Solving The "Impossible" Escape Riddle
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- Опубліковано 9 сер 2023
- Thanks to Rahul for the suggestion! A hospital has 16 rooms in a 4x4 grid. The rooms are numbered row by row and rooms and connected vertically and horizontally. Each room has exactly one patient. The patient in room 1 wants to meet all patients and then exit from room 16, but cannot meet another patient twice. How can it be done?
(Repost due to small mistake in original proof - the nxn grid needs to be even n)
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So no one else had the idea to...um....terminate the patients in each room so you don't meet them again? Just me?
Same. Not sure what that says about us that "m*rder everyone we meet" was our first thought 😂
Nope
@@KyleRDentThat youre both chaotic neutral, clearly
@@ohitsrusher842they are full chaotic evil
@@ohitsrusher842it’s a hospital. Perhaps some patients could be giving birth, so by spawnkilling, you would be chaotic evil
There are other "outside the box" solutions to a puzzle like this. Besides, say, having Patient 12 walk into Room 16 and visiting both patients at once, you can also note that a hospital laid out like this is a fire hazard. Simply pull the fire alarm, and you can meet all the patients at once after everybody files out through room 16.
your second solution is awesome lol
i had the idea of ending on room 12 after visiting everyone, then have 5 move to 1, 9 move to 5, etc until room 16 is empty, then leave. i had the thought that there could only be one patient per room at a time, which is why i came up with the solution that i did. wonder if adding the “other patients can’t move” clause brings you to the intended solution easier…
@tremen151 technically with that strategy, you can go to room 16 and exit while never meeting any patients
@@Kyle-nm1kh You have to visit every patient exactly once and zero is not equal to one.
Well played 😊
@@egwenealvereiscool7726 I know I was just stating that it can be done, not that it was a solution
So this isn’t just a math puzzle, but a word puzzle.
Math problems can be described either with words or numbers/symbols. But even numbers/symbols are words when you pronounce them. So actually every math puzzle is a word puzzle.
The patient in room 13 is probably a terminal case anyway, so if you just wait until they kick the bucket you don't need to go through that room.
Huh, 'cuz 13 is an unlucky number. I get it. o3o
No need to wait, a hospital is a great place to find potentially lethal drugs.
Your solution is not valid because you HAVE to visit all patients.
That hospital can't possibly not be in violation of the fire code.
Plus, in the US there's HIPAA to consider.
Privacy for the patient in room 16 must be nonexistent too…
its so hard for patient 1 to exit too
Another solution is just to open the door to another room and greet the next patient, but you haven't left the current patient yet. This is the problem with converting mathematical pattern problems into "humans do this" puzzles - there's all sorts of ways to solve the problems with a bit of lateral thinking.
lol
Or how about the fact that you can't "meet" a patient more than once, because the 2nd time you wouldn't be strangers
@@Kyle-nm1khlol, you can use the word "meet" even with acquaintances. You can say "i meet my friends in the park this afternoon" ;-)
@@maximepaccalet4449 which still would be only once. If you go to the park and meet with your friends, then go for a walk and run into them again, it's still only a total of once
@@acidbong204all it says is that patient 1 wants to “meet” them, not visit them
In addition to all the other "outside the box" solutions, you could literally leave the box by exiting through one of the windows then back in through another.
if there ARE any windows
You might as well teleport through the walls
There are no windows
It s the 50th floor
@@nicrevenif there are no windows, then make one
Just have the patient in each room come with you the rest of the way after you visit them. Then you’ll see them the whole trip, but only once.
but they're sick and can't walk much
@@Macieks300At least one of them should be able to walk, that's more than enough
maybe the real friends were the patients we met along the way
@@Macieks300murder ought to do it
Escape from a nuthouse. That is what I was thinking about this question.
I actually concluded that it was impossible without visiting some room twice just by brute force lol, and then the solution appeared immediately and painfully obvious.
Same, haha.
Same. Once I decided it was impossible, the solution was staring at me
I was about to declare it impossible before I payed attention to the wording. That’s when I solved it.
Use a time machine to make the whole thing never have happened and then technically you haven't visited any of them.
