wonkyth
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Ok, now I'm REALLY confused.

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Wit penetrates it;
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But Nonsense Transforms it"
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w.r.t. Aristophanes

brian_s
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Rabscuttle wrote:
I adapted it from a puzzle in a game I was playing recently. You could say I localised it?


0, 1, 2, 3, 1, 1, 2, 1, _, 2, 1, 3, 1, 4, 1, 5

If I am going to have a chance of getting this, I think I will need to know what game you were playing.

Rabscuttle
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Rabscuttle wrote:

An easy tricky one:

What is the missing number?

3 1 1 2 1 _ 2 1 3 1

TripleM
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As soon as you posted the problem I was thinking of those counting sequences (three 1s, one 2, ..) where you describe the previous term, and I'm sure you're intending that with your reference to historical events, but it doesn't match any of the standard counting sequences I can find.

Rabscuttle
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TripleM wrote:
As soon as you posted the problem I was thinking of those counting sequences (three 1s, one 2, ..) where you describe the previous term, and I'm sure you're intending that with your reference to historical events, but it doesn't match any of the standard counting sequences I can find.

No, that's not what I mean by the reference to "historical events."

==

Here's some more of the sequence (with hintful formatting)

3 0 1 2 3 1 1 2 1 _ 2 1 3 1

The game I was playing was Professor Layton and the Curious Village - the puzzle is essentially the same as one from that.

kie
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Rabscuttle wrote:

An easy tricky one:

What is the missing number?

3 1 1 2 1 _ 2 1 3 1

The missing number is 1.

The sequence is '31/12', '1/1', '2/1', '3/1'
i.e. the dates from New Year's Eve to 3 January



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cheers, kie

Rabscuttle
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correct!

stigant
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Bump... Kie - do you have a puzzle? I have one if you don't.

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kie
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Oops, sorry, I completely forgot about this thread.
Go ahead, stigant, thanks!

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cheers, kie

stigant
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Ok, there's four political dissidents in a jail. The warden has all of them in solitary confinement in sound proof, light proof, odor proof, touch proof, taste proof rooms (ie there is no way for them to communicate in any way with each other). The dictator dies and control passes to his successor who offers the prisoners the following deal. The jail contains a perfectly cylindrical room with a single light and single light switch. The bulb of the light is unreachable (ie it cannot be altered except by turning it on or off using the light switch). Each night at midnight, the warden will choose one prisoner at random and escort them to the room, where they will stay for one minute and then be escorted back to his cell. The room with the light will be cleaned thoroughly, but the light and the switch will not be touched. Then the prisoner will be asked the question "Have all four prisoners been in the room?" He may answer "Yes" or "I don't know". The process will continue until one of the prisoners answers "yes". If the first prisoner to answer "Yes" is correct, then all four will be released. If the first prisoner to answer "yes" is incorrect, all four prisoners will be shot. The prisoners are given one hour together to plan a strategy. After the planning session, they will not be able to communicate in any way (other than by turning the light in the room on or off). What strategy will guarantee their (eventual) release?

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calamarain
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Just a technicality - if the switch is not touched, they could leave a scratchmark on it to say they've been there, A, B, C and D.

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RoboBob3000
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Do the prisoners know the initial state of the light?

Edit: Or, alternatively, do the prisoners know when the first night is?

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[Last edited by RoboBob3000 at 06-03-2008 10:14 PM]

zex20913
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I believe this question was posed here before. But with 100, rather than 4. Anyway...

With 4, it's a lot less time consuming to use this process, as follows:

Label 3 onners, and one offer.

Onners role: If they see a light that is off, turn it on. Turn on the light only once (never twice). If they see a light that is on, do nothing.

Offers role: If they see a light that is on, turn it off. Count how many times they turn the light off. Once the count reaches three, say "yes".

Edit: If the switch is initially on, and the offer goes in, he should turn it off, and not count it.

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[Last edited by zex20913 at 06-03-2008 10:21 PM]

RoboBob3000
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Man, Zex, that was my answer almost exactly. I'm not convinced yours works entirely though. What happens when an Onner is first chosen and sees the light in the on position?

Here's my solution for comparison:



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zex20913
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Let's assume that they don't know when it started.

If the light starts off, then an onner seeing an on light would change nothing.

If the light starts on, then the count should be increased to 4. This throws off my theory, which assumed that the light started in an off position.

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stigant
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Well, as I can't think of a way to solve the problem if you don't know the original state of the light, and as Zex posted essentially my solution, I award him the points and anoint him puzzle master.

