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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)
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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)
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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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EPT
