**psst** Are you enjoying Scam School? You've gotta check out our *brand-new series* "The Modern Rogue" at ua-cam.com/users/modernrogue. If you've seen it, reply and tell me what you think. If not, then get on over there. I'll bet you a dollar you love it. -Brian
The numbers one is really easy, surprised they didn't get that one. Just knowing that 1 and 8 are on the ends and only can't touch one number means they go in the middle, which then means 2 and 7 have to go on the ends, and the other number are really easy from there. Interestingly I got the answer shown but with the middle flipped horizontally.
first one was easy: since the middle 2 need to have the next number as the opposite outside box, you need to start and end with a middle box. For the second puzzle, theres always that one wall you dont see
I did the ghost one! Before everyone goes its impossible read this through. The ghost one is possible. You just need to be a t-rex ghost and when you walk through all 5 rooms at once, you only go through each wall once! To better explain, the ghost you always were, was a small line from a marker. Get a big paint brush and paint over the whole thing in one swoop. Then bam you win. Do i get my free beer?
well while you are at the bar and someone challenged you.. are you going home and come back just to take a brush? and they gave you max.5mins to solve. what are you gonna do?
I was able to solve the numbers one pretty easily but I tried the ghost one like twice and said I dont have the patience for it right now. The numbers one was easy because for the two center numbers you just figure two numbers you are using that only have one number each they cant touch each which would be the first and last numbers, the 1 and the 8. Then you figure you place the two numbers that cant touch the 1 and 8, which would be the 2 and 7, and you place them to where they cant touch, which is on the side ends opposite on the side they are on. From there everything else was easy to put in place, just fill in the blanks with the remaining for numbers
+Scam School Ive decided to pick up magic as a hobby recently which brought me to your channel and im glad I came across your channel. Love your videos and love the scams. Your channel has shown me alot of cool new tricks and scams, now I just need to apply the techniques to scam some people
+Scam School I watched the video where you put the cards in chased order to find out the top card and I couldnt remember if chased ran from the top of the cards down or the opposite so I did the trick with the top then I flipped the cards face up and did the trick and it worked both times. With that being said I noticed if I do the trick the way you showed it that the d on matching card will always be the matching card to the bottom so now when I do the trick I can tell them the card they have from the top of the deck and the card from the bottom of the deck
Yup it's so easy... Just write down numbers and group every 2nd number so you get basic numbers for a side. The one space suggests that one side has to have high numbers and the other low... obviously having that guides you to set middle numbers counter the side numbers thus giving you a little mirror of numbers: high-low... easy..
the puzzle of the 8 numbers is the possible one, probably you will not belive in me, but I do it in my frist try. I just think before put the numbers and I realized the middle squares touch in all square except for one, so if is possible, the numbers 1 and 8 have to be in the squares of the middle, becouse this is the only numbers who doesn't have a predecessor and a sucessor, then you put the number 7 and 2 in the only possible place, after that the rest is easy.
the first one was really easy i did it in 10 seconds XD you just have to realize that 1 and 8 have to go in the middle, because they only have 1 neighbor (2 and 7). everything else has two numbers next to it, ex. 3 is next to 2 and 4.
In an attempt to explain the impossible one (since it wasn't done in the video)Count the number of walls around each room.Three of them have 5 walls. A, B, C. At the start, if the ghost is NOT in room A then it must end in room A. (In, out, in, out, in) the same goes for B and C.So the start point or end point must be in rooms A B and C,As there is only 1 start and 1 end it is therfore impossible.
On the numbers riddle: The two squares in the middle are each adjacent to _all other squares except one_. Each number has a number before and after it, except the first and the last, i.e. 1 and 8. So 1 and 8 must go in the middle. Their neighbouring numbers, 2 and 7, must go in the outer squares so that you have a row like 7 1 8 2 or 2 8 1 7. The 3 has to go above or below the 1 so that it isn't touching the 2. Along the same lines, the 6 has to go above or below the 8. But you can't put them _both_ above or both below, because then the two final squares for the 4 and 5 are adjacent. So if you put the 3 above the 1, then the 6 goes below the 8, or vice versa. Then the 4 goes beside the 6 and the 5 goes beside the 3, and you're done. On the walls riddle: Of the five rooms, three of them have 5 walls, which is an uneven number. If you start your line outside any of the rooms with with 5 walls, and ignoring all other rooms for the moment, your line will eventually go (1) in - (2) out - (3) in - (4) out - (5) in, and then you're stuck in the room. That means you can start your line within a 5-walled room and you can end it in one, but the third 5-walled room will not have all its walls crossed, otherwise you would have become stuck in it.
I paused the video, the number puzzle was easy. 1 and 8 go in the middle because they only have 1 number they can't touch and one space they aren't touching. 7 and 2 go at the sides with 3 above 1 and 4 below, 5 above 8 and 6 below. I haven't seen the answer yet and I am 12.
2 things: First, I like how Brian gave a hint to the solution of the number square puzzle Second, the Euler Trail Puzzle as Brian described it technically has a solution because it involves a "Ghost passing through walls only once" and not "walking through each door only once". Technically, under Brian's rules, the ghost doesn't have to pass through a doorway, just through a wall, and you can pass through all 16 walls only once if you do a combination of Corners and single walls. Again, this is "Technically" and under *Brian's Rules*, not the actual rules.
I knew 1 and 8 had to be in the middle because they are only adjacent to one other number. So 2 and 7 have to be on the ends and then I solved from there.
Red Eyes Sorry to say, but it is 100% impossible, no matter how hard you try. It is proven by mathematics, I was even studying this entire topology topic, you can't beat it.
the number one is realy easy (solved it in 30 sec ) because all u have to do is but 1 and 8 in the middle because they are the numbers that only have 1 number they cant be next too rest of the puzzle does it by his own
Scam School no it is the answer look 1 and 8 in the middle then you have to put 7 and 2 in the opposite corners because those are the only places where they can go. Then there are 2 places left where 6 can go and 2 places left where 3 can go the only error you can do now is put them both up or down but if you think ahead you will know that you cant do that becasue you have to put 5 and 4 then next to each other. So you but 3 and 6 diagonally away from eatch other and fill in 4 and 5 in the remaining spots.
For the number trick, think about it like this. The two middle boxes touch every other box except one--the ones on the end, respectively. Most of the numbers can't be touching two other numbers and so they CAN'T be placed in the middle (e.g. 3 can't touch 2 or 4; 5 can't touch 4 or 6). I put 1 & 8 in the middle boxes because each number could only NOT be touching one other number (1 can't touch 2 & 8 can't touch 7). So as soon as the middle row was laid out (7 | 1 | 8 | 2), the rest of it was easy.
Trickelodean enrage There can't be a proof. If you want to cross a room with five walls with one line then you have to start at the inside to end outside this room. Because you have 3 rooms with 5 walls and the line has only 2 ends wich with you can start on the inside you will definitly end in the third room having to get in and out with only 1 wall. (Hope you understand my bad English because it really sucks)
I also like the similarity of these problems. #1 is kind of the opposite of #2. In #1 you have to find a sequence of vertices (rooms) that are connected by edges (walls) such that every edge (wall) is crossed. In #2 you have to find a sequence of vertices (boxes) that are NOT connected by an edge (not adjacent) such that every vertex (box) is crossed.
VladKov36 - Clash and Craft um... It sounds a bit like you're celebrating that you solved them both. You definitely, definitely did not solve them both.
