Okay, I just hit the puzzle mother lode: XKCD Puzzles.
Blue Eyes is there, along with 69 others.
Wow. Puzzle heaven!
If you don’t already have a copy of Smullyan’s “What is the name of this book” I highly recommend it. My friends and I have spent several evenings working our way through the puzzles and though we’re only up to chapter 4, it’s been a heck of a lot of fun.
There’s no need for a special guru. Any brown-eyed person known to be completely honest could stand up and say “I see a blue-eyed person” and the same solution would follow.
When there’s only a very small number of blue-eyed people it is easy to see that the solution works, but in practical terms it seems very problematic with many blue-eyed people. For me this demonstrates the impracticality of rigorous logic: the slightest fuzziness or unreliability will upset the solution.
Nobody can talk except for the guru.
I agree with this to some extent. Do the islanders really have common knowledge of the guru’s statement? Suppose that Alice knows that Betty lost her hearing after eating some bad fish a few weeks ago, but that it came back. Cathy knew about the fish but Alice isn’t certain that Cathy heard about Betty’s recovery. So Alice doesn’t know that Cathy knows that Betty definitely heard the announcement. The more layers you go, the more tenuous your knowledge of whether the islanders definitely heard the announcement.
Yes. One of the brown eyes might be greenish and, even though I don’t consider it bluish, another might. Or another might worry that I do.
Or, more to the point, the puzzle requires that we all be perfect logicians, know that we’re all perfect logicians, and know that all the others know that. But isn’t it possible that
we all know that we all know that we all know that we all know that we all know that we all know that we all know that we all know that we are all perfect logicians,
but some of don’t know that
we all know that we all know that we all know that we all know that we all know that we all know that we all know that we all know that we are all perfect logicians.
Although I’ve never fully understand it, I 'believe" in fuzzy logic. Introduce even a tiny bit of fuzziness and this logic puzzle falls apart like a house of cards.
And my cakes turn out terrible when I put them in the oven upside down.
If you don’t want to follow the constraints of the exercise, then yes, you get different results. This isn’t surprising.
Forgot to post the solution chart:
1 1
2+1 3
3 3
4 3+1
5+3+1 9
6+3 9
7+3 9+1
8+1 9
9 9
10 9+1
11+1 9+3
12 9+3
13 9+3+1
14+9+3+1 27
15+9+3 27
16+9+3 27+1
17+9+1 27
18+9 27
19+9 27+1
20+9+1 27+3
21+9 27+3
22+9 27+3+1
23+3+1 27
24+3 27
25+3 27+1
26+1 27
27 27
28 27+1
29+1 27+3
30 27+3
31 27+3+1
32+3+1 27+9
33+3 27+9
34+3 27+9+1
35+1 27+9
36 27+9
37 27+9+1
38+1 27+9+3
39 27+9+3
40 27+9+3+1
“In PRACTICAL terms”?
For these kinds of puzzles you really have to figure out a way around that mindset…assuming you’d like to enjoy them, at least!
Looks good.
[spoiler]One way of arriving at a solution is using a “greedy” approach. We start with the empty set of weights and only add one when we can’t weigh something. Furthermore, we add the largest weight that we can while not leaving gaps.
Clearly, then, we need a 1 g reference weight to balance against the 1 g sample. How about a 2 g sample? Well, we know we can put the existing 1 g reference on the other side, and to balance against 2+1 we need a 3 g reference. From there we can count up to 4 g, but we can’t do 5.
To balance against 4+5 g, we need a 9 g reference. The pattern in adding/removing weights starts looking trickier, but there’s a simple solution: whatever order we added weights on in the first place, we reverse that order, but swap pans. Then swap again when we count up past the new reference. We counted up to 4 like this:
- S : 1
- S+1 : 3
- S : 3
- S : 3+1
Now we add the 9, and reverse the pattern for the 1 and 3 weights: - S+3+1 : 9
- S+3 : 9
- S+3 : 9+1
- S+1 : 9
- S : 9
And now back to the original pattern, except with the 9: - S : 9+1
- S+1 : 9+3
- S : 9+3
- S : 9+3+1
Now we’ve used all the reference weights, and need another. We have 13 g in weights and need to hit 14 g, so we can add a 27 g weight.
