I think the part where it shows you a square room with a door in the middle of each room is misleading since it makes me think that they have to be like that. If I were to just listen to the riddle, I'd probably think that there could be two doors on one wall with rooms side by side rather than the diagram that was shown...
How else would you reduce the problem? It's really the only way. If you say that you HAVE to connect to a new room, then for each room, you have to add two new rooms and then for those rooms, add two new rooms and so on exponentially, out of control. The number of rooms can only be reduced if you share
Hmm. In the condition "Each room has exactly three doors to other rooms on that floor", the word other implies that it is referring to three distinct rooms. You could argue and say that the condition should have been "Each room has exactly three doors to three other rooms on that floor." If you want to be pedantic over the way they phrase things and find loopholes inside of them, then you do you. But don't feel good about it for "thinking outside the box" and finding a solution that doesn't require as much logical thinking.
Yeah. I thought they were squares that were connected. Not circles. But it did say that there weren't any hallways, which wouldn't make it only squares, but in the video, it showed them like that, with them having hallways
If anyone is wondering, the size and shapes of the rooms are completely up to you to decide, and there can be more than one door on each wall. ( the solution doesn't even use squares and rectangles as it suggested)
Same I watched it and I was so confused I thought it would logically be the 4th if not the 5th but it turns out I was wrong. I still haven't grasped what the narrator is trying to say
I really miss the old riddles, which were more logic-orientated, of the increasingly mathematical ones which are hard to explain the answer to in person
My answer was the 4th level from the top. Never do they make it a requirement for two rooms to have only one door connecting them, so my imagination went wild.
Then the answer would be the top level. Cause just connect the control room to itself and you have everything they required. Thus even with this argument your answer is incorrect.
Rooms are normally rectangular in shape like depicted at the beginning of this riddle. The riddle said there are no hallways, meaning the rooms must be right next to each other. The answer shows circular rooms connected to each other with hallways which does comply with the information given at the beginning.
Air-Grip Biker, I think we all know that was a graph representation. Having said that, the representation is not congruent with the information given. Katie Hough claims to have drawn out square rooms that meet the criteria. I would like to see this drawing...
Paul Rezaei u said "The answer shows circular rooms connected to each other with hallways" but thats an incorrect claim. im sry but u dont know what a graph repr is. it is an abstraction, it just shows nodes and arcs, u gotta have to associate their meaning to them. nodes are rooms, arc are not hallways but doors here. they dont say anything about the implementation of the room shapes.. so no, you didnt know what a graph representation was. youre welcome
+Air-Grip Biker I can see how what I wrote was confusing. I knew it would be misinterpreted but I didn't feel like writing a long drawn out explanation ha ha. Maybe I'm wrong. Please explain how the solution can work with normal looking rooms.
Even after I understood the constraints, I got the wrong answer (4) because they didn't specify that two rooms could be connected by at most one door. There's a multigraph solution on just four nodes.
Your solution is wrong: You can have a solution with only 4 rooms. Note that the conditions don't say "Each room has doors to _exactly 3 other rooms_", but "Each room has _exactly 3 doors_ to other rooms". It's not said that the other rooms must be pairwise different ! In particular, if you connect at 1:28 the rooms B and C with _two_ doors, you have a solution where each non-control-room has exactly three doors to other rooms. qed In Ascii-art: CR | A / \ B=C Edit: Apparently they changed the wording in the video from "3 doors to other rooms" to "3 doors to 3 other rooms". So everything is fine now.
I tried a number of times but it turned out he'd messed them up or there was some hidden factor he forgot to tell us about, so now I don't even bother trying.
just try to provoke the guards by using stealth method, trigger false alarm, and see how many guards deployed in each level....the level with most number of guards is the control room...
Naman Agarwal wait for them to disperse, there's no time limit for this task, and if they found nothing threatening, I doubt they will stay there forever if they have specific post to begin with
I think it is implied though that a room is connected to another only through 1 door. Although to be honest they should have stated that explicitly in the rules.
+Shagas Heizenberg I think they meant for you to infer each wall only could have 1 door that is what all the diagrams had but yes they could have been more clear
hes saying that the information was intentionally misleading. their visualization even included rooms that were only connected at each compass direction.
It's not impossible, just a rectangular room with 3 doors on one side. Admittedly I found the animation to be confusing as I thought each side could only contain 1 door, but in the end it's not impossible..
@@amppi1236 hes saying that the information was intentionally misleading. their visualization even included rooms that were connected at each compass direction.
It's a shame that spy didn't find the control room which was in fact on the 4th Floor down. Where was it specified that two rooms couldn't have two doors between each other. So given what is shown at 1:20 B and C wrap around room a (and optionally the control room) with two doors between them. Not only does this provide a solution, but the rooms have only right angles between adjacent walls.
I too found 4 to be the answer. By nesting the control room inside of a room, I made two outside rooms that connected to each other twice, as well as the inner room that held the control room. Picture two brackets connected to each other on the outside edge and one room in between them. Both brackets connect to the middle room and to each other two times. The inner room contains the control room (making 3 doors) and that's that. Clean & feasible with square edges. The guidelines weren't clear on this one. I also found a solution that worked with 6 first without having 'nested' rooms that connected more than once but I figured that was the trick that most people would get so I tried again with less rooms.
Actually, using that same logic, which I agree with, the answer is the 3rd floor. Follow my description if you can. Draw the following One square room (control room) attached to another square room next to it on one side, with equal length sides as the control room. One door between these two rooms where the two sides are attached. For this example control room is on top and second room directly below it. There is ONE door between the two rooms. The third and FINAL room, on this 3rd floor, wraps around the three sides of the room that is NOT the control room so that the inside walls touch the other three walls of the NON control room, with a door on each of the three walls between the two rooms. For this example this third room would resemble the letter U when it wraps around the room that is not the control room. This give two rooms with three doors between each other and one room with only one door. THIRD FLOOR is the highest room in the pyramid possible that meets all criteria. Hope that is clear to you all as it is to me.
Wayne Robison doesn't that give the central room 4 doors (one to the control room and 3 to the one that wraps around) which breaks the rules? If it doesn't have 3 doors to the wrapping room but 2, then the outside room only has 2, breaking the rules - unless there can be a door facing out of the building (maybe a fire exit?), which i don't think was actually ruled out.
Well, this one can't be reduced to logic, since it's impossible to solve the way it's shown/explained when presenting the "riddle". ...but sure, let's math it up and pretend there's a correct solution.
Draw 4 rectangles stacked on each other horizontally then a vertical one on each side. Connect the bottom 3 horizontal to each room on the side and above it. That's your answer
I don't think the conditions eliminate the possibility of a room having two doors that lead to the same room. In which case the control panel was on the 4th floor down, you get caught cause you searched the wrong floor and the world is destroyed.
The conditions don't even eliminate a door from one room ( A suppose) leading to that every room( A itself). In that case it's the first floor. One room to the control panel and the other two to each other.
It does say other rooms, not another room. So I think it does eliminate the possibility of multiple doors to the same room. One just have to read it carefully.
@Forest Phantom Cat You can't use graph theory to solve a riddle about room. A node in a graph can have as many edges as it wants, but not physical room. Try drawing a floor plan with 5 rooms where each room connect to every other room. You can do it with a graph just fine, but you can't do it with physical rooms, no matter the shape.
