomes 4C. (To be honest, Day9 doesn't care how fast he's running and needed r & s for more important things.) As long as Day9 was inside the circle and the monster was chasing him, the monster's position was bounded by Day9 then imagined that he took the plunge and was out in no man's land beyond the safe circle of 1/4 and making a break for the shore. What was the optimum way to run for the shore? He knew he was bound by a parametric equation and the pythagorean theorem of the angular and radial differentials: Then he had to optimize outward movement over monster's angular gain. He wanted a path that gets him as close to the shore as possible while minimizing the angular catchup that the monster would achieve. The sign in the denominator has to be a minus, otherwise Day9 is going towards the monsters and should retry! Then subbed in the radial differential from the pythagorean theorem: Sanity check here. If Day9 chooses angular motion of 0, he should get a ratio of 1/4. He moves C out, the monster moves 4C around an arc. And it's true! I'm still sane. At this point, Day9 probably subbed in for angular movement, but frankly the following equation was beyond me. Day9 is hardcore: Then take the derivative with respect to the angular differential using chain rule and solve for zero. Some stuff fell out and the derivative looks like: (the denominator of the derivative doesn't matter as much but again good for sanity checks) plug that back into the pythagorean theorem and solve for radial differential and integrate to get the parametrized equation for radius. You know what that looks like? Day9 took one look and recognized it as the pythagorean theorem equation. What a smart guy. Reorganize and boom (pretend C0 is 0 for a bit)!!! He also knows that he's running Ct for distance so his parametric equation for path looks very much like a straight line. Day9 would make a beeline break for at his target point on the shore when it is tangent to the 1/4 circle that he was already running, checking once to see that monster was chasing the long way around. Solve for theta to prove that it is indeed a line, and it looks like a proper inverse tangent function for the line: Are we done yet? Nope! We still have to figure out how Day9 chose where on the shore to run for and how he decided to run after reaching the water. Dammit Day9. Why are you so smart.
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+ Show Spoiler +I would say to drop on floor 100, then 50. If it broke on 100, but not 50, try 75, otherwise 25. Basically this would be the interval halving method, and you would use the minimum amount of drops to find out where it breaks. Alternatively, pick a floor, drop the egg, and hope you've got it right
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+ Show Spoiler + Drop from 14. Assuming no breaks, drop form these floors until first break: 27 39 50 60 69 77 84 90 95 99
After first break, start at last known successful floor +1 then go by 1's up. This algorithm yields worst-case-scenario of 14 drops
Edit: Wooo i agree with flamewheel
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On November 15 2011 05:31 See.Blue wrote:+ Show Spoiler + Drop from 14. Assuming no breaks, drop form these floors until first break: 27 39 50 60 69 77 84 90 95 99
After first break, start at last known successful floor +1 then go by 1's up. This algorithm yields worst-case-scenario of 14 drops
Edit: Wooo i agree with flamewheel
Im so amazed at you guys just straight up doing the problem. My friend and I had to write a python script and when we discovered it is 14 we were so amazed and starred at it for 5 minutes for its beauty hahahahahaha
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On November 16 2011 19:10 evanthebouncy! wrote:Show nested quote +On November 15 2011 05:31 See.Blue wrote:+ Show Spoiler + Drop from 14. Assuming no breaks, drop form these floors until first break: 27 39 50 60 69 77 84 90 95 99
After first break, start at last known successful floor +1 then go by 1's up. This algorithm yields worst-case-scenario of 14 drops
Edit: Wooo i agree with flamewheel Im so amazed at you guys just straight up doing the problem. My friend and I had to write a python script and when we discovered it is 14 we were so amazed and starred at it for 5 minutes for its beauty hahahahahaha
+ Show Spoiler + Not sure if someone has mentioned yet, but for completeness, a way to show that 13 definitely doesn't work is:
Possible outcomes of the 13 drops If two breaks occur: 13C2 = 78 Only one break: 13 No breaks: 1
That's a total of 92 outcomes, which is not sufficient to differentiate the 101 possible outcomes of a 100 level building.
