Best...Puzzle...Ever....

[Richards]

Hmmm...how about

Spoiler

a tree fell over onto the cabin and crushed them?

Johnathan

 

[Pielorinho]

Nope --

Spoiler

the cabin is in a clearing, by which I mean there aren't trees nearby.

Daniel

 

[der_kluge]

My answers for for 1 and 2 are the same.

Riddle #3

Spoiler

The cabin is in the arctic. They froze to death.

 

[der_kluge]

As for the levers question:

It doesn't matter how you arrange the rooms, really. All you have to do is identify 9 levers with 3 states each. 1-A, 1-B, 1-C; 2-A, 2-B, 2-C; etc.

For every pair of levers, you have 6 distinct possibilities of configurations. Assign each single lever to two, non-adjacent, portals and put all the portals (doors, portcullises, etc.) in a straight line... Embed some levers in between doors for some extra fun.

Now divide the levers as follows:

1 & 2 is one pair, 3&4 is the second, 6&7 the third, 8&9 the fourth. Number 5 is by itself and controls 3 doors.

1&2 control doors A&B, 3&4 control B&C, 6&7 control C&D, 8&9 control D&E, 5 controls A,C, and E.

Place the doors in a randomly determined open/close pattern. Neglect to tell players how many doors there are, or what each lever does - only tell them precisely what they see/hear (doors opening/closing etc.)

Why would each lever need three states? Two doors only have 4 total states. up/up, down/down, up/down, and down/up.

 

[Greyline]

These puzzles are great, but for some puzzles with more of an adventure feel to them, try and find a copy of Palladium's Teenage Mutant Ninja Turtles Adventures--the Death Traps scenario has six puzzle/riddle rooms in it. You either solve the puzzle, or you die. Years back, we got through 3 of the rooms unscathed (and 3 scathed). Might be something you could mine for ideas.

Also, I would suggest situational problems as opposed to riddles--say a fight on a raft heading towards a waterfall--something that requires problem solving, but that doesn't have one right answer.

Greyline

 

[Pielorinho]

Die_Kluge, try my spoilered variation of the riddle; it's got the same answer, I'll wager, as the first one.

Here's another fun riddle. Although there are multiple answers, you're looking for the most elegant one.

Since his wife is on a business trip, Michael is staying at a friend's house, a beautiful old mansion. His friends give him the room with the gas fireplace. He's a little paranoid, so he locks the door and the windows of the room.

Next morning he doesn't come down for breakfast, so they bust down the door of the room to find him dead inside, the room filled with natural gas fumes. An inquisition finds that he died because the fire went out sometime during the night, and rule it accidental death.

Two weeks later, his wife having successfully got away with murder, she takes the insurance settlement to Tahiti and starts a new life with her boyfriend.

How did she kill her husband?

Daniel

 

[Wrath of the Swarm]

My guess is that she dumped a lot of dry ice down the chimney during the night. Between the cold and the liberated carbon dioxide, the fire was extinguished, and then the murder weapon evaporated.

 

[Vyper]

Corran:

The solutions that were posted by now are very good. But it's possible to solve the riddle completely without a computer or even a calculator. As you said, the main problem is the big solution space.

Well, you can reduce it significantly.

For example, every even number is the sum of two primes (Goldbach conjecture - this hasn't been proven, but it's certainly true for the small numbers we are talking about). So the sum must be odd.

If one of the numbers is a prime number >50, B would immediately know the numbers. So none of the numbers can be a prime >50, and since A must know this, too, the sum cannot be 53+x with 2 <= x < 100, neither can it be 97+x with 2 <= x < 100. Thus, the sum must be less than 55. (197 and 198 can be eliminated easily.)

So we are left with about 25 odd integers. We still need to find the ones that cannot be the sum of two primes. Every odd number is the sum of an odd and an even number. The only even prime number is 2, so the only remaining odd numbers that do not have the property P can be written as 2+p (with p prime).

So, if I didn't make a mistake in my explanations, only 11 numbers are left now.

btw, there is a complete (and quite elaborate) analytical solution of the problem that was posted to a German newsgroup (hence it's German): http://groups.google.de/groups?selm=9ualmd$7dcff$1@ID-58404.news.dfncis.de

However, I probably wouldn't use this puzzle in a real rpg, since it's much too difficult.

 

[tbitonti]

Every other prisoner ...

Math geek alert.

As an aside, the notes below are really a condensation of doing
the solution by a complete simulation. Even the chart at the
bottom is nearly so, as the numbers are a compression of the
number ranges and offsets.

