I want to get off 10 tile maze! I want to get off 10 tile maze! I want to get off 10 tile maze! I want to get off 10 tile maze! TH-cam 01 looks too intense for me! I want to get off 10 tile maze! I want to get off 10 tile maze! I want to get off 10 tile maze! I'm not going on TH-cam 01 - It isn't safe!
I remember as a kid that I would make very easy mazes and they would always walk in weird patterns and I'm like: "Wtf is wrong with these people" hahaha.
My favourite part about this maze is the bias against moving forward exists right to the end. Which means a guest that makes it all the way to the last couple tiles of the park-filling maze will likely turn around and slowly walk all the way back to the start.
@@coolkidsclub9943 Statistically for everyone that did solve it, the Earth formed and evolved for billions of years and we existed a second time as exact down to the atom clones of ourselves countless times. For every time that that maze is solved, I write this comment billions of times.
It's very, very unlikely they'd make it all the way back to the beginning. For every indent they pass, the chances increase that they'd enter it, at which point they have a 50-50 chance of turning around. Although it is pretty likely that a guest who made it almost to the end could still take several more days of real time to solve the maze, or even just get infinitely stuck despite literally being able to see the end of the maze.
@@professorracc.9780 No, poincare recurrence time for the universe vastly exceeds the expected solving time of this game. It's not even remotely close - it's not even large enough to be a rounding error in the exponent.
The 2018 Marcel: "You can charge a lot for this coaster, but the troughput is very low, making this a very inefficient ride". The 2020 Marcel: "This is how you make a maze with a 100% participation and 0 throughput"
@@Tomwesstein -- It's throughput is one guest every entire lifetime of a universe where one second is the length of the lifetime of another universe. Practically every guest coming out of the maze makes Xe-124 look like a toddler.
This explains why my mazes seemed to suck for the guests. I'd build a somewhat large, but easily solvable maze by our standards, but I watched guests spend over an ingame year before they got out.
"How many seconds in eternity?" The Shepard Boy says. "This is a labyrinth by Marcel Vos! It takes an hour to get past the first intersection, and an hour to go back where you came from! Every hundred years, a visitor comes and finishes it. And when every visitor finished the labyrinth once, the first second of eternity will have passed. You must think that's a helluva long time. Personally? I think that's a helluva maze."
After witnessing the fall of civilization, the extinction of the human race, our own Sun blacking out and eventually going supernova, the entirety of the Milky Way Galaxy being consumed by a supermassive black hole, the heat death of the universe, every single particle being attracted to a single point and, eventually, forming an entire new universe, over and over again, Guest #1422 had this to say: "Impossible Maze was great!"
That's not how the universe works. Our sun would not "black out" and would never go supernova(it's too small for that). It would instead blow up into a red giant possibly swallowing earth, but definitely mercury and venus. A few milion years later it would shed it's outer layers and become a white dwarf. After a really long time(in human timescales) it would eventually becoma a black dwarf(blacking out finally). The milky way galaxy would not be consumed by a black hole if our current predictions are correct. The heat death of the universe is literally every particle ceasing to exist due to entrophy. So they would not unite if heat death happens. And a new universe would not form. Guest #1422 would however finish before plance reocurrance time or something, which is the largest number that makes sense in out physical universe.
@@The360MlgNoscoper Oh, I'm fairly certain that an astrophysicist could point out every single thing that I got wrong, but I also didn't feel like peer researching my comment for the sake of a joke.
If he observed the heat death of the universe a trillion trillion times, it wouldn't even be equivalent to a single atom in the entire universe with regards to the [average] amount of time it would take to solve the maze.
The Spiffing Brit: "Marcel Vos is a guy who's trying to find the most optimal way of playing RC2." Marcel Vos: "Here's how to trap a guest in a maze until the heat death of the universe." Edit: Look. It's been three years. It's painfully obvious that this is an unserious joke comment. If you feel the need to make an 'um acksually' reply, I promise you that you're not as smart as you think you are.
Yeah, actually, when the heat death of the universe is happening, there will still be some park quests in the maze who hasn't even reached the 100th tile.
Whoa, the guest's strangely horrible "maze solving algorithm" might be similar to a gambling game stacked against you! It's like, "heads you win $1, tails you lose $1", but the coin is three times as likely to land tails than heads. And "beating the maze" means you've profitted $n overall, where n is the number of indents in the maze. I guess the benefit of this maze game is that you can't go into the negatives, so that's a nice plus...
