Right. Put another way, in your “3 of 10” scenario, the brownies think there are either 3 or 4 blues on the island- the three they see, or the three plus themselves. The blues think there are either 2 or 3- the 2 they see or those two plus themselves.
After day two, the blues realize that, in fact, there can’t be 2, so there must be 3- the two they see plus themselves and thus death is required. By killing themselves, they effectively announce that there are 3, so the brownies no longer think “3 or 4”…they have their answer.
The key is that neither the brownies nor the blues knew that the other opinion existed. Oh sure, they knew the opinion was different, but they don’t know in which direction. Everyone realizes “OK, the brownies think there are N+1 or N and the Blues think there are N or N-1” but they don’t know which opinion they hold. Knowing this rule, if your opinion was “there are either 3 or 4”, are you the N or N+1 statement or the N or N-1 statement? You don’t know, because you have no idea what’s going on in your cohabitant’s heads. If you did, you’d figure it out and know if you were a brownie or a blue.
No offense, but I will be glad when this thread falls off the first page of the forum. I keep seeing the “what’s your favorite lateral…” part of the title and thinking it’s a thread about wacky football plays.
Originally, no - of course they don’t. But the islanders are a 10 person set - not an incrementally increased group of people with a specific start date. It seems that the riddle only works if all the islanders come to the conclusion that the missionary’s talking about blue eyes is “Day 1”, and it works out logically from that.
Chessic Sense, you’re correct that any force whatsoever will make the ball go airborne, but juggliging requires synchronization of when each of the balls goes airborne. In fact, the minimum additional force needed to juggle 7 lb balls will be 3.5 lbs (so, 10.5 lbs total applied to a ball). But, if you use a force so low, you need to apply that force to both balls that you are holding (when one ball is in the air). If you apply at least 7 lbs of additional force, it needs to only be applied to one ball. So the peak forces will always be above the 171 lbs that would be required to carry them across the bridge without juggling.
The reason the days matter is because the islanders also gain new information every afternoon from learning who does and doesn’t kill themselves that noon.
I’m not sure what you’re saying about “not an incrementally increased group of people with a specific start date”. Who said anything about incrementally increasing the number of islanders?
You’re going to have to spell out your math better than that.
Where are you getting the 3.5 from? Who says I need to use half the weight of the ball?
Why am I applying force to 2 balls? I only hold one in my throwing hand. The other ball waits to be passed over from my resting hand to my throwing hand.
Even if the 10.5 is correct, how do you get 150+7+10.5 > 170? It’s 167.5.
Look, you’re holding 2 balls when you step onto the bridge. That’s a total of 157 plus whatever force you’re applying to the second ball. If it’s 7, that ball is holding still. If it’s 6, then you’re dropping the ball. If it’s anything greater than 7, then you can accelerate the ball into the air.
170 - 157 = 13 lbs. So you’ve got 13 lbs to work with, and you only need 7. So 9? Cool. 10.5? Fine. 12.9? That just gets the ball higher.
Again, you’re counting the weight of the ball twice. You’re saying it needs 7 to be held up and >7 more to be thrown. It doesn’t. You can either say it needs >7 total to be throw, or that it needs 7 + X to be thrown, where X is any positive number. You can’t do both. You don’t need 7 + (7+X) to throw the ball.
Browns think “I see N blues. So there are either N or N+1 (me) blues on the island.”
Blues think “I see N-1 blues. So there are either N-1 or N (me) blues on the island.”
Let me know if I’ve lost anyone.
Now, the only think keeping them alive is the fact that they can’t tell which formula their opinion is a reflection of. Let’s say you’re an inhabitant and you look out there and count 3 blues. So you think “There are 3 or 4 blues”, right?
Also, you know that either:
You’re a blue, have the N or N-1 opinion and the browns think “4 or 5”, where N is, in reality, 4.
You’re a brown, have the N or N+1 opinion and the blues think “2 or 3” where N is, in reality, 3.
Now suppose you found out one day that your neighbor thinks “There are either 4 or 5 blues”. What would you reason? Remember that each inhabitant is absolutely correct in their assessment.
So you know that there are “4 or 5” (his opinion) and also “3 or 4” (your opinion). It’s then easy to realize three things:
There are 4 blues.