@@noodleexpanding3407 Yes, except graphs are actually able to solve that issue, what they didn't mention is that you need a planar graph, that is a graph where it is possible to draw the lines such that they never cross. The issue you're describing would be represented by a K5 graph which is actually a non-planar graph hence it is not possible to connect the rooms, proved using graphs.
There was no restriction about multiple connections between rooms. You can put this scenario on the fourth floor if two of the rooms are connected by two doors each. And these two rooms are connected to a third room (one door for each of the two aforementioned rooms), which is then connected to the control room. This type of diagram also makes a nice rectangle.
There's another way of figuring out the solution using a drawn a map of the rooms. If you use square rooms you'll soon realise they'll just go to infinity, so the key is to make each room satisfy each other's condition in a way you won't need to add another room. For example, imagine that you have a square made of rectangles and an empty, smalller square spot, in the middle. If you start drawing the doors you'll see you can make each room conect to other 3, except one. That's the one connected to the controll room, so you'll have 4 rectangular rooms + 1 square room + controll room = 6. I hope that's not confusing.
I found another solution. You go on 1 path. You send 3 on another path You send 5 on yet another path You leave 1 path unexplored. On the group of 5, the majority will be honest, therefore you can see how many, if any cursed students were on that path and find out if the path is good or not. Therefore you know how many, if any cursed students were in the group of 3. And thus deduct if that path is good or not based on their answers. If all the paths were bad, then you know for sure that the unexplored one leads to freedom.
This doesn't work because the cursed students don't always lie, thus the number of cursed students cannot be accurately deducted based on the group of five.
+Hugo Fontes I'm not sure if this is what Frank was alluding to, but at 00:36 when it says "Each room has exactly 3 doors to other rooms on that floor," it shows a visual representation of a N-S-E-W type of doorway configuration. When I thought out the problem in my head, I was thinking rooms would have to be connected in one of two ways. The first, shown at 00:36, is strictly a N-S-E-W configuration with fairly rectangular rooms, almost like boxes placed on each other with holes in them leading to adjacent boxes. This seemed impossible because eventually a room must make up a corner of the floor, and each corner-room can only possibly connect to two other rooms, which breaks the 3-doors-or-1-room-with-1-door rule. The second, later described in the actual answer, could utilize rooms that diagonally cross and are not strictly rectangular. The question does not explicitly forbid this configuration, but it did not explicitly allow this either, and so the only thing I had to based the rules off of were what they explicitly said as well as implicitly showed us. What they explicitly say at 00:36 is supplemented by a visual that implicitly suggests a limitation of configuration to an only N-S-E-W setup. Past TED-Ed riddles are usually pretty good at defining limitations and permissions, but something about this riddle just seemed to leave a bit of a gray area. And as much as their riddles usually rely on thinking outside of the box, the (incorrect) implication here was pretty big. If the visual at 00:38 was accompanied by another room with say a non-rectangular room (like a pentagonal room with five sides and three doors), then it would have immediately made more sense to me, and I would have gotten it without any problems in a similar way that you probably did. A bit of a long-winded response, but I just wanted to back up Frank, whether or not this was his particular sentiment.
If there are no hallways how can diagonal rooms connect, say A to C and B to D. If it's A to C then A would have to be slightly skewed and no square, but that would mean B can't connect to C. Physically impossible.
No, it's NOT a hallway. It's a door that opens up into two rooms. There is no hall. Only three rooms. So no, the rest of you don't call it a hallway, unless you are a moron.
I drew a physical floor plan that satisfies the given conditions using only 4 rooms. Since this is worded a bit odd, I didn't follow the graph theory solution and drew room with different sizes and orientations. Let each letter symbolize a room, and the hyphens are doors: A A A-B B A C C C B A-C-D C-B A C C C B A A-B B B Room D is the control room and is completely surrounded by room C.
I suppose the answer was true for those graphs, but you showed cubic rooms, which means it was impossible unless you had diagonal doors that connected rooms which weren't even in touch
The rules are wrong! "Each room has exactly 3 doors to other rooms on that floor" does not equal "Each room has exactly 3 doors to other DIFFERENT rooms on that floor". Because of that 4 rooms are sufficient. You just have to have two rooms that are connected by two doors.
For those who find it hard to visualize it as an actual floor plan, try drawing a circular room and connect it to 4 other rooms (forming a square) and one of those rooms is connected to a separate control room. Or draw a square first, draw a circle within and extend 4 lines from the circle to the square. Then add an additional room.
And when you add that extra "control room" you no longer have a floor (level, elevation), that is square. This puzzle is imposable given the set up data of a pyramid with a square footprint and each room having 3 doors to other rooms
I must protest! The exact wording is "exactly 3 doors to other rooms" not "doors to 3 other rooms". On floor four if room A is the control panel and has one door to B, B has doors to C and D, and there are two doors between C and D this satisfies the criteria. The statement of the question didn't say you can't have more than 1 door between the same pair of rooms.
I came up with the same thing, then, upon changing my assumption to allow for the poorly worded riddle, I ran into another issue. If you try to make a floor plan of an actual building, not nodes and edges but rooms with doors, how could floor six be possible?
+Robert .Hartley Try drawing a square. Then connect the corners. make a point where the lines cross. Draw another square where the corners are on the lines. Remove the cross but keep the point. Connect two oppositting lines through the point. You should have two rectangles sorrounded by trapezes. One of the rectangles are the control room. Connect the trapezes to each other and each trapez to a rectangle. Done.
I blame my confusion on the graphic presented from 0:36 and on. The rooms are depicted as perfectly square. When I found out that the graphic shows out that diagonals are allowed, I was baffled as hell, since with that graphic, IT IS IMPOSSIBLE TO HAVE PERFECTLY SQUARE ROOMS, especially since the narrator also descriptively says that "there are no hallways".
Whaaaat? Wait, was there a part of the riddle that said the rooms had to be square/the same size and have one door per wall, or something? Because it could totally be the floor with four rooms. Here, imagine that the D's are doors: -------------------------------------------------- | | | | | | D D D | | | | | | | ----D----------D----- | | | | | | | --------------------------------------------------
Well, the riddle says that each room (except the control room) has three doors to other rooms. I assume they meant three doors to three distinct rooms, but the riddle is ambiguous on this. Brilliant solution!
You could also create a construct with two rooms. Consider: ________ | | | | D | | | | | | |_D__ __| |_______| The main room has a door that leads to its own room, so the door would be counted twice in graph theory.
"Each room has exactly 3 doors to 3 other rooms" There are only two doors and one doesn't lead to another room. Yours doesn't really follow the constraints of the question.
This is actually one I got by myself. Having figured out square rooms wouldn't work, I started messing around with other shapes and got one that makes more sense than the circles used (considering their doors have a hallway between the rooms). 4 trapezoidal rooms with a square room in the middle and the control room on the outside of one of the trapezoids. Just like the following: ______________ | \__1__/ | | 2 |__3__| 5 | | /___4__\ | |__6__|
OMG! I got it right (the first time I solved one of his riddles xD). Well in a more visualized way though xD My solution contained, that there are 2 rooms, that are 3 times larger than the rooms which were shown, but with 3 doors. He didn't say, that the rooms couldn't have a different size. I'll try to visualize it down here: A | C - B - D C - E - D C | D C - F - D Each character is one room (same number is one room without doors). The (-) are the doors between the rooms. Room (A) is the Controlroom with one door.