@@bensullivan5562 Ben, despite the incredibly long timespan it would take guest to complete the maze it's nowhere near infinity. Number marcel proposed is very large, but nothing compared to infinity
If you think about it, many things are terrible RNGs. If you hit a light switch, the light will turn on most of the time. If it doesn't, it usually won't no matter how often you hit the switch.
@@SuperAWaC That would take multiple RNG calls, which would have been computationally expensive and by an unpredictable amount as well. What if the guests all keep randomly selecting directions they can't go and have to pick again? This was a pretty reasonable solution IMO.
@@Graknorke things like xorshift are so fast (a few clock cycles) its not really different from simply adding 1 to the direction (1 clock cycle) considering every other aspect of the game needs more computational power than that.
You should see what the loop prevention algorithm looks like in large corporate networks. It's an impressive body of work under the hood - and far more reasonable to deal with at the configuration level. www.diva-portal.org/smash/get/diva2:1004743/FULLTEXT01.pdf
I'm just imagining some brave hero visitor trudging for an eternity and finally reaching the exit gate, just 1 tile away, and the RNG decides to send him into a backwards spiral that takes him all the way back to the beginning. I imagine all the children asking him what it was like out there at the end.
So I've had a nice scroll through the comments and I'm fairly certain no one picked up on the date being 6/9/420 at 6:24. Well played Marcel, well played.
So what you're saying is, roller coaster tycoon god has made a maze so long even he can't solve it before the heat death of the universe. How big would the maze have to be have the average solve time be the death of the sun, the heat death of the universe? I think this would be a neat way to represent galactic timescales.
The heat death of the universe is around 10^100 years according to wikipedia. To get there we would only need around 680 indents, which is less than 1% of the park filling maze.
@Mathew DeCoste To add, I don't know if this helps, or makes it even worse to visualize, but taking the avg solve time Marcel came up with divided by the heat death of the universe to find how many heat deaths would occur gives us 6.6 x 10^19,658 individual heat deaths added together, which is, extremely close to the avg solve time, showing off just the raw difference in scale between the two.
Actually, if you map out the random number generator function that the game uses, it's incredibly likely that there's no achievable sequence of RNG calls that makes the maze possible to complete at all.
I don't know the first thing about how RCT assigns RNG, but my initial reaction is that the large number of events in the park pulling RNG is going to create an amount of noise that means it should be possible.
I think you're right, but it is hard to say with certainty. Chris Sawyer did most of his programming in either C or Assembly, as I recall, and the rand() function in C has a maximum value of 32767. If the random number generator could produce a list of every possible permutation of numbers from 1 to 32767 (which it can't), then it could finish even much larger mazes. But because I don't know how random rand() is, I can't say what the largest possible solvable maze could be.
For some reason, I was laughing so hard when he was explaining just how large those numbers were. Imagining a guest being unable to complete that maze before the heat death of the universe is just too good. lmao
To follow the code of honour Die with a sword in your hand That is the way of the warrior Your forefathers left on the land To live by the sword you must die by the sword
@@BlueRaja Introductory statistics definitely covers things like calculating the minimum/maximum amount of time spent in such a maze, it's just some probability multiplied with itself, in this case "slightly" more often than with the typical examples with dice rolling etc. but the principle stays the same and a calculator won't care.
Due to almost all games using pseudorandom number generation, and since there is a finite set of seeds for said number-generation, this maze *may* be genuinely impossible. For example, the Mersner Twister PRNG has a period of just 10^6001. I have no idea how one might formulate a proof that this particular maze is impossible, as specifically testing every possible seed across every step of its period would be almost as impossible as the maze itself.
@@badidea6034 Err, no. Outside of what I discussed (i.e. the fact that this is pseudorandomness, and not randomness); Even if the chance of it occuring is on the order of 1/10^1000, mathematically speaking, it is not 0. Thus, it is only *nearly* impossible, unless it can be proven that no outcome possible from the PRNG would result in a completion.
@@tales9476 Great point, but RCT running on a different computer - say a NASA supercomputer may have a different implementation of RNG that has a longer period or is "true random" (based on quantum events)
@@kw4093-v3p Er, no. The algorithm RCT uses is deterministic. Start with the same seed, and get the same results. The period of the RNG algorithm is a property of the algorithm itself, not the hardware it is run on. If a computer were using true randomness (i.e. particle-detectors, etc.), instead of the PNRG algorithm, then that isn't vanilla RCT.
The park filling maze might actually be unsolvable by the guests' AI due to how random number generators aren't actually random. When you have to create that many random numbers the algorithm will eventually loop back on itself.