Scam School I got really excited...then as the video continued and saw how hard it was..i reviewed them...i failed the numbers one, kept re-trying and still haven't found a solution, but pretty confident in the ghost one, every wall was gone through once, but does the ghost have to end up outside the house again or any room? Lol, you replied right as i was gonna say something like this. :)
Scam School I just got the the part where you said the ghost rooms was impossible, ill tweet it to you,but i got through every wall once(or it looks that way). Both every wall once and every wall of a room once(I was a bit confused at the beginning lol)
Scam School You should have explained more about the math behind it. Perhaps not the reason, but a pointer on how to check if similar tasks are possible or impossible. It´s the Seven bridges of Königsberg problem of sorts. And the way to check if the task is possible is to count the walls of all the rooms. If there are 2 or less rooms with an uneven number of walls, it is possible. Otherwise it is not. The version you showed with the ghosts has 3 rooms with 5 walls. An easy way to think of it is, if you enter a room, you have to exit it as well, subtracting 2 walls each time. But if you do this with a puzzle with more than 2 rooms with uneven number of walls, you end up in one of the rooms with no way out. On the other hand, if there are 2 or less rooms with uneven numbers, you can just start in one of the rooms with uneven numbers and end in the other.
BonanaSquid It is impossible. 3 of the rooms have 5 walls. To draw a line going through each wall, one end gets trapped inside the room. Since the line only has two ends, you don't have enough for all three of the 5 sided rooms.
sap2002 BonanaSquid Actually, the 5 rooms one is mathematically proven to be impossible and has been attempted over and over but never successfully but it is ok, I lie sometimes too.
Well shit I missed one but it wasn't the middle segment ;-; main reason I said I didn't cous thats the one u said I got wrong. Didn't notice the one lurking around there :/
I don't believe he explained the rules correctly for puzzle 2. If the only rule is that you have one like that crosses every wall once. Than it is very solvable. If you can't cross the same room more than once than it's just a stupid puzzle because in order to intersect the 4 sides of a room you must enter it twice.
Point is that 3 rooms has an odd amount of walls (room 1,2 and 4). In order to "finish" such room you have to make an odd number of entrances/exits. In order to do that you have to start or end in that room. And since you can start/end in only two rooms you can never complete the third one.
He explained it correctly, he said you can't cross the same wall twice, not room, you can cross rooms more than once. But it's still impossible for the reasons mroz123 stated.
Only by going through a corner rather than a wall. Brian should have specified that they couldn't walk through a corner to count as entering from two walls.
In the real problem, you gotta start in one of the rooms and end up in one of the rooms (not necessarily the same one) making it impossible. But, the way he set it up in the video, you can start outside of the rooms and it's quite easily solvable.
Why am I looking at these old videos? Because puzzles are fun. :) The numbers were rather easy. The middle boxes just have one box that isn't adjacent each, and in a number sequence that is the first and last numbers, so 1 and 8 go into them. Following that, the numbers that are adjacent to them, 2 and 7, go into the far sides, as those are the only boxes they can go in. After that the remaining four are easy to place. The rooms with the ghost, unless I'm intepreting what counts as a wall wrong, have the three large rooms having five walls each (due to one wall in each room being separated by other walls; the three "vertical" ones). If the ghosts goes into a room, and out of it, that's an even number of walls. As this is an odd number of walls, this means the ghost has to either start in the room and end outside, or start outside and end inside. Since there are three of those, it's logically impossible to go through all walls only once, since there's only one starting point and one ending point. Unless you pull an outside-the-box strategy and fold the paper (which is the solution of some similar puzzles).
+Khryj Paca well you have the 3 and 5 at the wrong spot the 3 should be over the 1 and the 5 over the 8. so if you had them in the right spots and reversed them than the bottom right of the box with the 3 would be touching the top left of the box with the 2
I didn't get the right order for the 1 to 8 but it is easy to say the the ghost house is impossible, there are three rooms with an odd number of walls to cross, if you start outside you can't get out and if you start inside, you finish outside, meaning you are stuck inside the second one before the end.
Did the numbers in 5 seconds on the first try. Crossing the walls looked suspicious. I tried anyway and ended up with an art piece before I started counting the number of crossings. 3 rooms with uneven number of crossings leaves one hanging every time
Got both of them :) you cant start and end outside of a room with an uneven amount of walls, you can get around this by starting and ending your path in them, but there are 3 rooms w/ 5 wals in this puzzle, so you always have 1 wall left encrossed before you run out of legal moves.
With the ghost problem, I treated a couple corners as three walls -- the two left-right middle corners in particular. Start or end in the bottom middle room, and make sure you go through the corners to/from the bottom middle and the top ones. I have no idea whether it's allowed to treat corners like this.
I got the numbers one on my second try. The key is to notice which boxes have the most adjacent other boxes (there are two) and which numbers have the fewest adjacent numbers (there are two of those, too). Put those numbers in those boxes, and the rest is trivial.
I had to break out eight 8-sided dice to solve the number one. Once you realize that the two numbers in the two center squares touch every other number EXCEPT the one on the far left (or far right), you realize that those two center numbers must be 1 and 8, since they're the only two that are consecutive with just one of the other numbers (namely 2, and 7). The rest magically fell into place after that.
The 5 rooms one is impossible. A similar problem called "the 7 bridges of Konig burg" proves why. The gist is that once you enter a room you also have to get out so the number of adgesent rooms has to be even. The only exception is the room you started in and the room you ended in. Since the mansion has more than 2 rooms with an odd number of adgesent rooms, it is impossible
The room one was impossible, I thought I had solved it so many times, but I had either forgotten a wall or crossed a wall twice. The numbers one, however, was very easy if you ask me. I solved it in less than a minute. The logical thing is to put the numbers that only correspond with one other number in the middle, since the each of the squares in the middle is adjacent to all but one square. So 1 and 8 only correspond with 2 and 7. Put 1 and 8 in the middle and 2 next to 8 and 7 next to 1. Then the rest is simple. Put 3 and 4 in the top and bottom corner furthest from 2, put 5 and 6 in the top and bottom corner furthest from 7.
Per Brian's initial instructions on the 5 room puzzle, "The ghost can pass through each wall exactly one time," it is not impossible. If you go by his instructions toward the end of the video, when he says "pass through all 5 rooms exactly one time," then it is impossible.
The 8 square one is just logic. 1&8 only have 1 adjacent number, so you make them most vulnerable, now 2&7 only have one possible position. 3,4,5&6 are easy from there.
I'm glad I was able to solve the eighth one because it's symmetrical there's only like three options. It took me around five minutes to just go through cuz I knew certain things can be next to each other.
The number puzzle is the possible one...7 at the far left, 2 at the far right, 4 at the upper left, six at the upper right, 3 at the lower left, 5 at the lower right, 1 at the middle left, 8 at the middle right.
Second one is definitely impossible. Rooms 3 & 4 share a single wall with room 1 and rooms 4 & 5 share a single room with room 2. Therefore either side you go with room 4 (to room 1 or 2), you are automatically taking away from the ability to go through the other room. However, if you allow me to stop mid-wall at the very end, I can solve it. Coded RWRW Room started and the wall, which wall Room ended and the wall. for R - 0 outside, 1-5 stated in video W is the wall that the ghost goes through. 0 if the ghost is outside 1 for top wall, 2 for right wall, 3 for bottom wall, and 4 for left wall Solution: 0011, 1400, 0034, 3300, 0043, 4432, 3113, 1224, 2341, 4254, 5300, 0021, 2200, 0052, 5123 (but end in between the walls)
n one more thing....U R awesome.... thnx so much for making so much efforts..... I heartily Respect u man....!!!!Hats off...!!!! For making d channel SCAM SCHOOL....I literary love d channel... One of d Best channel ...!!!