But in the mean time you know all their personal details.
The problem at 0:32 asks you to find a solution without seen a patient twice, not visiting a room twice. You start by visiting patient in room 5 or room 2 and then you go back to room 1, depending on wich one you visited you take the other path and its easily done.
Or you see each patient once then when passing through you just don't look at them or blind them with mace or make them go inside of a box or behind a large room divider. Simple! Another way is to make all of them line up at the 2nd room to see you, then all leave into the 1st room when finished, then walk through all the other rooms towards your way out... Then don't forget have all of them leave their wallets in the last room to exit to cover the costs of hooking up all the gear necessary before leaving to start after you leave, and to buy snacks and drinks to come back and watch your own squidgames take place with those still in the building. 🤗
That's the solution I also came up with, as there is no patient in room 1.
Best solution@@mizum3458
Room 1 is also not "another" room - it is the starting room, so even if it uses the word "rooms", visiting room 1 twice is allowed.
refreshing to see someone that watched the video before commenting
This riddle gives me Ted-ED vibes with their virus riddle. Figured it out before playing the video.
its the same riddle lol
It's the same riddle with slight variation
lmao
yeah its the exact same riddle lol, its just different scenario
Same one lol
Well, I know there’s no way to end on 16 without doubling up on a room since you have to follow basically a line and there’s an even number of rooms. But since the rule is “cannot meet another patient” and you are patient 1, your room can be travelled in twice. You are able to exit you room and immediately come back since you have another door to leave from. But if you take more than once step before coming back, you’ll have to enter again from the only other exit. So, the solution must be to step out then back in, reducing the number of rooms to odd and from there it’s easy to get the rest.
Patient 1 is a very caring person :)
It was a murder spree…
Although it wasn't advised him to do that.
The uncaring version of patient 1 calls all to gather in room 16. Meets them all at once in the same room, and then exits. The less uncaring version just calls the patient in the neighbouring room to room 16 to meet at the door (without crossing over to their room) before exiting.
He injured all others 😂
Bro just wants to meet his mom :(
I was patient 10. It was a pleasure to meet you sir!
oh i was patient 12, i didnt meet you though
Well i was patient 7 he even got me chocolate😊
was patient 16, 1 just dashed through me so they could exit
I was patient 1, i did indeed give 7 a chocolate and ran straight for the exit, ignoring 16, now i am a free man which uhh, wait, why did i escape the hospital again?
I wanted to enter the hospital but they told me there were already 16, I died
Another plan:
1. Raid the room with the pharmaceuticals
2. Every time the patient visits a room, sedate the patient inside
3. Revisit rooms as often as you want, since the patient inside is unconscious and you're not "meeting" them anymore.
If you set the fire alarm off, all patients will be forced to leave the hospital and gather in the designated safe waiting place, you can then exit after them and meet them all at once, for added realism you can set your hair on fire just prior to exiting the building and joining the group.
Never realized it was THAT sort of hospital...
I doubt there's a working fire alarm. Considering how they designed the place, fire safety doesn't seem to be a consideration.
The real mystery to me is "If some of the patient rooms have 4 doors, wouldn't their bed block one of the doors🤔"
So my first thought is that it's not possible to go through every room exactly once, as both the start and end squares would be the same color if it were a checkerboard. Only moving cardinally, each move takes you from a black square to a white one and vice versa, but there would be 8 black squares to go through, and only 7 white ones, assuming we start on white. This means that a path through will always either miss at least one black square or end on one.
I may have misunderstood the question though
Edit: Yeah, I fell for the wording of it.
There are two solutions. Thanks for the exercise.
The solution: 1, 2, 1(back to his own room), 5, 6, 7, 3, 4, 8, 12, 11, 10, 9, 13, 14, 15, 16(exit)
There are 8 possible solutions.
@@TwistedSoul2002 yes my friend, I realised that later. Actually it took me 4 attempts using pen and paper. I am not very good with mazes and patterns. So when I solved it, I shared it out of excitement..😀
There are way more solutions.
