Oh good, I think this is the first time I've seen a puzzle before someone has solved it comment.

Just to be pedantic, Bogo-sort is an algorithm, just not an efficient one.

You could say "flip the cup in the North corner" every time and, given enough time, the probability of having a solution would approach 1.

I'll try to come up with an efficient solution now.

Oooh! We could have a *steal the prize* clause, where if you can come up with a more efficient algorithm, you can steal the honor.

For the record, it can definitely be achieved in a finite number of steps, although I don't remember what n is.

I've never played that game with a spinning table. Mostly, you just get all your teammates in a line with the alcoholics near the front. Then you drink and flip your cups and everyone shouts and cheers. I mean, if you want to practice flipping cups when you're sober, it is a useful skill to have anyway. But I've never known the compass directions to matter.

Besides, if you lose, your whole team drinks. So, no penalty. I'm not really understanding this puzzler of yours.

You know what would be awesome? If, some future Friday, I made up some bullshit puzzler like:

You have three people and you have to guess the number I'm thinking of, between one and two million, using only matchsticks and a can of beans.

I'm muddling around with "well, okay, these start scenarios are equivalent to each other, and for this one you can get it cleared out of the way like so..." thinking.

You could say "flip the cup in the North corner" every time and, given enough time, the probability of having a solution would approach 1.

I assume the goal is to have a definite upper limit on the number of steps involved.

Yes.

While I'm working on it, my current belief is that there is no solution that is guaranteed to be finite. I haven't thought of an algorithm that can't have loops in it.

I'm probably missing something, or there is always the possibility (however small) that the game will go on forever.

For the record, it can definitely be achieved in a finite number of steps, although I don't remember what n is.

Interesting, I'm have not found the correct track.

Hey, who am I to knock someone's faith? But 8 is not my belief.

While I'm working on it, my current belief is that there is no solution that is guaranteed to be finite. I haven't thought of an algorithm that can't have loops in it.

I'm wondering if the fact that four is composite is the key here, because I agree with you for the case of a two-position table. I'm working it out now for three.

I assume the goal is to have a definite upper limit on the number of steps involved.

NickS's solution is O(heebie's patience).

Okay, here's the strategy I would pursue, but I didn't check my work at all, so I may well have failed to work this out correctly:

First, if we aren't guaranteed that the start condition is not all face down, flip all.

Then:
1. Flip all
2. Flip any two opposite corners (say, E and W)
3. Flip all
4. Flip any two adjacent corners (say, N and E)

Repeat from 1 until satisfied.

I must be wrong though, or other people would have gotten it!

You have three people and you have to guess the number I'm thinking of, between one and two million, using only matchsticks and a can of beans

Tell me the number or I set you on fire, lady.

O(cⁿ), n being the number of beers you give Heebie.

Oh drat, mine isn't guaranteed to be finite either.

1. Flip all
2. Flip any two opposite corners (say, E and W)
3. Flip all
4. Flip any two adjacent corners (say, N and E)

I don't think this works.

13 has an obvious loop if the table never spins

(let + be face up and - face down)

If you start

---+

Your algorithm would generate
+++-
-+--
+-++
-+++
+---
(1 more loop gets to ---+ and you start over).

I have an appointment that I have to go to soon, so I'll see whether someone's solved it when I get back.

I don't think this works.

No, you're quite right, it doesn't. I'm no genius this evening. Also, I figured it would be more fun to be wrong in public than to say nothing while I checked my work in private.

Flip all
Flip adjacent corners
Flip all
Flip opposite corners
Flip all
Flip one
Flip adjacent corners
Flip all
Flip opposite corners
Flip all

Oh, dang, put another flip all in there after flip one.

I'm thinking maybe if you add a "flip three / flip all" or "flip one / flip all" in there...

Right! Like xyzzy is saying. I haven't been able to make a loop with that.

Don't forget to start with an extra "flip all" at the beginning, in case it started in the desired configuration.

Oh, dang it again, that doesn't work anyway.

The following flipping sequence will always work
1. NSEW
2. NW
2. NSEW
3. NS
4. NSEW
5. N
6. NW
7. NSEW
8. NS
9. NSEW

hint: start by assuming some number of cups in the wrong condition - 1,2,3 or 4

If you start with 2 cups in the wrong position
after step 1, there are still 2 cups wrong
after step 2, there are 0, 2 or 4 still wrong
after step 3, there are 0 or 2 wrong
after step 4, there are 0 or 4 wrong
after step 5, we're done.