Your opinion was the “N or N-1” variety and your neighbor is “N or N+1”
You’re a blue, both due to the formula and that you can only find 3 other blues out there.
Let me know if I’ve lost anyone by this point.
Now if someone were to just blurt out “I think there are x or y”, then you’d know that you either share the same opinion and have the same eye color (so you have to die) or that you differ in opinion and thus have the opposite eye color (and have to kill yourself). Everyone that heard this would know the same thing instantly.
When the missionary comes, he rules out “There are 0”. So the next day, anyone that had thought “there are 1 or 0” would know that there is in fact only 1, and they’re it.
When no one is dead, you know that no one though “There are 1 or 0” so the lowest opinion there could be is “there are 2 or 1”. That’s fine, though. You already knew that the only possible opinions are “4 or 5” and “2 or 3”.
By day two, when no one dies, you know that there are no “2 or 1” opinions. Again, this is not news. You knew this would happen. So did everyone else.
By day three, we learn that no one thinks there are “3 or 2” opinions. If there were, someone would have to die. Knowing that 3/2 doesn’t exist, all the 4/3s know that they’re the lowest opinion on the island that there is. So they know they’re the N or N-1 opinion and thus blue. Simultaneously, the browns realize they’re the high opinion on the island. Thus they have to die too.
Chessic Sense - that makes sense. But I still don’t understand why everyone assumes that the missionary ruling out “there are 0” is, effectively, Day 1.
I, for one, don’t. I’m calling the next morning Day 1. The announcement day is Day 0 for me. The day before that is D -1. We’re doing that for ease of communication. That way, N blues die on Day N. It’s just an arbitrary labelling so we can talk about it easily.
Why is this a stumbling block? What system would you prefer? I’m confused about your confusion.
Day 1, Day 0, Day A, doesn’t matter. Why is the missionary’s statement automatically assumed by the islanders to be the day to start counting? Like I said earlier - these types of puzzles typically lay that specifically out - the abbot in the abbey, the warden, etc. say “starting today _____ give me the right answer _________ or everyone dies.” Fill in your own blanks for the variables of the puzzle. But this puzzle is different - there is NO specific starting period, and it doesn’t seem logical to assume that all the islanders would come to the conclusion to start counting today/tomorrow/whatever.
It’s not that they start counting… it’s that the missionary’s statement gives them new epistemological properties (it ensures that everybody knows that everybody knows that everybody knows that… there is at least one blue-eyed islander, a state which they weren’t in and couldn’t have reached previously). Everything is set into motion by their gaining that new epistemological state.
At any point when he’s throwing a ball or catching a ball. If him plus all three balls weigh a total of 171 pounds, then that’s how much he’ll weigh on average, no matter what he does. If he juggles, then sometimes he’ll weigh less than that, but that just means that sometimes he’ll weigh more.
One thing you might be missing here is that it doesn’t take force to catch or throw a ball, it takes impulse, or force multiplied by the time the force is sustained for. If I catch a ball by exerting a force on it just barely greater than the ball’s weight, then I’ll have to spend a long time in the catching. But there’s a limit to how long a time I can spend catching a ball, since I’ve got to finish catching it and get it airborne again before the next ball comes down.
Gladly*. I’m going to construct a simple Newtonian model of juggling. First, let’s consider the arc of a single upward toss, say, from right hand to left hand. Suppose that the ball leaves the right hand and velocity v, travels through a arc in the air, and is caught in the left hand. By standard mechanics, the total time the ball spends in the air is T = 2v/g, where g is the gravitational acceleration**. For use later in the calculation, I will write this expression as v = Tg/2.
Now, let’s consider the actions involved in one “iteration” of juggling. Suppose Ball A is currently in you right hand, and Ball B is falling towards your right hand. To successfully juggle, you must:
[ol]
[li] You must accelerate Ball A from a stop to velocity v.[/li][li] You release Ball A from your grasp.[/li][li] Ball B falls into your hand.[/li][li] You now decelerate Ball B from velocity v (downwards) to a stop.[/li][/ol]
Denote the time it takes to accelerate or decelerate the ball by t. For simplicity, I assume that no time elapses between steps 2 and 3 (i.e., you catch Ball B immediately after releasing Ball A). Now, after releasing Ball A, it travels to your left hand, which then throws Ball C back towards your right hand. These two tosses take total time 2T. During this time, you must decelerate Ball B after catching it, and then accelerate Ball B to throw it again. The time required to do this is 2t. Therefore, in order to be juggling, it must be true that 2t <= 2T, or t <= T.
