If a train leaves Chicago at noon carrying 20 passengers, 5 of whom smoke, in 4 cars, what is the name of the conductor's dog?

July 30, 2009 1:39 PM

A nearly impossible logic puzzle.Other mindtwisting puzzles. More puzzles of varying difficulty (don't scroll down too far!). XKCD's take. (previously epic post Also: 1, 2)
posted by desjardins (47 comments total)

Well, I've given it some thought and really all I can come up with is that if I were Kratos I would just fucking kill all three of them.
posted by kbanas at 1:54 PM on July 30 [7 favorites]

The problem is that so many of these are just variations on the same logic problems that have been around for centuries.
posted by subaruwrx at 2:02 PM on July 30

If you managed to ask the True idol if the Random idol would answer your question correctly, you would automatically ID all three of them, as he wouldn't be able to answer - in that case, and that case only, it is not a yes/no question, as the answer is maybe.

Given that odd behaviour, I would imagine there are a lot of possible solutions, as there could be any number of imaginative ways to frame a question to force one of the other two to identify Random. So, nearly impossible this is not.
posted by mek at 2:08 PM on July 30

Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are ‘da’ and ‘ja’, in some order. You do not know which word means which.

This isn't so much a logic puzzle as it is seizure-inducing word salad cleverly disguised as some sort of narrative.
posted by Avenger at 2:09 PM on July 30 [6 favorites]

Well, I looked at the explanation of the nearly impossible puzzle. And I don't like it at all. I don't think it works (it's tough to tell, since it is poorly written). Semi-spoiler: the writer asserts that the Random Idol will be unable to answer a certain type of question, but that is not true.

Great website though!
posted by taliaferro at 2:11 PM on July 30

The first problem sounds like it from "What is the name of this book?" by Raymond Smullyan, or one of his sequels. The incremental increase in complexity of each problem did always seem a tad ocd to me.
posted by sammyo at 2:13 PM on July 30

nice mek - except that wouldn't that question produce the same non-answer in the False Idol?
posted by taliaferro at 2:14 PM on July 30

(My preacher always told me to beware of false idols)
posted by taliaferro at 2:14 PM on July 30 [2 favorites]

"Each question must be put to exactly one god." So the questions must be distinct, but we don't have to give each god a question? We can give multiple questions to the same god? The wording could be a little clearer.
posted by equalpants at 2:17 PM on July 30

If you managed to ask the True idol if the Random idol would answer your question correctly, you would automatically ID all three of them

Actually, if you ask that question of the False Idol, he's as unable to answer as the True Idol. Only the Random Idol would be able to answer it. But identifying the Random Idol doesn't help you differentiate between the other two.
posted by yoink at 2:17 PM on July 30

i-spoiler: the writer asserts that the Random Idol will be unable to answer a certain type of question, but that is not true.

Bah. If you're asking it indecidable questions, that's lame. "Is the integer, which when multiplied by itself equals 29, even?"

Hehe. "Take "P", the shortest (and lexicographically first) Common Lisp program which, when given a Common Lisp program as standard input, will print "yes" if the program will ever reach the end of its execution, and "no" if it does not. Is the least significant bit of the MD5 hash of program "P" equal to 1?"
posted by blenderfish at 2:19 PM on July 30 [2 favorites]

Oh lovely. Random guy doesn't answer true or false at random, but truely or falsely.
posted by Pronoiac at 2:25 PM on July 30 [1 favorite]

"I should ask the man whether he was a tree-frog."
posted by billysumday at 2:34 PM on July 30 [3 favorites]

Actually, if you ask that question of the False Idol, he's as unable to answer as the True Idol. Only the Random Idol would be able to answer it. But identifying the Random Idol doesn't help you differentiate between the other two.

Quite right, sorry. But identifying the Random idol simply reduces the problem to its original version, which requires only one question to solve. So if you ID the Random idol in two questions, you can use your third to finish the job. The only problem is in my original example, if you start by talking to the Random idol, your first question is wasted. Which means it only solves the puzzle 2/3rds of the time.
posted by mek at 2:36 PM on July 30

C'mon now, folks, there is really only one question needed to decode their identities.

"Is it Ghostbusters 2?"
posted by Diagonalize at 2:36 PM on July 30 [3 favorites]

Okay, so I cheated and looked at the answer.

Semi-spoiler: the writer asserts that the Random Idol will be unable to answer a certain type of question, but that is not true.

The writer actually gives two solutions; the second one doesn't rely on that. It's a pretty straightforward extension of the usual knights-knaves method; everything you need to know is in the problem statement.
posted by equalpants at 2:44 PM on July 30

Whenever you get more than one god in the same place they end up having sex and then killing one another, so I'd just stick around for that.
posted by turgid dahlia at 2:48 PM on July 30 [2 favorites]

The Monty Hall problem is a classic. This one vexes me. Sometimes I can see it, sometimes I can't.

