Best...Puzzle...Ever....
- Thread starter RithTheAwakener
- Start date
der_kluge
Adventurer
I worked up a map similar to the one Merak posted at the bottom of page 1. This one is a bit more complex.
There are 6 levers and 9 secret doors.
Each room 1-6 has a lever (numbered 1-6).
The levers open the following doors:
1. B
2. A,B,C,D,H
3. A,B,D
4. D,F,I
5. C,E,G
6. G,F
Find the sequence that will open the right-most room.
solution:
There are 6 levers and 9 secret doors.
Each room 1-6 has a lever (numbered 1-6).
The levers open the following doors:
1. B
2. A,B,C,D,H
3. A,B,D
4. D,F,I
5. C,E,G
6. G,F
Find the sequence that will open the right-most room.
solution:
according to my calculations, the following sequence will work: 1,3,4,6,5,3,2,1,3
Attachments
die_kluge said:
Nicely crafted, but as a puzzle not all that challenging -- until near the very end of the sequence you don't have any nontrivial choices. You can either return to where you came from, or flip the lever and go through the only open door. And returning to the last room is pointless, because you'll never be able to do anything other than go back or go forward again, retracing your footsteps in either direction.
Now if you changed it so that in the first room you had a choice of two open doors, then you could get a real maze out of it. One choice will ultimately lead to a dead end, the other won't but will make you feel like you're going around in circles. The solver, if on the right path, has to resist the temptation to give and go back to try the other path.
Tiefling
First Post
Zander said:
I got the same answer as everyone else, using the following algorithm:
Equation of Death - # of Killings to Go at the End - Highest Remaining # at the End - Calibration Number
(Calibration Number is 0 if the first remaining number in the next round will be killed, 1 if the second remaining number in the next round will be killed.)
1+n*2 - 499 - 1000 - 0
2+n*4 - 249 - 1000 - 0
4+n*8 - 124 - 1000 - 0
8+n*16 - 61 - 992 - 1
32+n*32 - 30 - 976 - 1
48+n*32 - 0 - 976 - N/A
In which n is and integer on [0,Previous Remaining Killings Minus One]. Each line can be arrived at by taking the previous line and performing some simple logic and arithmetic. Crude, I know, but it works and takes about 5 minutes.
(Calibration Number is 0 if the first remaining number in the next round will be killed, 1 if the second remaining number in the next round will be killed.)
1+n*2 - 499 - 1000 - 0
2+n*4 - 249 - 1000 - 0
4+n*8 - 124 - 1000 - 0
8+n*16 - 61 - 992 - 1
32+n*32 - 30 - 976 - 1
48+n*32 - 0 - 976 - N/A
In which n is and integer on [0,Previous Remaining Killings Minus One]. Each line can be arrived at by taking the previous line and performing some simple logic and arithmetic. Crude, I know, but it works and takes about 5 minutes.
Concerning the 52 card problem-
Take 13 cards from the deck to form the second pile, and flip them all over. If you subtract x turned cards from the first pile it has 13-x turned cards. Since the second pile had 13-x rightside-up cards, it now has 13-x turned cards. Equal.
Concerning the Evil Villain, Secret Numbers problem-
Poor villain. He managed to A) imprison two lightning-fast number theorists and B) pick exactly the wrong pair of numbers.
This problem requires that a few ground rules be laid out to determine the validity of a pair of numbers. But first, let me define a few terms for this problem: a "sumoid" is any pair of numbers whose sum is a desired number. A "productoid" is any pair of numbers whose product is a desired number. A sumoid of a productoid is a sumoid of the number that the productoid pair adds to. A productoid of a sumoid is a productoid of the number that the sumoid pair multiplies to. Whew!
1. A and B cannot both be prime, for obvious reasons.
2. None of the sumoids of A can be both prime, as Person A could not be sure that it was a false sumoid and/or that Person B could not factor it.
3. [For every false productoid of B, there must be at least one sumoid of that productoid that is a pair of prime numbers. This cannot apply to the true productoid of B.] This allows Person B to eliminate those false productoids upon hearing that Person A knew he could not factor B, leaving only the true productoid, (X,Y).
4. Rule 3 does not apply to any of the false sumoids of A. That is, if you have a false sumoid (c,d) then either zero or two or more of its productoids lack any prime sumoids, but never exactly one. This allows Person A to locate the false sumoids of A when informed of Person B's success.
This could also be written as follows:
1. X and Y are both integers on [2,99]
2. X /= prime OR Y /= prime
3. X+Y = A
4. X*Y = B
5. A-q /= prime OR q /= prime where q is any integer on [2,A-2]
6. B/C = D where C is any integer on [2,B/2] EXCEPT X or Y
7. C+D-q = prime AND q = prime for at least one integer q on [2,C+D-2]
8. E(A-E) = F where E is any integer on [2,A-2] EXCEPT X or Y
9. F/G = H where G is any divisor of F on [2,F/2]
10.G+H-q = prime and q = prime for either ZERO or TWO OR MORE possible integers q on [2,G+H-2]
So the sequence goes like this:
1. Person A finds that none of the sumoids of A are both prime, meaning that Person B could not possibly factor B into its true productoid without more information.