I used a similar method and was able to do it with only 4 rooms. D is the control room and is completely surrounded by room C A A A-B B A C C C B A-C-D C-B A C C C B A A-B B B
@TactiGrill Each Room must be connected to 3 different rooms. In your example A is connected to B and C but no other 3rd room. A minimum solution must have 6 rooms, and one has been given by @SternenblutDE (which is a one of many physical representation of the Graph modelization in the video)
1. Draw a big squar (mark each side A/B/C/D) 2. Draw 4 mini squares in a line inside of the big squar from side A to C ( roughly in the middle, so that you kinda "halve" the big squar) 3. Now you sould have 6 rectangle( "The Rooms") inside the big square. 4 of them are the mini squares and 2 larger rectangels 4. The larger rectangels on side B and D have each 3 doors to 3 of the mini squares. Now each of those 3 mini squars have 2 doors. One mini squar hasn't a door yet. 5 Link two of the mini squares(that have already 2 doors) so they have each 3 doors and make one door between the other two mini squars. Voila each room has 3 doors except one
it's not that clear in the question, that you can't connect two rooms with more than one door. if you can do that, the correct answer would be the fourth floor, you connect the rooms A and B with two doors, A to C, B to C, and C to the control room with one door each.
Each room has to have 3 doors (except the control panel). The fourth floor doesn't work as B and C can only has 2 connections (one less then the required amount).
O O O O O O O O O O I got the bottom floor with ten rooms, 3 rows of 3 rooms, and then the Control room just connected to one of the other 9 rooms...I was wrong...
I was limited by my preconceptions of geometry on this one. I assumed each room was square, equal size, and could only connect on their sides. Basically just thought of it as a square grid system. By this logic, I don't actually believe there is any way to solve the puzzle. Assuming we start with A and have it connect to the CR, the two branching rooms would need two other doors, which, by the nature of room A being stuck between them, means that they need to create two new rooms each. Then those rooms need new rooms of their own, and so on. Even when connecting the two branches together, there is one room that still needs to be created, and in the process you've made dozens of other branches.
guys the rooms arent actually circles that are connected by corridors, the circles are just there to represent each room and the lines represent a door so that it can be easier to visualise.
Among other things, I find it implausible that you will have time to search exactly one floor, no matter the number of rooms on that floor. You're saying I will be able to search a floor with, let's say 10 rooms, but were I to look at the one room on the top floor, I won't have time to search two more rooms downstairs?
abstract thinking is an index of intelligence. I suggest you to watch ccp gray video about 'you are two' . You might guess which part of your brain made you mumble that feel good explanation
Zen Waichi I'm not incapable of abstract thinking. Do you want to guess which part of your brain made you comment that "feel good" attempt at a smack down? I suggest you to get over your superiority complex. If I'm presented with a riddle taking place in the real world, I expect it to make sense. "feel good explanation"... Jesus...
Aῳ сυвεɾ ...and I prefer to stick to my own opinion that the riddle failed itself. Thank you for your contribution though. Sure showed me there! Not that I in this comment mentioned me failing the riddle... To be honest, I did kind of fail the riddle. Not because of what was mentioned here, but because I read the text and saw the illustrations, and got stuck up on how every single floor would be impossible, if it was made up of the rooms we were shown. I dare say it's not a tough riddle, but I find it flawed in several ways.
3:16 dang, he really predicted the corona virus 4 years before it happened, he even showed it going from asia to europe avoiding africa and into the U.S.
Please make better explanations for your videos. It wasn't clear at any point that you could just connect any room to any other (because in the real world a room has to have a shape) and the fact that you said "ignore hallways" made it even more confusing, as it could be interpreted as "no hallways". I'm getting tired of not being able to solve the riddles because I have to pay close attention to every interpretation of whatever you say in order to have a chance at glancing the problem you're trying us to solve.
Here is a link to a better diagram of the solution to this riddle. photos.google.com/share/AF1QipO18neYA0m2pt9Y_N4u6WWdcxnKWOQoYAU7irXrxJlQeFqlmLuh5K8KY8eVR2Ks2A?key=MUZFT1pDSHJLN2ZmRi1KbE0xZEFRa0J0R2hPemRn
I got the same result as well, because a door links two rooms, and more than one door can, in theory, link the same two rooms. But if each room has only one door to every other room it links to, then I would indeed have 6 rooms.
You cannot have ”multiple connects“. The video states “Each room has exactly 3 doors to 3 _other_ rooms” at 0:37. Granted, it would be more explicit to say _“distinct_ other rooms”.
is it level 6? the control room only has to connect to one room, but that room has to connect to three, so you just add more rooms until each has exactly enough to connect to eachother.
i was correct! yay! im a very visual thinker so i found it not as challenging as the other riddles. i even challenged myself to not draw it and instead to visualise it in my mind :D
I visualized it instantly and decided that it was impossible, since some floors didn't have enough rooms to make the connections, and no floor would be able to make the connections in the corner rooms.
Gabzo Avro It is impossible, if you go by the illustrations. Square rooms with doors leading to other rooms in 3 of the walls. With that it's impossible to make the corner rooms, and floors like the top floor can't be made (Can't be made, even if you don't go by the illustrations.).
1. On the conditions list it did not say you cannot have multiple doors to the same room (essentially making level 2 possible as three doors lead to the control room, site "except the control panel room which only CONNECTS to 1" it does not say through a single door). This still fulfils the first rule "Each room has exactly 3 doors to other rooms on that floor" as the rule does not say 3 DIFFERENT rooms. 2. On the conditions it did not say the control room is in the highest possible room making the suggested level 6 the control room just a guess as even using the Graph Theory logic provided there are many other floors that mathematically work to contain the control room.
They didn't say that two rooms couldn't have multiple doors between them. If so, you can do it on the fourth floor: A and B have two doors between each other, and one each to C. That leaves C with one free door to connect to the secret plan room
Unclear conditions! The solution you presented applies if each room must have 3 doors to 3 DIFFERENT rooms If lets say room A has 2 doors both connecting to room B and 1 door connecting to room C then the correct answer is floor 4 It doesn't say anywhere that it have to be different rooms. Even though it makes the most sense (and I believe many didn't even question it) you can't make up information and take it for granted. There is no restriction on the arrangement of doors and rooms among the conditions so your solution is incorrect!
I had the same solution. I drew non-square rooms which connected such that room A connects to the CR, room B, and room C. Room B connects twice to room C, and you have 3 doors on each room except the CR.
Not seen the answer yet. I say floor 4. Just have 2 rooms have 2 doors leading those 2 rooms. Control room 1 door leading to main room. Main room 3 doors, 1 to control room and 2 doors to 2 other side rooms. Both side rooms have 1 door each to main room and 2 doors each leading to the other side room. Total 3 doors each.
I still validate my answer. If you're going to add rules to the riddle as you explain the answer (no more than 1 door per edge) then what's the point in giving us riddles -_- . I thought you guys were smart.
Each room has three connection to a different room (except the control room which only has one). 4 doesn't cut it as B and C only connects to two rooms (which is one short).
the riddle didn't specify that only one door could be between rooms. So the answer I got was the 4th floor. 1 connected to 2, 2 connected to 1, 3, and 4, and then 3 connected to 4 twice and 2 once. 4th room being connected to 2 once and 3 twice.
I cannot calculate it! You said each room has 3 doors to 3 more rooms. But since a pyramid is square then the corners will only be able to have 2 doors. If i still include those i would assume it's the tenth floor. Let's see if i was right or if i again was terribly wrong. I WAS WRONG AGAIN, DARNIT!!