You cannot SEE a patient twice. Just close your eyes if you go to a room where you’ve already seen a patient 😉
My solution was to have a patient in the next room come over. Nowhere does it say that the patients are bound to their own rooms, or that Patient 1 can meet only one other patient at a time.
Why stop there? Have all 15 other patients meet you at Room 1, then exit before having the patients return to their respective rooms.
That's mainly a wording issue. If you consider the word "meet" it's entirely possible, but some parts of the video use the word "visit", which implies that you're stepping in their room.
I have not watched the solution yet, but I just found out a way the technically answers the question. The patient can't visit ANOTHER patient twice, but can go back to room 1. So go to room 2, back to room 1, and now there are many easy ways to solve it.
i was thinking the same thing, nice one i guess.
Same for me.@@sqw0t
Reminds me of the Euler trail puzzle.
There can be many ways.
It says you have to visit each patient but doesn't say that you have to enter every room.
Pausing at the pause point in the video, here is my read.
The instructions on this one are different than what I have seen before - namely in this instance you are a patient, not the doctor. I originally heard this - one patient in each room, must visit every room exactly once & leave from room 16. In that case, I do not believe it is solvable. In this case you are a patient and must visit every patient exactly once, and exit through room 16. The obvious and seemingly unimportant difference is that you are a patient and not a doctor, but that does actually make a meaningful difference because it means there is one room you can reenter, your own room.
So you go into room 2 or 5, immediately turn around and go through your room into the other. From there, any reasonable path will be successful.
Now watching to see if this is the same solution
Edit: immediately after pressing play they explained the patient vs room thing haha. Nice video
Has this hospital never heard of hallways?
No privacy fr lol😂😂😂
My solution was a little different. As you visit each patient, move them to the room you were in just before. That also results on only visiting each patient once. Though doing so winds up with a sort of mirrored version of the solution where the back track is at the end instead of at the start.
That only works if 15 of the 16 rooms are occupied at any given time. There are 15 patients, but there's also you so it's 16/16. That means there's no room to shuffle around like that without coming into contact on a transfer or without double packing rooms.
@@Kyle-nm1khthe solution doesn’t have to be practical, especially considering the fact that the hospital apparently isn’t OSHA compliant (according to another comment)
As you visit each patient, you tell them to come along
so then when you revisit their room, you dont "visit the patient" in that room anymore cuz they're not there; they're with you.
This way you can freely move around any room
I love spatial puzzles. I paused a long time. The ultimate answer was very validating lol. Also makes sense now why it was worded the way it was instead of having a care provider rounding between the rooms, which would otherwise make a lot more sense XD
Another morbid solution: Just walk back and forth, then self terminate when all rooms are visited. The coroner will take the remains out via 16, and patient 1 will not be meeting anyone during that exit path.
Others have already discussed the opposite, terminating each patient as you pass through their room so you don't meet them again.
Not very moral solutions, but solutions nonetheless.
Oh man, I saw this one from the thumbnail and assumed it was a doctor trying to visit each room without repeatedly seeing a patient, that seemed impossible, it's so much easier when it's a patient just trying to visit other patients without repeating.
I originally figured you could just peek your head in the door to meet a patient, now I like the idea of bringing all the patients with you as you meet them so you never meet anyone twice.
With a (2n)^2 board you necessarily need (2n)^2 moves to get there from the diagonal. Which is impossible if you touch every single square once, you'll always have (2n)^2-1 moves. You need one more.
But you can still use your own room as many times you want. 1-5-1-2 or 1-2-1-5 and so on.
Easy! I guessed from thumbnail😬
is this like a perfect "social anxiety" example puzzle? I can imagine a doctor walking in on a patient they already has been to and is like "oh... yeah.. sorry... i'll just leave... you sure you good? no, ok, i'm.. seriously leaving this time... erm... yeah... just, like... bye... don't forget your medicine :) ... ok, bye"
There's actually a solution: and it's to start at a different spot. This also poses a problem with the chess explanation. The chess explanation assumes that you cannot move yourself through all other 15 spaces without touching another once. However, when you move starting at box 2, 3,5,6,9,12,14, or 15 the solution quickly becomes apparent. Not going to spend too much time pondering why that algorithm is; so that starting at a different spot works, but it works. If someone has a logical explanation please let me know. I'm very curious! :)
You can imagine the layout as a 4x4 chess board with alternating black and white cell. If 1 is white then 16 must also be white. Each time you move you alternate between black->white or white->black. After 16 moves (starting in a white cell) you have to be landing on a black cell, so if you start at 1 (white) you can never land on 16 (also white) after 16 moves without going back to some cell a second time.