2. Person B finds that one of B's productoids has a sum with no prime sumoids, while all the other productoids have at least one such sumoid.
3. Person A tells Person B that Person B could not have found the answer.
4. Person B deduces that only the productoid with a sum producing no prime sumoids could be correct, as all the others would lead to uncertainty in Person B. He announces his success.
5. Person A figures out what Person B deduced, and finds that it only works on the sumoid of A that correspond to with Rule 3, not those that correspond to Rule 4. He determines what the true sumoid is, and also announces his success.
The rules allow a program to be written for or a bored mathematician to be prodded into examining each of the 9604 possible combinations in turn until finding one that works. But I'm not about to try.
This problem requires that a few ground rules be laid out to determine the validity of a pair of numbers. But first, let me define a few terms for this problem: a "sumoid" is any pair of numbers whose sum is a desired number. A "productoid" is any pair of numbers whose product is a desired number. A sumoid of a productoid is a sumoid of the number that the productoid pair adds to. A productoid of a sumoid is a productoid of the number that the sumoid pair multiplies to. Whew!
1. A and B cannot both be prime, for obvious reasons.
2. None of the sumoids of A can be both prime, as Person A could not be sure that it was a false sumoid and/or that Person B could not factor it.
3. [For every false productoid of B, there must be at least one sumoid of that productoid that is a pair of prime numbers. This cannot apply to the true productoid of B.] This allows Person B to eliminate those false productoids upon hearing that Person A knew he could not factor B, leaving only the true productoid, (X,Y).
4. Rule 3 does not apply to any of the false sumoids of A. That is, if you have a false sumoid (c,d) then either zero or two or more of its productoids lack any prime sumoids, but never exactly one. This allows Person A to locate the false sumoids of A when informed of Person B's success.
This could also be written as follows:
1. X and Y are both integers on [2,99]
2. X /= prime OR Y /= prime
3. X+Y = A
4. X*Y = B
5. A-q /= prime OR q /= prime where q is any integer on [2,A-2]
6. B/C = D where C is any integer on [2,B/2] EXCEPT X or Y
7. C+D-q = prime AND q = prime for at least one integer q on [2,C+D-2]
8. E(A-E) = F where E is any integer on [2,A-2] EXCEPT X or Y
9. F/G = H where G is any divisor of F on [2,F/2]
10.G+H-q = prime and q = prime for either ZERO or TWO OR MORE possible integers q on [2,G+H-2]
So the sequence goes like this:
1. Person A finds that none of the sumoids of A are both prime, meaning that Person B could not possibly factor B into its true productoid without more information.
2. Person B finds that one of B's productoids has a sum with no prime sumoids, while all the other productoids have at least one such sumoid.
3. Person A tells Person B that Person B could not have found the answer.
4. Person B deduces that only the productoid with a sum producing no prime sumoids could be correct, as all the others would lead to uncertainty in Person B. He announces his success.
5. Person A figures out what Person B deduced, and finds that it only works on the sumoid of A that correspond to with Rule 3, not those that correspond to Rule 4. He determines what the true sumoid is, and also announces his success.
The rules allow a program to be written for or a bored mathematician to be prodded into examining each of the 9604 possible combinations in turn until finding one that works. But I'm not about to try.
Last edited:
der_kluge
Adventurer
orsal said:
Yes, that is true. So, if lever 1 opened both doors B, *and* A, that would be a more interesting challenge. I'll have to play with it to see what kind of nastiness I can create if I do that.
I know one thing, my players are gonna hate this when I spring it on them!
Pielorinho
Iron Fist of Pelor
Corran, I do hope you're right, because your reasoning (minus the computer work) were as far as I was able to take the problem over the weekend. I'm not a programmer, so I couldn't figure out how to throw a CPU at the problem; short of figuring out every number P, I was going to have to throw up my hands.Iron Sheep said:
I'll make one minor adjustment to the way you stated property P: a number with property P is neither the sum of two primes NOR the sum of a prime and its square. Twelve, for example, is the sum of three and nine, and is thereby disqualified; if A had been given twelve, then B could have been given 27, and 27 may only be factored into 3 and 9. Even though nine isn't a prime number, there's no other two integers greater than 1 that can be multiplied to equal 27.
This is a very minor adjustment, for two reasons. First, there's only four numbers that are thereby disallowed: 6 (2 and 4), 12 (3 and 9), 30 (5 and 25), and 56 (7 and 49). Second, all these pairs can be disqualified by other means: 6 is disqualified by 3 and 3 (assuming X and Y can be the same number), 12 by 5 and 7, 30 by 13 and 17, and 56 by 43 and 13.
Still, it's an interesting quirk to the problem.
