How is this a lateral thinking puzzle? This seems to be a rote exercise in straightforward - and utterly unrealistic - logic.
I agree that it’s fundamentally a simple math problem and not an “Aha!”-moment puzzle, beyond the extent to which simple math problems can have surprising results. However, the thread has long been largely about what are ultimately simple math problems with surprising or non-intuitive results. I raised this point earlier in the thread, in post #13, but ultimately, that’s the way the thread has gone pretty much from the start and there’s no way to change it now.
Has everyone agreed that if there are n people they die on day n? You don’t need to go through the thought process to prove that.
If there is 1 blue, he dies on day 1.
If there are 2 blues, then nobody dies on day 1, which tells them there can’t be just 1 blue, so there must be 2, and they die on day 2.
If there are 3 blues, then nobody dies on days 1 or 2, which tells them there can’t just be 2 blues, so there must be 3, and they die on day 3.
If there are 4 blues, then nobody dies on days 1, 2, or 3, which tells them there can’t be just 3 blues, so there must be 4, and they die on day 4.
This is the easy way to prove it. Because of this easy proof we know they are getting information from the original statement. There is no question about it.
The hard part is going through the thought process and explaining what information they are getting. It has been explained many ways already, so I won’t repeat it. But in the end it is irrelevant to the riddle because there is a much easier, less confusing proof.
Yes, that is the easiest way to present the matter. Did we actually forget to put it that way originally? This has gone on for so long I no longer remember the path the dialectic traced. But perhaps we should stick to focusing the argument on Carmady’s presentation, since it has the advantage of such simplicity. (But, hell, I don’t really know. I’m pretty sure no one in this thread has changed their minds about anything, like I said, so I wonder what we’re all doing. Maybe we should just go back to presenting puzzles and not worry about trying to explain whether their presented solutions are right or wrong. :))
If it is any consolation, I’ve heard this puzzle many times and I’ve understood the inductive proof Carmady presented but I never really grasped the “A knows that B knows..” logic. But between you and Chassic Sense I now have a much better idea of how it works. I still haven’t completely internalized it, but I can see the logic path now. In other words, I have learned something and I really appreciated you and Chassic Sense taking so much time to explain it. Thanks for that.
Actually, I think the thought process is the most interesting part of the riddle.
The easy proof isn’t that interesting, but at least it does prove that the original statement provided new information.
Once you are convinced of that, it may be easier to understand the (more interesting) explanations of exactly how it provided new information.
What do you mean? I already said what C is writing down. She’s writing what she thinks could be on D’s paper on day 2, which happens to be the same as it was on Day 1 and Day 0.
D wrote about possible realities. C wrote about possible D papers. B wrote about possible C papers. A wrote about possible B papers.
Why are there 3 in the first group and only one in the second group? Because the first group is “If I, B, am a blue” while the second is “If I’m a brown.” Naturally, C never wrote down the second group (0100) because B isn’t brown, but B doesn’t know that. Anyway, in the “If I’m brown” second group, 0101 isn’t possible because it has 2 blues and they would have been dead already. 0110, same reason. 0111 was deleted on day 1 when no one died.
Why did C write either one group or the other? Because C knows the answer to the second slot, which is 0. But B doesn’t know that, and she doesn’t know if when C looks at her, she sees a 1. If that were the case, then C would have written down the latter group (now only consisting of 0100).
It’s Chessic. Because I play chess. And I have a sense about what to do on the chess board. It’s a chessic sense.
Agreed. I wasn’t trying to grump about it.
Well, he can catch 2 without exceeding the max weight. But if he has to catch 3, he’s had it. IOW, he needs either to throw one ball all the way across, make two trips, or settle for making off with just 2 of the three.
No complex analysis is needed: the average downward force will be equal to Max plus the total weight of whatever balls he’s transporting. Tossing and catching introduces transient variation, but doesn’t change the average.
As an aside, if you toss a ball gently (using a maximum force not much above its weight) you’ll find that it’s airborne for only a short percentage of the total time. Most true juggling requires that a ball’s time in the air be reasonably long, so more aggressive tosses are necessary
I wasn’t accusing you of a mistake in that presentation, Chessic Sense; I just had trouble understanding. I missed the part where this was all taking place on day 2, which led to some confusion as to why certain things were left out.
Sorry about that.
Here’s an old puzzle. Several centuries old, in fact, but still tricks people.
