This one's late, and these answers should have been here yesterday. But, as Sunday-Wednesday is my work week, and I do the whole school in the mornings thing, that didn't happen. So here are your answers, and then if I can find my digital camera by the end of writing this post, I'll take a picture of my scribbly work at solving HLPE (I still do not have an answer).
First off, how about I just know out the one in the footnote to last time. The answer is fairly well known, for which we might blame Bowie and Henson. You ask either guard if, say, the other guard would tel you that the road to your left will lead you to paradise. Then choose whichever road the guard indicates that the other guard would tell you doesn't lead to paradise. Or watch this.
Gnomes Buried to Their Gnecks
All right, for this one, lets start right away and clear up a strategic question. A good way to figure out the answer is by asking "What would make this puzzle easy?" and then moving from there if we find anything helpful at that point. Again we have:
Now hand-drawn! (And yeah, I always make A smoke a cigarette for some reason.) |
Again, terrible drawing. |
Blue, Brown, and Green
Here's my favorite one. It's hard but it's damn satisfying. This one requires yet another counterfactual like above, but then the critical move is abstraction from the simple case to the much more complex one. It's also most rigorously shown with mathematical induction, but I'll sketch that after demonstrating the answer.
First, imagine the counterfactual: It would be simple if only she said this to a group, and only one member of which had blue eyes. The blue-eyed islander would look around and not see anybody else on the island that had blue eyes. He'd figure it must be him. Now here's the crucial step to get the solution, and it seems so simple in retrospect: What if there were two people? Why, they would look around, see one other person and realize that this could be the only person with blue eyes. After the ferry man came and went though, these blue-eyed islanders would of course see their other blue-eyed island mates not having left. They would realize that this person must have seen someone else, but since they only saw one other person with blue eyes, it must be them, and indeed on the second night, these two blue-eyed people would leave. The case of three runs exactly the same way, and they leave the third night. This is iterable indefinitely. Thus all and only blue-eyed people would leave on the 100th night. Everyone else will never leave the island.
A quick sketch of the answer by Mathematical Induction: So we've already done the case of 1, that a single blue-eyed person will leave on night 1. Lets assume n blue-eyed people will leave on night n. Now we show that if n+1 people were on the island, they cannot leave on night n because they will see n people but they know that those people would leave on night n if there are n people. Excluding other people, they know there must not be n people, but that there can be at most one more person with blue eyes, them. And so it follows from the n case that the n+1 case. Q.E.D.
Don't Get Stuck with Random
...And a segue into the HLPE which will be posted with solution and commentary next week.
I'm going to be fairly lazy with this response, since it is hard to come up with and I can't easily reduce it to a simple case like I have done with the others. First we have to notice though, that essential to the strategy this one uses is to cut out as a possibility the unreliable person you are asking. If we never pick the person we question we don't have to worry about asking random and not being able to decide if the person who answered us was random. Whatever answer she gives, we won't be picking her this way. So the best question to ask if random isn't who we're talking to is (lets imagine we're gesturing to one and then the other, and if that seems like cheating, imagine phrasing the question another way): Is that sister older than that sister? Then, to guarantee you pick only the eldest or youngest, pick whichever sister she indicates is younger based on her answer.
If you're talking to the eldest, who only tells you the truth, she'll indicate random is older and marrying the younger of these two puts you with the youngest. If you're asking the youngest, she'll lie and indicate the middle (random) girl is older, and so marrying the one she indicates to be younger gets you the eldest.
Relation to the Hardest Logic Puzzle Ever:
So we might use this and reformulate the HLPE. We might reduce it to asking a similar question and either determining all but the answer on the second question where we could rule out who we're asking as random, and then just determine our one last bit of information. The ja and da thing is making this difficult though. You can't waste any of these questions, so the idea is to ask about as many things as possible with one question. I've been assuming answers and drawing trees based around possible questions like "Is it true that Not Both ja is yes and C is more likely to tell me the truth than A?" said to B... It's yielded no results, so another thing is, and I'll accept help: If you can get an answer in 5 questions or less for this problem, I want your sequence of questions.
An upcoming post, I think, will be a post about craft beer and style put next to traditional. Not in a value based way, but maybe just determining qualities for them.