@@nguyentanhao6545 that is true, but my solution proposes that we don't start on square 1. Within 16 moves you can solve the problem whilst touching all squares without touching the same twice. Now it really seems like the problem is very dependent on where your starting and ending point is
@@edwardpan57615 moves to reach all rooms. That's an odd number. That means you have to start in a square that can land on #16 in an odd number of moves. Room 1 cannot reach room 16 in an odd number of moves. If you go to 2, 6, 7, 11, 12, 16 - that is an even number of moves (6). Any path you take is even. So 1 is an invalid starting room unless you do the backpedal trick.
@@edwardpan576 If you start on square 4, it's a black square - and it satisfies the alternating pattern he's talking about correctly.
After having played Zelda Oracle of Ages a lot in my life, I can assure you that this puzzle reignites many traumas =)
A great puzzle and certainly doable. Thanks for the idea for the proof regarding paths in 2px2p grids
Since the question doesn't require the other patients stay still, you could also move them around. If Patient 5 visits room 1 first, then you can travel through all the other rooms without repeating a patient or a room.
Or, you can just every patient you do meet back to room 1, which frees up every room you've visited so you can pass through it as many times as you want.
It also didnt give you permission to move them out of the roomsss ;-;
If the difference between the available moves (total rooms-1) and amount of inner straight walls (note:doors don't separate walls; walls extend from one end of the layout to the other end) is odd, then the layout is unsolvable (when no room can be visited more than once).
Each inner wall (of which there are 6 in the 4x4 room example; 4-1 + 4-1) has to be crossed an odd number of times to end up on its other side. Let m be the amount of available moves and w the amount of inner walls. So each wall has to be crossed at least once, leaving us with m-w moves; that is, the difference I talked about in this comment's first sentence. The remaining moves have to be allocated in pairs to the walls to maintain which side you end up in, which is impossible if there are an odd amount of said moves left.
Patients: all go to room 1
Patient 1: goes to room 16
I totally forgot my graph theory knowledge from discrete mathematics😂
I think that you can write a computer program in which you assign initially a value of 0 to 15 to all 64 (16 rooms each have 4 possible values.) of the 2-dimensional array values which come from rooms belonging to 16 patients including yourself (Your room is Room 0.). You then link each of the patients to all the patients that are next door to that patient (This linkage would require the aforementioned 2-dimensional array. The maximum possible number is 4.). You then start with a for loop of say i = 1 to 4 and run a recursive subroutine that calls itself for each of the possible patients that you are next to. If there are less than 4 patients then you simply put a value of "-1" for those positions that are not able to be used. If the value of the patient is "-1" then you don't run that recursive subroutine. If the 2-dimensional array's value is anywhere from 1 to 15 then you set all the affected 2-dimensional array values involving the room that you just left equal to "-1" (You simply take the rooms that that room normally leads to and find which of those destination rooms' paths lead to that room and set those values equal to "-1". For example, if room 5 were to lead to rooms 8, 9, and 10 then you would set R(8, J), R(9, K) and R(10, L) all equal to "-1" with J, K, and L being the "doors" that lead to that room from rooms 8, 9, and 10 respectively.). You then build the sequence that you are on by adding a character that represents the room that you just entered to the string that has been accumulated so far as you are traversing the rooms. You then save any sequences that made it to the end and have all 15 patients mentioned in a text file to read later. If I were to run this program with your example then I would end up with no sequences. If I were to convert the program so that room "0" (your room) never has any paths to it set equal to "-1" then I would get the sequences that you mentioned.