If any one didn't understood the question : 1) Total 10 floors are there 2) On 1 st floor 1 room, 2nd floor 2rooms ,... and so on 10 th floor 10 room , here floor are from top to bottom. 3) find the highest floor satisfying the conditions. 2 type of rooms are there 1) simple room 2) control room simple rooms are connected to 3 other rooms.(both rooms possible) control room is connected to only 1 other room.(simple room) Find on which floor this arrangement is possible thats it !!!
Shouldn't it just be the topmost level? Since in a floor with just the control room, all rooms other than the control room(none) have the correct number of doors.
They mention to not include stairs as a doorway. So at the topmost floor, the only room should have no door. The room with the control panel needs one door to satisfy the conditions of the riddle.
I'm getting dumber and dumber after each riddle that now I couldn't even understand the question. 😂😂
AHHH WAHTS HE TALKING ABOUT
-Me by the 4the riddle video
exactly....same here
lol same
I feel you
Thought I was the only one..
"or the spread of an epidemic though different locations". Well I guess that technique was useful.
Too bad no one understands how it works
@@kerchow1837 Except for South Korea maybe lol
I literally was about to comment and the top comment was this. I’m disappointed :(
Did they edit it so it said that or something? I swear I never remembered that bit
Yes VeRY HelPuL
TED-Ed: Graph Theory can even be used to find out the spread of a pandemic
Me in 2020: *laughs quietly with mask on*
Slime Kiddo Funny. You made my day. Thanks
SLIME RANCHER
...
amazing u r awsome
Well I guess ted ed didnt make the graph. If he did we wouldnt be in this situation
I was thinking actual rooms, like on a grid. I think the riddle could be explained a bit better.
Yeah i was thinking of square rooms, there's usually not any other shape room...
Rectangle rooms :P
Cause you know I didn't think doors could be in corners
I think the part where it shows you a square room with a door in the middle of each room is misleading since it makes me think that they have to be like that. If I were to just listen to the riddle, I'd probably think that there could be two doors on one wall with rooms side by side rather than the diagram that was shown...
Actually there is a solution with all rooms being square.
I can't fly to the top of the pyramid and destroy the death ray?
Who says it wont destroy you?
Flames Gamez the possibility that the death ray needs cords to fire and radar intelligence to spot him if he is in sight view.
Well of course they would see you 'cause of surveillance and destroy you 'cause they are "EVIL"
It'll destroy you first
Oh yeah
3:34 “Now, to solve the mystery of why your surveillance team always gives you cryptic information.”
It was a test of strategy all along
They probably lazy.
Or they just doing it to spite me…….
Maybe it’s a test…
I may never know.
They are goin’ to betray you!
I didn't even understand the question...
lol me 2
same here
Yea.....
Ikr!
I had the "Oooooooh I understand now!" moment the 5th time I watched it xD
Well it never said that you couldnt have 2 connections between the same two rooms
In that case it would be floor 4!
How else would you reduce the problem? It's really the only way. If you say that you HAVE to connect to a new room, then for each room, you have to add two new rooms and then for those rooms, add two new rooms and so on exponentially, out of control. The number of rooms can only be reduced if you share
David Boucard Or you could just say "Only one door between the same two rooms allowed" XD
Hmm. In the condition "Each room has exactly three doors to other rooms on that floor", the word other implies that it is referring to three distinct rooms.
You could argue and say that the condition should have been "Each room has exactly three doors to three other rooms on that floor." If you want to be pedantic over the way they phrase things and find loopholes inside of them, then you do you. But don't feel good about it for "thinking outside the box" and finding a solution that doesn't require as much logical thinking.
And on the image shown rooms were squares with doors in the middle of the walls but they are not obviously.
“As your country’s top spy-“
Well there’s the problem. I’M the top spy. My country is doomed. R.I.P U.S.A
AND YAY FOR ZE MOTHERLAND!!!! >:P
Your country is doomed anyway
magnus bennett We really are
*sips tea in corner*
ITS COLONISING TIME
But you arent the top spy
I actually didn't understand what the puzzle was about until I saw the solution. Anyone else?
Same
Yeah. I thought they were squares that were connected. Not circles. But it did say that there weren't any hallways, which wouldn't make it only squares, but in the video, it showed them like that, with them having hallways
They arent circles, its a graph
i think the straight lines are not hallways... I think those are the doors and its pathway
I didn't understand anything, even after the solution...
If anyone is wondering, the size and shapes of the rooms are completely up to you to decide, and there can be more than one door on each wall. ( the solution doesn't even use squares and rectangles as it suggested)
Step 1: Confirm you have green eyes
Step 2: Ask the alarm system to allow you to search another floor without it going off
madlad
@@mathguy37 pog
haha youre funny
Yes.
hmmmmmm ulu
.......what......? Is anyone else really confused? Maybe it's just because I'm tired, but still.
It's kinda dumb because those connections make impossible rooms unless the rooms are super weird shaped...
Same I watched it and I was so confused I thought it would logically be the 4th if not the 5th but it turns out I was wrong.
I still haven't grasped what the narrator is trying to say
TheAztecGamer It's not your fault. The narrator got the puzzle wrong.
+Evan Smith Exactly like I'm so confused
+TheAztecGamer I thought it was number 4 too, but the all the rooms were connected to each other and it didn't make sense..
Am I the only one that watches these to do them on my friends
Same
Tdx Yt I do them on my teachers
Tdx Yt same
Me to
I do it to 😄😄😄😄😄
Can you solve this control room riddle? No I can not, but I can watch and pretend I understand.
I mean, you could, it would just take a bit of time
Lol
@@cookiecakeeater6340 I don't think thats possibla.
I really miss the old riddles, which were more logic-orientated, of the increasingly mathematical ones which are hard to explain the answer to in person
womp womp
Why is he blue
Thunder Hollow simple...
because...
he's blue daba dee da ba da ba da ba dee daba da daba fee da ba da
Thunder Hollow Why not?
he is jake sully
Mike Q. why did you sing like Ella Fitz Gerald?
Because hes a smurf
I wasn't listening I was just reading the comments.😬
Exactly what I'm doing now
Same
same here
Smaw
Uh huh
My answer was the 4th level from the top. Never do they make it a requirement for two rooms to have only one door connecting them, so my imagination went wild.
I did the same
Then the answer would be the top level. Cause just connect the control room to itself and you have everything they required. Thus even with this argument your answer is incorrect.
+Gonenc Mogol the video does say that the doors must lead to other rooms. I'm in team 4.
Gonenc Mogol "Just connect the control room to itself". That doesn't make any sense. Are you high?
Levi Bailey Even if it didn't, what the hell does it mean to "connect a room to itself"? We need GladOS to explain this to us.
Rooms are normally rectangular in shape like depicted at the beginning of this riddle. The riddle said there are no hallways, meaning the rooms must be right next to each other. The answer shows circular rooms connected to each other with hallways which does comply with the information given at the beginning.
those werent rooms, it was just a graph representation, it is not how the actual rooms look like.
I did the same thing they did while solving
I just drew them out and got the right answer, I had square rooms and one L-shaped room.
Air-Grip Biker, I think we all know that was a graph representation. Having said that, the representation is not congruent with the information given. Katie Hough claims to have drawn out square rooms that meet the criteria. I would like to see this drawing...
