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 [quote="bonanova"][quote="DejMar"]Does breaking across a line that has already been broke count as a single break? If yes, then does "stacking" include the re-arrangement of the rectangular broken sections for further breaks?[/quote] Prohibiting stacking is meant to preclude counting simultaneous breaks as a single break. Pick a piece. Break it. Repeat. Count the steps. But I'm not sure that addresses your first question.[/quote]
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Posted: Fri Dec 07, 2012 1:23 am    Post subject: 1

Duke Gnome wrote:
Mark of Death
A remote island tribe is unusual in that every single citizen is a perfect logician.

Is there any chance we could convince the entire tribe to run for government office?
bonanova
Posted: Fri Oct 12, 2012 3:05 pm    Post subject: 0

DejMar wrote:
Does breaking across a line that has already been broke count as a single break? If yes, then does "stacking" include the re-arrangement of the rectangular broken sections for further breaks?

Prohibiting stacking is meant to preclude counting simultaneous breaks as a single break. Pick a piece. Break it. Repeat. Count the steps. But I'm not sure that addresses your first question.
DejMar
Posted: Tue Oct 09, 2012 10:53 am    Post subject: -1

Does breaking across a line that has already been broke count as a single break? If yes, then does "stacking" include the re-arrangement of the rectangular broken sections for further breaks?
bonanova
Posted: Tue Oct 09, 2012 4:19 am    Post subject: -2

The Hershey Bar

A Hershey Bar is a rectangle of chocolate, scored to delineate m smaller squares along its length and n squares along its width. You desire to break it down completely into smaller squares. If stacking is not allowed and all breaks are along straight lines, what is the minimum number of breaks required?

Every break produces one more piece. mxn-1 breaks are required.
Scurra
Posted: Mon May 28, 2012 5:15 pm    Post subject: -3

That's why I specifically noted in my comments that there are only two "common" words, meaning words that people would used in day-to-day conversation. The wording of the version I used was careful to emphasise the trick nature of the question as well.

For me, this is one of those puzzles that falls into the same category as Find a word that rhymes with orange to which the "right" answer is still there isn't one, even though there are some technically correct answers (mostly place names and the like.)
groza528
Posted: Mon May 28, 2012 4:22 pm    Post subject: -4

Actually there are more than three; this puzzle was on the front page once and the solution contains a pretty extensive list.
Amb
Posted: Sun May 27, 2012 11:59 pm    Post subject: -5

Going back to Scurra's Gry question. The answer is "Gry": http://www.onelook.com/?w=gry&ls=a

It isnt a trick question as he states. There are three words ending in Gry, and Gry is one of them.
groza528
Posted: Sun May 27, 2012 9:30 pm    Post subject: -6

Big square, small square

Given a square

Construct, using a compass and a straightedge, a square with exactly 1/5 the area of the original square.

[Bisect all four edges. Connect the SW corner to the midpoint of the N edge, the NE corner to the midpoint of the S edge, the NW corner to the midpoint of the E edge and the SE corner to the midpoint of the W edge. In the center will be the small square.]
Zag
Posted: Wed Sep 28, 2011 3:20 pm    Post subject: -7

English Words with Special Properties.

These are just the ones I can think of off the top of my head and with very quick Google searches. PM me others and I'll add them here.

Shortest common word with all five vowels: Sequoia
Shortest common word with all five vowels appearing in order AEIOU: facetious Adding the Y: facetiously
Allowing uncommon words: aerious
Longest word with only 2 different letters (several of each): deeded

On a QWERTY keyboard ...
... the longest common word you can make from just the top row of letters: Typewriter
... the longest common word that can be typed using only letters from the middle row: alfalfa.
... the longest common words using letters in typewriter order: weigh, quips, quash, quaff, quill.
... the longest common words using letters in reverse typewriter order: soiree, sirree.
... the longest common words using just the left hand on the typewriter: aftereffects, stewardesses, reverberated, desegregated.
... the longest common words using just the right hand on the typewriter: polyphony, homophony.
... the longest common word using alternating hands: dismantlement.
... the longest common word using one finger: deeded.
... the longest common words from adjacent keys: assessed, reseeded.
ctrlaltdel
Posted: Sat Feb 26, 2011 12:14 am    Post subject: -8

4 Triangles Using 6 Matches

How to form 4 triangles using 6 matches? (no breaking, etc. etc.)

