Delightful Puzzles

December 16, 2009 12:20 PM

A gathering of puzzles including many old chestnuts but also perhaps one or two you haven't met before.
posted by Wolfdog (29 comments total)

1. Cool post. I'm looking forward to trying some of these.

2. Weird background effect. At first I was worried something was wrong with my eyesight.
posted by Jaltcoh at 12:24 PM on December 16, 2009 [1 favorite]

Excellent stuff, I havent added a blog feed for over a year but I added this one.

Thanks
posted by therubettes at 12:36 PM on December 16, 2009

I must be dumb -- I don't get #1 even after reading the solution. And that's in the "easy" category.
posted by Jaltcoh at 12:37 PM on December 16, 2009 [1 favorite]

Thanks Wolfdog!

Jaitcoh: Walk through an example. The 10 cards he sets aside contain 2 cards face up for example. He flips them all over. Now he has 8 cards face up which is what the second pile also contains, right?
posted by vacapinta at 12:52 PM on December 16, 2009

Delightful indeed; thanks for this.

First I couldn't see the background effect. Then I too thought my eyes were going funny. Apparently Wordpress has a snowstorm plugin.

On preview - that's quite the spoiler there, vacapinta.
posted by motty at 12:54 PM on December 16, 2009

My brain hurts. I'm going to go find some pictures of cats and reboot. I'm bookmarking this to come back to when I'm ready to think.
posted by Slack-a-gogo at 12:58 PM on December 16, 2009

Ah, thanks, vacapinta. Got it. Very clever.

I wish the blog post were more precise in spelling out the exact circumstances of the set-ups. For instance, we're told only that the blind man in E1 "is handed" a 52-card deck with 10 face-up cards, not that he knows this. Also, in E6, we're not told (but are supposed to assume) that the person mentioned first makes the first move; also, I'm not clear on how strategically the players in that problem are supposed to play.
posted by Jaltcoh at 1:00 PM on December 16, 2009

Also I was hoping to see the Monty Hall Problem. ;)
posted by Jaltcoh at 1:02 PM on December 16, 2009

Monty Hall is in there.
posted by Wolfdog at 1:05 PM on December 16, 2009

Oh! Perhaps I am the one who is blind.
posted by Jaltcoh at 1:07 PM on December 16, 2009

Puzzle: A prison has 1000 cells. Initially, all cells are marked with + signs. From days 1 thru 1000, the jailor toggles marks on some of the cells: from + to – and from – to +. On the i-th day, the signs on cells that are multiples of i get toggled. On the 1001-th day, all cells marked with + signs are opened. Which cells are these?

Source: I remember hearing this puzzle sometimes in 1987, when I was in grade 9, from Samir Khosla. And I remember solving it in the football field in school.

Solution: A number has an odd number of divisors iff it is square. So cell numbers 1, 4, 9, 16, 25, and so on are opened.

This solution is so far over my head that it appears to me to be the answer to a different question entirely, except for that it mentions cells.
posted by Navelgazer at 2:23 PM on December 16, 2009

Awesome.
posted by alligatorman at 2:28 PM on December 16, 2009

Navelgazer: Same here. I just posted a comment (pending approval by a moderator) to ask for clarification. I've subscribed to the comments on that post and will update here if I get an answer.
posted by Jaltcoh at 3:00 PM on December 16, 2009

Navelgazer: "This solution is so far over my head that it appears to me to be the answer to a different question entirely, except for that it mentions cells."

Essentially, they're working from a mapping between cells and numbers (cell numbers, duh). Cell 8 gets toggled on the 1st, 2nd, 4th and 8th day, because 8 is a multiple of all of them. Or to state that another way, the divisors of 8 are 1,2,4, and 8. Simple so far.

For each cell, there's only two states, on and off, and the state is toggled as many times as it has divisors. Usually, divisors come in pairs. 1*8, 2*4. So at first brush you might imagine that all cells are toggled an even number of times. However, consider the number 9. For that you have 1*9 and 3*3. 1,3,9. These squares (3*3) are where the odd number of divisors comes from.