Of course, you have to make sure when you do the converted version of the program that you don't allow the program to visit your room more than twice if you include starting there as a visit. Otherwise, the program will never be able to stop running since it would keep going back to your room. To prevent that from possibly happening, you do set those paths equal to "-1" the second time that your room is visited assuming that your starting there counts as a visit.
It's not impossible. The challenge stated that Patient 1 cannot meet another patient twice. Nowhere in the statement did it mention anything about the room. By that logic, Patient 1 can revisit Room 1 as they themselves are the Patient in Room 1.
Edit: totally didn't spend an hour thinking it all out
This is and escape puzzle wrapped up inside a riddle 😮
I usually try to solve these from the thumbnail. But that doesn't say the human is the patient in room 1, I assumed it was a doctor visiting all rooms, so I didn't figure it out.
Very nice task. So easy and nicely done
He comes back through his own room dosent he.
Nailed it. Now I feel like a cool boy.
Right when I heard that I was the patient in room 1, I immediately knew all 8 solutions.
Literally that one Ted-ed riddle way back with that virus thing
You need a lot of patients to solve something like this.
I saw the thumbnail ad took a good moment trying to figure it out but the second I heard the wording I got it immediately
The problem is impossible if you cannot visit the same room twice (forget the patients). It becomes possible if the number of lines or rows is odd.
Once you've met a patient, you've met them. Next time through the room: "good to see you again"
Very clever and satisfying resolution!
It was very satisfying when it clicked before the explanation.
theoretically, if we're going by the challenge rules? every patient could simply meet patient 1 in room 1. then congregate in room 2. to which patient 1 can simply walk in an L. exiting through room 16, having met every patient once.
with dozens of other permutations of that solution available.
I just sat here, though about it a bit, and found a way. Since just going down won't give you anything, you should find the other ways. If you don't go to 2 or 5 at the beginning, you won't ever get back to them. But then I thought I found the way(in fact I didn't notice how I stepped on the same point twice) and thought "so this is how we meet 15 of patients", but then "why 15?" and it became so obvious
カンタベリーパズルに似たようなものが書いてありました
Got it. :))
The first step was to prove (or intuit) that the obvious reading of the problem leads to a dead end. Then work out how to fiddle it by looking for some special case that is not entirely covered by the wording...
Imagine patients with high level of anxiety in the middle rooms: each room has 4 doors and from each door another patient might come in. Damn, that's some horror scenario
I was hoping that you could somehow turn this into a problem for Euler tours. It would be, if your task was to visit every ENTRANCE exactly once. But oh well, maybe in another video.
It took me a second from seeing the thumbnail, but when I heard that "you" were patient 1, I immediately figured it out.
I like both the first part and second part.
Cannot meet another patient twice. Room one is the lynchpin. The sequence, starting in room 1:
1, 2, 1, 5, 6, 7, 3, 4, 8, 12, 11, 10, 9, 13, 14, 15, 16
Didn’t get it, at first, but I think “impossible”, because we all get too hung up on the rooms, rather than the patients.
Examples like this are the reason I think the details are so important. How can we problem solve in real life if we don’t sheer our logic with specific examples to help us recognize the difficult stuff
I get so much disapproval from people who say they want to solve a problem or are angry about a misunderstanding and they want to leave it alone because there’s “no other way to see it” and I’m literally being stopped from helping them see it.
We’re human and we have emotional waves and our egos and are pride can be bruised, but dammit
Easy. Visit 14 of the cells, leaving the exit cell and one adjacent. Pop a ceiling panel and crawl through the overhead plenum to get back to the other two and leave.
The bad thing about this sort of puzzles is that it is primarily an exercise in semantic pedantry, not in actual problem solving.
knew i had seen this somewhere, found it on a ted ed video. ironically that was the only one i managed to figure out
When problem seems impossible look at the wording. Good one
i saw this concept on a TedEd video! so i solved the riddle without even watching the video yet! It's brilliant.
Solved it with the 3rd path!
Too easy.😀
Problem was nice but i managed to get it! Thx for these though. It turns my brain off after im watch youtube shorts all day.
I had a feeling that it required you to return to the starting room, but didn’t consider the idea of doing that right away.
Решил, достаточно быстро. Просто надо держать в уме, что в свою палату возвращаться можно.