[Position them into a tetrahedron.]

no idea how to make it invisible.

I did it for you. --Zag (Edit or quote it to see how.)
PuzzleScot
Posted: Wed Jan 26, 2011 11:08 am    Post subject: -9

Quote:
Two men working at a construction site were up for a challenge...

I was going to say a hand-grenade!
bonanova
Posted: Thu Jul 29, 2010 6:32 am    Post subject: -10

Four Ants on a Square

Four ants occupy the corners of a square one meter on a side; each faces a nearest neighbor, but no two face each other. Beginning at the same time each ant walks, with the same speed as the others, in the direction of the ant he faces. Eventually they meet at the square's center. How far will each ant have traveled?

[By symmetry the ants' positions remain the corners of a square, and the paths of pursuing and pursued ants remain perpendicular. The distance each ant travels is thus the original separation, one meter.]
Duke Gnome
Posted: Thu Jun 24, 2010 9:03 am    Post subject: -11

Mark of Death

A remote island tribe is unusual in that every single citizen is a perfect logician. Every citizen knows that every other citizen is a perfect logician and they all know everybody else knows etc.

They also have a strange condition that sometimes a strange mark will appear on their forehead. There are currently 100 tribesmen with this mark. They are all superstitious and it is widely known that anybody who realises that they have this mark will kill themselves at exactly noon on the first day after learning about it. Because of this all mirrors and reflective surfaces are banned, as is discussing whether or not someone has the mark.

An explorer visits the tribe, learns the language and is made to be very welcome. On his last day every single tribe member is gathered together to see him off. The explorer talks about how much fun he has had, and finshes by saying "as I was saying to a man with a strange mark on his head, this is the best place I have ever visited"

What will happen to the tribe?
[The ones with the mark will all kill themselves in 100 days. The explorer's comment has introduced the fact that there is at least one person with a mark into the collective conciousness. (The fact that they all knew this anyway is irrelevant). So if there was only one person with the mark he would die tomorrow. When this doesn't happen, in a hypothetical village with 2 people with the mark, they would realise that the reason nobody died was that everybody saw someone else, so the 2 people with the mark realise that each saw the other and so didn't kill themselves. They then kill themselves on day 2. In another hypothetical village with 3 guys with the mark, at the end of day 2, all 3 wonder why the other 2 are not dead and work out they must have the mark etc up to the 100th day where hypothesis catches up with reality and they all die.]
Antrax
Posted: Wed Jun 23, 2010 5:33 am    Post subject: -12

That was actually a question on my combinatorics exam. I wish I saw this chestnut before that
Zag
Posted: Tue Jun 22, 2010 11:35 pm    Post subject: -13

Combinations of Patio Tiles in a Path

I have a number (N) of patio tiles which are all 1 foot by 2 feet. I am going to use this tiles to make a path that is 2 feet wide, as long as I can with the tiles I have. (So the path will always be a rectangle, 2 ft. X N ft.) The question: Considering all tiles to be interchangable, how many unique ways can I lay them out? For instance, if I had exactly three tiles, there are three ways I could lay them out:

1. = |
2. | =
3. | | |

(where the equals sign represents two bricks side-by-side going along the path, and the OR bar represents one brick laid across the path)

The answer is the Nth fibonacci number, starting with fib(1) = 1 and fib(2) = 2.

