Riddle me this?

The French version of the KFC website is great. I love the Super-Croque

Aww man! I thought this was going to be a Gravel Puzzleworth thread full of puzzles. :(
― marianna lcl (marianna lcl), Thursday, April 7, 2005 8:53 PM (5 years ago) Bookmark

Absolute fave recent puzzle: http://jig.joelpomerantz.com/fun/dogsmead.html

fantastic!

<3

thank fuck i've finished it so i can go to bed!

Two classics that were in the movie Fermat's Room

-----------------------

A candy-store owner receives three opaque boxes of mislabeled candy. One contains sweet candies, one contains mints, and the third contains a mix of sweets and mints. The three boxes are, correspondingly, labeled “Sweets,” “Mints,” and “Mix,” but none of the labels is on its correct box. How many pieces of candy must the shopkeeper remove to determine the actual contents of the three boxes and relabel them properly?

-----------------------

The professor has 3 daughters. He tells a student "the product of their ages is 36, and the sum of their ages is my house number."

The student says "I'm missing a piece of information."

The professor says, "you're right. The eldest plays piano."

What are their ages?

2nd one of those has multiple solutions?

supposedly, but the solution is the one that most makes sense
this riddle has been told in different ways but this way is good enough

36, 1, and 1, the younger two are twins

"A student asks his teacher, 'How old are your three daughters?' The teacher replies, 'If you multiply their ages you get 36. If you add their ages you get my house number.' 'I am missing a detail,' protests the student. 'Oh yes,' says the teacher, 'the older one plays piano.' How old are the 3 daughters?"

I don't get how the sum = house number helps in any way? A house number could be anything? There are several solutions, surely? (I played piano at a young age).

you have to assume that the eldest isn't 35 because the teacher isn't old or something...

72, 1/2, 1, the youngest one is adopted

typo 36*

also, the house number line is important

but that's all I can say

so does the student already know the house number...?

18, the square root of -2, i, two of them are fabrications invented to make mathematicians' sums neat and tidy

there is no piano

does the student know the house number?

yes and that is why you can figure there is only one piece of information missing

actually that is also why 36, 1, 1 doesn't work

lol I googled the answer, captainlorax you messed up in your retelling - you need to use the comparative & not the superlative form

puzzles whose solutions involve nitpicking the semantics of puzzle-people talking the way nobody actually does = dud

I told the problem twice dayo

yeah, there was semantics involved. you had to figure out that the student knew the teacher's home number because he only need one detail

xxxxxp that's surely important to mention then?!
9,2,2

Use of "older" was telling. It's not clear the student knows the house number, my o/h thought this was what the 'extra info' he needed was.

oh now I C - now I have one more trick up my sleeve when I'm at the bar

that must be some nerdy bar

apparently the real answer is that if the student knows the house number, then he is deciding between two possibilities (6, 6, 1 & 9, 2, 2,) that adds up to the house # and knowing that there is only one elder 'solves' it

assuming that the twins in the 6, 6, 2 were born by c-section at the same time so neither one is 'older' than the other

it's a classic riddle supposedly

I don't know why I'm trying to do that Dog's Mead one, but I am. On first look it doesn't seem right, particularly clue 9 Down, and also 7 Down being 5 squares doesn't seem to fit with 14 Across being 3 squares. Is there anything stupid like blank squares to make it work?

Surely with the crappy student puzzle 2, 3 & 6 is another solution?

9 down is elegant when it all fits together! No blank spaces.

10a/10d trouble me - they seem to have multiple possible solutions.

oh wait unless martha comes into it. damn, can't remember how old she was.

i think i've proved to my satisfaction that the squares (^2 i mean) aren't 4 and 8. so they must be 1 and 1.

now instantly disabuse me of this notion.

what? 8 isn't 4^2. idiot.

Is there an answer online? I may have reached one, but I don't trust it. (ilxmail works if you don't want to spoil it onthread)

I took a leap of faith at one 2-choice junction and don't believe I hit a contradiction, so I suppose I should go back to the other option and make sure it doesn't work...

(a solution is here, if you want it... http://jaxwebster.wordpress.com/2010/06/19/self-referential-number-square/)

Interesting. I have that answer and I did (more or less) those steps but in a completely different order. I like a puzzle which doesn't have to be done in the one true order - you might think that something as self-referential as this would be pretty rigid about ordering, but it seems not.

Thanks, GP!

Haha I finally finished this and really liked it.

