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 THE SECRET CHAMBER LOGIC - not adequate for real life situat Goto page 1, 2  Next
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Quailman
His Postmajesty

 Posted: Mon Feb 12, 2001 10:56 pm    Post subject: 1
mathgrant
A very tilted cell member

 Posted: Mon Feb 12, 2001 11:32 pm    Post subject: 2 OK, here's what I think. Reach your hands into any two opposite holes and make sure they are the same direction, let's say down. Take both hands out and let it spin. If the chamber opens, you're finished. Otherwise, reach into any two opposite holes and push them down if they aren't already. If they are, you're reaching into the same holes as before. Repeat until you manage to find the other two holes and the chamber will open. There's probably a better solution, but that's what I think. ------------------ [This message is being edited by mathgrant (being edited right now).]
Zephyr
Daedalian Member

 Posted: Tue Feb 13, 2001 12:19 am    Post subject: 3 I think it should take no more than 4 tries. Try 1: Stick your hand in adjacent holes and flip the switches both up and both down. If the chamber opens, we are done. If the chamber doesn't flip one up and flip one down and remove your hands to randomize. Try 2: Since the chamber did not open in the first try, the other switches must be 1 up, 1 down so the configuration must be either U D U D or U D D U So stick your hands in OPPOSITE holes and if both switches are the same, flip them and we're done. Otherwise: Try 3: We know the configuration must be UDDU so stick your hands in ADJACENT holes. If the switches are the same, flip them and we're done. otherwise, flip them and: Try 4: Since the configuration must now be UDUD, stick your hands in OPPOSITE holes and flip them both. Anyone see any faster method?
borschevsky
Chessnut

 Posted: Tue Feb 13, 2001 12:38 am    Post subject: 4 Similar to Zephyr's method: Try 1: Stick your hands in opposite holes, try both switches up and both switches down. If the chamber opens, we're done. If not, the other 2 switches are 1 up and 1 down. Switch one of your switches up, the other down. The configuration is now UDDU for sure, so go to Zephyr's step 3. We'll get it open in at most 3 tries.
babydoc
Guest

 Posted: Tue Feb 13, 2001 2:32 pm    Post subject: 5 maybe i'm missing something, but i think everyone's assumption is that after the column spins you wouldn't be able to tell which of the holes you had just worked on. but considering that there is writing on the column (which the puzzle said you could read) why not just remember which holes you just worked on by remembering the words above or below it? if this was the case you could finish in just 2 moves.... (i looked back at the picture and it looks as if the writing and the holes would all spin together - i.e. the column is one piece)
mwf
Daedalian Member

 Posted: Tue Feb 13, 2001 10:26 pm    Post subject: 6 Does anyone have a piece of chalk or marking pen or something I can mark the hole with? Then again that would take all the fun out of it.
Ghost Post
Icarian Member

MAT
Icarian Member

 Posted: Wed Feb 14, 2001 2:08 am    Post subject: 8 --Try One-- Take a marker and mark the spot where you put your hands in then put your hands in those holes and fip the switches down. --Try Two-- Then find the two holes you didn't mark and put you hands in and flip the two switches down. And boom you're done. --Mat-- --The speed of the leader, determines the rate of the pack.-- [This message has been edited by Mat (edited 02-14-2001).]
MAT
Icarian Member

 Posted: Wed Feb 14, 2001 7:03 am    Post subject: 9 Ok... Im serious now Here's what you need to do, there's writing on the column right? Yes. So use the writing as a reference point. This is pluasible because the writing IS mentioned in the story, so it IS a variable. I don't know if someone already mentioned this, I don't always read all the posts. Sorry if I copied you. --Mat-- --Risk always involves progress, you cannot steal second and keep your foot on first.-- [This message has been edited by Mat (edited 02-14-2001).] [This message has been edited by Mat (edited 02-14-2001).]
MAT
Icarian Member

 Posted: Wed Feb 14, 2001 7:17 am    Post subject: 10 The Writing on the Column... See, there is writing both above and below the holes so you CAN use the writing as reference points. --Mat-- --Genius is the ability to reduce the complicated to the simple.-- [This message has been edited by Mat (edited 02-14-2001).]
Dave from Ireland
Guest