Consider that you can only see the very last portion of the path which contains N tiles. It could be a single cross-wise tile, in which case the number of combinations of the remaining tiles is f(N-1) -- the total combinations possible without this brick. Or it could be the ends of two bricks side-by-side, in which the number of combinations is f(N-2) -- the total combinations possible without both these end bricks. So, for any N, the total combinations is

f(1) = 1 (by "exhaustive" analysis)
f(2) = 2 (by "exhaustive" analysis)
f(N) = f(N-1) + f(N-2)

Which is the definition of the fibonacci series.
Duke Gnome
Posted: Tue Jun 22, 2010 2:27 pm    Post subject: -14

One I've started to see around quite a bit lately.

First Alphabetic Odd Number

If all integers were written out in full in American English (no 'and') and the spaces were ignored, what would be the first odd number alphabetically?

[8,018,018,885. Eight Billion Eighteen Million Eighteen Thousand Eight Hundred Eighty-five]
BraveHat
Posted: Sun Apr 25, 2010 5:50 am    Post subject: -15

Ah yes, that is true. I forgot to consider the winner of heat 6.

I think I have an alternate solution to the lining up hat puzzle:

Zag wrote:
100 of you are locked in a room by the king of Puzzleania and hear the following: "In 15 minutes, knockout gas will be released. When you all wake up, you will each have a red or black hat on, such that you can see everyone else's color but not your own. With NO communication, you must form yourselves in a queue with all the red hats on one side and all the black hats on the other. Any communication after you awaken or any deviation in the line and you will all be killed. You have 13 minutes remaining in which to agree on a strategy." What is the strategy you propose (and, hopefully, the idiots in the room listen to)?

Before getting knocked out, tell everyone that when they first come to, they must count the number of people they see wearing black hats. In fact tell them to count three times and make sure they get the same number each time before doing anything. Designate one side of the room for people who count an odd number, and the other side of the room for people who count an even number (zero being even). They will automatically be congregating with like colors.
Zag
Posted: Sun Apr 25, 2010 2:51 am    Post subject: -16

Since we are only interested in the top three, we know that 3rd behind #2 of heat 6 is, at best, fourth. He was beaten by two horses in his own race, and the better of those was beaten by yet another horse (the winner of heat 6). So it doesn't matter that he might possibly be better than #3 of heat 6, it only matters that he can't possibly be in the overall top 3. That is, he might be better than #3 of heat 6, but only if that horse is, at best, 5th.
BraveHat
Posted: Sun Apr 25, 2010 1:18 am    Post subject: -17

bonanova wrote:

1-5: Race 5 groups of five horses.
Eliminate all fourth and fifth place horses.

6: Race all five first place horses.
Eliminate the fourth and fifth place horses.
Eliminate the horses they defeated in the first round.
Eliminate the horse who [in 1-5] was 3rd behind horse 2 of heat 6.
Eliminate both horses who [in 1-5] finished behind horse 3 of heat 6.
Six horses remain; heat 6 winner is known to be fastest.

7: Race the other 5 to get 2nd and 3rd fastest.

I think the answer is actually 8. I don't see why we are able to safely eliminate the horse who finished 3rd behind #2 of heat 6. That horse could very well be faster than #3 of heat 6.
Zag
Posted: Fri Jan 29, 2010 6:33 pm    Post subject: -18

Ant on an Elastic

Imagine you have an infinitely stretchy elastic attached to a fixed point and a car's rear bumper. The car is traveling away from the fixed point in a straight line at 10 mph, and the elastic stretches uniformly as the car pulls away. Assume an infinitely long, perfectly straight road.

When the car is 0.1 miles away from the fixed point, you place an ant on the elastic at the fixed point. It travels along the elastic towards the car. At all times it is moving at 0.1 mph relative to the point of elastic upon which it is standing. Of course, the elastic under its feet is moving (once it is off the starting point) so it is actually moving faster than that.

Does the ant ever reach the car?