Paul Rezaei
u said "The answer shows circular rooms connected to each other with hallways"
but thats an incorrect claim. im sry but u dont know what a graph repr is.
it is an abstraction, it just shows nodes and arcs, u gotta have to associate their meaning to them. nodes are rooms, arc are not hallways but doors here. they dont say anything about the implementation of the room shapes.. so no, you didnt know what a graph representation was.
youre welcome
+Air-Grip Biker I can see how what I wrote was confusing. I knew it would be misinterpreted but I didn't feel like writing a long drawn out explanation ha ha. Maybe I'm wrong. Please explain how the solution can work with normal looking rooms.
I'm using this to take revenge on my math teacher, who's joining me?
Don't kid yourself, your math teacher is still smarter than you.
Aaaand, he solved it. Frick
Me
Potato_from.midgard, I 100% with you. I use these to make me look smarter than the teachers so I don’t look like I need tests.
@@vampirevideos3460 But you're not.
I love these vids, but this time I've found it quite confusing, I haven't really understood the riddle until I've seen the explanation.
Even after I understood the constraints, I got the wrong answer (4) because they didn't specify that two rooms could be connected by at most one door. There's a multigraph solution on just four nodes.
+Jonathan Sharman Same here
i just watch for answer
Your solution is wrong: You can have a solution with only 4 rooms.
Note that the conditions don't say "Each room has doors to _exactly 3 other rooms_", but "Each room has _exactly 3 doors_ to other rooms". It's not said that the other rooms must be pairwise different ! In particular, if you connect at 1:28 the rooms B and C with _two_ doors, you have a solution where each non-control-room has exactly three doors to other rooms. qed
In Ascii-art:
CR
|
A
/ \
B=C
Edit: Apparently they changed the wording in the video from "3 doors to other rooms" to "3 doors to 3 other rooms". So everything is fine now.
1) good idea
2) theres the other videos?
You aren't dead!!!
Or you could just nuke the whole pyramid
but then they would know that your attacking them and use it before it is destroyed
I actually thought the same thing
Screw that, taking out the Death Star was easier.
Anyone else too lazy to try to solve these riddles?
Yea... But im 11 😂
Same
+Anna Benita Pal im 10 and i am from Denmark😇
This one was hard for me even with explanations so sometimes I guess
I tried a number of times but it turned out he'd messed them up or there was some hidden factor he forgot to tell us about, so now I don't even bother trying.
just try to provoke the guards by using stealth method, trigger false alarm, and see how many guards deployed in each level....the level with most number of guards is the control room...
Yeah it is the best way!
But then how will you fight with the maximum gaurds on that floor??
Naman Agarwal wait for them to disperse, there's no time limit for this task, and if they found nothing threatening, I doubt they will stay there forever if they have specific post to begin with
You are right!
Good :D
Or you could throw a nuke destroying the plan and the pyramid.
Science Power no
Tbh i'd rather quit my job as a spy than attempt that 😂😂
Not with the pay you'd be getting if they trusted you alone to go in and do that.
same
I'm so early the comment section is respectful and orderly.
lol
WHAT DID YOU CALL ME???!?!?!?!
jk
somebody call a racist already!
+Drizz that's discrimination 😂😂😂
Yeah, keep deleting my comments because I'm saying something mean on the Internet.
Visualizing it probably took you so long the alarm probably went off......
People watching in 2020:
make sure you listen to 3:16
😐😐😐😐😐😐😐😐😐😐😐😐😐
4 rooms are enough. Control room connects to room A, room A connects to rooms B and C, and rooms B and C are connected to each other with two doors.
Room B and C can't connect without a hallway which violates the requirements
I think it is implied though that a room is connected to another only through 1 door. Although to be honest they should have stated that explicitly in the rules.
+Shagas Heizenberg I think they meant for you to infer each wall only could have 1 door that is what all the diagrams had but yes they could have been more clear
There is no reason B and C cant be joined without a hallway.
That doesn't work because it doesn't satisfy the 3 door requirement for rooms B and C.
oh i thought we were accepting that these rooms could exist physically... in which case the rules make the entire building impossible
And what's impossible about having rectangle or L shaped rooms?
hes saying that the information was intentionally misleading. their visualization even included rooms that were only connected at each compass direction.
It's not impossible, just a rectangular room with 3 doors on one side. Admittedly I found the animation to be confusing as I thought each side could only contain 1 door, but in the end it's not impossible..
Dude I can even draw you what the floor would look like it's not impossible at all lmao
@@amppi1236 hes saying that the information was intentionally misleading. their visualization even included rooms that were connected at each compass direction.
It was the dog all along
ShadowHunter lol
It was trump all along
Judy Hopps this is probably what would happen if Trump saw this
Lol
And yet no one gets the Silent Hill reference.
It's a shame that spy didn't find the control room which was in fact on the 4th Floor down. Where was it specified that two rooms couldn't have two doors between each other. So given what is shown at 1:20 B and C wrap around room a (and optionally the control room) with two doors between them. Not only does this provide a solution, but the rooms have only right angles between adjacent walls.
I too found 4 to be the answer. By nesting the control room inside of a room, I made two outside rooms that connected to each other twice, as well as the inner room that held the control room. Picture two brackets connected to each other on the outside edge and one room in between them. Both brackets connect to the middle room and to each other two times. The inner room contains the control room (making 3 doors) and that's that. Clean & feasible with square edges. The guidelines weren't clear on this one.
I also found a solution that worked with 6 first without having 'nested' rooms that connected more than once but I figured that was the trick that most people would get so I tried again with less rooms.
Actually, using that same logic, which I agree with, the answer is the 3rd floor.
Follow my description if you can.
Draw the following
One square room (control room) attached to another square room next to it on one side, with equal length sides as the control room. One door between these two rooms where the two sides are attached.
For this example control room is on top and second room directly below it. There is ONE door between the two rooms.
The third and FINAL room, on this 3rd floor, wraps around the three sides of the room that is NOT the control room so that the inside walls touch the other three walls of the NON control room, with a door on each of the three walls between the two rooms.
For this example this third room would resemble the letter U when it wraps around the room that is not the control room.
This give two rooms with three doors between each other and one room with only one door.
THIRD FLOOR is the highest room in the pyramid possible that meets all criteria.
Hope that is clear to you all as it is to me.
Wayne Robison doesn't that give the central room 4 doors (one to the control room and 3 to the one that wraps around) which breaks the rules? If it doesn't have 3 doors to the wrapping room but 2, then the outside room only has 2, breaking the rules - unless there can be a door facing out of the building (maybe a fire exit?), which i don't think was actually ruled out.
+Fabrizio Lungo nope, clearly says "doors to other rooms on that floor"
Yes it does, If i am standing in one room and open a single door, that door leads to a different room.
This is not a riddle, this is a mathematical problem -.-
Shhhh...Don't let the casuals know..
lol
All of the riddles from this channel reduce to logic or math problems.
Bad day?
Well, this one can't be reduced to logic, since it's impossible to solve the way it's shown/explained when presenting the "riddle".
...but sure, let's math it up and pretend there's a correct solution.
3:16 Me seeing this during COVID-19: oh well i guess this is the right time to learn about nobes and edges
I got the 8th floor because I imagined square & rectangle rooms, not circle rooms. No one would waste so much space.
They don't need to be circles
Draw 4 rectangles stacked on each other horizontally then a vertical one on each side. Connect the bottom 3 horizontal to each room on the side and above it.
That's your answer
Uhohhotdog But wouldn't stacked rooms make an extra floor?
Jeyna LSC
I'm talking about a 2D drawing.