I saw a peacock with a fiery tail
I saw a comet rain down hail
I saw a cloud with ivy wrapped around
I saw a mighty oak tree creep along the ground
I saw a spider swallow up a whale
I saw the ocean full of ale
I saw a glass thirty foot deep
I saw a well full of men’s tears that weep
I saw their eyes all in a flame of fire
I saw a house as big as the moon and higher
I saw the sun out in the middle of the night
I saw the man who saw this wondrous sight.
How is this possible?
It’s a totally fair puzzle and the solution is reasonable. Think about it, try and puzzle it out. Then see the solution.
I saw a peacock.
With a fiery tail, I saw a comet.
Rain down hail, I saw a cloud.
With ivy wrapped around, I saw a mighty oak tree.
Creep along the ground, I saw a spider.
Swallow up a whale, I saw the ocean.
Full of ale, I saw a glass.
Thirty foot deep, I saw a well.
Full of men’s tears that weep, I saw their eyes.
All in a flame of fire, I saw a house.
As big as the moon and higher, I saw the sun.
Out in the middle of the night I saw the man who saw this wondrous sight.
At a point 10 + 5/pi miles away from the South Pole (11.59 miles). If you walk 10 miles south (i.e., towards the pole), you’ll be 1.59 (5/pi) miles from the SP. Then, walking east is basically walking 10 miles around the SP. (radius if 1.59, diameter is 3.18, and circumference is 3.18pi or 10 miles.) Then 10 miles north is back to the beginning point.
That’s not quite exactly right, since on the curved surface of a sphere, the distance around all the points at distance r from the South Pole (or anywhere else) is not 2πr, but rather, in fact, generally less than this. But there is a line of lattitude close to the South Pole where the distance around is 10 miles and any point 10 miles north of that does work as a solution, as I mentioned previously (for that matter, 10 miles north of anywhere where the distance around the line of lattitude is any divisor of 10 will work); it’s just that these points won’t be distance 10 + 5/π miles from the South Pole, but rather will be a tiny bit more to the north.
My solution:
I was dreaming.
That’s wonderful. Thank you!
No, I didn’t get it without looking at the answer. I thought it might be about the shape of the letters in each word. Wrong.
That reminds me of this big “aha!” moment I had a few years back. For the life of me, I couldn’t figure out what the hell Martina McBride was talking about in “I love you”
I’m like “what the hell does ‘desperately sure’ even mean”? Physically acting? Is there another way to do it? Ok… Then I realized that I was punctuating it wrong. The real lyrics are:
Now it makes total sense!
I am trying to come up with true “lateral thinking” examples of solutions to the Max/golden eyes/rickety bridge problem.
[ul][li]The obvious solution is “drop one golden eye before getting onto the bridge”. The ability to be flexible enough to realize that alive and with two golden eyes is better than dead with three is likely to be more valuable than any amount of effort helping islanders avoid suicide because of their eye color - although not as much fun. [/li][li]My solution is for Max to throw one of the eyes as far ahead onto the bridge as he can. Then run up to it, throw another eye, pick up the first, and repeat. [/li][li]If that solution allows the natives to catch up to him, no problem, as long as he is on the bridge. None of the natives can get onto the bridge as long as Max is on it - if they do, the bridge collapses and the golden eyes are lost. So the natives have to wait until Max is off the bridge. Max then reaches the other side, waits until a native gets onto the bridge, and puts one foot on the bridge (holding onto the pilings with one hand). When the native reaches the center, Max leans hard enough on the bridge to make it collapse, jumps back, and escapes at his leisure.[/ul][/li]
Regards,
Shodan
Except that all the more interesting explanations are wrong.
notation: b = Blue; r = bRown; , = “or”, A.x means “A thinks x”
If a villager sees 4 blues, then he knows that nobody sees less than 3 blues, and he thus knows that everybody knows that nobody sees less than 2 blues. That means that everybody knows, right off the bat, before the visitor says anything, that nobody thinks anybody sees:
brrrr,rbrrr,rrbrr,rrrbr,rrrrb,rrrrr.
(If there are any browns in town besides the 5 blues, then they know that nobody sees less than 4 and thus that nobody thinks that anybody sees less than 3, eliminating bbrrr,brbrr,brrbr,brrrb,rbbrr,rbrbb,rbrbr,rbrrb,rrbbr,rrbrb,rrrbb as well.)
These cases are a priori eliminated, for exactly the same reason that the rock falling a thousand feet is eliminated by the ten-foot rope; they can be eliminated by a different and separate method.
So:
E.bbbbb,bbbbr.