Weirdly once I had what ledge posted I could do the rest!

I didn't realise that the "new language puzzle" upthread uses (or maybe is just loosely based on, I haven't looked into it for more than 20 seconds here) a real "conlang", if there can be such a thing as a "real" "artificial language".

But anyway, this is a thing, which actually gets used in the world outside puzzles, apparently:
http://en.wikipedia.org/wiki/Blissymbolics
http://www.blissymbolics.org/pfw/

Blissymbols have become popular as a method of augmentative and alternative communication (AAC) for non-speaking people with cerebral palsy or other disorders, for whom it can be impossible to otherwise communicate with spoken language. However, Bliss did not approve of his language being used as an AAC; he sued the celebral palsy centers who employed it in this way, and they settled out of court

what a twat!

I kinda feel I lost some credit on this thread for the bad conlang puzzle upthread - anyway here's a conlang puzzle I think is absolutely brilliant, I ended up ordering a photocopy of the original book from the French national library.

http://twitpic.com/3vmtc/full

(not my desktop fwiw)

think i fail at the first hurdle for trying to click on the scrollbar/extra pages on the right of that pic.

and for saying "fail at the first hurdle"

I was looking for a languages thread but there are some articifial language puzzles itt so I'll put it here:
http://www.lrb.co.uk/blog/2015/09/03/helen-dewitt/tashu-duset-sekar/

and if you don't want to do the puzzles I agree with the article underneath too fwiw (modulo whiny caveats about opening things up w/o letting the new incomers be worn down by 3 years of being sneered at); hopefully the article got circulated round some teenager-populated nerd tumblr communities and such

I'd be curious to know how kids who haven't done Latin etc get on with the above but short of kidnapping some I will probably never know

surely you want someone who has done Latin as a control, right

(I could do them)

(I sometimes run a course that involves inventing a language - it's good fun)

Ooh, those were fun. Really not too hard, either.

xp I guess I was the control, though I am not a kid

(I could do them unless we were supposed to detect any nuances to the word order - I took the "unfixed" from the rubric to mean it didn't matter, though I spent a short time checking for patterns. But then I did Latin long ago and have spent time this very week swearing at German adjective declension, though studying German doesn't seem to rule applicants out of taking Test 1)

Tell me about your course, imago! Who is the target audience, how long is it, what do they do?

It's a day-long thing that forms part of either a weekend or a summer/easter week, generally aimed at 13-18 year-olds. I'll show them basic Esperanto and its mechanisms, discuss why it succeeds or fails as a language, and then run an exercise by which we write down letters in piles of vowels and consonants, turn them over, and then go around the table selecting patterns of vowels and consonants which are then turned over (by me) to make randomised new words. (For example, consonant-consonant-vowel-consonant-vowel-consonant could give us fbopad, or equally mjisev). Then I split up the group into four: nouns, verbs, adjectives and 'other' - each new group gets to claim in turn the words we've created and give them meaning. Finally all the groups collaborate to write a formative constitution for the new language. As I said, great fun :D

The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must leave the room exactly as he found it and is permitted no further communication with the others.

The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed. Find a strategy for them which which has probability of success exceeding 30%.

I assume once names are found the boxes/names are not removed and are left exactly as they were?

Yep!

This is somewhat harder than the monty hall problem. I thought the old 'simplify!' would help, got a solution for two people but I can't generalise it. Hint needed or I will have to cheat.

Having a ... glimmering... of what to do, but can't turn it into a concrete plan of action yet.

Slowly I turned, step by step

Okay, I think I know what the strategy is in the case of four prisoners.

This is a hard puzzle derived from a computer science topic. Reading the solution, I first thought "why would THAT work?" then had an epiphany. I like it.

Yeah, I figured it was related to something like that, which is where I came up with my sketch of a strategy. Can't quite get the calculation to go through yet though.

Something I don't get about this. Seems like the very first guy has a 50% chance of finding his own name and then the second guy has a tiny bit higher than 50%(third guy a little bit more than that etc.). Already based on the first two 1/2 * (1/2 +epsilon) = 1/4 + delta = 25% + something small is already less than 30%.