Does anyone know a non-iterative way to solve this using elementary
functions (or even using Fibonacci numbers, or a base 2 representation
and a summation)? I put a mini-formula at the bottom, but it's just
a start.

Here are my notes (they read best with a fixed width font):

from 1 .. 1000 [by 1] --> take out 1, 3, 5 .. 999 --> leaving 2 .. 1000 [by 2]
[count == 1000] [count == 500]
from 2 .. 1000 [by 2] --> take out 2, 6, 10 .. 998 --> leaving 4 .. 1000 [by 4]
[count == 500] [count == 250]
from 4 .. 1000 [by 4] --> take out 4, 12, 20 .. 996 --> leaving 8 .. 1000 [by 8]
[count == 250] [count == 125]
renumber by dividing by 8.
from 1 .. 125 [by 1] --> take out 1, 3, 5 .. 125 --> leaving 2 .. 124 [by 2]
[count == 125] [count == 62]
from 2 .. 124 [by 2] --> take out 4, 8, 12 .. 124 --> leaving 2 .. 122 [by 4]
[count == 62] [count == 31]
from 2 .. 122 [by 4] --> take out 6, 14, 22 .. 118 --> leaving 2 .. 122 [by 8]
[count == 31] [count == 16]
from 2 .. 122 [by 8] --> take out 2, 18, 34 .. 114 --> leaving 10 .. 122 [by 16]
[count == 16] [count == 8]
from 10 .. 122 [by 16] --> take out 10, 42, 74 .. 104 --> leaving 26 .. 122 [by 32]
[count == 8] [count == 4]
from 26 .. 122 [by 32] --> take out 26, 90 --> leaving 58, 122 [by 64]
[count == 4] [count == 2]
from 58, 122 [by 64] --> take out 58 --> leaving 122 [by 128]
[count == 2] [count == 1]

Leaving 122.

Backing up: 122 * 8 == 976. The 976'th prisoner is last.

Model:

Prisoner numbers:
a(i) + (k - 1) * 2^i

i gives the generation number

a(i) is the number of the first prisoner

k ranges from 1 .. K(i)

K(i) is the number of prisoners remaining

a(i) ... (K(i) - 1) * 2^i gives the prisoner number range

The initial generation is generation 0, before any prisoners are removed.
The initial prisoner number, a(0), is 1.
The initial prisoner count, K(0), is 1000.

Then:

For each generation 'i' there is a parity of '0' or '1', where
parity '0' means that the first prisoner will be removed,
and parity '1' means that the second prisoner will be removed.

P(i) == 0, 1

In the given problem, P(0) == 0.

For a given 'i', 'P(i+1)' is given by:

P(i+1) = P(i) if K(i) is even;
P(i+1) = P(i) + 1 (mod 2) if K(i) is odd.

For a given 'i', 'a(i+1)' is given by:

a(i+1) = a(i) + 2^i if P(i) is 0;
a(i+1) = a(i) if P(i) is 1.

For a given 'i', 'K(i+1)' is given by:

K(i+1) = K(i) / 2 if K(i) is even;
K(i+1) = (K(i) + 1) / 2 if K(i) is odd and P(i) is 1;
K(i+1) = (K(i) - 1) / 2 if K(i) is odd and P(i) is 0.

Then:

i a(i) P(i) K(i) 2^i
0 1 0 1000 1
1 2 0 500 2
2 4 0 250 4
3 8 0 125 8
4 16 1 62 16
5 16 1 31 32
6 16 0 16 64
7 80 0 8 128
8 208 0 4 256
9 464 0 2 512
10 976 0 1 1024

Note that 976 == 1024 - 48 == 1024 - 32 - 16,
that is:

a(10) == 976 == sum { 0 .. 9 } of ( 2^i * (1 - P(i)) ) + 1

But, how does P(i) relate to the initial number?

 

[Pielorinho]

My guess is that she dumped a lot of dry ice down the chimney during the night. Between the cold and the liberated carbon dioxide, the fire was extinguished, and then the murder weapon evaporated.

Interesting, but even if her climbing on the roof didn't wake the guy up, surely the clatter of dry ice falling down his chimney would do the trick. In any case, fireplaces normally have baffles to prevent things from falling down them.

This is one of those puzzles with an elegant solution: once you figure it out, you'll see that it's the only possibility. That means things like hypnotizing him and giving him a post-hypnotic suggestion are unlikely to be the correct answer .

Daniel