The answer is yes. Imagine drawing 199 uniformly spaced marks along the elastic (dividing it into 200 equal parts). When the ant is standing at the starting point, the closest of those marks is moving away from him at 10mph/200 or 0.05 mph. Since he moves at 0.1 mph, he clearly will catch it.

Once he is standing at the first mark, the next one is moving away from him at the same 0.05 mph, the difference between the speed the mark he is on moves and the next one. Clearly, he will reach that one, too.

Etc.
3iff
Posted: Thu Jan 21, 2010 3:45 pm    Post subject: -19

Lilies in a pond

A lily in a pond will double every day. Starting with one lily, in 50 days the pond will be full. In how many days will it be half full?

Solution.

The pond is half full in 49 days...not 25 as many people say!
Nsof
Posted: Fri Jan 15, 2010 10:39 pm    Post subject: -20

100 ants on a stick

100 ants are placed on a one meter stick. The location as well as the direction of each ant is randomly chosen. Once all ants are placed they start walking at 1 meter per minute.
Note
• There are only two possible directions on the stick - the two ends
• Ants cannot pass next or over each other
• If two ants collide they immediately (zero time) turn around and continue walking
• If an ant reaches the end of the stick it falls off
How much time in worst case until all ants fall off of the stick?

After at most one minute.
The way to see this is to imagine that in every collision instead of turning around the ants pass through each other. In other words, think of each original ant as a virtual ant that replaces its host after each collision and always continues walking in the same direction.
bonanova
Posted: Sun Nov 02, 2008 3:14 am    Post subject: -21

How many horse races?

The annual Horse racing tournament is a week away.
You own 25 horses, and you're allowed to enter up to three.
You decide to hold a series of races to find your three fastest horses.
You have access to a track that permits five horses to race against each other.

[1] No two of your horses are equally fast - there will be no ties.
[2] You have no stopwatch - you can't compare performances from different heats.
[3] You do not want to tire the horses needlessly - the fewer heats, the better.
[4] You want to know which horse is fastest, which is next fastest and which is third-fastest of the 25.

How many heats are needed? Only seven.

1-5: Race 5 groups of five horses.
Eliminate all fourth and fifth place horses.

6: Race all five first place horses.
Eliminate the fourth and fifth place horses.
Eliminate the horses they defeated in the first round.
Eliminate the horse who [in 1-5] was 3rd behind horse 2 of heat 6.
Eliminate both horses who [in 1-5] finished behind horse 3 of heat 6.
Six horses remain; heat 6 winner is known to be fastest.

7: Race the other 5 to get 2nd and 3rd fastest.
ralphmerridew
Posted: Mon Aug 25, 2008 9:09 pm    Post subject: -22

The version I heard was that the prisoners had to flip one switch or the other each time they entered the room.
lostdummy
Posted: Mon Aug 25, 2008 9:07 pm    Post subject: -23

Duke Gnome wrote:
All prisoners not sent in on the first day...

you are correct, I missed that text state prisoner must be called on specific first day. if problem stated that "random prisoners on random days were called" it would be more suitable.

ralphmerridew wrote:
There's a way to alter the method to get around that problem

yes, I was thinking along that line, but with variant that could count only up to 99:

One elected prisoner will always flip switch#1, and reset switch#2 to off. Starting with his second visit, he will count how many times he found switch#2 "on" and will add one count to counter depending on state of switch #1 (means he will keep two separate counters).

Other prisoners will turn switch#2 "on" only in case it was off when they entered and they didnt turn it on previously when switch#1 was in current state. That means everyone else will turn switch#2 on only twice - once when switch#1 was on, other time when switch#2 was off.

That way elected prisoner can declare completition when first of his two counters reach 99. In worst case it will be after 197 counts, but in best case it will be half that, thus on average saving some time.