The eight-number problem was easy. You only have to notice that the middle two blocks are connected with every block except one. Therefore, 1 and 8 must go in those two blocks, forcing 2 and 7 in the appropriate left and right end blocks. 3, 4, 5, and 6 must go above and below -- say, the odds above and the evens below. Make sure 3 and 6 are away from 2 and 7. No problem.
I don't like how brian doesn't explain how the puzzle was done, just fills in the solution. For the number puzzle the first thing you need to figure out is that the 2 central numbers touch 6 numbers (out of 7) so that means they can only be occupied by 1 and 8 (since 0 and 9 aren't going to be included they only have 1 adjacent number). once you figure that out there is only one location 2 can go and one location 7 can go, once you do that there is only 1 locations 3 and 6 can go and so on until its done. Filling the boxes in isn't really going to annoy people as much as telling them what they missed. And in scam school that seems to be one of his goals :)
The way you described it, it is possible. I'm so confused because I think I'm doing something wrong, yet I'm passing through each wall. I think it's possible.
As many have said, the numbers one can be solved logically. It didn't take me long. The second one, however, CAN be solved strictly MATHEMATICALLY, as the host kept claiming that it couldn't be. There are several solutions, but the main thing to remember is the point where all line segments intersect is a part of each line segment. It is fairly simple to do most of the puzzle, leaving 3 line segments at the end which intersect. By drawing a line through that intersection, you have met all the conditions of the puzzle. By the way, one of the conditions of the original puzzle IS that the line you're drawing cannot cross itself. If you could it would be an easier puzzle to solve.
The rooms puzzle can be solved ;) Here's the instructions from the video verbatim: "Your goal [...] is to fly through each of the walls in the castle in one continuous run, passing through each wall exactly one time, no more, no less." The problem being, of course, that the long stretch of wall dividing rooms 1/2 from rooms 3/4/5 has two rooms on one side and three on the other. So while you would have to pass 3N ("North" wall of room 3), 4N and 5N once, for a total of 3 times, you can only pass through 1S ("South" wall of room 1) and 2S once each, for a total of 2 times. HOWEVER... The instructions did NOT specify that you have to pass through each wall at an angle of 90 degrees. So there is one solution, where you pass through one wall length-wise instead. Picture: imgur.com/LDGr1Vm Following the syntax of the rooms numbered as in the video, and using N/E/S/W for North/East/South/West, here's the path: Start in room 1. Pass through 1W. Turn around. Pass through 3W, 3E/4W, 4E/5W, 5E. Turn around the SE outside corner. Pass through 5S, 5N/2S, 2N. Turn around. Pass through 1N, 1S/3N, 3S. Turn around. Pass through 4S, 4N right through the wall the separates rooms 1 and 2 (1E/2W), and out the other end. Turn around the NE outside corner. Pass through 2E.
There's a whole branch of mathematics this problem created where it is clearly defined you would need to pass from one room to an adjacent room or inside to outside in order for it to count. There are any number of conditions they did not explicitly forbid here, but it's not a philosophical exercise and I'm sure with further query they would have clarified moving along a wall is not passing through. It's often presented as travelling to islands on bridges over water and was an actual practical problem studied by Euler known as the Seven Bridges of Königsberg, but it can be generalized as abstract concepts of nodes connected by lines where these are the only moves permitted. Walking along a wall is not a connection between nodes. This solution also doesn't include the paths moving through room 4 on the top left to room 1 or on the top right to room 2. It's impossible because you would always have to enter at least 2 rooms with an odd number of walls. Once you enter room 1, 2, or 4, there are only 4 valid walls left for that room, and whichever wall you choose to exit leaves only 3 options. Go back in and out of the room again, and there's only 1 wall left, which means you'd always have go through that wall last since it leaves you inside the room with no paths to get out. You can't finish in 2 different places at once, which would be necessary in order to complete it. You could start and end in an odd-numbered room so it could work with 2, but not with 3.
@@KalOrtPor The issue is that in mathematics it's obviously one way, in these type of pub question videos often the solution lies in a trick of how the question is worded. As such this should count as a solution because of the failure of the person asking the question. Marleen very likely does not believe they have outwitted the very old Seven Bridges problem but I would argue Marleen did outwit our spikey haired host due to his failure to be sufficiently specific in his question.
Going top row goes: 57 Middle row goes: 3142 Bottom row goes: 68 Honestly, i was starting to think that half way threw nicks time in doing that puzzle, but i needed to rewind it back before he gave the answer so that I could see the format because I forgot
I’m curious, for the room one, what happens if you pass through a corner? Will it count as moving through every wall connected to said corner? If so, does that present a solution?
Second one is a Euler trail, which is impossible. The first one is pretty easy once you put 1 and 8 in the centre 2 squares, then find places for 2 and 7, 3 and 4, then 5 and 6.
If you want to prove that the ghost problem is impossible: Call the outside part “room zero” Draw six boxes, on in the middle with the other five around it spaced out (you don’t have to do it like that, but it’s easier) Label the center box 0, and the other 5 are 1-5. Doesn’t matter which order. Draw lines connecting each pair of boxes which you can move between (0 connects with all the others, which is why you put it in the middle). A couple of colors is useful because you’ll have some crossing over. Now, the puzzle of traveling around this network going along every line exactly once is mathematically the same as the original problem, because the lines are the walls. Stop and think. Ignoring the very first and very last moves, each move into a room must be followed by a move out again along a different path. This has to remove two lines from the network. You can work out from this that any room with an *odd* number of connections must be either the start or the end, because that’s the only way to take care of the odd connection. The network for this puzzle has three rooms with odd numbers of connections (rooms 3, 5 and 0). You can’t do it: you only have two moves - the first and last - which can deal with an odd number, but you have three odd numbers.
The line is possible if you go through one of the corners. Aside from that there are squares with an odd number of points meaning that you cannot have three ends of one line so it is impossible if you don’t use corners
Simple reason why #2 is impossible: if a room has an odd number of walls, any valid path that goes through every wall either starts inside that room and ends outside, or vice versa. If there are 3 or more rooms like that, it's impossible. You have to start outside of at least two of those rooms and you can't end in both of those two (you can only be in one room at any given time). Thus the valid path condition about odd walls is broken for any path, so there are none. This is actually the same sort of problem as The Seven Bridges of Königsberg, solved by Euler in the 18th Century.
the ghost rooms puzzle is impossible because the 3 wide rooms all have an odd number of walls. If you start inside you can only can be outside after all walls are crossed and vice versa. Therefore there can only be at most 2 rooms with an odd number of walls to complete this puzzle: start in one of the two rooms and if there are 2 end in the other. The number puzzle can be tackled the following way: From the numbers 1 through 8 there are only two numbers that only have one adjacent number: 1 and 8. Put those in the middle squares (since they have the most adjacent squares). Now there is only one possibility for both 2 and 7 (the outer squares). Then there are two possible outcomes: put the 3 either above or below the 1. This leaves 4, 5, 6 and an open square above AND below the eight and a square above OR below the 1. This means that either there are two empty adjacent squares at the top or at the bottom. That means that 5 can not be in one of the adjacent squares, as that would mean there would either be a 4 or 6 next to it. So the 5 will be next to the 3. The 4 and 6 now only have one position to go as the 6 can not be near the 7.
If you only use the inner walls in the ghost rooms puzzle, then it is possible to solve as there are now only 2 rooms with an odd number of walls: the bottom 2 wide rooms: Start in one of the bottom wide rooms, go to the small room, go to the upper wide room, go to the same initial bottom wide room, go to the other bottom wide room, go back to the upper wide room, go to the remaining small room and then finish in the second bottom wide room.