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Rabscuttle
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If you don't know the original state of the light, you can get the non-counters to turn from on to off twice. Then the counter has to count 6 times that they turn the light from off to on.

If the light originally was on, then the non-counters have each been in twice.
If the light originally starts off, then one non-counter has been in once, the others have been in twice.

--

There's some interesting discussion of this problem on various places around the web. If the prisoners know that one of them will enter every night and they know how many nights it has been since the beginning, then you can use some more complicated strategies that can get them released faster.

zex20913
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Heard this one today.

A cruise line has many routes. We will focus on two--the Miami to Montreal and the reverse.

Each day, at noon, a ship leaves Miami for Montreal, and one from Montreal to Miami. The trip takes exactly 6 days--144 hours. How many Montreal to Miami ships can the passengers on the Miami to Montreal ship see during their cruise?

Because the number could be guessed randomly, an explanation will be necessary.

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NiroZ
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Well 6, because in the 6 days that it takes to travel that route, 6 boats would have left and theoretically one boat traveling the route must pass all of them.

Tuttle
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If you're on a boat leaving Miami, at the moment your boat leaves there are 7 Montreal-Miami boats on the water -- one leaving Montreal, one arriving at Miami and 5 in the middle somewhere. Your boat sees all of those.

In addition, 6 more boats will leave Montreal during the journey -- one each day, including the one that leaves at the moment you arrive. Your boat sees all of those too.

So 13 all up.

zex20913
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The Tuttle's have it.

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Tuttle
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You have 25 roaches in a box. They're distinguishable (probably just numbered, but if you want to name one Stanley I won't stop you). Each roach travels at a constant, distinct speed when moving. The challenge is to figure out which of those roaches is the fastest, the second fastest and the third fastest out of the lot.

You have a race track with five lanes and a little starting mechanism, and the roaches are smart enough to run from one end to the other without turning around or climbing over the walls. You can use that to compare the speed of up to five roaches at once. Photo finish equipment is available to record the results accurately.

You do not have access to a radar gun, stopwatch or any other method of directly measuring a roach's speed -- you can only compare the placings within a race.

How many races are necessary to determine the three fastest roaches in order? (And like the previous puzzle, you'll need to explain why.)

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[Last edited by Tuttle at 06-08-2008 01:04 PM : Clarification: Results from each race can always be determined accurately.]

Rabscuttle
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I can do it in 7.

Let's letter the roaches from A-Y

First do 5 groups of 5.
Next, do one race with the winner of each set.
Order the sets by the order in the sixth race.

So we get a partial ordering like so:

A < B < C < D < E
                ^
F < G < H < I < J
                ^
K < L < M < N < O
                ^
P < Q < R < S < T
                ^
U < V < W < X < Y

where < means "is slower than."

The roaches that don't have more than 3 faster than them are
O, S, T, W, X, Y.

Y is clearly the fastest, so finally race
O, S, T, W, X.
First and second of that race are second and third fastest overall.

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[Last edited by Rabscuttle at 06-08-2008 01:26 PM]

Tuttle
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That's it.

Rabscuttle
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I can't think of anything. If anyone wants to jump up with a puzzle, go for it.

Vike91
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Try to solve this:

       2
     2 2 1   2 4 6 1 2 2
     3 1 1 1 2 1 2 7 5 1|
    3                   |
  6 2                   |
2 4 1                   |
  1 4                   |
  1 5  [grid goes here] |
1 1 3                   |
1 1 2                   |
1 1 2                   |
  1 2                   |
  1 2                   |
------------------------+

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[Last edited by Vike91 at 06-08-2008 05:54 PM]

Wolfcastle
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Vike91
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Oh, you can do Japanese puzzle?

Your turn.

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Wolfcastle
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When the seven ravens of witch Axabra came back from a journey they said:"We were sitting next to each other on a power supply line today."

Draaks:"Either to my right sat Braks or to my left Crangs."

Quraksch:"To my left sat either Prahgz or Crangs."

Krachz:"Either to my left sat Braks or to my right Crangs."

Prahgz:"To my right sat either Crangs or Draaks."

Crangs:"Either to my left sat Grooks or to my right Krachz."

Grooks:"To my left sat either Draaks or Crangs."

Braks:"Either to my right sat Prahgz or to my left Grooks."

The witch is confused:"Not possible. I bet some of you are lying."

Quraksch:"We are not lying. But not everyone of us was looking north like I did."

In what order did the ravens sit?

zex20913
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Before I get started, does this mean "immediate left" and "immediate right"? That is, no ravens between others if they say "to my left"?

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