Qball
In the Quorner Pocket

 Posted: Wed Feb 14, 2001 1:24 pm    Post subject: 12 I'm going to have to agree with mathgrant on this one. The chamber will open when all the switches are in the same position. Depending on your guess on where to put your hands for the second spin, the chamber could open as soon as the first try, or if you're very unlucky, you could be there all day. But I think the opposite hand method is definitely the way to go. WTG mathgrant. ------------------ This sig available in audio cassette
mwf
Daedalian Member

 Posted: Wed Feb 14, 2001 1:36 pm    Post subject: 13 1) Put hands opposite holes and feel the switch if they are the same goto step 3 and remember wether they are up or down. 2) Since the are not the same push the switchs in the same dirrection. If this is your second time at step 2 the set the switch the same way as the first time. Then remove hands quickly. if the door did not open go back to step 1. 3) Stick your hand in the opposite holes not used and set the switch to the same setting they were in step 1 and the door should open.
Evil Empire
Soopy's Favourite

 Posted: Wed Feb 14, 2001 5:10 pm    Post subject: 14 I think writing on the column defeats the purpose of the puzzle. Why not just have you friend(the one with the marker) put his hands in the holes so all four switches can be pushed down at the same time? Done in one move, 2 with a marker and no friend. Neither of these answers if very sporty. It looks like borschevsky solved this with a minumim of 3 moves (Zephyr gets an assist)
Ghost Post
Icarian Member

 Posted: Wed Feb 14, 2001 5:40 pm    Post subject: 15 a thought: if you look at the column from above, the holes are (lets say) on the N(orth), E(ast), S(outh) and W(est). so, first you ensure that W and S switches are in DOWN position. then, you check W and E and make sure that both of them are DOWN. now you have 3 switches out of 4 DOWN (worst case scenario). all it takes now is to find the fourth switch. how? as fermat had put it, not enough space here... S
mwf
Daedalian Member

 Posted: Wed Feb 14, 2001 6:05 pm    Post subject: 16 All 4 of the switch can not be in the same position or the door would be open. So 1 or 2 of the switchs must be out of place. So feel what the position of each switch is. If 1 is different that the other 3 then move it to the position of the other 3. If you have 2 up and 2 down. Then pick 2 that are the same and move them to the other position. Door should open on the first try.
mathgrant
A very tilted cell member

 Posted: Wed Feb 14, 2001 7:21 pm    Post subject: 17 Gee, thanks QBall. But that was the first opinion anyone had. After I read the other, I started thinking that borschevsky is right. His method takes no longer than 3 step, while mine could take 1,000,000,000,000 steps. Also consider the expected number of steps. Let's modify my method so that the first step is similar to Zephyr's first step. Then there's 1/2 a chance that it will be done in one try. Otherwise there's 1/4 a chance that it will be done in two tries, 1/8 a chance that it will be done in three tries, . . . 1/2^n chance that it will be done in n tries, . . . . So the expected number of tries is: code:``` 1*1/2 + 2*1/4 + 3*1/8 + . . . +n*1/2^n+ . . . = (1/2 + 1/4 + 1/8 + . . .) +(1/4 + 1/8 + 1/16 + . . .) +(1/8 + 1/16 + 1/32 + . . .) + . . . =1 + 1/2 + 1/4 + 1/8 + 1/16 + . . . =2 ``` But with borschevsky's answer, the expected number of tries is 1/2+2/4+3/4=1 3/4. borschevsky's answer is much better than mine. [edited to add emphasis] [This message has been edited by mathgrant (edited 02-14-2001).]
Quailman
His Postmajesty

 Posted: Wed Feb 14, 2001 7:43 pm    Post subject: 18 On the first try, won't his likelihood of opening the chamber be 3/7 rather than 1/2? There are 14 possible configurations, of which 2 involve alternately UDUD so he will get in for sure, and 4 are arranged UUDD, so he won't get in for sure. In the other 8 cases there is only one U or only one D, and he has a 50% chance of finding it and getting in. The other two potential configurations UUUU and DDDD are not possible because the door would be open.
mwf
Daedalian Member