Uhohhotdog Ah, ok.
I don't think the conditions eliminate the possibility of a room having two doors that lead to the same room. In which case the control panel was on the 4th floor down, you get caught cause you searched the wrong floor and the world is destroyed.
I agree. The 4th floor could have the 3rd and 4th rooms connected to each other with 2 doors.
The conditions don't even eliminate a door from one room ( A suppose) leading to that every room( A itself).
In that case it's the first floor.
One room to the control panel and the other two to each other.
Manaswini Ayala They do. The condition is "Each room has exactly 3 doors to other rooms on that floor."
IAmProcrastinatingRightNow Oh yes 😅
It does say other rooms, not another room. So I think it does eliminate the possibility of multiple doors to the same room. One just have to read it carefully.
there is a difference between connected doors and flippin hallways
@Forest Phantom Cat You can't use graph theory to solve a riddle about room. A node in a graph can have as many edges as it wants, but not physical room. Try drawing a floor plan with 5 rooms where each room connect to every other room. You can do it with a graph just fine, but you can't do it with physical rooms, no matter the shape.
@@noodleexpanding3407 Yes, except graphs are actually able to solve that issue, what they didn't mention is that you need a planar graph, that is a graph where it is possible to draw the lines such that they never cross. The issue you're describing would be represented by a K5 graph which is actually a non-planar graph hence it is not possible to connect the rooms, proved using graphs.
@@noodleexpanding3407 can you draw it in graph?
@@noodleexpanding3407 Here's how I did it with physical rooms. i.imgur.com/XxYJgQd.jpg
@@noodleexpanding3407 you can
xxx
x x
xxx
x
That's just a theory. a graph Theory. thanks for learning
Ayyyyy! I see your reference! (Game theory is awesome)
+Grace bergeon haha ikr😂 GT ftw
#MatPatForPresident2016
+TheMonkeyMage #matpatforworldleader2016
+Lauren Burnett Nice to see some other GT fans
There was no restriction about multiple connections between rooms. You can put this scenario on the fourth floor if two of the rooms are connected by two doors each. And these two rooms are connected to a third room (one door for each of the two aforementioned rooms), which is then connected to the control room. This type of diagram also makes a nice rectangle.
I feel like TED-ED spends more time figuring out the scenario than the question itself
When you make it this complicated anything can be a riddle
Wait..... so floor 10 was the lowest floor?
*facepalm*
10th floor from the top.
There's another way of figuring out the solution using a drawn a map of the rooms. If you use square rooms you'll soon realise they'll just go to infinity, so the key is to make each room satisfy each other's condition in a way you won't need to add another room.
For example, imagine that you have a square made of rectangles and an empty, smalller square spot, in the middle. If you start drawing the doors you'll see you can make each room conect to other 3, except one. That's the one connected to the controll room, so you'll have 4 rectangular rooms + 1 square room + controll room = 6. I hope that's not confusing.
I found another solution.
You go on 1 path.
You send 3 on another path
You send 5 on yet another path
You leave 1 path unexplored.
On the group of 5, the majority will be honest, therefore you can see how many, if any cursed students were on that path and find out if the path is good or not.
Therefore you know how many, if any cursed students were in the group of 3. And thus deduct if that path is good or not based on their answers.
If all the paths were bad, then you know for sure that the unexplored one leads to freedom.
Wrong vid
LMAO
Before typing such a long comment you should have at least looked at the title of this video.
Gagan Deep it was a joke
This doesn't work because the cursed students don't always lie, thus the number of cursed students cannot be accurately deducted based on the group of five.
I love these riddles make more!!!
I KNOW RIGHT?!?!?!?!?!
and so they did
“Or the spread of an epidemic through different locations.”
Well, this did not age well.
wow, poorly explained question
Agreed...
what was poor about it? I got it, no problems at all
+Hugo Fontes thanks
thanks no thanks
+Hugo Fontes I'm not sure if this is what Frank was alluding to, but at 00:36 when it says "Each room has exactly 3 doors to other rooms on that floor," it shows a visual representation of a N-S-E-W type of doorway configuration. When I thought out the problem in my head, I was thinking rooms would have to be connected in one of two ways. The first, shown at 00:36, is strictly a N-S-E-W configuration with fairly rectangular rooms, almost like boxes placed on each other with holes in them leading to adjacent boxes. This seemed impossible because eventually a room must make up a corner of the floor, and each corner-room can only possibly connect to two other rooms, which breaks the 3-doors-or-1-room-with-1-door rule. The second, later described in the actual answer, could utilize rooms that diagonally cross and are not strictly rectangular. The question does not explicitly forbid this configuration, but it did not explicitly allow this either, and so the only thing I had to based the rules off of were what they explicitly said as well as implicitly showed us. What they explicitly say at 00:36 is supplemented by a visual that implicitly suggests a limitation of configuration to an only N-S-E-W setup.
Past TED-Ed riddles are usually pretty good at defining limitations and permissions, but something about this riddle just seemed to leave a bit of a gray area. And as much as their riddles usually rely on thinking outside of the box, the (incorrect) implication here was pretty big. If the visual at 00:38 was accompanied by another room with say a non-rectangular room (like a pentagonal room with five sides and three doors), then it would have immediately made more sense to me, and I would have gotten it without any problems in a similar way that you probably did. A bit of a long-winded response, but I just wanted to back up Frank, whether or not this was his particular sentiment.
If there are no hallways how can diagonal rooms connect, say A to C and B to D. If it's A to C then A would have to be slightly skewed and no square, but that would mean B can't connect to C. Physically impossible.
No, a door could open to two rooms that share a single wall and you are able to enter one room to the left and the other on the right, not impossible.
Chris Harrington the rest of us call those hallways.
No, it's NOT a hallway. It's a door that opens up into two rooms. There is no hall. Only three rooms. So no, the rest of you don't call it a hallway, unless you are a moron.
CallOnTheCountryBalls look three comments up
I drew a physical floor plan that satisfies the given conditions using only 4 rooms. Since this is worded a bit odd, I didn't follow the graph theory solution and drew room with different sizes and orientations. Let each letter symbolize a room, and the hyphens are doors:
A A A-B B
A C C C B
A-C-D C-B
A C C C B
A A-B B B
Room D is the control room and is completely surrounded by room C.
Your solution doesn't work
due to the fact you have a and b connected twice which isn't allowed. It says each room is connected to three other rooms
I suppose the answer was true for those graphs, but you showed cubic rooms, which means it was impossible unless you had diagonal doors that connected rooms which weren't even in touch
The rules are wrong! "Each room has exactly 3 doors to other rooms on that floor" does not equal "Each room has exactly 3 doors to other DIFFERENT rooms on that floor". Because of that 4 rooms are sufficient. You just have to have two rooms that are connected by two doors.
Yes! I came to the same soultion
Why can't there team mates give them the information and not do it
It's Their not There
lps cooki3 idk ( i wahgt your videos i like them
Today we’re gonna teach you bout a common writing error.the difference is amongst the form of there and their and the’re/jacksfilms
For those who find it hard to visualize it as an actual floor plan, try drawing a circular room and connect it to 4 other rooms (forming a square) and one of those rooms is connected to a separate control room. Or draw a square first, draw a circle within and extend 4 lines from the circle to the square. Then add an additional room.
wow that really helped :D
And when you add that
extra "control room" you no longer have a floor (level, elevation), that is square. This puzzle is imposable given the set up data of a pyramid with a square footprint and each room having 3 doors to other rooms
I must protest! The exact wording is "exactly 3 doors to other rooms" not "doors to 3 other rooms". On floor four if room A is the control panel and has one door to B, B has doors to C and D, and there are two doors between C and D this satisfies the criteria. The statement of the question didn't say you can't have more than 1 door between the same pair of rooms.