Whaddaya mean "walk forward"??!1? I CAN'T use "just walk forward algorithm" when "in maze"-flag is TRUE! Ugh, stupid malfunction people don't even know their algos...
"If you want to solve a maze, just keep going left" "Uhh are you sure? This one just looks like a straight path with a bunch of tiny dead ends." "Shhh just trust me it works"
In a maze like this, you'd actually be done relatively quickly, as you'd move towards the exit every single time. You'd end up stepping on every single tile of the world, but only once (in the fingers) or twice (in the main pathway) each.
@@zgthor No, all you need is at least one single loop within the maze. Think a simple straight wall that can be circled. Or in other words, a maze that consists of not only dead ends.
I'm surprised no one has done this yet - I'll give you the exact math for how much each left indentation gives. Because walking backwards has different behavior depending on where you start walking backwards, let's track the current indentation with a variable: x. For the first indentation, x=0. For the second, x=1, etc. We have the first indentation at x=0 because, if the NPC enters the indentation and exits to the right (so they are facing the entrance), they make zero choices before getting back to the first indentation again. Right now, our goal is to calculate the expected number of units an NPC moves to reach position x+1 from position x. Let's say the total number of indentations is n. Now we are ready to begin: let b be a NPC, and let b start at position x, facing forward. As mentioned in the video, there is a 3/4*1/2=3/8 chance they end up at position x-1, facing backwards, when they make their next choice not at position x. Similarly, there is a 1/4+3/4*1/2=5/8 chance they end up at position x+1, facing forwards, when they make their next choice not at position x. B moves 4 units when they end up facing backwards at position x-1 (1 into the indentation, one out of the indentation, and two to get to x-1), and has a 1/4 chance of reaching x+1 in just 2 moves, and a 3/4*1/2=3/8 chance of reaching x+1 in 4 moves. Now let's say b is at position x, facing backwards. As mentioned in the video, there is a 1/4*1/2=1/8 chance they end up at position x+1 facing forwards when they make their next choice not at position x, and a 3/4+1/4*1/2=7/8 chance they end up at position x-1 facing backwards. Upon ending at position x+1 facing forwards, b has moved 4 units. There is a 3/4 chance b ends at position x-1 facing backwards in 2 moves, and a 1/4*1/2=1/8 chance b ends at position x-1 facing backwards in 4 moves. This is all we need to begin the hard part: calculating the expected number of moves (aka units moved) to reach position x+1, facing forwards, from position x, facing forwards. I will use the notation Y_x to mean the random variable denoting the number of moves to reach position x+1 (facing forwards) from position x (facing forwards), and Z_x to mean the random variable denoting the number of moves to reach position x+1 (facing forwards) from position x (facing backwards). Note that it is impossible to reach position x+1 facing backwards without first reaching it facing forwards, so this sort of notation comes naturally from asking the question. Let's start with a simple case: starting the maze. In this case, b is at x=0 facing forward, so we are talking about the random variable Y_0. There is a 3/8 chance b enters the first indentation, does a 180, exits the first indentation, turns right toward the start, makes one move, then does a 180 and makes one move back to where they started (4 moves). There is a 3/8 chance b reaches position x=1 in 4 moves, and a 1/4 chance b reaches position x=1 in 2 moves. Therefore we can write the expected number of moves Y_0 as E[Y_0]=3/8*(4+E[Y_0])+3/8*4+1/4*2, which we can rewrite as 5/8*E[Y_0]=28/8, so E[Y_0]=28/5. We can also calculate the expected number of moves Z_0 as E[Z_0]=1/8*4+1/8*(4+E[Y_0])+3/4*(2+E[Y_0]) = 5/2+7/8*E[Y_0]=5/2+49/10=37/5. We can confirm these values by compiling and running the C code I uploaded to pastebin here: pastebin.com/kcxugAhB. dist is the distance to travel (if, for example, this is set to 2, a round in the maze ends when x=2 is reached for the first time), startdir is 0 for forwards and 1 for backwards, and startpos is the current x value (x as used in this comment). Now, note that, for x>0, we can calculate E[Y_x]=3/8*4+3/8*(4+E[Z_{x-1}]+E[Y_x])+1/4*2=7/2+3/8*E[Z_{x-1}]+3/8*E[Y_x] so E[Y_x]=56/10+3/5*E[Z_{x-1}], and