Now, let’s consider the forces involved in accelerating or decelerating a ball.
Let m be the mass of the ball, and let F be the force your hand exerts on the ball. When you are accelerating or decelerating the ball, notice that F > mg (the weight of the ball). When you are holding the ball steady, F = mg. If F[sub]L[/sub] and F[sub]R[/sub] are the forces of your left and right hands, respectively, my claim is that F[sub]L[/sub] + F[sub]R[/sub] >= 3mg (the weight of 3 balls), at least during a portion of the time spent juggling.
Newton’s Second Law says that ma = F – mg, where a is the acceleration of the ball. Integrating this expression through t, i.e., the time required to accelerate the ball from a stop to v, we get: mv = (F – mg)t. We can rewrite this expression to get F = mv/t + mg. Then using the expression v = Tg/2 (which we found above), this simplifies to F = mg(T/(2t) + 1).
F = mg( T/(2t) + 1 ) is an important expression, so I’m going to focus on it for a minute. In the example you gave (7 lb balls, pushing on them with 9 lbs of force), we can see that mg = 7, F = 9, so T/(2t) = 2/7. However, since t <= T, we can see that T/(2t) >=1/2. Since 2/7 = 0.285…, you simply cannot juggle by exerting the forces that your example requires. By substituting T/(2t) = 1/2, you we quickly see the answer to your first question. That is, the minimum force you can juggle with is F = 3/2 mg. This seems to suggest that F[sub]L[/sub] + F[sub]R[/sub] >= mg + 3/2 mg = 2.5mg. This is the calculation you did in question 3 (total force = 7 + 10.5). I will now show you why it’s mistaken.
There is one final complication to the model. Suppose that Ball A is in the left hand, Ball B is in the right hand, and Ball C is in the air, and is caught by the right hand at time 0. The Ball B is also thrown at time 0, and Ball C is brought to a stop at time t. Ball B is caught by the left hand at time T, but before that happens, the left hand must accelerate Ball A, which it begins to do at time T-t (and releases Ball A at time T). Notice that, if t > T/2, there is a period of time that the right hand is decelerating Ball C and the left hand is accelerating Ball A simultaneously. In this case, then T/(2t) >= 1/2. During this time, the total force on the balls is
and, again, the total force exerted on the bridge is at least the weight of all three balls. This should answer your questions 2 and 3, in particular, the weight on the bridge when the balls are moving simultaneously is 150 + 10.5 + 10.5 = 171.
In conclusion, this is all just a long-winded way of agreeing with what Malacandra said (on preview, also what Chronos said):
I wanted to get that last post onto the board before I had to go to a meeting. Now I’m back from the meeting, so I have time to explain.
** Here, I’m making the assumption that the ball is caught at the same height that it was thrown. I think this is a reasonable assumption.
Because the missionary released information. If someone had come out and said “I think there are 6 or 7 blues”, then everyone that agrees will have the same eye color as him and everyone that disagrees will be the opposite. If you think it’s 7/8, then you’re a brown and he’s a blue. If you had though 5 or 6, then you’re a blue and he’s a brown.
It’s not a requirement that anyone tells anyone else who has what. The only important thing is that another person’s opinion to known. Since he can see you, once you learn his opinion, you can compare it to yours and from the difference, you can deduce what he’s counting YOU as - blue or brown?
Think of it this way:
The blues always have the low guess. But they don’t know that! They look at all the other blues and say to themselves “We either share an opinion or they’re one lower than mine”. This rule is well-known to everyone.
So when blue A looks at blue B, he knows it’s possible that blue B has a lower opinion than himself. And what does A think about B’s opinion of C? It’s possible (not guaranteed) that it’s one lower than that. A knows that B and C share the same opinion only because A knows that B and C are in the same group. But B doesn’t know that. So A thinks that B thinks that C has an opinion one lower than B.
And what would A say about B’s reasoning about C’s reasoning about D’s guess? Well it’d be even lower. Again, C thinks “D could be lower”. B knows that and so he thinks “C could be one lower than me, so C thinks that D could be two lower than me.”