I once saw Teded mention a riddle like this
Imagine being patient 16, everyone coming in and going out the hospital need to pass his/her room.
Nice puzzle!
Now I’m interested in working out how such parity rules could be applied to n dimensional matrixes.
Obviously for n=0 you start at the end as it’s a point. As for n=1, every cell may be traversed once for all starting and ending cells and for all lengths m. This video further solves n=2 of the 2-dimensional solution. But we still have the 3 dimensional matrixes unsolved as well as those of hyper-matrixes at this point.
I just came up with a puzzle with a twist. I am guessing this can be solved for a person inside 3 by 3 by 3 matrix such as a rubix cube and I am guessing you can go from various different starting positions to different ending positions. If I am correct the twist to the puzzle would be to then solve for if the rubix cube is in an initial unsolved state and at every time step. Ie: every time the person changes his position to a new cube either 1 or 2 faces of the rubix cube are rotated along an axis 90 degrees. The axis of rotation is determined by shortest or minimum path to solve the rubix cube given its random starting condition. The person in the cube must first determine if it is possible to traverse every cube within the rubix cube and if it is possible he must correctly visit every location in order while it is rotating.
Ted Ed posted the same video but the funny part is how differently they posed the problem.
Myd: feel free to pause the video and try to solve the problem
Meanwhile Ted Ed: if you don't solve the problem, the virus breaks out and everyone DIES
Sometimes you have to "take a step back" to pick the right path.
when 2nd smartest student in the class cheats
Go into room 5, then back into your own room (your not visiting anyone). Then you can go 2,6,7,3,4,8, then it's a clear shot to the end
I figured out you needed to visit the first room at least twice but didn't know how that could be used until like 10 seconds after MindYourDesiscions in videotime told you to stop the puzzle.
Problem solved:
Go 1-2-3-4 than 8-7-6-5 and by going on 9-13-14-10-11-15-16 you can just go to the door of number 12, say hello without entering the room and just exit at 16.
Job done...
Seen this on TedED.
But when did it state that you are the patient in room 1?
Okay, it does, right at the start. I completely missed that it was a patient going to the other rooms. In my mind it was a doctor or nurse doing the rounds, and there was a patient in the starting room they visited.
Clever little riddle, I'd say, more than a logic problem.
I couldn't solve this from just the thumbnail, but after hearing the full riddle, it was easy.
This is absolutely genius!!
Nice, actually solved this one. I did it by going into room 5 and doubling back.
The moment I heard that room 1 was empty (patient 1 wants to meet), it became really easy to solve. Literally the first thing I tried after that worked.
Play the Italian strategy to gas the room so that they don’t greet you
this is like that one TED riddle where the rooms you go in fill up with concrete.
I'm actually proud of myself. I did figure out the way to do so before the solution, entering only your own room twice
On the assumption that patient 1 has to go into the other patients respective room to 'meet' them and they can't move. My solution is starting at 1, go down to 5, back up to 1 and then you can do 2,3,4,8,7,6,10,9,13,14,15,11,12, and finally 16 and 'patient 1' will never "meet another patient twice"
Edit: I love it when I am actually am able to figure out the answers before I watch the video.
I know this one. Since the pov character IS the patient in room 1 he can still return there. So all one needs to do is go to one room, return to your own, and go through the rest of them.
Well, if you’re using wordplay to solve this riddle, you can argue that you can really only “meet” a person once. So you can take any path you want as long as you’ve met a particular patient already.
Other out of the box solutions:
-demand all other patients move (for example, all into room 2, then you can simply exit from 16)
-tear down the walls (floor and ceilong too if you want) for a literal out of the box solution
-unplug the life support of patients as you go, not technically visiting them again if they arnt alive...
-cure other patients as you visit them, they then arnt patients, and you can now visit their room as many times as needed
4D entities will definitely help us in this situation.
I actually had to deal with this while making minecart patterns for my farms
I was thinking just take every new patient you meet with you that way you can go wherever you want and never meet anybody twice
Took me 40 seconds. I had to check that not possible, then read the task again. And got it.