Oh, dang it again, that doesn't work anyway.

It doesn't?

Whoops, left out step 5.5 NSEW

Oh my god, trying to read these solutions is making me dizzy. Hang on.

Repaired 28 is the same as repaired 22.

32

True enough - we're cross posting. 22/23 are faster than I am

Problem being that it doesn't always work if you start out with

xo
ox

Ignoring the flip alls as noise, the adjacent flip can result in

ox
ox

then the next diagonal flip

xo
ox

which is no good. That algorithm isn't guaranteed to resolve the 2 case for both adjacent and diagonal starts.

which is no good. That algorithm isn't guaranteed to resolve the 2 case for both adjacent and diagonal starts.

The flip-one will jar it out of being a two case, though, right?

Okay - I'm sure I'm making errors, too, but I believe there is no correct solution yet.

Okay I think this may work.

Flip all
Flip opposite corners
Flip all
Flip adjacent corners
Flip all
Flip opposite corners
Flip all
Flip one
Flip all
Flip opposite corners
Flip all
Flip adjacent corners
Flip all
Flip opposite corners
Flip all

34 is wrong

A diagonal flip starting from

OX
OX

results in either
XX
OO

or
OO
XX

not
XO
OX

then the next diagonal flip

Wait, nuh-uh, that would be with two adjacent flips in a row.

ox
ox

and a diagonal flip would give you

oo
xx

or

xx
oo

Hi, I should stop cross-posting now.

1. Smack it up.
2. Flip it.
3. Rub it down.
4. Move to the Jacuzzi.
5. Oooh that booty.

Yes, I'm pretty sure about that on a second glance. This portion:

Flip opposite corners
Flip all
Flip adjacent corners
Flip all
Flip opposite corners
Flip all

should correctly resolve all 2/2 cases.

Starting w/

xo
ox

You win on the first pair.

On

xx
oo

The first flip pair switches it to

xo
xo

Second pair either resolves it or switches it to

xo
ox

And third resolves.

The opening flip all covers the 4/0 case and the flip one in the middle should convert any 3/1 to a 2/2 that will be correctly resolved by repeating the other sequence.

37 IS THE BIG WINNER!!

xyzzy, relish in your hard-earned smug satisfaction!

Repaired 28 is the same as repaired 22.

22 isn't an algorithm. "Flip opposite corners" isn't a step, unless it means flip NSEW (ie, all opposite corners). 28 has you flip one set of specific opposite corners, then flip all, then flip the other set, then flip all—you can't get that from 22.

37 suffers from the same problem as 22. What do I do when I get to the step that says "flip opposite corners"?

Hooray!

Interpret "flip opposite corners" as flip any arbitrary pair of corners opposite one another. The spinning makes the labels irrelevant.

Ohhh, no wonder I was tempted to keep cycling between opposite and adjacent at first. I forgot there was actually a reason for that. Foolish me! Good puzzling, people who are not me.

Although 41 wins the contest to first answer the question, "Which verse begins The time was 6 o'clock on the swatch watch. No time to chill! Got a date! Can't be late! Hey, the girl is gonna do me..."

You can't forget the J, the I, the M, the M, the Y. I need a body bag.

Start: (1,2adjacent,2diagonal,3,4)
1. NSEW -> (1,2adjacent,2diagonal,3)
2. NS -> (1,2adjacent,3,4)
3. NSEW -> (1,2adjacent,3)
4. NW -> (1,2diagonal,3,4)
5. NSEW -> (1,2diagonal,3)
6. NS -> (1,3,4)
7. NSEW -> (1,3)
8. N -> (2adjacent,2diagonal,4)
9. NSEW -> (2adjacent, 2diagonal)
10. NS -> (2adjacent,4)
11. NSEW -> (2adjacent)
12. NW -> (2diagonal, 4)
13. NSEW -> (2diagonal)
14. NS -> (4)
15. NSEW

48 -- YEAH! I knew I could win at something, as long as it did involve logic. U-S-A! U-S-A! U-S-A!

Don't these all need a do nothing step at the beginning in case the tester is sneaky enough to start with them all down?