Daniel
Pielorinho
Iron Fist of Pelor
This problem as stated has too many elegant answers. Add in the clause about being wet, and there's at least two answers, one more elegant than the other. The elegant one has been stated; the less elegant one would be more elegant except for the phrase "on the nightstand":Wippit Guud said:
The man was sleeping in a life raft; his companion in the raft, driven insane by weeks at sea, punctured the raft with the scissors. The bed he lies in is the ocean bed. As for the nightstand, well, you gotta carry something with you when you abandon ship!
Daniel
Tilla the Hun (work)
First Post
Right...
I'll jump in with an old riddle (but still quite good) or two, then a comment on the levers question from Merak
Riddle#1: Picture a bedroom with two beds, two dressers, two of everything, even two young girls fish bowls. Speaking of young girls, Sally and Jane lay dead on the beds. Both died of drowning, yet there are no marks of struggle, and the beds are quite dry. How did they die??
Riddle#2: John was found hanging 5 feet off the floor from a rope tied around the rafters of a 20 foot high ceiling in a concrete floored warehouse. There's a large spreading pool of water below him. He's quite dead, and there's quite literally nothing else in the warehouse. How did he die?
Like I said - oldies, but goodies
Oh yeah - one more:
Riddle #3: Eddie and Aaron are dead in a cabin in the woods. It's over 500 miles to the nearest living being, and no on else was here when they died. If they didn't kill each other and didn't die from natural causes, nor other animals, what killed them?
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.)
Stand back and wait for the howls of anguish as they -slowly- work their way through it.
Alternatively, try this one on for size:
6 levers on the wall, six rooms. Each room is shaped like a D6, with round, open holes connecting to the next room. Each wall contains the same number of holes as a corresponding D6. Each lever on the wall connects to a single room and rotates the room in 2 dimensions. The goal is to find a configuration wherein the holes in the walls line up with holes in the wall of the next room...
Just don't tell them it's a d6 room Erase holes on the ceiling and floor to hide it better. Also, you can add 3 positions to each lever - the first position rotates the room to the right, the second to the left, and the third switches the roof with the floor (an engineering marvel to be sure)
For more fun with either variation, included teleportation traps that are moved by the levers, that teleport the players back to start. Make one such trap be a total 'reset' of all the doors/rooms.
I'll jump in with an old riddle (but still quite good) or two, then a comment on the levers question from Merak
Riddle#1: Picture a bedroom with two beds, two dressers, two of everything, even two young girls fish bowls. Speaking of young girls, Sally and Jane lay dead on the beds. Both died of drowning, yet there are no marks of struggle, and the beds are quite dry. How did they die??
Riddle#2: John was found hanging 5 feet off the floor from a rope tied around the rafters of a 20 foot high ceiling in a concrete floored warehouse. There's a large spreading pool of water below him. He's quite dead, and there's quite literally nothing else in the warehouse. How did he die?
Like I said - oldies, but goodies
Oh yeah - one more:Riddle #3: Eddie and Aaron are dead in a cabin in the woods. It's over 500 miles to the nearest living being, and no on else was here when they died. If they didn't kill each other and didn't die from natural causes, nor other animals, what killed them?
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.)
Stand back and wait for the howls of anguish as they -slowly- work their way through it.
Alternatively, try this one on for size:
6 levers on the wall, six rooms. Each room is shaped like a D6, with round, open holes connecting to the next room. Each wall contains the same number of holes as a corresponding D6. Each lever on the wall connects to a single room and rotates the room in 2 dimensions. The goal is to find a configuration wherein the holes in the walls line up with holes in the wall of the next room...
Just don't tell them it's a d6 room
Erase holes on the ceiling and floor to hide it better. Also, you can add 3 positions to each lever - the first position rotates the room to the right, the second to the left, and the third switches the roof with the floor (an engineering marvel to be sure)For more fun with either variation, included teleportation traps that are moved by the levers, that teleport the players back to start. Make one such trap be a total 'reset' of all the doors/rooms.
Riddle #1:
Riddle #2:
Riddle #3:
I’m pretty sure of the first two; the third’s a guess.
Johnathan
Sally and Jane are goldfish, who flopped out of their respective bowls and onto the beds, where they drowned in the air.
Riddle #2:
He stood upon a block of ice to hang himself. The block of ice has since melted.
Riddle #3:
Umm, Eddie and Aaron are potted plants that dies when nobody watered them?
I’m pretty sure of the first two; the third’s a guess.
Johnathan
Pielorinho
Iron Fist of Pelor
Richards, you have the same guess for the first one as I did, and the second one I'd heard before. The third one I've heard told differently; while I think the version I've heard is harder, I'll put it in spoilers in case it has a hint:
Daniel
Frank and Tina are hiking in teh woods when they come to a clearing. There's a cabin ahead of them in the clearing. They've never seen the cabin before, and it's not on fire, and they can't see anyone inside of it, and there's nothing moving near it, and they weren't expecting to see it, but Tina immediately says, "You stay here, I'll run and get help." How come?
Daniel
Similar Threads