Do the same thing as before: add the new 27 g reference but reverse the order of the 1/3/9 pattern so that we are subtracting from it. Then go in the normal order once you pass the reference weight.
I’m pretty sure you figured this out already, but others might be curious…[/spoiler]
Not sure if you “mined” this from some other source - the math puzzle with the number 6…
0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6+6-6 = 6 (example)
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6
You can use any math function between the numbers, and it has to equal 6.
I have composed a wide variety of puzzles, including some for paid publication in logic puzzle magazines. Puzzles for paid publication even included puzzles in the Knight/Knave genre. I am well aware of the assumptions. 
And I do enjoy these puzzles.
I didn’t intend to demean the puzzle. I did think it interesting to point out that, while solutions to Monty Hall or Sultan’s Dowry have real-world applications, it might be hard to construct rule-world usage for Hundred Blue Eyes. Try it if you wish (and obviously you could simulate the conditions with digital programming), but please report if you have a real-world emulation in mind.
BTW’s:
[ul][li]French logic puzzle magazines circa 1980 took care to impose “Every statement uttered or read in the chamber is either true or false.” The need for this admonition is sometimes overlooked.[/li][li] I mentioned a card-play probability problem in the on-going Cafe Society Monty Hall thread. Why doesn’t one of you go post a spoiler for it, please.[/li][li] … and explain if you will – if it’s even true – that the card-play puzzle is closely related to Monty Hall. (This card-puzzle is very well known to expert bridge players, but even many mathematically inclined intermediate players get it wrong until the solution is shown.)[/li][/ul]
Glad you do enjoy these kinds of puzzles; good for you for constructing them.
My point is simply that for puzzles like these to work, there need to be any number of qualifications and other “rules” that have to be spelled out.
In the Monty Hall problem, for instance, you have to specify (perhaps among other things) that:
*There is always a prize;
*There is always exactly one prize;
*Monty always knows where the prize is;
*The prize will not be moved at any point after it is placed, whether on Monty’s orders or by some minion acting alone;
*The prize will not be removed after it is placed;
*Monty always opens one door to reveal that the space behind it is empty, regardless of whether the original choice is correct or not;
*Monty always asks the guest if he or she would like to switch…
Sometimes all of these are spelled out; sometimes they are not.
To my mind, that’s an awful lot of qualifications, which taken together severely reduce any “real-world” applications to the puzzle. To me, that makes the MH problem very like the assumptions in the blue eyes problem that everyone on the island can hear and that everyone on the island perceives eye color the same way. Your mileage obviously varies, and that’s fine.
He doesn’t need to know where the prize. He only needs to know one door that the prize isn’t behind, and only after the contestant has selected a door.
Okay, let me amend that: SOMEbody on Monty’s staff has to know where the prize is, and must be able to secretly communicate that information to Monty while the game is being played out. Probably simpler, all in all, if Monty already knows where the prize is before the game begins, but you’re right that the other way works too.
The main point really being that Monty isn’t in the dark–that is, he does not open a door at random.
Here are solutions to a few of them:
[spoiler]2 + 2 + 2 = 6
(3 x 3) - 3 = 6
sqrt4 + sqrt4 + 4 = 6
5 + (5/5) = 6
7 - (7/7) = 6
cbrt8 + cbrt8 + cbrt8 = 6
sqrt9 x sqrt 9 - sqrt9 = 6
But it kind of seems like cheating to use square roots and cube roots, as if you’re allowing that it’s hard not to allow irrational powers to be used as well, which would make the problem trivial (assuming I am correct that there is always a number n such that x^n = y, where x and y are both rational. But maybe I am wrong about that). Also, I don’t know how you can get the 1 and 0 cases to work.[/spoiler]
One I’ve noticed before, that it occurs to me makes for a good riddle:
For most verbs in English, if you want to form a noun that means “person or thing that does that”, you can append the suffix -er to the end. For instance, a walker is a person who walks, a maker is a person who makes, and a primer is a thing that primes. But there’s at least one verb in English that follows the exact opposite pattern: To form a noun that means a person or thing that does it, you remove -er from the end of it. What is the verb?
A pest pesters.
[spoiler]I don’t know any way that’s not “cheap”, but these work:
floor(sqrt(tan(tan(cos(0))))) - cos(0) - cos(0) = 6
floor(sqrt(tan(tan(1)))) - 1 - 1 = 6[/spoiler]