I got puzzle one in under one minute and sat on two for about five minutes until I did the math on needing to have a start and end in each of the five sided rooms and then gave up.
only took me one try to get the number puzzle :D did the 5 rooms puzzle 3 times, and that last time I found out you just need 1 more wall and I would've had it I knew that was mathematically impossible at that point :D
I messed up the numbers one on the first try but figured 1 and 8 had to be the middle and 2 and 7 had to be on the far ends. Then just played with the rest until they fit.
The number one was quite easy. I always look for patterns and unique characteristics so the first observation I made was that every number is adjacent to two numbers except the starting and ending numbers in the range of numbers given (1 and 8), which are only adjacent to one number each. Then I looked to see which squares are the most restricting, which are the middle two as there's only 1 square not touching each. That means 1 and 8 have to be in the middle as they're the only numbers who can satisfy that. And also the numbers adjacent to each have to be in the only square they're not touching, which let's us fill the entire middle row with 7-1-8-2. From there filling the rest is straightforward, put 3 and 4 in the squares not touching 2, and put 5 and 6 in the squares not touching 7. The second one I quickly gave up on though because it was too tedious lol. XD
There is mathematical proof that can be done to show that the five rooms problem is impossible. What we essentially have are three rooms (1,2,4) which have 5 walls each and two rooms (3,5) which have 4 walls each. Because the rooms 1, 2, and 4 have an odd number of walls, then if you begin inside the room, in order to pass through all of the walls, you must end up outside the room and vice versa. Therefore, in order to complete a 5-walled room and continue to another room, you must start inside the room. Similarly, in order to complete a 5-walled room and finish the puzzle inside that room, you must start outside. You can set you start and end points in two of the 5-walled rooms, let's say 1 and 4. However, there is a third 5-walled room to complete. Since you began inside room 1 and must end inside room 4, then you must both start outside and end up outside room 2, which is impossible. A form of this puzzle can only work either if all of the rooms have an even number of walls or if exactly two of the rooms have an odd number of walls and the rest have an even number.
the easiest way to solve these, and prove the impossibility of the one is to realize the duplicity, both are series that must work both ways. Knowing that, it only takes a few minutes of testing to "solve" them.
**psst** Are you enjoying Scam School? You've gotta check out our *brand-new series* "The Modern Rogue" at ua-cam.com/users/modernrogue. If you've seen it, reply and tell me what you think. If not, then get on over there. I'll bet you a dollar you love it. -Brian
Scam School For the second one could you overlap one of the lines and have it still be counted as if it were to cross it?
I did the ghost one easily, it's not impossible. I have it done
Scam School I solved the 8 square one
Scam School i solved both legitimately.
Scam School I did the second puzzles, it is actually possible.
The numbers one is really easy, surprised they didn't get that one. Just knowing that 1 and 8 are on the ends and only can't touch one number means they go in the middle, which then means 2 and 7 have to go on the ends, and the other number are really easy from there. Interestingly I got the answer shown but with the middle flipped horizontally.
MizukiTHPS ikr? i took more of a visualization approach of all impossible places and worked it out from there
I solved it with the same way, 1 and 8 have to be in middle and 2 and 7 have to be on edges. That one was pathetically easy.
MizukiTHPS Yeah the numbers one was very easy, but the other one I didn't even try because I knew it was impossible.
Casey Lambert I did it and it's not impossible took me about a minute
Diane Kao Did you not watch the video? It is mathematically impossible to do the ghost one, but the numbers one was very easy.
numbers was ez af... took me 15 seconds
You're super smart.
Aw stop it you. You're making me blush 😊
+Master BatesTM or is he making you brush
wood
Lol same
Same
8:18* in the background*
When you forget why you entered a room
I did it in less than 10 seconds. it was obvious that at the middle you put 1 and 8.
ikr that one was so easy
Amirite I figured the number puzzle out in the first try and the second puzzle in 3 cuz that one was a lil sh*t
Same it was simple
Ikr
Same
first one was easy: since the middle 2 need to have the next number as the opposite outside box, you need to start and end with a middle box. For the second puzzle, theres always that one wall you dont see
I did the ghost one! Before everyone goes its impossible read this through. The ghost one is possible. You just need to be a t-rex ghost and when you walk through all 5 rooms at once, you only go through each wall once! To better explain, the ghost you always were, was a small line from a marker. Get a big paint brush and paint over the whole thing in one swoop. Then bam you win. Do i get my free beer?
Better start carrying a paintbrush out at bars
MandEParty OMG GENUISSS!!! HEYY Scam School HE FIGURED IT OUT YAYYY!!!
MandEParty It's called Houdini's Last Trick, it's literally impossible.
400lobo Well its impossible if you a sharpie like I said. If you use a paint brush its not impossible. To win you gotta bend the rules.
well while you are at the bar and someone challenged you.. are you going home and come back just to take a brush?
and they gave you max.5mins to solve. what are you gonna do?
I was able to solve the numbers one pretty easily but I tried the ghost one like twice and said I dont have the patience for it right now. The numbers one was easy because for the two center numbers you just figure two numbers you are using that only have one number each they cant touch each which would be the first and last numbers, the 1 and the 8. Then you figure you place the two numbers that cant touch the 1 and 8, which would be the 2 and 7, and you place them to where they cant touch, which is on the side ends opposite on the side they are on. From there everything else was easy to put in place, just fill in the blanks with the remaining for numbers
Glad I gave up on the ghost puzzle now that I see its impossible
+Robby Mejia **whew**. me, too.
+Scam School Ive decided to pick up magic as a hobby recently which brought me to your channel and im glad I came across your channel. Love your videos and love the scams. Your channel has shown me alot of cool new tricks and scams, now I just need to apply the techniques to scam some people
+Robby Mejia awesome. thanks!
+Scam School I watched the video where you put the cards in chased order to find out the top card and I couldnt remember if chased ran from the top of the cards down or the opposite so I did the trick with the top then I flipped the cards face up and did the trick and it worked both times. With that being said I noticed if I do the trick the way you showed it that the d on matching card will always be the matching card to the bottom so now when I do the trick I can tell them the card they have from the top of the deck and the card from the bottom of the deck
I did the numbers real quick, like less than a minute
same
...Jk
Took me 5 tries, but I got it
liar
Yup it's so easy... Just write down numbers and group every 2nd number so you get basic numbers for a side. The one space suggests that one side has to have high numbers and the other low... obviously having that guides you to set middle numbers counter the side numbers thus giving you a little mirror of numbers: high-low... easy..
the puzzle of the 8 numbers is the possible one, probably you will not belive in me, but I do it in my frist try. I just think before put the numbers and I realized the middle squares touch in all square except for one, so if is possible, the numbers 1 and 8 have to be in the squares of the middle, becouse this is the only numbers who doesn't have a predecessor and a sucessor, then you put the number 7 and 2 in the only possible place, after that the rest is easy.
I did it in the 2nd try :) It was quite easy. On the 2nd puzzle I didn't even bother because it's impossible. :D
+Pedro Paulo I didn't think of that. Damn.
+Pedro Paulo took me 8 goes to realise you had to put 8 and 1 in the middle, and then i got it
it didn't even took me a minute
+Pedro Paulo i realized this too on my second try
the first one was really easy i did it in 10 seconds XD you just have to realize that 1 and 8 have to go in the middle, because they only have 1 neighbor (2 and 7). everything else has two numbers next to it, ex. 3 is next to 2 and 4.
I knew the 8 one already, my uncle showed me that one in Scotland, over 30 years ago.