However the problem sails straight over one important thing: initial state. If everything starts +, it takes an even number of toggles to remain in that state.
posted by pwnguin at 3:05 PM on December 16, 2009

thanks pwnguin - and it's worth noting that the initial state is different on the front page than it is inside.
posted by Navelgazer at 4:16 PM on December 16, 2009

Great post!
posted by storybored at 6:19 PM on December 16, 2009

His choice of language is occasionally ambiguous, but most puzzles I've seen do a poor job of setting out the assumptions clearly. I'm having a lot of fun with this. So far I've gotten the first five or so right, but I felt bad about the marble one, which I guessed instead of proving.
posted by BrotherCaine at 6:22 PM on December 16, 2009

So, that snowfall effect got old pretty quickly, eh?
posted by primer_dimer at 4:50 AM on December 17, 2009 [1 favorite]

So, that snowfall effect got old pretty quickly, eh?

Between a family history of detached retinas, macular degeneration and diabetic retinopathy, I really don't like things that make me think my eyes are going wonky, so yes.
posted by BrotherCaine at 5:28 AM on December 17, 2009

In response to my comment on the "1000 cells" question, a commenter named DevilsAdvocate responded:

On the first day, the status of every cell is flipped. On the second day, the status of every even-numbered cell is flipped. On the third day, the status of every cell whose number is divisible by three is flipped. And so forth.

So at the end, the status of every cell will have been flipped a number of times equal to the number of divisors the cell’s number has, including 1 and the cell number itself. For example, 24 has 8 divisors (1,2,3,4,6,8,12,24), so its status was flipped 8 times. Since 8 is even, the status of the cell is the same as when it started (“-” with Sreeram’s correction). Cells whose numbers have an even number of divisors have the same symbol at the end as when they started; cells whose numbers have an odd number of divisors have the opposite symbol.

Every non-square number has an even number of divisors, because the divisors can be paired off to give products that equal the original number (1×24, 2×12, 3×8, 4×6). Square numbers have an odd number of divisors, because there’s one divisor, the number’s square root, which doesn’t get paired with a different number. E.g., 36 has 7 divisors (1,2,3,6,12,18,36) – if you try to pair them off you get 1×36, 2×18, 3×12, 6×6…but oops, you don’t count 6 twice among its divisors. So cells with non-square numbers end up with the same symbol they started with, and cells with square numbers end up with the opposite symbol.

posted by Jaltcoh at 6:25 AM on December 17, 2009

Hi!
posted by DevilsAdvocate at 12:42 PM on December 17, 2009

And I missed 36=4x9, so 36 actually has 9 divisors, not 7, but the basic reasoning still applies.
posted by DevilsAdvocate at 12:46 PM on December 17, 2009

Oh, hello
posted by Jaltcoh at 1:09 PM on December 17, 2009

Yeah, I clicked through to the link yesterday without reading the comments in this thread, lost a couple of hours there, including seeing your comment there (and recognizing you as a MeFite, or at least someone with the same username as a MeFite) so responded there, but by then I was running late for something (thanks, Wolfdog) and didn't get around to reading the comments in this thread until today, whereupon I saw that you had also posted your comment (and subsequently my response) here.
posted by DevilsAdvocate at 3:11 PM on December 17, 2009

Cool!
posted by Jaltcoh at 12:13 PM on December 18, 2009

His choice of language is occasionally ambiguous, but most puzzles I've seen do a poor job of setting out the assumptions clearly.

It seems like all the problems are ambiguous. M6 ("Cap Colors") is particularly confusing to me -- wouldn't there need to be more limitations than just a "bunch" of gnomes with an unspecified number of blacks and white hats for anyone to be able to guess the color of their own hat?
posted by Jaltcoh at 12:27 PM on December 18, 2009

It seems like all the problems are ambiguous. M6 ("Cap Colors") is particularly confusing to me -- wouldn't there need to be more limitations than just a "bunch" of gnomes with an unspecified number of blacks and white hats for anyone to be able to guess the color of their own hat?

The question is worded very poorly. He should have said "Each gnome can see everybody that is shorter than him" rather than "the tallest gnome....".

Also, the question should be something like "How many gnomes survive if they use an optimal strategy?"

Besides that, "bunch of gnomes" is enough to come up with answer.
posted by alligatorman at 4:06 PM on December 18, 2009

OK, got it now. For some reason I always have to read the solution about 10 times before I understand what it's describing.
posted by Jaltcoh at 11:12 PM on December 18, 2009

Here's another very good collection (maybe slightly better edited, too).
posted by Wolfdog at 5:25 AM on December 29, 2009 [1 favorite]