 Posted: Wed Feb 14, 2001 8:12 pm    Post subject: 19 If you some how tell if what a switch is set to when you stick you hand in. Then it is easy to get the answer right on the first try, since no more than 2 switch can be out of place at any one time. If there is no way of telling how the switch are set (they return to a nuetral position after you press them). Then using opposite holes and pressing the switch up (or down if you like) after every spin is the only the best thing to do. It will still require a little luck, but it is your best odds.
Evil Empire
Soopy's Favourite

 Posted: Wed Feb 14, 2001 8:27 pm    Post subject: 20 mwf All the information needed is given in the puzzle. You can tell what position the switches are in and there is no nuetral position. The problem is once you remove ONE hand from a hole the column spins. If you leave your other hand in a hole you lose it. Granted, your hand would make for a great marker but you are still screwed because you now only have one hand to work with. [This message has been edited by Evil Empire (edited 02-14-2001).]
mwf
Daedalian Member

 Posted: Wed Feb 14, 2001 9:03 pm    Post subject: 21 Your right. I will have to take up some reading classes again. Since removing hand triggers rotation. Then there is no way of knowing what hole is which after rotion. So it is keep trying opposite holes. Darn!!! Why is it that when you bring 50+ porters along, no one remember a pen, pencel or any to write with.
borschevsky
Chessnut

 Posted: Wed Feb 14, 2001 9:59 pm    Post subject: 22 Quailman is right about the odds. There are 14 possible configurations to start. There are 2 cases of UDUD, they take 1 step for sure. = 1/7. There are 4 cases of UUDD, they take 2 steps half the time and 3 steps half the time. = 2/7*(2*1/2 + 3*1/2) = 5/7. There are 8 other cases, they take 1 step half the time, 2 steps 1/4 of the time and 3 steps 1/4 of the time. = 4/7*(1*1/2 + 2*1/4 + 3*1/4) = 1 So the expected number of steps is 1 + 5/7 + 1/7 = 13/7.
Alfie
Bovine Member

 Posted: Wed Feb 14, 2001 11:43 pm    Post subject: 23 Ok, why put your hands in at all? Use a coat hanger or the like. You could even use four of them so you didn't have to take them out. Or you could train a monkey to reach into holes and turn switches down. He could do two and you two.
Zephyr
Daedalian Member

 Posted: Thu Feb 15, 2001 2:52 am    Post subject: 24 I don't know whether I'd trust a monkey not to take his hands out before I take out mine .
Evil Empire
Soopy's Favourite

 Posted: Thu Feb 15, 2001 3:54 pm    Post subject: 25 Instead of spending all that time training a monkey you could just have your friend Hank turn 2 of the switches.
Greymorn
Guest

 Posted: Thu Feb 15, 2001 8:55 pm    Post subject: 26 If you want to make the puzzle more iron clad, have a stationary pillar with the switches spinning inside and out of sight. (A cylander within a cylander.) Marking the pillar becomes useless. To remove any possability of of help from a friend, a trained monkey, coathanger, etc. put covers over the holes, press a button under the hole to open the cover, and coordinate the buttons so that only two covers can be open at once. Just a thought.
Ghost Post
Icarian Member

 Posted: Thu Feb 15, 2001 9:25 pm    Post subject: 27 I mistakenly made a new topic with this post, so this is a repost in the correct place. Sorry. I am assuming I can flip a switch only once per try, I have to place both hands in at the same time, and nothing happens until I remove my hands. 1. Try opposites: If same, flip both down (up). The chamber will open if the other pair are both down (up). The chamber will not open if the other pair do not match. (I will assume for the remaining steps that if the switches were the same on the first try, they were up, and I switched them down.) If different, flip up to down. The chamber will open if the unknown pair are both down. The chamber will not open if the other pair are both up or do not match. 2. Try opposites: If different or both up (must be unknown pair from step 1), flip so both are down and the chamber will open. If both down (must be original pair again), flip both up. The chamber will open if the unknown pair were both up (original pair mis-matched). The chamber will not open if the unknown pair were mis-matched, and the resulting pattern must be UUUD. 3. Try opposites: If different, flip down switch to up and the chamber will open. If both up (original pair yet again), flip one down, and the resulting pattern must be UUDD. 4. Try adjacents: If they match, flip both switches, and the chamber will open. If they do not match, flip both switches, and the resulting pattern must be UDUD. 5. Try opposites: flip both switches and the chamber will open. [This message has been edited by Rob (edited 02-21-2001).]
Kmeson
Icarian Member