In an attempt to explain the impossible one (since it wasn't done in the video)Count the number of walls around each room.Three of them have 5 walls. A, B, C.
At the start, if the ghost is NOT in room A then it must end in room A. (In, out, in, out, in) the same goes for B and C.So the start point or end point must be in rooms A B and C,As there is only 1 start and 1 end it is therfore impossible.
er.. i paused the video after it explained the 8 squares trick.... took a piece of paper, solved it in ~30 seconds... so... lets do the next one!
Alex Mueller my pencil broke while i was trying the 5 rooms, so whatever..
Alex Mueller same here -.- easiest thing ever bet the 5 rooms thing had me fooled for a bit before realization that its impossible so...
Well I didnt have time to find paper and pen before I solved it. And I already knew the other one. And proved it to be wrong. And I'm 15 by the way XD
Albert Arvad Ottosen
i was 13 when i solved the first one
Wouldn't surprise me. Its not that hard. As long as puzzles can be solved by pure mathematics, they're pretty straight forward.
9:57 "huge double plus thanks" nice to see newspeak is finally catching on. Brian are you a member of the party?
Once you see that only 1 and 8 can go in the middle, the rest falls into place.
kilroy1964 good point.
M1sterW1zzz The Bottom-Middle room has a wall in the top left corner that has not been passed through
Nick B I wouldn't say crap, more like easy. The other one is easily proved impossible as well.
On the numbers riddle:
The two squares in the middle are each adjacent to _all other squares except one_. Each number has a number before and after it, except the first and the last, i.e. 1 and 8. So 1 and 8 must go in the middle. Their neighbouring numbers, 2 and 7, must go in the outer squares so that you have a row like 7 1 8 2 or 2 8 1 7. The 3 has to go above or below the 1 so that it isn't touching the 2. Along the same lines, the 6 has to go above or below the 8. But you can't put them _both_ above or both below, because then the two final squares for the 4 and 5 are adjacent. So if you put the 3 above the 1, then the 6 goes below the 8, or vice versa. Then the 4 goes beside the 6 and the 5 goes beside the 3, and you're done.
On the walls riddle:
Of the five rooms, three of them have 5 walls, which is an uneven number. If you start your line outside any of the rooms with with 5 walls, and ignoring all other rooms for the moment, your line will eventually go (1) in - (2) out - (3) in - (4) out - (5) in, and then you're stuck in the room. That means you can start your line within a 5-walled room and you can end it in one, but the third 5-walled room will not have all its walls crossed, otherwise you would have become stuck in it.
The one with the eight squares was very easy to solve.
Daniel Davidsen lol same :D
I paused the video, the number puzzle was easy. 1 and 8 go in the middle because they only have 1 number they can't touch and one space they aren't touching. 7 and 2 go at the sides with 3 above 1 and 4 below, 5 above 8 and 6 below. I haven't seen the answer yet and I am 12.
Surfboarder4
Okay and now what
The ghost took me literally 2 seconds off the top of my head, EZ you are a giant ghost.
I'm the first one took me about 15 seconds to get it.
2nd one is imposible, Euler Game dude.
I did it once and I'm 8 years old the impossible one Eulers trail
+nathanjv30 haha draw it in paint and send link to pic, I wanna laugh when I see you forgot 1 wall
+Kids in africa little Kids think they can beat maths. This is not a video game, this is the real live kid
2 things:
First, I like how Brian gave a hint to the solution of the number square puzzle
Second, the Euler Trail Puzzle as Brian described it technically has a solution because it involves a "Ghost passing through walls only once" and not "walking through each door only once". Technically, under Brian's rules, the ghost doesn't have to pass through a doorway, just through a wall, and you can pass through all 16 walls only once if you do a combination of Corners and single walls. Again, this is "Technically" and under *Brian's Rules*, not the actual rules.
Same I got the corner trick too
I knew 1 and 8 had to be in the middle because they are only adjacent to one other number. So 2 and 7 have to be on the ends and then I solved from there.
Michael Hardy well done.
Please... we all know that the second one is impossible. That puzzle is called Euler's Trail and I can easily prove that it is impossible.
Gary Lu I completed the ghost one. If you start outside and end up inside, it is completely possible.
nataleragemusic I know you did not solve it.
Red Eyes Sorry to say, but it is 100% impossible, no matter how hard you try. It is proven by mathematics, I was even studying this entire topology topic, you can't beat it.
+Red Eyes (Darkness Dragon) I think I got them
+Kainan Woodard go to twitter and search Warlock_Exo
the number one is realy easy (solved it in 30 sec ) because all u have to do is but 1 and 8 in the middle because they are the numbers that only have 1 number they cant be next too rest of the puzzle does it by his own
close, but I don't think that gets you there.
Scam School no it is the answer look 1 and 8 in the middle then you have to put 7 and 2 in the opposite corners because those are the only places where they can go. Then there are 2 places left where 6 can go and 2 places left where 3 can go the only error you can do now is put them both up or down but if you think ahead you will know that you cant do that becasue you have to put 5 and 4 then next to each other. So you but 3 and 6 diagonally away from eatch other and fill in 4 and 5 in the remaining spots.
***** yep did the same thing took me a minute
Or. 7
352
184
6
Daniel Spollin 4 and 5 are adjacent there
Excellent. A variant on the Seven Bridges of Königsberg demonstrating the impossibility of finding an Eulerian path in certain scenarios.
Glad you liked it!
The first one was easy.
I've seen the second one before. It's literally impossible.
I found four different solutions to the first one.
+Joshua Moyer there's only 2 solutions. Please explain the other 2?
*****
Not sure which two you already got, but:
64
2817
53
53
2817
64
46
7182
35
35
7182
46
Ahh flipping the 1 and the 8. I see, I guess I didn't exhaust all the possibilities. Thanks for clearing that up man.
if you switch 3&5 7&1 8&2 4&6 you have another answer which is correct
major gamer No that puts 2 and 3 next to each other as well as 7 and 6 next to each other
I swear the GOD i solved the first one
Show me.
_ | 1 | 3 |_
5 | 7 | 6| 8
| 2 | 4 |
im bad at this keyboard art ting
before watching it i swear plus this one is a different way
blindshot. 6 and 7 are next to eachother
For the number trick, think about it like this. The two middle boxes touch every other box except one--the ones on the end, respectively. Most of the numbers can't be touching two other numbers and so they CAN'T be placed in the middle (e.g. 3 can't touch 2 or 4; 5 can't touch 4 or 6). I put 1 & 8 in the middle boxes because each number could only NOT be touching one other number (1 can't touch 2 & 8 can't touch 7). So as soon as the middle row was laid out (7 | 1 | 8 | 2), the rest of it was easy.
Pfft, did the ghost one in my first try, too easy.
Same here and the number one took me 10 min.
That ghost one is the impossible one.
So no, you got it wrong and probably don't understand the game to call it 'too easy'.
645akz No I got it right and I aint stupid, and yes it is way too easy (for me).
Proof?
Trickelodean enrage There can't be a proof. If you want to cross a room with five walls with one line then you have to start at the inside to end outside this room. Because you have 3 rooms with 5 walls and the line has only 2 ends wich with you can start on the inside you will definitly end in the third room having to get in and out with only 1 wall. (Hope you understand my bad English because it really sucks)
The first one was easy, come on!
This guy must REALLY love beer.
hooked on this math pub challenge puzzles. keep up the good work
Second 1 is impossible
First one is
*35*
7182
*46*
the second one is possible
+Sohan Brar ni it isnt
+Cat119 The second one is possible and can be solved if your smart
it is mathematically impossible. Have you watched the video?
