OK, that’s an interesting twist… I’ll have to think about that for a while to figure out if it works.
Well, you know, presumably the rules of the game disallow random murder… That having been said, even if you change the rules to allow it, if such murder can be distinguished from mass suicide upon realization of eye color, then it wouldn’t free the islanders from their burden; the inductive proof still goes through. It’s still the case that if mass suicide doesn’t occur by day n, then you know there are at least n blue-eyed islanders, and it’s still the case that once you reach the point that you know there are more blue-eyed islanders than you see, you commit suicide.
(I know, I know, I already made my last post, but this is not continuing the previous argument but just adding a new note on a related question)
Actually, maybe not, hold on, let me think too…
Certainly a more noble act would be to commit suicide the first day in order to save the lives of the other 99…expecially as there is a fifty percent chance you’d eventually have to do it anyway. That, it seems to me, would stop the whole thing…unless it turned out you had brown eyes to begin with…
Wait…scratch that. If there are multiple eye colors and only 10% blue eyes, the only right method is to fake the suicide of another blue-eyed member on the first day. Killing one to save a hundred. Too risky to just kill self without knowledge of one’s own eye color.
If there were some way to get the visitor, while still on the first day, to retract his statement by saying: “I lied. I don’t see anyone with blue eyes. I was just messin’ with you.”
Indeed, couldn’t the stranger say that any day before the mass suicide and prevent it…
Wouldn’t that be enough to save the villagers?
It would be quite odd, since every villager knows that the missionary does see someone with blue eyes. If they were already the sort of people who would blindly trust the missionary (as implicit in the problem), then they should be the sort of people who react to “I don’t see anyone with blue eyes” by blindly trusting him and deriving a contradiction. And thus deriving that they have both brown, blue, green, and purple eyes, then killing themselves with extra passion.
After hurting my brain a bit, I think the missionary would have to retract his statement on the first day for it to work. Something along the lines of, “I thought I saw blue eyes, but they could have been green.”
Though everyone would know the missionary was lying, they wouldn’t all know that everyone knows that everyone knows…(etc.)…that the missionary was lying.
Probably be best at this point to just kill the missionary.
Killing a blue-eyed villager by the first noon would indeed prevent any more deaths. You need not be subtle about it either, just pull out a shotgun and blast the sucker in full view of everyone.
Alternatively, the noble thing to do would be to ask “do I have blue eyes?” If the answer is no, you live. If the answer is yes, you die but everyone else lives.
(Allowing killing is really no different than allowing discussion)
This, however, would not work.
The required body count goes up each day. If you make your kill by the first noon, one body suffices. If you wait until the 98th noon, you have to kill 98 people.
If you kill fewer than the required body count, all 100 still die on noon 100.
Okay, how about these ideas:
-
Once the missionary blabs, you get everyone rip-roaring drunk so that they can’t think logically for a few days and thus can’t infer from other’s actions/non-actions.
Or would it just start all over again after everyone sobered up…? No, I think this would work… -
Let’s say everyone can see what is coming and wants to prevent it. Suppose the decision is made to isolate several hundred islanders so that they cannot see the actions of others or know if the suicides took place. By isolating various islanders and releasing them at various times, couldn’t you take away the necessary consensus day count? So if everyone knows that everyone knows …etc…that there is at least one blue eye, but you remove a standard timeline for the process…doesn’t that break the process and as long as no one says anything again it’s OK. If you did this after the third day, wouldn’t everyone be left with…everyone knows that everyone knows…etc…that there are at least 3 blue eyes, but everyone does not know that everyone knows…etc…whether there are 4 or more blue eyers?
Excepting as I’ve noted (assuming anybody ever listens to me) the outcomes of the first 98 days can all be extrapolated nanoseconds after the missionary makes his statement on day 0. (Well, nanoseconds plus however long it takes the villagers to calculate the 1,267,650,600,228,229,401,496,703,205,376 possible initial answers to A.B.C.[…].CV.CW.CX, and then repeatedly go through the list and cross off the ones for each day until they get down to the final 101. But we assume that they’re pretty fast.)
Given that, anything that they do before noon on the 99th day should be exactly as effective (or ineffective) as anything they do on the afternoon of day 0.
Espectially since every villager will arrive at the same solution (assuming it’s the logically correct one) and everyone trying to murder a random blue is better than everyone in the tribe committing suicide on day 1.
Though they need to be careful not to be certain that all the blues are dead, because then they’d all know that the survivors are all brown and thus must all die. Hence my suggesting that everyone pick and try to kill a random blue, rather than everyone killing all the blues they see or lining up in a row and killing the first blue to their left.
Would you guys take these endless discussions to new threads, please? They’re completing bogging this one down.