 Posted: Thu Feb 15, 2001 9:41 pm    Post subject: 28 I agree with the 3 step solution of borschevsky. However, on my first reading of the puzzle I assumed that either flipping a switch, or removing hands caused the column to spin. This makes step 1 of Zephyr/borschevsky unfeasable and the problem more interesting: I have a 5 step solution, but haven't tried too hard to minimize. There is an additional challenge at the bottom for those interested. There are 4 possible initial states, S1-4, that differ from the solved state, S0: S0 : 1111 0000 S1 : 1000 (+ cyclical permutations) S2 : 1010 S3 : 1100 S4 : 1110 Now let's state our goal slightly differently: Let S_i be the set of possible states after i flips. Let F_i be the operation done on the ith flip. S_0 = {S1 S2 S3 S4} S_i --F_i--> S_(i+1) we want to find a set of maps F = {F_1, F_2, .. , F_i} s.t. S_i = {S0} (note that S0 --F_i--> S0 becomes part of the problem definition) ---- What is the space of operations, F_i? First, one can either pick adjactent or opposite holes. Next one can switch based on the current state of the switches switches 00 01 10 11 newstate 00 01 10 11 For all possible F_i there exists a 9 bit number, n_i which defines its operation. bit 1 is on for adjacent, off for opposite, bits 2-3 define the new state if the old state is 00, bits 4-5 for 01, bits 6-7 for 10 and bits 8-9 for 11. Now, my solution: F_1 = 0,00,00,00,00 S_1 = {S0, S1, S2} F_2 = 0,11,00,00,00 S_2 = {S0, S4} F_3 = 0,01,11,11,01 S_3 = {S0, S3} F_4 = 1,11,10,01,00 S_4 = {S0, S2} F_5 = 0,11,10,01,00 S_5 = {S0} In english: F_1 = pick opposite holes, flip switches down F_2 = pick opposite holes, if they are different flip switches down else flip both switches F_3 = pick opposite holes, if they are different flip switches on else flip one switch F_4 = pick adjacent holes, if they are different flip both switches else flip both switches F_5 = pick opposite holes (they will be the same), flip both switches ------------- Here is an interesting extention (no solution yet): Starting with all switches up, your gambling mathematician buddy offers a wager: He observes that if the switches were randomly positioned then S0+S2:S1:S2:S3:S4 = 2:2:2:2, and 3:1 odds would be a fair bet on any of these combinations. He offers you the opportunity perform N switch flips (telling him what you will flip depending on what you find but not what you find. He knows the F_i.) and then, allowing him to pick the state, wager 2:1 on the outcome (he pays a dollar if wrong, and gets 2 if right). For what N would you accept, and how would you randomize the column? (Let's assume for now that the secret chamber door is disabled so doesn't give away state S0).
Ghost Post
Icarian Member

 Posted: Fri Feb 16, 2001 12:01 am    Post subject: 29 How bout this? Reach into any two holes and flip the switchs down. Before removing your hands, wrap your legs around the pole. When you pull your hands out of the hole, hang on to the pole for dear life while it spins. When it stops spinning, you're still positioned in the same position you were before and you can reach in the other two holes and flip the switches down.
Ghost Post
Icarian Member

 Posted: Fri Feb 16, 2001 6:30 pm    Post subject: 30 Man, I don't know whats in this room, but it had better be worth it ------------------ You are not thinking. You are merely being logical. - Neils Bohr to Albert Einstein
waterdelph
Guest

 Posted: Sat Feb 17, 2001 3:19 am    Post subject: 31 How about inserting hands first in opposite holes and flipping both switches up, take them out (column spins), then insert hands in adjacent holes and flipping them up? After these two steps, 3 out of the four switches are definitely up. The fourth switch, however, is still a guessing game that can take any number of tries to find. Any ideas?
Ghost Post
Icarian Member

 Posted: Sun Feb 18, 2001 1:37 am    Post subject: 32 How's this? http://www.seankell.com/puzzle/puzzle.htm I've manage to it down to no more than two spins... [This message has been edited by sfkell (edited 02-19-2001).]
art begotti
Guest