+Cat119 Its possible
there is a different solution for the possible pne and it was really easy
86
1324
56
+Virginia Webster the 2 and the 3 are next to each other and you left out a 7
+Virginia Webster Here is a hint. 1 and 8 have to be in the middle squares.
Wait why is everyone saying the possible one is the numbers? I failed badly on the numbers, but on the ghost I got it first try. What?
+MinecraftTestSquad oh nvm, missed two, I thought I did it, but I did solve the numbers after
I solved both of them lol
+Jose Martinez no you didn't. Check your ghost one again.
+Scam School on the ghost one are you able to go around the outside of the house? Cause thats what i did.
+Jose Martinez sure you can. It's still impossible to hit all the walls. en.wikipedia.org/wiki/Five_room_puzzle
+Scam School i can send you a picture tell me if it's wrong. Ill dm you on twitter?
+Scam School can you go threw all of the walls and end up inside or do you have to start and end on the outside?
I also like the similarity of these problems. #1 is kind of the opposite of #2. In #1 you have to find a sequence of vertices (rooms) that are connected by edges (walls) such that every edge (wall) is crossed. In #2 you have to find a sequence of vertices (boxes) that are NOT connected by an edge (not adjacent) such that every vertex (box) is crossed.
THEY WERE BOTH SO FUDGING SIMPLE!!! HAHA! Got them bot on one try :P
But Coollll! :D
VladKov36 - Clash and Craft um... It sounds a bit like you're celebrating that you solved them both.
You definitely, definitely did not solve them both.
Scam School I got really excited...then as the video continued and saw how hard it was..i reviewed them...i failed the numbers one, kept re-trying and still haven't found a solution, but pretty confident in the ghost one, every wall was gone through once, but does the ghost have to end up outside the house again or any room?
Lol, you replied right as i was gonna say something like this. :)
Scam School I just got the the part where you said the ghost rooms was impossible, ill tweet it to you,but i got through every wall once(or it looks that way). Both every wall once and every wall of a room once(I was a bit confused at the beginning lol)
VladKov36 - Clash and Craft send me a picture, but the odds are that you missed one of the two center walls. (those are the ones everyone misses)
Scam School You should have explained more about the math behind it. Perhaps not the reason, but a pointer on how to check if similar tasks are possible or impossible.
It´s the Seven bridges of Königsberg problem of sorts. And the way to check if the task is possible is to count the walls of all the rooms. If there are 2 or less rooms with an uneven number of walls, it is possible. Otherwise it is not.
The version you showed with the ghosts has 3 rooms with 5 walls.
An easy way to think of it is, if you enter a room, you have to exit it as well, subtracting 2 walls each time. But if you do this with a puzzle with more than 2 rooms with uneven number of walls, you end up in one of the rooms with no way out.
On the other hand, if there are 2 or less rooms with uneven numbers, you can just start in one of the rooms with uneven numbers and end in the other.
both it is possible
+Dhruv Aiyer nope.
+Scam School I did them both tho and i'm only 14 XD
+Wylie Wyant Ikr except I'm 13
thank you thay both are possible
+Wylie Wyant its literally mathematically impossible. Pretty sure VSauce did a video on it.
The 5 rooms one ain't impossible
BonanaSquid easy it's just diagonal lines through the boxes
oops forgot one my bad it is impossible
BonanaSquid It is impossible. 3 of the rooms have 5 walls. To draw a line going through each wall, one end gets trapped inside the room. Since the line only has two ends, you don't have enough for all three of the 5 sided rooms.
There both possible
sap2002 BonanaSquid Actually, the 5 rooms one is mathematically proven to be impossible and has been attempted over and over but never successfully but it is ok, I lie sometimes too.
Awesome vid, was trying both for ages! Finally got the number one but did it slightly differently though!
I did the 5 rooms one and I'm confused y he says its impossible
Nope. You missed one of the two middle wall segments (just like everyone else)... Check again
+Scam School hate to brake it to ya, but it's very much possible, I found multiple ways of doing it
+Scam School I got em all it is possible
show me. Because if either of you is right, you have a nobel prize for mathematics waiting for you: en.wikipedia.org/wiki/Five_room_puzzle
Well shit
I missed one but it wasn't the middle segment ;-; main reason I said I didn't cous thats the one u said I got wrong. Didn't notice the one lurking around there :/
I don't believe he explained the rules correctly for puzzle 2. If the only rule is that you have one like that crosses every wall once. Than it is very solvable. If you can't cross the same room more than once than it's just a stupid puzzle because in order to intersect the 4 sides of a room you must enter it twice.
No, he explained it correctly. Post a link to a pic of the solution if your so sure.
Point is that 3 rooms has an odd amount of walls (room 1,2 and 4). In order to "finish" such room you have to make an odd number of entrances/exits. In order to do that you have to start or end in that room. And since you can start/end in only two rooms you can never complete the third one.
He explained it correctly, he said you can't cross the same wall twice, not room, you can cross rooms more than once.
But it's still impossible for the reasons mroz123 stated.
+Bartek Strukowski I've done it tho
Only by going through a corner rather than a wall. Brian should have specified that they couldn't walk through a corner to count as entering from two walls.
I did the one they said was impossible on my second try
Tell us how then...
In the real problem, you gotta start in one of the rooms and end up in one of the rooms (not necessarily the same one) making it impossible. But, the way he set it up in the video, you can start outside of the rooms and it's quite easily solvable.
Why am I looking at these old videos? Because puzzles are fun. :)
The numbers were rather easy. The middle boxes just have one box that isn't adjacent each, and in a number sequence that is the first and last numbers, so 1 and 8 go into them. Following that, the numbers that are adjacent to them, 2 and 7, go into the far sides, as those are the only boxes they can go in. After that the remaining four are easy to place.
The rooms with the ghost, unless I'm intepreting what counts as a wall wrong, have the three large rooms having five walls each (due to one wall in each room being separated by other walls; the three "vertical" ones). If the ghosts goes into a room, and out of it, that's an even number of walls. As this is an odd number of walls, this means the ghost has to either start in the room and end outside, or start outside and end inside. Since there are three of those, it's logically impossible to go through all walls only once, since there's only one starting point and one ending point. Unless you pull an outside-the-box strategy and fold the paper (which is the solution of some similar puzzles).
Before watching, the number one is possible
13
8675
42
+Jackson DeStefano 6 & 7 together in the middle
35
7182
46
Thats the answer you can reverse two numbers above( 3 and 5) and the lower part.
+Khryj Paca well you have the 3 and 5 at the wrong spot the 3 should be over the 1 and the 5 over the 8. so if you had them in the right spots and reversed them than the bottom right of the box with the 3 would be touching the top left of the box with the 2
tlcd2006 This is the correct format
35
7182
46
+Khryj Paca yeah that's what i was saying
i could do the five rooms puzzle
+oliver clausen doubtful. Show me.
wait i found out i mad a mistake. But cool channel i love what you do.
+oliver clausen i did the number easily but im stuck on the wall one D:
+oliver clausen watched the rest of the video and no wonder :L
check my link !
Hey Brian. Normally I struggle with your puzzles. These, I solved in less than ten seconds. Feels had man.
I didn't get the right order for the 1 to 8
but it is easy to say the the ghost house is impossible, there are three rooms with an odd number of walls to cross, if you start outside you can't get out and if you start inside, you finish outside, meaning you are stuck inside the second one before the end.