E[Z_x]=1/8*4+1/8*(4+E[Z_{x-1}]+E[Y_x])+3/4*(2+E[Z_{x-1}]+E[Y_x])=5/2+7/8*E[Z_{x-1}]+7/8*E[Y_x]. This is a complex recurrence relation, but using the base cases and Wolfram Alpha, we can solve it, giving E[Y_x]= 1/10*(-55 + 111*(7/5)^x*UnitStep(x-1) + 111*UnitStep(-x)). The UnitStep(n) function is 0 when n0. I think it's easier to see what's going on if we instead write it casewise: for x>1, we have E[Y_x]=11.1*(7/5)^x-5.5, and E[Y_1]=10.04 and E[Y_0]=5.6. We don't really care about E[Z_x] but Wolfram Alpha solves it as well, giving E[Z_x]=37/10*(-5+7*(7/5)^x). We are almost done now: we have calculated the expected number of moves between each value of x, so we just need to sum up the expected values from E[Y_0] to E[Y_n]. Assuming n>0, this gives the expected number of moves required to reach the nth position in the maze as f(n)=38.85*(7/5)^n-5.5n-33.25. In asymptotic notation, this is in Θ((7/5)^n), which means that, as n gets arbitrarily large, adding an additional indentation multiplies the time required by 7/5=1.4, which is pretty close to what you got in the video! Neat! I think I may be slightly inaccurate by assuming the hedge is as wide as the indentation is deep (i.e. hedges are placed in a unit grid and each takes up a 1x1 space).
Dreams luck is significantly easier to achieve than this lol. What would take longer, getting struck by lightning twice in a row, or picking up every atom in the universe and waiting one Google amount of years in-between? The universe hasn't even been alive for one TRILLIONTH of the time that each of those atoms would take to pick up XD
Imagine being so petty you'd lie to your nine-year-old fans about cheating and still get fifth place in a speedrun. I just don't understand people like Dream.
Probability: The chance that one of the NPCs finishing the maze is so vanishingly close to 0 that any system that simulates it will invariably break down and decay completely into radiation before any of them can complete the maze. Marcel: *So you're saying there's a chance*
The stupid thing is the guests will jump to peek over the wall when they go through a maze, but even then they can't see that they just need to go straight lol
It is a common strategy for mazes, but it no longer works if there is a loop in the maze (unless of course you realize you've been walking in circles. I have actually designed mazes in the past purposefully designed to not make this strategy work regardless if you are a right or left hugger)
I like how you can see the guests going from one maze to the next, thinking "oh that was easy let's check out the other one!" And then get trapped for the rest of their existence.
when you get to this levels of random, you might want to consider looking into how RCT handles random number generation, as you might go long enough to create an RNG loop
These ain't rookie zeros. 10^19758 goes beyond all physical meaning. For you see, a plank length is 1.6*10^-35 meters. A plank cube would be 2.5*10^-103 cubic meters. The radius of the observable universe is 4.4*10^26 meters. The volume of the observable universe is 3.6*10^80 cubic meters. There are about 1.4 * 10^183 plank volumes in the observable universe. Now let's have a million universes, each a hundred times larger than our own. We get 1.4*10^191 Hm...still 19,567 orders of magnitude short. Let's say each of these plank volumes blinks every microsecond, from the dawn of time until our sun burns out. Across all of our million universes, there will be 2.2*10^214 blinks. Still 19,544 orders of magnitude short *We haven't even made a dent* Now let's say a god is writing a book for his 10^214 friends. To be specific, he is writing the complete dictionary of all 470,000 words, with an extensive definition for each 500 words long. But this god can only write a letter every 500 billion years.When he finishes, he begins translating the book into every language that has ever existed (about 10,000), one character at a time. By the time he is done creating a complete set of dictionaries for all his friends, 2*10^239 years have passed. Still not even a dent. Now don't get me long, it's certainly possible to conjure up a much larger number. But only if you resort to abstract math. 10^19758 goes beyond all physical meaning.
@@tomc.5704, But see that's when you need to start bringing in a larger scale. Just because the amount of zeros goes beyond all physical meaning to humans, doesn't mean that it has no meaning. Because on the universal scale, humans are as small as, if not smaller, than a higgs boson particle.