BTW, one advantage of RM solution is that it is doable with only one switch.
ralphmerridew
Posted: Mon Aug 25, 2008 4:37 pm    Post subject: -24

There's a way to alter the method to get around that problem:

Create one counter who flips switch A off whenever he sees it on. After switching it off 197 times, he announces everyone has visited the room.
Everyone else flips switch A on the first two times they see it off.
Duke Gnome
Posted: Mon Aug 25, 2008 3:47 pm    Post subject: -25

lostdummy wrote:

yes, but I think both original and my version of puzzle already suppose that, with "prisoner...selected at random without regard for the past" - prisoners don't know if they are first/only to enter so far. Having them more often than one per day does not influence problem, just speed it up.

If the king sends for prisoners exactly once a day, then the first prisoner will know he's the first prisoner, can treat switches like they're set at Off. All prisoners not sent in on the first day now know the starting position and may treat it as a simplified 1 switch solution, just faster. If the king sends for prisoners when he feels like it then when a prisoner waits 5 years before being sent in the first time, he still doesn't know if he's the first one or not.

lostdummy wrote:

I think that can work, but it has (small) probability of never finishing. Lets suppose one prisoner (A) will first time see position off/off , and then he is not called in untill all other prisoners already used their "set #2 once". That means after elected prisoner set #2 back to off, if even number of prisoners are called between each call to PrisonerA, he (A) will always see off/off , and will never flip his #2 switch.

There is solution less dependant on call order, and thus with chance to release prisoners sooner. But above solution is also ok, unless puzzle text is made more restrictive.

Actually it has 0 chance of "never finishing". Eventually it will give a result. Of course we have to assume that the prisoners live indefinitely long, but that's the same with all solutions as there is always a diminishing chance that 1 of the prisoners will never be called.

I won't derail the thread with this discussion any further though.
lostdummy
Posted: Mon Aug 25, 2008 3:23 pm    Post subject: -26

Duke Gnome wrote:
I believe the situation you're thinking of is one where .... first prisoner doesn't know he's the first prisoner

yes, but I think both original and my version of puzzle already suppose that, with "prisoner...selected at random without regard for the past" - prisoners don't know if they are first/only to enter so far. Having them more often than one per day does not influence problem, just speed it up.

Duke Gnome wrote:

One solution: ...

I think that can work, but it has (small) probability of never finishing. Lets suppose one prisoner (A) will first time see position off/off , and then he is not called in untill all other prisoners already used their "set #2 once". That means after elected prisoner set #2 back to off, if even number of prisoners are called between each call to PrisonerA, he (A) will always see off/off , and will never flip his #2 switch.

There is solution less dependant on call order, and thus with chance to release prisoners sooner. But above solution is also ok, unless puzzle text is made more restrictive.
Duke Gnome
Posted: Mon Aug 25, 2008 12:28 pm    Post subject: -27

Your 100 prisoners with 2 lightswitches puzzle is trivial, as it is just a simpler version of the 1 lightswitch puzzle. I believe the situation you're thinking of is one where the king calls prisoners in whenever he feels like it, not once per day. This means that the first prisoner doesn't know he's the first prisoner, and all further prisoners are never entirely sure whether they are the first/only prisoner to ever enter until they see the switches change.

One solution:

[Elect one prisoner counter. The counter will flip the first switch every single time he enters. He will set the second switch to Off the first time he enters. After that every time he sees the second switch on he will turn it off. After turning the second switch off 99 (or 100 if it was originally On) times he declares.

Everyone else does nothing until they see a different arrangement to the last one they saw. After that they flip the first switch every time they enter. They turn the second switch on if they have never done so before .
]
dethwing
Posted: Mon Aug 25, 2008 12:09 am    Post subject: -28

How old are my children?

A census taker asks a lady how old her three children are.
She says that the product of their ages is 36.
The man of course says that's not enough information.
She then tells him that their ages sum to the house adress.
The man goes to see the adress, but returns claiming that's still not enough information. The woman adds, "My youngest loves strawberry ice cream." The man says, that's enough and leaves. How old are the children?