D.E.bbbbb,bbbbr,bbbrb,bbbrr.
C.D.E.bbbbb,bbbbr,bbbrb,bbbrr,bbrbb,bbrbr,bbrrb,bbrrr.
B.C.D.E.bbbbb,bbbbr,bbbrb,bbbrr,bbrbb,bbrbr,bbrrb,bbrrr,brbbb,brbbr,brbrb,brbrr,brrbb,brrbr,brrrb BUT NOT brrrr.
A.B.C.D.E.bbbbb,bbbbr,bbbrb,bbbrr,bbrbb,bbrbr,bbrrb,bbrrr,brbbb,brbbr,brbrb,brbrr,brrbb,brrbr,brrrb,rbbbb,rbbbr,rbbrb,rbbrr,rbrbb,rbrbr,rbrrb,rrbbb,rrbbr,rrbrb,rrrbb BUT NOT brrrr,rbrrr,rrbrr,rrrbr,rrrrb,rrrrr.
You’ll note that no villager is incorrect in what they think the other thinks; they allow for extra cases but never miss any actually considered ones. And nobody even considers the case that anyone thinks that there are no blue-eyed people in town.
So you can go on all you want about exponentially increasing ignorance, but you’re simply wrong; these villagers are not restricted to only one method of deducing each other’s thoughts. So when the visitor opens his big mouth, he’s not telling anyone anything they didn’t know, because they already had eliminated the possibility long ago.
But when the visitor opens his mouth the villagers all are doomed to die anyway.
Not because they learn anything about what anybody thinks anybody thinks anybody etcetera. So if he’s not imparting information, why do they die?
Because he is imparting information - the information that the game is afoot. Carmandy is right that the induction dictates the result, and all the villagers know it. So, as soon as the visitor says it, all the brilliant villagers say to themselves:
“Oh crap we’re all dead. Not because anybody learned anything today; they didn’t, any more than I did, and we all know it because there’s too many blue-eyed people in this village for anybody to be overlooking them all.”
“The problem is the induction - we all know the day we’ll die. We just don’t know why we’ll be dying yet. Take me for example - I see 4 blue eyes, so I’ll be dying on day 5. All those blue-eyed people in town? They’re either slated to die on day 5 like me, or day 4 if I’m brown - and whichever they are, they already know it (though they too don’t know why yet). Only on day 4 will I learn if they saw 3 or four, which will be the first new piece of information I get since game start. Unfortunately that piece of information directly tells me my eye color and then I’ll have to die - right on schedule, on day 5.”
So I was wrong when I said that they could avoid their fate - but pretty much every single one of you have been wrong when you attempted to rationalize why. The actual reason that they all die is simply and exclusively that they are aware of this logic puzzle and know that everyone will follow the script. Disappointingly simple, perhaps, but that’s just the way things are sometimes.
begbert2, let’s play a two-player game. The game has different named pieces called “islanders”, which can be placed in one of two states at any moment: blue-eyed or brown-eyed.
There are also cards in this game. They come in two flavors: atomic and connective. One of the atomic cards is labelled “Someone has blue eyes”. And for each islander X, there’s a connective card labelled “Islander X can be certain of:”. These cards also have instructions on the back, which I’ll explain in a second.
By the “proposition deck”, I mean a particular stack of cards with an atomic card at the bottom and some number of connective cards on top of it, which is used during the game. (These, of course, represent propositions.)
The way the game works is like this: on every turn of the game, you have to point at the proposition deck and say “I believe the rules of the game will validate this proposition in the current scenario”.
What happens next depends on the top card of the proposition deck, which will have instructions on the back saying how to proceed.
Specifically, the instructions for “Someone has blue eyes” say “The game is now over. If any islander has blue eyes, begbert2 wins; otherwise, Indistinguishable wins”.
And the instructions for “Islander X can be certain of:” are as follows: I get to say “Since Islander X doesn’t know his own eye color, but is still so certain, he must be able to remain certain no matter what his eye color is.” Then I get to optionally change Islander X’s eye color to whatever I want. Then we remove this card from the top of the proposition deck and keep playing the next turn of the game with the rest of the proposition deck.
At the very start of the game, you get to set all the islanders’ eye colors to whatever you want, and you get to set up the proposition deck however you like. Then we start playing from the first turn. At the end, if you win, we’ll say that the rules of the game validated the initial proposition in the initial scenario, and if you lose, we’ll say that the rules of the game failed to validate the initial proposition in the initial scenario.
Ok? Are you willing to play this game?