Oh wait. Maybe not

heh, the prisoners one must be making the rounds cuz someone just posed it to me the other day. still haven't looked it up but i have the impression it's not reasonable to solve and impossible to calculate

some feasible riddles

1a. (easy, elegant solution) 9 gold coins, 8 of equal weight, 1 dud coins that weighs less. you have 2 weighs of a scale to find the dud coin
1b. (harder, inelegant solution) same as 1a but 12 gold coins, 3 weighs, and you don't know if the dud is lighter of heavier than the rest

2. you have 25 horses, and you want to pick the fastest 3 horses out of those 25. In each race, only 5 horses can run at the same time because there are only 5 tracks. What is the minimum number of races required to find the 3 fastest horses without using a stopwatch?

3. (fun algebraic* spin on a classic balance puzzle) i pick an arbitrary polynomial in one variable p(x) = a + bx + cx^2 + dx^3 + ... yyou can give me values of x and i will evaluate them, p(x). our goal is to figure out what it is; i.e, to tell me a, b, c, d, ... in the fewest number of evaluations. which numbers do you evaluate at, and how many evaluations does it take? (note: solution has to hold for ANY polynomial, of arbitrary (finite) degree)

*don't let that scare you! anyone can solve this one ;-)

Thanks!

Back to prisoners. Feel like the comp sci approach has to do with binary numbers and checksums maybe, but not sure about the latter. Studying the case of 4 prisoners now. The only thing you know going in is that the boxes are ordered and that other guys have all chosen correctly, presumably. (But is this last actually of any use? Looks like not) There are 4 choose 2 ways equals 6 for anyone to pick. Which 4 of the six should they use? You don't want any two guys to use the same exact choice, based on some kind of symmetry argument, and furthermore you don't want to favor one particular box, so each box should be opened by exactly two prisoners. One possible set would be
P1 chooses B1,B2
P2 chooses B1,B3
P3 chooses B2,B4
P4 chooses B3, B4

The probability of any one prisoner's success is 1/4 but you can't just multiply the probabilities since the successes are not independent. Need to get a pencil and do the calculation. (Also note that the chances of overall success with this approach should be the same as that of any "conjugate" approach, such as one that has P1: B1,B4

Probability still way too low, unless there is some form of communication inside or outside the room I am missing, like they do something to the paper with the names on them to leave a binary encoded message.

I cheated. V elegant, still tough to intuit *how* it works, even for only four prisoners. Wouldn't have come up with it myself in a million years.

Yeah this is advanced set theory stuff, right? I feel like I get the principles at work but no way I could prove one method of search could have a better probability of success than another. Permutations is an album title to me.

Okay. Just took a look, almost seems like some kind of diagonal argument variant.

No, not quite.

advanced set theory stuff, right?

Psst, Tombot, over here

Anyway, really cool that a probability problem is solved with a group theoretic approach.

What I like best about this problem is that even when you solve it correctly all 100 prisoners are executed more often than not. Imagine them putting in all that work and planning, then the first prisoner sent in doesn't find his name.

B-b-but they have a moral victory. They converted a vanishingly small probability of survival into a 30% chance! Socrates would have blushed.

For the case of four prisoners I got a probability 1/3 of failure, 2/3 of success. Furthermore, after the first two prisoners make it, it is over, they have succeeded. The same as after the first 50 prisoners in the 100 prisoner case.

Also seems like the first prisoner reduces his own personal probability of failure from 1/2 to 1/4.

No, that's not quite right, I left something out.

No, first guys probability by his lonesome does not change.

But simply by surviving, he has already eliminated most of the cases the strategy doesn't work for.

By my count, the 4 prisoner problem has 14 bad cases and 10 good cases, so overall probability of success is 10/24 = 5/12. The first prisoner doesn't not find himself in 12 of the 14 bad cases, leaving two bad cases for the second prisoner to either encounter or to miss in order to survive, in which case victory is assured. Calculating the two step probability gives 1/2 * 10/12 = 5/12 once again.

Ah no, still not right, first guy does not eliminate as many cases as that.

i have the impression it's not reasonable to solve and impossible to calculate

I solved it! I couldn't calculate it until I understood it a bit better, but:

(spoilers)
Lbh'er rffragvnyyl orggvat gung gurer'f ab 'punva' nobir yratgu 50. Vs gurer vf, lbh ybfr*, vs abg, lbh jva. Lbh pna pnyphyngr gur punaprf gung n punva bs yratgu a rkvfgf naq vg'f fhecevfvatyl arng - sbe vafgnapr gur punaprf bs gurer orvat n punva bs yratgu 100 ner.... 1/100!

*: oneevat n fhcre jrveq beqrevat rssrpg

Excited to try flopson's tonight!

solved the prisoners one last night... god damn it that's a fantastic riddle