And so on. So if A thinks it’s 5/6, then he reasons “B thinks that C thinks that D thinks that E thinks that F thinks it’s 0/1”, even though it’s clear that no one thinks it’s 0/1. That’s not the point.
By day 1, when F doesn’t kill himself, A thinks “B knows that C knows that D knows that E knows that F never thought it was 0/1, because otherwise he’d know it was in fact just 1 and he’d have offed himself. So the lowest F could possibly think is 1/2.”
A then says “So B knows that C knows that D knows that E knows that F is not alone in his blueness. E and F are now in the same boat.” Problem is, E and F don’t know that they are in fact E and F. Those labels were just attached to them in A’s mind, and that’s important to understanding this.
The next day, when E and F don’t kill themselves, everyone realizes that those two blues (whoever they might be) don’t realize that they’re the only two blues there are. So there must be a third. Of course, everyone knew there was a third (and more) blue already, but it wasn’t until just now that it was widely known that this info was common knowledge.
In other words, after the second day where no one dies, A thinks “B now know that C knows that D knows that he’s with E and F in their pool of blueness. This is becuase A thinks that B thinks that C is clueless and not counting himself in the pool. The poor chap-tomorrow he’ll realize that they were counting him too and he’s a sucker. Then they’ll all die together.”
But that doesn’t happen. And so on, and so on. To understand why this is so, try not to think of each person’s guess at N. Try instead to think of people’s assessment of people’s assessments of people’s assessments of the two possible Ns.
This whole chain of events kicks off because the missionary stopped the chain of logic leading all the way down to someone thinking someone else thought that someone else thought that someone else thought that someone else though it possible for there to be 0 blues on the island.
We could have skipped this whole process at any time if someone had just TOLD someone else what they thought were the two possible values of N. Or even what they thought someone else’s opinion could be. The whole island runs around going “Don’t tell me how many blues you see! Don’t! Lalalala, I can’t hear you!”
I was going to say that you’re thye one missing the impulse. The impulse in catching the ball is the same as the impulse in throwing it. If I send it up at .5 m/s, it’s going to come down at -.5 m/s, right? So to stop it, I have to spend the same amount of time and force catching it as I did throwing it.
Look, let’s say that every time I launch a ball, I impart an extra 2 lbs for a total of 9 lbs, OK? So when I launch or catch a ball, it effectively weighs 9 lbs. At rest, it’s of course still 7.
I’m carrying A, B, and C. I start out at 171, then launch A so I’m 173. Then I step onto the bridge. I now weigh 166. I launch B at 168. Then I weigh 157. Then Ball A lands and I’m 168 on the scale while I decelerate it. I’m at 166 for a while before launching C at 168 again. After C goes airborne, I’m at 157. Repeat.
You failed to show your work, so no points for you. When come back, bring time line with weights. At no point am I holding three balls, or holding two balls and catching/tossing a third. Neither throwing nor catching take 13 lbs of force at any time.
If you want to challenge this, tell me what each ball is doing and how much force it’s applying to my/Max’s body and what that totals to. Nowhere will you find it more than 170 unless you’re throwing or catching in excess of 13 lbs, which isn’t required by the problem.
Chessic, if you hold a 7lb ball steady you have to apply 7lb force to it all the time. If you want to arrange matters so that you’re only holding it two-thirds of the time, you’ll find you have to apply an average of 10.5lb force to it while you’re holding it. Or to look at it another way, gravity is accelerating the ball downwards at 32ft/sec/sec continuously, so you have to arrange to accelerate it upwards at the same rate over time. If you’re not accelerating it continuously, you’ll have to accelerate it more while you do hold it to compensate for when you’re not.
How much work do you need, for the statement that if something’s below average some times, it must be above average at other times? That’s all that the calculation comes down to.
Right, but those times are when Max is off the bridge. It’s higher when he launches the first ball and also when he catches it at the end. He doesn’t need to average under 170, he just needs to be under it for the dash across the bridge. And Max doesn’t need to launch balls at 9 lbs of thrust. He can do it at 13. And the first ball can go up as hard as he can throw it. It’s only restricted by his arm, not a weight limit.
If necessary, the OP can shorten the bridge to 50 ft or whatever. It just needs to be far enough that you can’t toss the orb over.