Here is a brute force solution that I have not tried.
posted by Xoebe at 2:59 PM on July 30

The title of this post reminds me of one of my favorite puzzles:

Bob: What are the ages of your three children?
Ted: The product of their ages is 36.
Bob: That's not enough information to tell their ages!
Ted: The sum of their ages is your apartment number.
Bob: That's still not enough information to tell their ages!
Ted: Okay, the oldest child plays the tuba.
Bob: Now I know their ages!

What are the children's ages?

This puzzle actually does have a solution. Solution
posted by twoleftfeet at 3:09 PM on July 30 [10 favorites]

Is there a word for being able to figure out the answer but unable to explain how it works? Even when I read the explanation, it still doesn't make sense to me.

Do I still get my little brother back though? Mom's gonna be pissed if I let Bowie turn him into a goblin.
posted by Sova at 3:13 PM on July 30 [2 favorites]

Well, I've given it some thought and really all I can come up with is that if I were Kratos I would just fucking kill all three of them.

Agreed; killing three guards is wasteful, though. Kill two of them in rapid succession, and then threaten the third with your bloodied sword, and I imagine you'd have your answer right quick, though.
posted by misha at 3:14 PM on July 30

However, the fact that the oldest child plays the piano implies that there has to be an oldest child, which means that (1, 6, 6) is eliminated.

Why couldn't the two oldest children both be 6, nine months apart?
posted by East Manitoba Regional Junior Kabaddi Champion '94 at 3:35 PM on July 30 [10 favorites]

twoleftfeet: That's a great puzzle. But it would be even better if there were two or more sums that could be arrived at in multiple ways, only one of which had exactly one possible child-set without eldest twins. Then you would also be using the fact that Bob was able to eventually find out. (As given, you don't even need his last line--you could stop after "oldest child plays the tuba" and the puzzle would still be solvable.)

...But I haven't been able to come up with a product and number-of-children that has that property. Anyone?
posted by equalpants at 3:49 PM on July 30

However, the fact that the oldest child plays the piano implies that there has to be an oldest child, which means that (1, 6, 6) is eliminated.

Why couldn't the two oldest children both be 6, nine months apart?

Or they're twins. Twins don't come out at the same time.
posted by billysumday at 4:06 PM on July 30 [1 favorite]

Why couldn't the two oldest children both be 6, nine months apart?

Totally valid point there. There's an implicit assumption that "older" means "having an age that is a larger integer", which is unsupported by biology. There's nothing in the statement of the puzzle about twins.

Actually, the puzzle is usually stated "the oldest child plays the piano", but given the same implicit assumption one could rephrase this as a statement about the youngest child and still have a workable puzzle.

Oh, and tubas are more fun than pianos.

equalpants, I'm not sure what you mean. When you say "two or more sums" do you mean "two or more children" or two or more sums of three numbers?
posted by twoleftfeet at 4:22 PM on July 30

To make it more precise, the riddle could say something like, "the oldest enjoys playing his tuba for the twins." That might give too much away, though.
posted by billysumday at 4:27 PM on July 30

the oldest enjoys playing his tuba for the twins

Hey, this is a family website!
posted by yoink at 4:29 PM on July 30 [2 favorites]

twoleftfeet: I got hung up on the idea of non-integer ages. Makes the problem more difficult. Also, I couldn't for the life of me figure out what the hell it meant to have a number add up to a date, because I kept thinking that the month had to be involved somehow. In short, if I ever meet a sphinx, I'm just going to try to shit my pants quickly, so that it doesn't enjoy eating me.
posted by Humanzee at 4:31 PM on July 30

twoleftfeet: Yeah, that wasn't a very clear explanation, I reckon. I meant two or more sums, each of which can be arrived at by two or more arrangements of children.

For example, let's temporarily suppose that 3+3+4 equals 11 instead of 12; then the possible child-sets are:
1+1+36 = 38
1+2+18 = 21
1+3+12 = 16
1+4+9 = 14
1+6+6 = 13
2+2+9 = 13
2+3+6 = 11
3+3+4 = 11 (*for now)

Ted tells Bob the sum, and Bob still doesn't know the kids' ages. In the original puzzle, we also knew the sum at this point. But this time, we don't know the sum yet, even though Bob does--we only know it's either 11 or 13, since those are the sums that don't give Bob the answer immediately. Now Ted tells Bob (and us) that the oldest age is unique; that doesn't help us, because it only rules out 1+6+6; we still have three possible answers.

Now Bob says that he knows. If the sum had been 11, then Bob still wouldn't know the ages, since both sets of ages that sum to 11 have a unique oldest age. Since Bob does know, we now know that the sum must have been 13.

(Still haven't thought of an actual set of numbers that would work like that, though.)
posted by equalpants at 4:45 PM on July 30

I got hung up on the idea of non-integer ages.