Did the numbers in 5 seconds on the first try. Crossing the walls looked suspicious. I tried anyway and ended up with an art piece before I started counting the number of crossings. 3 rooms with uneven number of crossings leaves one hanging every time
Got both of them :) you cant start and end outside of a room with an uneven amount of walls, you can get around this by starting and ending your path in them, but there are 3 rooms w/ 5 wals in this puzzle, so you always have 1 wall left encrossed before you run out of legal moves.
With the ghost problem, I treated a couple corners as three walls -- the two left-right middle corners in particular. Start or end in the bottom middle room, and make sure you go through the corners to/from the bottom middle and the top ones. I have no idea whether it's allowed to treat corners like this.
I got the numbers one on my second try.
The key is to notice which boxes have the most adjacent other boxes (there are two) and which numbers have the fewest adjacent numbers (there are two of those, too). Put those numbers in those boxes, and the rest is trivial.
I had to break out eight 8-sided dice to solve the number one. Once you realize that the two numbers in the two center squares touch every other number EXCEPT the one on the far left (or far right), you realize that those two center numbers must be 1 and 8, since they're the only two that are consecutive with just one of the other numbers (namely 2, and 7). The rest magically fell into place after that.
When I was 10 it took 1hour...now I'm 32...almost 22yrs ago... I still give my friends this task... I really love it
The 5 rooms one is impossible. A similar problem called "the 7 bridges of Konig burg" proves why. The gist is that once you enter a room you also have to get out so the number of adgesent rooms has to be even. The only exception is the room you started in and the room you ended in. Since the mansion has more than 2 rooms with an odd number of adgesent rooms, it is impossible
Hey proud of me! Did the first chalenge on secund try! The 1-8 numbered pussle!
The room one was impossible, I thought I had solved it so many times, but I had either forgotten a wall or crossed a wall twice. The numbers one, however, was very easy if you ask me. I solved it in less than a minute. The logical thing is to put the numbers that only correspond with one other number in the middle, since the each of the squares in the middle is adjacent to all but one square. So 1 and 8 only correspond with 2 and 7. Put 1 and 8 in the middle and 2 next to 8 and 7 next to 1. Then the rest is simple. Put 3 and 4 in the top and bottom corner furthest from 2, put 5 and 6 in the top and bottom corner furthest from 7.
Per Brian's initial instructions on the 5 room puzzle, "The ghost can pass through each wall exactly one time," it is not impossible. If you go by his instructions toward the end of the video, when he says "pass through all 5 rooms exactly one time," then it is impossible.
The 8 square one is just logic. 1&8 only have 1 adjacent number, so you make them most vulnerable, now 2&7 only have one possible position. 3,4,5&6 are easy from there.
I got the exact same thing, only it was flipped horizontally. (In the 7 box I put 2, in the 3 box I put 5, in the 8 box I put 1, etc.)
so far I've only seen about 3 or 4 of your videos and have not failed a single puzzle yet!... yes I am a bit proud...
moving on!
I'm glad I was able to solve the eighth one because it's symmetrical there's only like three options. It took me around five minutes to just go through cuz I knew certain things can be next to each other.
The number puzzle is the possible one...7 at the far left, 2 at the far right, 4 at the upper left, six at the upper right, 3 at the lower left, 5 at the lower right, 1 at the middle left, 8 at the middle right.
Second one is definitely impossible. Rooms 3 & 4 share a single wall with room 1 and rooms 4 & 5 share a single room with room 2. Therefore either side you go with room 4 (to room 1 or 2), you are automatically taking away from the ability to go through the other room.
However, if you allow me to stop mid-wall at the very end, I can solve it.
Coded RWRW Room started and the wall, which wall Room ended and the wall. for R - 0 outside, 1-5 stated in video
W is the wall that the ghost goes through. 0 if the ghost is outside 1 for top wall, 2 for right wall, 3 for bottom wall, and 4 for left wall
Solution:
0011, 1400, 0034, 3300, 0043, 4432, 3113, 1224, 2341, 4254, 5300, 0021, 2200, 0052, 5123 (but end in between the walls)
n one more thing....U R awesome.... thnx so much for making so much efforts..... I heartily Respect u man....!!!!Hats off...!!!! For making d channel SCAM SCHOOL....I literary love d channel...
One of d Best channel ...!!!
I wish youtube did pictures because I have a picture of this mathematically impossible one
The eight-number problem was easy. You only have to notice that the middle two blocks are connected with every block except one. Therefore, 1 and 8 must go in those two blocks, forcing 2 and 7 in the appropriate left and right end blocks. 3, 4, 5, and 6 must go above and below -- say, the odds above and the evens below. Make sure 3 and 6 are away from 2 and 7.
No problem.
I don't like how brian doesn't explain how the puzzle was done, just fills in the solution.
For the number puzzle the first thing you need to figure out is that the 2 central numbers touch 6 numbers (out of 7) so that means they can only be occupied by 1 and 8 (since 0 and 9 aren't going to be included they only have 1 adjacent number).
once you figure that out there is only one location 2 can go and one location 7 can go, once you do that there is only 1 locations 3 and 6 can go and so on until its done.
Filling the boxes in isn't really going to annoy people as much as telling them what they missed. And in scam school that seems to be one of his goals :)
Man, knowing simple graph theory makes this pretty easy.
The way you described it, it is possible. I'm so confused because I think I'm doing something wrong, yet I'm passing through each wall. I think it's possible.
As many have said, the numbers one can be solved logically. It didn't take me long. The second one, however, CAN be solved strictly MATHEMATICALLY, as the host kept claiming that it couldn't be. There are several solutions, but the main thing to remember is the point where all line segments intersect is a part of each line segment. It is fairly simple to do most of the puzzle, leaving 3 line segments at the end which intersect. By drawing a line through that intersection, you have met all the conditions of the puzzle. By the way, one of the conditions of the original puzzle IS that the line you're drawing cannot cross itself. If you could it would be an easier puzzle to solve.
I actually figured out one of your puzzles for once!
The rooms puzzle can be solved ;) Here's the instructions from the video verbatim:
"Your goal [...] is to fly through each of the walls in the castle in one continuous run, passing through each wall exactly one time, no more, no less."
The problem being, of course, that the long stretch of wall dividing rooms 1/2 from rooms 3/4/5 has two rooms on one side and three on the other. So while you would have to pass 3N ("North" wall of room 3), 4N and 5N once, for a total of 3 times, you can only pass through 1S ("South" wall of room 1) and 2S once each, for a total of 2 times.
HOWEVER...
The instructions did NOT specify that you have to pass through each wall at an angle of 90 degrees. So there is one solution, where you pass through one wall length-wise instead.
Picture: imgur.com/LDGr1Vm
Following the syntax of the rooms numbered as in the video, and using N/E/S/W for North/East/South/West, here's the path:
Start in room 1. Pass through 1W.
Turn around. Pass through 3W, 3E/4W, 4E/5W, 5E.
Turn around the SE outside corner. Pass through 5S, 5N/2S, 2N.
Turn around. Pass through 1N, 1S/3N, 3S.
Turn around. Pass through 4S, 4N right through the wall the separates rooms 1 and 2 (1E/2W), and out the other end.
Turn around the NE outside corner. Pass through 2E.
There's a whole branch of mathematics this problem created where it is clearly defined you would need to pass from one room to an adjacent room or inside to outside in order for it to count. There are any number of conditions they did not explicitly forbid here, but it's not a philosophical exercise and I'm sure with further query they would have clarified moving along a wall is not passing through. It's often presented as travelling to islands on bridges over water and was an actual practical problem studied by Euler known as the Seven Bridges of Königsberg, but it can be generalized as abstract concepts of nodes connected by lines where these are the only moves permitted. Walking along a wall is not a connection between nodes. This solution also doesn't include the paths moving through room 4 on the top left to room 1 or on the top right to room 2.