 Posted: Sun Feb 18, 2001 3:05 pm    Post subject: 33 i was thinking that the idea of wrapping the legs around the pole was a good idea, but then i read the puzzle again. he said that there has to be a FINITE number of moves. i figured it must be only one move. if you had it all figured out in a definite two moves, there may be a chance that you may "accidentally" get it on the first try. and you can't just say " if A doesn't work than try B" Keyword: FINITE. so my solution is this. take your flashlight and shine it down into the holes AT AN ANGLE. therefore you will either see the switch with your light at a direct angle, or, by again changing the position of the flashlight, by reflection off of one of the inner walls. you can then see the position of all of the walls. so there are three possibilities. 1. U U U D, 2. U U D D, or 3. U D D D. so if it is case one simply switch the one up. if its is case two switch either the two ups or two downs. if its case three switch the one down. and the chamber should open!
art begotti
Guest

 Posted: Sun Feb 18, 2001 3:07 pm    Post subject: 34 correction: see the position of the switches in P2
Captain Psquire
Guest

Icarus
Guest

borschevsky
Chessnut

 Posted: Wed Feb 21, 2001 2:01 am    Post subject: 37 But it's not the same as having a bag with 4 balls in it, because the switches are in a certain order. If you use two opposite switches or two adjacent switches, you gain information.
Acer
Guest

 Posted: Wed Feb 21, 2001 5:50 am    Post subject: 38 Well, I am new here, but I think the answer to the riddle is to mark the first two holes flip the switches up or down which ever, then let the collum spin, and which ever two holes are not marked is the two you switch up or down to open the door. what do you think??
Ghost Post
Icarian Member

 Posted: Wed Feb 21, 2001 10:18 pm    Post subject: 39 My solution assumes that (1) you can't tell which hole is which after the column spins, and (2) as soon as all four switches are aligned, either up or down, the door opens. The problem can be solved in a maximum of four tries, fewer if you're lucky. First Try: Put your hands in two holes opposite each other. If they are in the same position, flip them both. If the door opens, you're done. Otherwise, leave both switches in the down position and go to Second Try. If they are in opposite positions, flip one so they are the same. If the door opens, you're done. Otherwise, flip them both. If the door opens, you're done, Otherwise, leave both switches in the down position and go to Second Try. Second Try: Since the door didn't open on the first try, you now must have three switches down (the two you tried in First Try plus one of the others) and one switch up. Let's call the four holes 1, 2, 3, and 4, and let's assume that 1 is up and 2, 3, and 4 are down. Put your hands in two adjacent holes. If one of the switches is up, switch it to down and you're done. Otherwise, you've either put your hands in holes 2 and 3 or in holes 3 and 4. Flip the one with your left hand up and proceed to Third Try. Third Try: In Second Try, you flip either switch 2 or switch 3 up; switch 1 was already up, and switch 4 is still down. So now the configuration is either 1 and 2 up, 3 and 4 down, or 1 and 3 up, 2 and 4 down. Now put your hands in two holes opposite each other. If the switches are in the same position, switch them both and you are done. Otherwise, you know that the configuration is 1 and 2 up, 3 and 4 down, but you don't know whether your hands are in 1 and 3 or 2 and 4. It doesn't matter. Just switch both switches. Now the switches opposite each other are in the same position. Proceed to Fourth Try. Fourth Try: Put your hands in two holes opposite each other. They will both be in the same position. Switch them both and you are done.
Ghost Post
Icarian Member

 Posted: Thu Feb 22, 2001 6:47 am    Post subject: 40 I have to agree with NoCloset's logic but there there is one flaw in step one. It was previously mentioned that when one switch was flipped, the post would turn even if your hands weren't removed. Therefore there is no flipping the switch one way and if the door doesnt open flip it the other way. Instead, if both swithes are the same flip them both and if the door doesn't open then go to step two. However if they are different, it calls for an extra step. Flip them so they are both down for instance. If the door doesn't open then you either have two down and two up that are opposite each other, or you have three down and one up. To find out which is the case, now put your hands in two adjacent holes and flip both down if they arent' already. IN this case either the door will open or you will know for sure that three switches are down and you can proceed to step two. Therefore I believe that the smallest finite number of times is 5 versus 4. I give NoCloset credit for this as it is based entirely on his conjecture.
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