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Sanya12364 Posts
On November 16 2011 22:17 Ivs wrote:+ Show Spoiler + Not sure if someone has mentioned yet, but for completeness, a way to show that 13 definitely doesn't work is:
Possible outcomes of the 13 drops If two breaks occur: 13C2 = 78 Only one break: 13 No breaks: 1
That's a total of 92 outcomes, which is not sufficient to differentiate the 101 possible outcomes of a 100 level building.
Nice brute force counting way of proving it! A little more more elegance you have: + Show Spoiler +But I think it's easier to omit no breaks and think in number of levels rather than number of breakage results. It is only adding 1 to each side of the equation. Generalized for any number of eggs:
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On November 13 2011 17:50 Warble wrote: I also have another one. It is similar to a level in Chip's Challenge. I don't know which came first, whether Chip's Challenge originated the idea or whether it was adapted from the idea.
Day9 is on an perfectly circular island of radius r surrounded by water. The water monster from Amnesia is in the water waiting for Day9 to leave the island. Day9 must eventually leave the island to continue the game, or else his viewers will scorn his manhood. There is gate some distance away in the water - the specifics here don't matter. We all know that he must head for that gate.
Day9 runs at speed s both on land and in water (this is the shallow water area he knows and loves) and has unlimited stamina. The water monster swims at speed 4s, clearly outpacing Day9. This monster is tougher because it always knows where Day9 is, even when he's not in the water, and it will always take the fastest route to his position, even while he is on land. (The original monster only moved when Day9 was in the water. This monster has no such restriction. It moves all the time - but can only move in water.)
However, Day9 is awesome at maths and immediately works out how to best prolong his life so that we can all enjoy the maximum amount of pants-wetting terror.
What strategy gives Day9 the best chances of reaching that gate? Can he safely enter the water? Can you prove (mathematically or otherwise) that this is the best strategy?
Ha! I loved Chip's Challenge. You mean this level, right?
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Sanya12364 Posts
About Day9's race to the gate we're off the island, but it's really hard to prove. Here's what I think is the best solution though:
+ Show Spoiler +Day9's lead in arcs at point of entry into the water is about .58 (compare this to .14 when doing the native perpendicular way) and that means that he's got an upper bound on the gate's distance from his point of entry into the water at about .19. If the gate is pictured according to scale at C, Day9 knows he's doomed. It's easier to prove that aiming for F is better than aiming for B.
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Sanya12364 Posts
On November 13 2011 15:05 rotinegg wrote: You got 13 bottles of wine, one of which is poisoned. You got 4 lab rats, which you can test the wine on. Once poisoned, the rat will die after 23 hours 58 minutes + or - 1 minute. You are having guests in 24 hours, and must prepare 12 bottles of un-poisoned wine. What is your strategy of testing out the wine bottles?
+ Show Spoiler + Another counting problem. You do this by feeding each bottle to a unique combinations of rats. When that combination of rats die you know you'll which bottle was the culprit (no rats is a possibility). For 4 rats there is 2^4 or 16 possible combinations and more than enough for the 13 bottles.
On November 13 2011 17:04 rotinegg wrote: You got a 95 by 85 by 2000 cube, made up of little 1x1x1 cubes, just like a rubiks cube. Your buddy paints the outside of the 95x85x2000 cube black, and smashes it into 95x85x2000 little pieces of 1x1x1 cubes. He then puts all the cubes in a bag. He shakes the bag up, then pulls one out at random, and rolls it like a die. What's the probability that the top face is black?
+ Show Spoiler + First we'll consider the cube along 3 axis (x for the 2000 thick, y for the 85 thick, z for the 95 thick). We'll partition the outcome into 3 equally possible subcategories. The cube's top face could be aligned along the x axis, the y axis, or the z axis.
For each of the of these axis subcategories, the probability that the side is painted 1/n where n is the thickness of the cube along that axis. There are 2 painted sides and 2 * n total sides.
The probability of rolling a painted side is 1/3(1/85+1/95+1/2000)
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