Solution:
. At first glance, it seems that not knowing what the numbers sum to, this is impossible. However, the MAN knows, so we can use this information. First, make a table of the possible factorization of 36, and note the sum.
Code:

36 1 1 : 38
18 2 1 : 21
12 3 1 : 16
9 4 1 : 14
9 2 2 : 13
6 6 1 : 13
6 3 2 : 11

If the adress was anything but 13, the man would have enough information and be satisfied. (For instance, if it's 16, he knows it must be 12-3-1). Since it's NOT, we can rule out everything but 13. The man can't know whether it's 9-2-2 or 6-6-1. The strawberry information is irrelevent, the point is the woman HAS a youngest. Therefore, he concludes that it can't be 9-2-2 [Would not have a youngest]. He leaves knowing it must be 6-6-1.

Edit: I can't seem to make it invis the table. Oh well.
Victoria Silverwolf
Posted: Sat Aug 23, 2008 5:28 am    Post subject: -29

That reminds me of this:

Letter Sequence

What's next in this series?

O, T, T, F, F, S, S

The next letter in the sequence is E. This is just the first letters of the counting numbers. One, Two, Three, Four, Five, Six, Seven, Eight.
/dev/joe
Posted: Fri Aug 22, 2008 7:21 pm    Post subject: -30

1, 11, 21, 1211

What is the next term in the sequence 1, 11, 21, 1211, 111221 and how is the sequence continued?

The next term is 312211. This is the audioactive sequence, extensively studied by John H. Conway. It is also called the Look-and-say sequence and the Morris number sequence.

Each term is created by reading the previous term as a sequence of groups of like digits. It begins with 1, and then next term states that there is one 1, making 11. The next term states that there are two 1s, so 21. This number has one 2 and one 1, so 1211, so so on.

The last given term, 111221, is read as three 1s, two 2s, and one 1, thus giving the answer 312211.
groza528
Posted: Fri Aug 22, 2008 7:17 pm    Post subject: -31

I almost had the chance to pull that one, but the strong man himself was so heavy I wasn't sure I could do it.

The Plate Game
Two men are playing a game involving a perfectly circular table and a number of plates that are all of the same diameter. The two men alternate placing plates on the table until they are unable to fit any more. A plate may hang off the edge of the table, but it my not be resting on or partially on any other plate. Whoever places the last plate wins.
Assuming they both use perfect strategy and have perfect hand/eye coordination and balancing skills, which player will win?

The first player will win. He should put his first plate in the exact center of the table, then mirror his opponent's plays after that.
MNOWAX
Posted: Fri Aug 22, 2008 6:38 pm    Post subject: -32

and my other favorite one

Two men working at a construction site were up for a challenge, and they were pretty mad at each other. Finally, at lunch break, they confronted one another. One man, obviously stronger, said “See that wheelbarrow? I’m willin’ to bet \$100 (that’s all I have in my wallet here) that anything you can wheel to that cone and back, I can wheel twice as far. Do we have a bet?”

The other man, too dignified to decline, shook his hand, but he had a plan formulating. He looked at the objects lying around: a pile of 400 bricks, a steel beam, the 10 men that had gathered around to watch, and a stack of ten bags of concrete mix; he thought for a while, and then finalized his plan.

“All right,” he said, and revealed his object.

That night, the strong man went home thoroughly teased and \$100 poorer. What was the weaker man’s object?

-He looked the strong man right in the eye and said, "get in." -
MNOWAX
Posted: Fri Aug 22, 2008 6:37 pm    Post subject: -33

Mathgrant had been captured by a Spanish general and sentenced to death by his 50-man firing squad.

Mathgrant cringed, as he knew their reputation for being the worst firing squad in the Spanish military. They were such bad shots that they would often all miss their targets and simply maim their victims, leaving them to bleed to death, as the general’s tradition was to only allow one shot per man to save on ammunition. The thought of a slow painful death made Mathgrant beg for mercy.