Car Talk often has great puzzles along these lines. Last week’s was pretty good: what’s so special about the number 8,549,176,320?
the digits are in alphabetical order
I have a non-lateral one similar to that - what’s so special about the number 333667?
Take its first and last digit to get 37, and multiply the two: you get 12345679. Note that there’s no 8 - so to try to rectify that, multiply it in too. You get 98765432.
Along similar lines is 142857. Double it, triple it, etc., and you just re-order the digits
142857
285714
428571
571428
714285
857142
12345679 times anything, modulus 999999999 (which it evenly divides), is either of the form of the same three digits repeated three times, or has no repeated digits, depending on whether or not the number you’re multiplying it by is divisible by 3.
(any integer, anyway.)
An account of the phenomena pointed out by begbert2:
The first phenomenon:
Let onestring(b, n) = what’s denoted by a string of n many 1s in base b. [E.g., onestring(10, 4) is 1111]. Clearly, onestring(b, n) * onestring(b[sup]n[/sup], m) = onestring(b, nm).
Also, note that 37 = onestring(10, 3)/3 and that 333667 = onestring(10[sup]3[/sup], 3)/3, which makes the whole thing tick.
Accordingly, 37 * 333667 = onestring(10, 3)/3 * onestring(10[sup]3[/sup], 3)/3 = onestring(10, 9)/9.
But why should onestring(10, 9)/9 come out to 12345679?
Well, now let’s define sweepright(b, n) as what you get by writing out a string of n many increasing digits in base b, starting with 0 at the very left and moving to the right. [E.g., sweepright(10, 4) is 0123].
Observe that sweepright(b, n) * b + n = sweepright(b, n + 1) = sweepright(b, n) + onestring(b, n). Accordingly, onestring(b, n)/n = sweepright(b, n) * (b - 1)/n + 1. Setting n to b - 1, we obtain that onestring(b, b - 1)/(b - 1) = sweepright(b, b - 1) + 1.
And so, when b = 10, we have that onestring(10, 9)/9 = sweepright(10, 9) + 1 = 012345678 + 1 = 12345679.
The second phenomenon:
Of course, if we multiply the number we just obtained by b - 1, we end right back up with onestring(b, b - 1), from the left side of the equation. So to multiply it by b - 2 instead, we can simply subtract it from onestring(b, b - 1); thus, according to the right side of the equation, multiplying the number from the first phenomenon by b - 2 causes us to end up with onestring(b, b - 1) - 1 - sweepright(b, b - 1) = b * onestring(b, b - 2) - sweepright(b, b - 1) = the result of taking each digit of sweepright(b, b - 1) except the initial 0, and subtracting it from b. That is, a number with b - 2 digits, decreasing from left to right, starting with b - 1.
In the particular case of b = 10, this is, of course, 98765432.
The third phenomenon:
As we saw previously, our magic number from the end of the first phenomenon, when multiplied by b - 1, gives us onestring(b, b - 1). Since it evenly divides a string of b - 1 ones, it of course evenly divides a string of b - 1 repetitions of any digit. [In particular, it produces b - 1 repetitions of b - 1 when multiplied by (b - 1)[sup]2[/sup]]
As for the rest of it, clearly, nx mod mx is equal to (n mod m)x. Thus, it suffices to only consider the values of nx where n is in [0, m). (For our purposes, x is the same magic number as always, and m, as mentioned above, is (b - 1)[sup]2[/sup], just high enough to make mx as large as possible for a number consisting of (b - 1) digits.)
Recall that our magic number is onestring(10, 3)/3 * onestring(10[sup]3[/sup], 3)/3, with the first factor being an integer (indeed, both are). Thus, when multiplied by a multiple of 3, we will of course get the corresponding multiple of onestring(10, 3)/3 * onestring(10[sup]3[/sup], 3). The second factor (along with the constraints on how large n gets) ensures that this will take the form of three digits repeated three times. The first factor ensures that the specific three digits form a multiple of onestring(10, 3)/3.
As for the case when n is not a multiple of 3, with all digits showing up except -n mod 9, science is powerless to explain this.
Indistinguishable, you are a math god.
Though you’re still wrong about villagers bothering to organize and incrementally cull one and a quarter nonillion possible cases.)
Thank you (I’ll take what I can get. :)).
Incidentally,
You stress the anything here as though in contrast to Chronos’s example, but his has a very similar property, so long as you take the result modulo 999999; to wit, it will either produce zero or a permutation of the original digits, according as to whether you multiplied it by a multiple of seven or not.
Of course, the two work on very similar grounds. As we saw, 12345679 = onestring(10, 3)/3 * onestring(10[sup]3[/sup], 3)/3 = 9 * onestring(10, 3 * 3)/3[sup]4[/sup], while 142857 = 9 * onestring(10, 3) * onestring(10[sup]3[/sup], 2)/7 = 9 * onestring(10, 3 * 2)/7.