If you don't assume that ages have to be integers then the problem isn't uniquely solvable. This is easier to see with two children than three. If a pair of numbers has a certain product we get an equation like xy = p, which gives the hyperbola y = p/x. If x + y = s cannot be determined then y = s - x cannot be determined and you get a family of infinitely many lines (all parallel to -x) which will intersect the given hyperbola in infinitely many places.
posted by twoleftfeet at 4:59 PM on July 30

Damn! By "equals 11 instead of 12", of course I meant "equals 11 instead of 10"...
posted by equalpants at 5:05 PM on July 30

There is some truly appalling grammar on that website.
posted by doobiedoo at 5:06 PM on July 30

This problem inspired me to solve a problem at university on a worksheet on multiple integration without doing any integration. I was told it was clever, but not to be so clever again. Make of that what you will.
posted by edd at 5:08 PM on July 30

"names in boxes" is my favorite Very Difficult Puzzle. it's the first one listed in peter winkler's amusing collection "Seven Puzzles You Think You Must Not Have Heard Correctly".

it's great because there really does not appear to be any solution at all. and yet there is a reasonable answer.
posted by bruceo at 5:16 PM on July 30 [4 favorites]

we don't know the sum yet, even though Bob does... Now Bob says that he knows.

OK. I get it now. Interesting! A solution might take some time...
posted by twoleftfeet at 5:24 PM on July 30

Metafilter: let's temporarily suppose that 3+3+4 equals 11 instead of 12
posted by 256 at 6:17 PM on July 30 [1 favorite]

Diagonalize: C'mon now, folks, there is really only one question needed to decode their identities.

"Is it Ghostbusters 2?"

There are two other options, I believe:

"This puzzle; it vibrates?"

and

"Is this something I would need a television to understand?"
posted by tzikeh at 6:17 PM on July 30

Okay, it works for 5-10 kids with product 96, or 4 kids with product 192. (And probably many more; I only checked the ranges 2-10 kids and product 4-200.)
posted by equalpants at 6:25 PM on July 30

Conan the Destroyer was a dreadful movie, except for the part where Conan yells "Enough talk!" and then starts killing people. These puzzles remind of that scene for some reason.
posted by Brandon Blatcher at 6:52 PM on July 30 [1 favorite]

Xoebe said: The Monty Hall problem is a classic. This one vexes me. Sometimes I can see it, sometimes I can't.

Think of it like this: You have three doors, A, B, and C. The car is behind door C. Monty will always open a random non-prize door after you choose your door.

If you choose A, he has to open B. Switching wins.
If you choose B, he has to open A. Switching wins.
If you choose C, he can open either A or B. Switching loses.

Whether or not switching wins depends entirely on what sort of door you picked initially. If you pick an empty door, switching always wins. If you picked the prize door, switching always loses. 2/3 of the doors are empty doors.
posted by SemiSophos at 7:07 PM on July 30

Faced with the situation described, I would put all the idols on an airplane atop a conveyor belt moving at the exact same speed as the airplane but in the opposite direction, and see if the airplane takes off.
posted by notswedish at 8:54 PM on July 30 [2 favorites]

And when I woke up, my pillow was gone!
posted by DoctorFedora at 9:10 PM on July 30

I got hung up on the idea of non-integer ages.

Whereas I immediately figured that a six-year-old wouldn't be strong enough to hold up a tuba. Different focus, I guess.
posted by kittyprecious at 4:52 AM on July 31

equalpants, the smallest number having the property you mention is, I think, 288:
2 + 12 + 12 = 4 + 4 + 18 = 26 (incidentally, this just doubles all the ages from the "36" version of the problem)
1 + 9 + 32 = 2 + 4 + 36 = 42
Of course, not too many people have both infants and kids in their 30s, but this is math, not reality. And I haven't thought too hard about whether my program would generate all the solutions.

90 and 144 also have some interesting factorizations. For 90, you have 1 + 9 + 10 = 2 + 3 + 15 = 20, 2 + 5 + 9 = 3 + 3 + 10 = 16, so there's probably a puzzle in there that hinges on the existence of a pair of twins. Similarly for 144, which has basically the same thing with different numbers: 3 + 4 + 12 = 2 + 8 + 9 = 19, 3 + 6 + 8 = 4 + 4 + 9 =17;.
posted by madcaptenor at 6:18 AM on July 31

How did we get this far without mentioning this?
A little convoluted, though completely logical. (No thinking out of the box solutions)
posted by fizzzzzzzzzzzy at 9:30 AM on July 31

How did we get this far without mentioning this?

I think someone did (or maybe one of the links upthread took me to it on further clicking?). Anyway, it hurt my brain for a long time until I went through the hypothetical "what if there was just 1 blue-eyed person...what if there were 2...3...4...5...etc." It does make sense in the end.
posted by yoink at 9:43 AM on July 31

Bob: Forget it. I don't care about your damn kids. I was just trying to be polite.
posted by ODiV at 3:24 PM on July 31