“Very well, I have some compassion. You may choose where the men stand when they shoot you and I will add 50 extra men to the squad to -ensure someone will at least hit you. Perhaps if they stand closer they will kill you quicker, if you’re lucky,” snickered the general. “Oh, and just so you don’t get any funny ideas, they can’t stand more than 20 ft away, they must be facing you, and you must remain tied to the post in the middle of the yard. And to show I’m not totally heartless, if you aren’t dead by sundown I’ll release you so you can die peacefully outside the compound. I must go now but will return tomorrow and see to it that you are buried in a nice spot, though with 100 men, I doubt there will be much left of you to bury.”

After giving his instructions the general left. Upon his return the next day, he found that Mathgrant had been set free alive and well. “How could this be?” demanded the general. “It was where Mathgrant had us stand,” explained the captain of the squad.

Where did Mathgrant tell them to stand?

-Mathgrant told them to form a circle around him. All the squad was facing in at Mathgrant, ready to shoot, when they realized that everyone who missed would likely end up shooting another squad member. So no one dared to fire, knowing the risk. Thus at sundown he was released.-
mtc32
Posted: Fri Aug 22, 2008 6:37 pm    Post subject: -34

Slamming Locker Problem

You have a room with 100 students with 100 lockers numbered 1-100. Each student stands in front of his/her locker. The principal comes in and asks everyone to open his/her locker. He then asks every second person to close his/her locker starting with locker 2. He then asks every third person to close or open his/her locker (depending on whether or not it was currently open or closed) starting with locker 3.

The principal continues this sequence asking every fourth, fifth, sixth etc. locker to open/close until he reaches the hundreth person where only locker 100 is closed/opened.

After all of that slamming, which lockers are now currently open?

It helps to figure out that people are opening and closing their lockers when the number given is a factor of thier locker number. An open locker at the end of the sequence would indicate an odd number of factors for the locker number. The only numbers that have an odd number of factors are perfect squares. Therefore the open lockers should be 1,4,9,16,25,36,49,64,81, and 100.
Zag
Posted: Fri Aug 22, 2008 2:49 pm    Post subject: -35

Identifying Light Switches

You are at the top of the stairs and you see three light switches. You know that they control the three bulbs at the bottom of the stairs (around a corner so you can't see them from here). You would like to know which switch controls which bulb. What is the minimum number of trips down the stairs to glean this information?

Answer: Just one. First, turn on switch #1 and wait a couple of minutes. then turn it off and turn on switch #2 and go down the stairs. One bulb will be on -- that is switch two. Feel the two that are off -- the one that is hot will be switch #1.
ralphmerridew
Posted: Fri Aug 22, 2008 2:01 pm    Post subject: -36

Variation on /dev/joe's puzzle:

1: The wells no longer contain the glasses directly; instead, they contain a button. Pushing the button will cause the glass to flip. You can push at most two buttons between spins. You win if all glasses are pointing the same way when the table is spun. How many spins are necessary & sufficient to win?

2: Like 1, except that you must get all glasses pointing up, but you may press as many buttons as you want each spin. How many spins are N&S to win?

3: Like 2, except that there N wells at the vertecies of a regular N-gon table. For what values of N can you guarantee a win? How many spins are N&S in such a case?

4: Like 3, except there are two tables, left and right, each with N wells. You may only push buttons on one table per spin. In addition, pushing any buttons on the left table will cause the right table to re-randomize. You win by getting all 2N glasses pointing up. Now for which values of N can you win, and how many spins are needed.

1: 7 spins are necessary and sufficient. Label the wells A,B,C,D clockwise. Flip AC, AB, AC, A, AC, AB, AC. 7 spins are necessary because there are 14 possible initial cases, and each spin eliminates at most 2.