I came up with the same thing, then, upon changing my assumption to allow for the poorly worded riddle, I ran into another issue. If you try to make a floor plan of an actual building, not nodes and edges but rooms with doors, how could floor six be possible?
One door has to be a cat flap at floor level, the other a serving hatch above it. I think that's the only way it can work.
+Robert .Hartley Try drawing a square. Then connect the corners. make a point where the lines cross. Draw another square where the corners are on the lines. Remove the cross but keep the point. Connect two oppositting lines through the point. You should have two rectangles sorrounded by trapezes. One of the rectangles are the control room. Connect the trapezes to each other and each trapez to a rectangle. Done.
+Robert .Hartley Hope it helped you.
EXCATLY! I thought I was the only one who saw that
Step 1: Confirm you have green eyes
Step 2: Ask the death ray to deactivate
madlad
So if I'm my country's best spy, why do I have a fucking surveillance team where one of them is a cat?
Must be a very poor country
I blame my confusion on the graphic presented from 0:36 and on. The rooms are depicted as perfectly square. When I found out that the graphic shows out that diagonals are allowed, I was baffled as hell, since with that graphic, IT IS IMPOSSIBLE TO HAVE PERFECTLY SQUARE ROOMS, especially since the narrator also descriptively says that "there are no hallways".
"Or to map the spread of a pandemic"
*Well that aged well*
Whaaaat? Wait, was there a part of the riddle that said the rooms had to be square/the same size and have one door per wall, or something? Because it could totally be the floor with four rooms. Here, imagine that the D's are doors:
--------------------------------------------------
| | | | |
| D D D |
| | | | |
| ----D----------D----- |
| | |
| | |
--------------------------------------------------
Well, the riddle says that each room (except the control room) has three doors to other rooms. I assume they meant three doors to three distinct rooms, but the riddle is ambiguous on this.
Brilliant solution!
Indeed. He is saying that the riddle doesn't rule out doing so.
"Each room has exactly 3 doors to 3 other rooms". Here, the second room from left has 3 doors to 2 other rooms, so it's ruled out.
You could also create a construct with two rooms. Consider:
________
| | |
| D |
| | |
| |
|_D__ __|
|_______|
The main room has a door that leads to its own room, so the door would be counted twice in graph theory.
"Each room has exactly 3 doors to 3 other rooms"
There are only two doors and one doesn't lead to another room. Yours doesn't really follow the constraints of the question.
This is actually one I got by myself. Having figured out square rooms wouldn't work, I started messing around with other shapes and got one that makes more sense than the circles used (considering their doors have a hallway between the rooms). 4 trapezoidal rooms with a square room in the middle and the control room on the outside of one of the trapezoids. Just like the following:
______________
| \__1__/ |
| 2 |__3__| 5 |
| /___4__\ |
|__6__|
“Can you solve the control room riddle?”
I CANT EVEN SOLVE THE SOLUTION
You should make it clear that there are no repeated arcs in the graph. i.e. Rooms cannot be connected by more than one doorway.
OMG! I got it right (the first time I solved one of his riddles xD).
Well in a more visualized way though xD
My solution contained, that there are 2 rooms, that are 3 times larger than the rooms which were shown, but with 3 doors. He didn't say, that the rooms couldn't have a different size. I'll try to visualize it down here:
A
|
C - B - D
C - E - D
C | D
C - F - D
Each character is one room (same number is one room without doors). The (-) are the doors between the rooms. Room (A) is the Controlroom with one door.
yes so it should be 5th floor
Wow, you got it right. And your floor plan is physically practical too!
poojitha *6*, look again
I used a similar method and was able to do it with only 4 rooms. D is the control room and is completely surrounded by room C
A A A-B B
A C C C B
A-C-D C-B
A C C C B
A A-B B B
@TactiGrill Each Room must be connected to 3 different rooms. In your example A is connected to B and C but no other 3rd room. A minimum solution must have 6 rooms, and one has been given by @SternenblutDE (which is a one of many physical representation of the Graph modelization in the video)
1. Draw a big squar (mark each side A/B/C/D)
2. Draw 4 mini squares in a line inside of the big squar from side A to C ( roughly in the middle, so that you kinda "halve" the big squar)
3. Now you sould have 6 rectangle( "The Rooms") inside the big square. 4 of them are the mini squares and 2 larger rectangels
4. The larger rectangels on side B and D have each 3 doors to 3 of the mini squares. Now each of those 3 mini squars have 2 doors. One mini squar hasn't a door yet.
5 Link two of the mini squares(that have already 2 doors) so they have each 3 doors and make one door between the other two mini squars.
Voila each room has 3 doors except one
my answer would be to not undertake the mission until further information was acquired.
Rinoa Super-Genius but then they will use the death ray before you destroy it
You could have explained the riddle better
Am i the only one that has actually drew it, and solved it that way lol
Yup lol
I usually love these but this one was very confusing due to the wording
it's not that clear in the question, that you can't connect two rooms with more than one door. if you can do that, the correct answer would be the fourth floor, you connect the rooms A and B with two doors, A to C, B to C, and C to the control room with one door each.
exactly!
Each room has to have 3 doors (except the control panel). The fourth floor doesn't work as B and C can only has 2 connections (one less then the required amount).
O
O O O
O O O
O O O
I got the bottom floor with ten rooms, 3 rows of 3 rooms, and then the Control room just connected to one of the other 9 rooms...I was wrong...
Me: doesn’t understand the question nor answer
Also Me: *talking to friends* btw there is this riddle and can u do it it’s super ez
this makes no sense, it's a horribly designed building, such a waste of space.
wow, you'd do great as a scientist /sarcasm
haha
+Khorps /s
When you find an evil person whose happen to have a good plan you tell me.
I was limited by my preconceptions of geometry on this one. I assumed each room was square, equal size, and could only connect on their sides. Basically just thought of it as a square grid system. By this logic, I don't actually believe there is any way to solve the puzzle. Assuming we start with A and have it connect to the CR, the two branching rooms would need two other doors, which, by the nature of room A being stuck between them, means that they need to create two new rooms each. Then those rooms need new rooms of their own, and so on. Even when connecting the two branches together, there is one room that still needs to be created, and in the process you've made dozens of other branches.
guys the rooms arent actually circles that are connected by corridors, the circles are just there to represent each room and the lines represent a door so that it can be easier to visualise.
Welcome back to another episode of: "How Shitty Can We Make People Feel About Themselves"
How would you connect room B and C if they were separated by room A?
Peterolen i think i'm to dumn for this
I Can't even spell
Plattfisken IDK. At first I thought it was connected by hallways but room A is inbetween
ducktastic cat yeah someone understands what i mean
ducktastic cat plus there are no hallways
Spy: do you have the floor plan?
The surveillance team who played solitaire for 5 hours: ozo
I was confident on 4, I should've known it wasn't that easy
Among other things, I find it implausible that you will have time to search exactly one floor, no matter the number of rooms on that floor. You're saying I will be able to search a floor with, let's say 10 rooms, but were I to look at the one room on the top floor, I won't have time to search two more rooms downstairs?
abstract thinking is an index of intelligence. I suggest you to watch ccp gray video about 'you are two' . You might guess which part of your brain made you mumble that feel good explanation
Zen Waichi I'm not incapable of abstract thinking. Do you want to guess which part of your brain made you comment that "feel good" attempt at a smack down?