It's impossible because you would always have to enter at least 2 rooms with an odd number of walls. Once you enter room 1, 2, or 4, there are only 4 valid walls left for that room, and whichever wall you choose to exit leaves only 3 options. Go back in and out of the room again, and there's only 1 wall left, which means you'd always have go through that wall last since it leaves you inside the room with no paths to get out. You can't finish in 2 different places at once, which would be necessary in order to complete it. You could start and end in an odd-numbered room so it could work with 2, but not with 3.
@@KalOrtPor The issue is that in mathematics it's obviously one way, in these type of pub question videos often the solution lies in a trick of how the question is worded. As such this should count as a solution because of the failure of the person asking the question. Marleen very likely does not believe they have outwitted the very old Seven Bridges problem but I would argue Marleen did outwit our spikey haired host due to his failure to be sufficiently specific in his question.
For the grid you can do a diffrence thing
2.
6
2
There is a diffrence in between the answer.And it always goes from bigger to least
Going top row goes: 57
Middle row goes: 3142
Bottom row goes: 68
Honestly, i was starting to think that half way threw nicks time in doing that puzzle, but i needed to rewind it back before he gave the answer so that I could see the format because I forgot
I’m curious, for the room one, what happens if you pass through a corner? Will it count as moving through every wall connected to said corner? If so, does that present a solution?
Second one is a Euler trail, which is impossible. The first one is pretty easy once you put 1 and 8 in the centre 2 squares, then find places for 2 and 7, 3 and 4, then 5 and 6.
If you want to prove that the ghost problem is impossible:
Call the outside part “room zero”
Draw six boxes, on in the middle with the other five around it spaced out (you don’t have to do it like that, but it’s easier)
Label the center box 0, and the other 5 are 1-5. Doesn’t matter which order.
Draw lines connecting each pair of boxes which you can move between (0 connects with all the others, which is why you put it in the middle). A couple of colors is useful because you’ll have some crossing over.
Now, the puzzle of traveling around this network going along every line exactly once is mathematically the same as the original problem, because the lines are the walls.
Stop and think. Ignoring the very first and very last moves, each move into a room must be followed by a move out again along a different path. This has to remove two lines from the network. You can work out from this that any room with an *odd* number of connections must be either the start or the end, because that’s the only way to take care of the odd connection.
The network for this puzzle has three rooms with odd numbers of connections (rooms 3, 5 and 0). You can’t do it: you only have two moves - the first and last - which can deal with an odd number, but you have three odd numbers.
The line is possible if you go through one of the corners. Aside from that there are squares with an odd number of points meaning that you cannot have three ends of one line so it is impossible if you don’t use corners
I solved the 5 rooms puzzle, but only based on Brian's description.
Simple reason why #2 is impossible: if a room has an odd number of walls, any valid path that goes through every wall either starts inside that room and ends outside, or vice versa. If there are 3 or more rooms like that, it's impossible. You have to start outside of at least two of those rooms and you can't end in both of those two (you can only be in one room at any given time). Thus the valid path condition about odd walls is broken for any path, so there are none.
This is actually the same sort of problem as The Seven Bridges of Königsberg, solved by Euler in the 18th Century.
the ghost rooms puzzle is impossible because the 3 wide rooms all have an odd number of walls. If you start inside you can only can be outside after all walls are crossed and vice versa. Therefore there can only be at most 2 rooms with an odd number of walls to complete this puzzle: start in one of the two rooms and if there are 2 end in the other.
The number puzzle can be tackled the following way: From the numbers 1 through 8 there are only two numbers that only have one adjacent number: 1 and 8. Put those in the middle squares (since they have the most adjacent squares). Now there is only one possibility for both 2 and 7 (the outer squares). Then there are two possible outcomes: put the 3 either above or below the 1. This leaves 4, 5, 6 and an open square above AND below the eight and a square above OR below the 1. This means that either there are two empty adjacent squares at the top or at the bottom. That means that 5 can not be in one of the adjacent squares, as that would mean there would either be a 4 or 6 next to it. So the 5 will be next to the 3. The 4 and 6 now only have one position to go as the 6 can not be near the 7.
If you only use the inner walls in the ghost rooms puzzle, then it is possible to solve as there are now only 2 rooms with an odd number of walls: the bottom 2 wide rooms: Start in one of the bottom wide rooms, go to the small room, go to the upper wide room, go to the same initial bottom wide room, go to the other bottom wide room, go back to the upper wide room, go to the remaining small room and then finish in the second bottom wide room.
The first time i actualy beat BOTH puzzles. I am soo happy
I got puzzle one in under one minute and sat on two for about five minutes until I did the math on needing to have a start and end in each of the five sided rooms and then gave up.
i solved the box with the 5 and 7 in the outside squares. then 8 and six in the middle and then 1, 2, 3 and 4 in the rest
this was posted on my birthday!!!! My 6th Birthday! Awwwe the good ole days
I watched a video before this and new the second one wasn't possible
only took me one try to get the number puzzle :D did the 5 rooms puzzle 3 times, and that last time I found out you just need 1 more wall and I would've had it I knew that was mathematically impossible at that point :D
Somehow, my social studies teacher took out a corner and magically made it possible.
I messed up the numbers one on the first try but figured 1 and 8 had to be the middle and 2 and 7 had to be on the far ends. Then just played with the rest until they fit.
Hey! I actually beat this one! If I ever run into you at a bar, you owe me a beer Brushwood!
I literally figured it out instantly after hearing the problem
The number one was quite easy. I always look for patterns and unique characteristics so the first observation I made was that every number is adjacent to two numbers except the starting and ending numbers in the range of numbers given (1 and 8), which are only adjacent to one number each. Then I looked to see which squares are the most restricting, which are the middle two as there's only 1 square not touching each. That means 1 and 8 have to be in the middle as they're the only numbers who can satisfy that. And also the numbers adjacent to each have to be in the only square they're not touching, which let's us fill the entire middle row with 7-1-8-2. From there filling the rest is straightforward, put 3 and 4 in the squares not touching 2, and put 5 and 6 in the squares not touching 7.
The second one I quickly gave up on though because it was too tedious lol. XD
There is mathematical proof that can be done to show that the five rooms problem is impossible. What we essentially have are three rooms (1,2,4) which have 5 walls each and two rooms (3,5) which have 4 walls each. Because the rooms 1, 2, and 4 have an odd number of walls, then if you begin inside the room, in order to pass through all of the walls, you must end up outside the room and vice versa. Therefore, in order to complete a 5-walled room and continue to another room, you must start inside the room. Similarly, in order to complete a 5-walled room and finish the puzzle inside that room, you must start outside. You can set you start and end points in two of the 5-walled rooms, let's say 1 and 4. However, there is a third 5-walled room to complete. Since you began inside room 1 and must end inside room 4, then you must both start outside and end up outside room 2, which is impossible. A form of this puzzle can only work either if all of the rooms have an even number of walls or if exactly two of the rooms have an odd number of walls and the rest have an even number.
I solved the number one in EXACTLY the same order that Brian did! Maybe the ghost is communicating between us...
the easiest way to solve these, and prove the impossibility of the one is to realize the duplicity, both are series that must work both ways. Knowing that, it only takes a few minutes of testing to "solve" them.
Took me longer than I care to admit to realize 1 and 8 had to go in the middle but once I realized that it was pretty easy.
In my 5th grade math class we had a while I tire class of doing puzzles like the 2nd one and etc, I solved all of them
For the ghost one, you don't need to cross your own path. Instead of starting from the outside, start from the inside. I figured it out