2: 15 spins are N&S. Flip ABCD, AC, ABCD, AB, ABCD, AC, ABCD, A, ABCD, AC, ABCD, AB, ABCD, AC, ABCD. 15 spines are needed because there are 15 possible initial cases, and each spin eliminates at most 1.

3&4: It's possible to win iff. N is a power of 2. Problem 3 takes 2^N - 1 moves in such a case, and problem 4 takes 2^(2N)-1 moves.

- Suppose N = (2m+1)*2^k, and m>1. On any move, there are two glasses distance 2^k such that both are flipped or neither are. (If not, then any two glasses of distance (2i+1)*2^k have one flipped and one not, and any two (2i)*2^k are both flipped or neither. But each glass is distance (2m+1)*2^k from itself, and cannot be one flipped and one not. Contradiction.

- If you can win problem 3 with N wells, you can win problem 4 with N wells. Alternately make the series of moves that would win the right table, and make one move on the sequence that would win the left table. Eventually you will get the left table all pointing up, and then you will win the right table before making another move on the left table.

- If you can win problem 4 with N wells, you can win problem 3 with 2N wells. Imagine a virtual left & right table. Define glass #i on the left table as being up if glasses #i and #(i+N) on the real table are both up or both down. Define glass #i on the right table as being up if glass #i on the real table is up. To make a move on the right table that flips #j on the right table, flip #j and #(j+N) on the real table. (Note that this doesn't affect the virtual left table.) To make a move that flips #j on the left table, flip #j on the real table.

- It is possible to win problem 3 with 1 glass. (Left as an exercise.)
lostdummy
Posted: Fri Aug 22, 2008 1:47 pm    Post subject: -37

100 Prisoners and Two Light Switches

After seeing above post, I remembered this one too. It has exactly same problem setup as one above, except for:
- there are two light swithces in room
- prisoners do not know initial position of those switches

Since this post ended on new page, here is full text of altered problem:

The King of Puzzlania has gathered 100 of his prisoners in the prison exercise yard. He explains that there is a room in the palace that is completely bare except for two light switches. The light swiches are currently in unknown position. The prisoners will have 20 minutes to discuss strategy after which they will be taken to solitary confinement. Every day thereafter at exactly noon one prisoner, selected at random without regard for the past, will be taken to the palace where she may switch the light or not at her discretion.

At any time any prisoner may declare that everyone has been to the palace at least once. If she is correct then everyone will be set free, otherwise they will all be subjected to mathgrant's music. Since nobody wants even the slightest chance of having to listen to mathgrant's music they all agree that they won't make any declaration unless they are 100% percent sure. Given that they are all in good health and can be expected to survive indefinitely, give a method that can ensure release eventually.
Duke Gnome
Posted: Fri Aug 22, 2008 11:33 am    Post subject: -38

100 Prisoners and a Light Switch

The King of Puzzlania has gathered 100 of his prisoners in the prison exercise yard. He explains that there is a room in the palace that is completely bare except for a light switch. The light swich is currently set for Off. The prisoners will have 20 minutes to discuss strategy after which they will be taken to solitary confinement. Every day thereafter at exactly noon one prisoner, selected at random without regard for the past, will be taken to the palace where she may switch the light or not at her discretion.

At any time any prisoner may declare that everyone has been to the palace at least once. If she is correct then everyone will be set free, otherwise they will all be subjected to mathgrant's music. Since nobody wants even the slightest chance of having to listen to mathgrant's music they all agree that they won't make any declaration unless they are 100% percent sure. Given that they are all in good health and can be expected to survive indefinitely, give a method that can ensure release eventually.

[The simplest method is to elect one person as counter. She will turn the light off when she sees it on, and do nothing when it's off. Everybody else will turn the light on if they have never turned it on before and do nothing if the light is on. The counter may declare after she has turned the light off 99 times. There are better methods that can cut average time down but this is the simplest.]