I suggest you to get over your superiority complex.
If I'm presented with a riddle taking place in the real world, I expect it to make sense.
"feel good explanation"... Jesus...
cover up the fact that you failed the riddle dumb###
Aῳ сυвεɾ
...and I prefer to stick to my own opinion that the riddle failed itself.
Thank you for your contribution though. Sure showed me there!
Not that I in this comment mentioned me failing the riddle...
To be honest, I did kind of fail the riddle. Not because of what was mentioned here, but because I read the text and saw the illustrations, and got stuck up on how every single floor would be impossible, if it was made up of the rooms we were shown.
I dare say it's not a tough riddle, but I find it flawed in several ways.
k
3:16 dang, he really predicted the corona virus 4 years before it happened, he even showed it going from asia to europe avoiding africa and into the U.S.
Please make better explanations for your videos. It wasn't clear at any point that you could just connect any room to any other (because in the real world a room has to have a shape) and the fact that you said "ignore hallways" made it even more confusing, as it could be interpreted as "no hallways". I'm getting tired of not being able to solve the riddles because I have to pay close attention to every interpretation of whatever you say in order to have a chance at glancing the problem you're trying us to solve.
Here is a link to a better diagram of the solution to this riddle. photos.google.com/share/AF1QipO18neYA0m2pt9Y_N4u6WWdcxnKWOQoYAU7irXrxJlQeFqlmLuh5K8KY8eVR2Ks2A?key=MUZFT1pDSHJLN2ZmRi1KbE0xZEFRa0J0R2hPemRn
Gerardo Teniente I understood the riddle and the solution. But only after having to see through to the answer, and that's what annoys me.
"Info from your team"
Me: You mean from a granny
Bullshit. I can just hide under a cardboard box and check every room.
Nice Reasoning
Agreed
I got 4 by using non-square rooms with multiple connects, as it never states that that is not possible in the stipulated rules.
I got the same result as well, because a door links two rooms, and more than one door can, in theory, link the same two rooms. But if each room has only one door to every other room it links to, then I would indeed have 6 rooms.
ah2190
Yep. Vague wording leads to multiple solutions.
You cannot have ”multiple connects“. The video states “Each room has exactly 3 doors to 3 _other_ rooms” at 0:37. Granted, it would be more explicit to say _“distinct_ other rooms”.
is it level 6? the control room only has to connect to one room, but that room has to connect to three, so you just add more rooms until each has exactly enough to connect to eachother.
i was correct! yay! im a very visual thinker so i found it not as challenging as the other riddles. i even challenged myself to not draw it and instead to visualise it in my mind :D
I was ablr to do the same.
I visualized it instantly and decided that it was impossible, since some floors didn't have enough rooms to make the connections, and no floor would be able to make the connections in the corner rooms.
it was 6 it's not impossible
Gabzo Avro It is impossible, if you go by the illustrations. Square rooms with doors leading to other rooms in 3 of the walls. With that it's impossible to make the corner rooms, and floors like the top floor can't be made (Can't be made, even if you don't go by the illustrations.).
Step 1: Flip switch
Step 2: confirm you have green eyes
Step 3: ask the guards to leave
Remember when these riddles made sense?
It still makes sense...but just after they explain it lol.
+Angelle W yah, but they're not as interesting
3:17 Ted-Ed had guessed the COVID Pandemic from 2016:D
1. On the conditions list it did not say you cannot have multiple doors to the same room (essentially making level 2 possible as three doors lead to the control room, site "except the control panel room which only CONNECTS to 1" it does not say through a single door). This still fulfils the first rule "Each room has exactly 3 doors to other rooms on that floor" as the rule does not say 3 DIFFERENT rooms. 2. On the conditions it did not say the control room is in the highest possible room making the suggested level 6 the control room just a guess as even using the Graph Theory logic provided there are many other floors that mathematically work to contain the control room.
Ted-Ed: *_breathes_*
My brain: *_s t r o k e_*
Didn't understand the wording of the question at all
They didn't say that two rooms couldn't have multiple doors between them. If so, you can do it on the fourth floor: A and B have two doors between each other, and one each to C. That leaves C with one free door to connect to the secret plan room
Unclear conditions!
The solution you presented applies if each room must have 3 doors to 3 DIFFERENT rooms
If lets say room A has 2 doors both connecting to room B
and 1 door connecting to room C then the correct answer is floor 4
It doesn't say anywhere that it have to be different rooms. Even though it makes the most sense (and I believe many didn't even question it) you can't make up information and take it for granted.
There is no restriction on the arrangement of doors and rooms among the conditions so your solution is incorrect!
That was my solution and logic as well.
I had the same solution. I drew non-square rooms which connected such that room A connects to the CR, room B, and room C. Room B connects twice to room C, and you have 3 doors on each room except the CR.
Not seen the answer yet.
I say floor 4. Just have 2 rooms have 2 doors leading those 2 rooms.
Control room 1 door leading to main room.
Main room 3 doors, 1 to control room and 2 doors to 2 other side rooms.
Both side rooms have 1 door each to main room and 2 doors each leading to the other side room. Total 3 doors each.
I still validate my answer. If you're going to add rules to the riddle as you explain the answer (no more than 1 door per edge) then what's the point in giving us riddles -_- . I thought you guys were smart.
+ShadowLynx777 I thought floor 4 too! I feel like this riddle didn't specify the conditions enough.
I agree with the insufficient data. My answer was 4 as well.
Each room has three connection to a different room (except the control room which only has one). 4 doesn't cut it as B and C only connects to two rooms (which is one short).
the riddle didn't specify that only one door could be between rooms. So the answer I got was the 4th floor. 1 connected to 2, 2 connected to 1, 3, and 4, and then 3 connected to 4 twice and 2 once. 4th room being connected to 2 once and 3 twice.
I cannot calculate it! You said each room has 3 doors to 3 more rooms. But since a pyramid is square then the corners will only be able to have 2 doors.
If i still include those i would assume it's the tenth floor.
Let's see if i was right or if i again was terribly wrong.
I WAS WRONG AGAIN, DARNIT!!
Who told us that all the space has to be occupied or that the rooms are rectangles?
Forterosen First assumption one gets.
Forterosen really? I dont remember that ó.0
Best way to destroy your brain.
"Alright, it's-!"
*Alarm system reactivates*
If any one didn't understood the question :
1) Total 10 floors are there
2) On 1 st floor 1 room, 2nd floor 2rooms ,... and so on 10 th floor 10 room , here floor are from top to bottom.
3) find the highest floor satisfying the conditions.
2 type of rooms are there 1) simple room 2) control room
simple rooms are connected to 3 other rooms.(both rooms possible)
control room is connected to only 1 other room.(simple room)
Find on which floor this arrangement is possible
thats it !!!
Shouldn't it just be the topmost level? Since in a floor with just the control room, all rooms other than the control room(none) have the correct number of doors.
They mention to not include stairs as a doorway. So at the topmost floor, the only room should have no door. The room with the control panel needs one door to satisfy the conditions of the riddle.
The door that leads outside.
There has to be atleast a single door to the outsie of the death ray pyramid.
It says in the requirements that the door connects to another room.