The lame joke in 50 could have been funnier if I'd remembered to include a key "not." To be clear: I suck at both logic and editing, but I like booty.

1. NSEW
2. Replace E with a google image search for "ben w-lfs-n"
3. NSFW.

QED.

53: Ever since you first called our attention to this fact, WS, I have been very concerned about our Benjamin's blind-dating prospects.

What are you talking about? Attractive, intelligent oven-ready turkeys can't get enough of him.

Did the powers that be not activate heebiegeebie@unfogged.com or heebie-geebie@unfogged.com when they made h-g a front pager? Can someone, heebie or othereise, e-mail me heebie's e-mail? I have a clarifying question about the puzzle but don't want to look at this thread.

I e-mailed you. The @unfogged ones haven't been created yet.

In case anyone else wants to send me super puzzlers or has questions, I'll leave my e-mail address on my name for this comment.

The @unfogged ones haven't been created yet.

You have the technology. Don't let Ben tell you otherwise. Go for it. Also… mouseover text, abouts, and the blogroll. ANARCHY!1!l!

The puzzle in mathy terms is that you've got a random unknown 4 bit binary number (representing the initial start position of each cup, up or down), and you want to find another 4 bit number (representing which ones to flip) such that they xor to 0000 (representing the state of all cups being down). So you have to test every number from 0001 to 1111, the only complication being that the state is saved after each operation. So call the original state A1, the second A2, etc., and you'd want to test

A1 xor 0001
A1 xor 0010 (equivalent to A1 xor 0001 xor 1100, or A2 xor 1100)

etc.

Solution guaranteed in 2^4 = 16 operations, no table rotating needed.

Oh, damn, I just realized the table rotating wasn't user-controlled. Ignore that answer then.

53: Ever since you first called our attention to this fact, WS, I have been very concerned about our Benjamin's blind-dating prospects.

You and me both, lady.

Also, I just did a GIS for my name and wtf? I wonder when that was taken.

Jeez, you look about 16 in that.

62: ben, it seems those folks are looking for you, too:

Ben w-lfs-n
N/A
???

You should contact them. Perhaps the mothership is returning soon.

No way, man. Ever since dag right-square-bracket-gren snubbed me when I went to Finland, I'm through with that crowd.

I do wonder what's up with plorkwort, though.

And CRGRE!!!

i'm bad at solving puzzles
yesterday i read the post and like felt physically how all the receptors're running out of mediators
always happens when i read something and can't concentrate, for example you guys always refer to MY and i tried to read his blog, a very rare state of mediators, i'll try again maybe when i'll get interested in politics
in the morning it's better and i actually read it and my algorithm would be just start from any corner, turn 90 degrees, flip the cup - repeat it 3 times each until all are flept
then i don't see why it's a puzzle
it's an impossible to carry out plan if the table just spins and stops at wherever degrees of angulation
but if it spins and stops at the corner exactly then it could stop however many times at the already flept cup corner, so chaos and when all the cups will be done depends on gods only and how much push you'll apply to the table
now when i got what the puzzle's about i'm waiting for the correct answer like pretty impatiently

The way I see it, there are 16 possible starting conditions, and each condition has 4 rotations, so one might think at any time there could be one of 64 results.

But some rotations don't result in unique positions and there are actually only 16 possible results. The 0000 result would end the game so I think the following positions are possible:

1111, 0001, 0010, 0100, 1000, 1110, 1101, 1011, 0111, 1010, 0101, 1100, 0110, 0011, and 1001.

My algorithm would be to use these 15 templates for the flips in sequence repeatedly until one works. My solution requires that the table rotations are random so, for example, it won't simply rotate 4 or a multiple of 4 every single time. My algorithm does not guarantee a win but it does make a win very very likely over time.

Comment 41 was sponsored by Ronnie DeVoe Real Estate Inc.

2. You issue a command, of the form: "Flip the cups in the ____ positions",
and you can specify exactly which corners you want, as many as you want.

Can I have 5! flips or does as many as you want max out at 4 ?

You can have 5!!!! flips! But I won't tell you if you've succeeded or not until I've finished carrying out your tall order.

I'd like to order a tall drink of water, please.

Coming right up! Do you have a sufficiently tall straw?

You can have 5!!!! flips!

If that's ((((5!)!)!)!) flips, that would be a lot of flips.

No, it's 5(!(!(!!))).