My question is, how does this affect the rating of the maze, if at all? Does increasing the difficulty of it increase it's excitement rating, or does it only count towards the amount of tiles used? How many tiles of left indents do you need for guests to pay $20.00 for the maze?
the stats of a maze are determined by the amount of tiles used - each additional tile adds a very small fixed amount to the excitement and intensity. however, there's a cap at one hundred tiles, so any maze bigger than that will have identical stats despite being more expensive and having worse throughput
My guess was "an absolutely meaninglessly, incomprehensibly large number" so I'd say my guess was basically spot on. If anything it was a bit smaller than I was expecting. Also to all the people talking about the finite period of RNGs and so on: While it's true that the PRNG in games like this has a finite (albeit very large) period, just because it has a small period doesn't make it trivial to prove that it's impossible to use that PRNG to generate the necessary sequence to let a guest escape. There are many, many other things in the game that will advance the PRNG by a most likely unpredictable amount, which will lead to an incredibly difficult to predict system. So while the PRNG is easy to predict, the behavior of the system is not. If you could see state of the RNG and manipulate it through actions in the game, to cause it to be incremented a different number of times (for example, how tool assisted speedruns work), you could probably trivially make a single guest walk all the way to the end of the maze without ever turning into a single alcove.
Me: But you can't buy things while on a ride. Six Flags Manager: Mod in food stalls. Six Flags Employee: Sir, we don't need mods, we can just cut the hedges. Six Flags Manager: Even better!
There's a news story about a couple at a corn maze, they called 911 two hours after the maze had closed because they were lost and unable to find the exit... Or think about simply stepping through the corn to go to the building that they could clearly see from where they called 911 from...
@@veggiet2009 Protip: Any maze that doesn't have a 4 way intersection can be solved by turning the same way at every intersection, though it might take a while if it's particularly complicated. Even if it does have a 4 way intersection, it can still be solved by picking a different (cardinal) direction every time you reach it. It only starts to get complicated when the maze has multiple 4 way intersections.
@tommyjay_97 "Viviene S." literally just said that with a grumpy face, prompting me to try to figure out how to end her maze misery. 🥺😅...worse case, I'll just delete the maze entirely 🤦🏾♀️😂
Imagine being citizen when Attilla the Hun was starting his campaign in Europe. You escape into a maze and wander around it till you find the exit. When you exit you find out this thing called WWII was happening.
Cern: Yeah, six sigma is a high enough confidence interval to determine the mass of subatomic particles. Marcel Vos: Plebs, grahams number sigma or GTFO
Note that the end of the universe is not really agreed upon and 10^100 (a googol) years is the assumption for heat death with proton decay. Without proton decay the equivalent would be slightly higher with 10^(10^120) years. And it still isn't entirely dead after that. Probably. What I'm saying is: Start the computers, there is still a chance.
I've never played this game, and it doesn't matter - I literally enjoyed this video more than anything else on TH-cam to date! A computer game, analysed to the extreme, dripping with snark.... brilliant!
you mean day 529292518473829572719384726274928263728283726174939172538492726393917374829204750602963318204017274920274829472949282747493010373838263636
Being a mathematician, I really like your scientific approach. Your materials on longest ride time records on RCT2 (the synchronised coasters with your deterministic analysis and this one with your probabilistic analysis) could probably make a publishable scientific article.
I guessed that it would take 400 million real-life years to solve. I was wrong by approximately 6.6 x 10^19758 years. 400 million years isn't even noticeable on that timescale. It would be roughly as noticeable as a single atom being removed from our universe. If our universe was multiplied by itself 233 times.
I love these vids. Just a guy torturing virtual people in an ancient game. It would be hilarious if once a person finished the maze, they went back in it.
my guess was "heat death of the universe" and it turns out that was actually lowballing i can't- (had completely forgotten this video until binge watching the whole channel and oh WOW do i not envy you when building that maze. the dedication sure, but the actual process? surprised you didn't go insane during it. well, any more than you already might've been at least ;p)
HP printer factory worker: 'Sir, we've run out of printer ink!' manager: 'How is this possible?!' worker: 'Marcel Vos accidentally printed out his impossible maze calculations'
11:10 "There is a chance a guest will walk from the entrance to the exit without going into a single indent." Maybe with a true random number generator, but some rough math indicates you'd need your RNG to have a state size of on the order of at least 8 kiB for there to be even a 50% chance for this to be possible. Even a _top-of-the-range_ Mersenne Twister has a state size of only around 2.5 kiB, so even if it used that there'd still only be about a 1 in 1ᴇ48836 chance of there even _being_ a configuration that results in a guest traversing the maze in a single go. With the (at best) 32 bit state size of the linear congruential generator RCT probably uses, the chances are more like 1 in 1ᴇ54992. I guess you could argue that the states of the thousands of guests traversing the maze are in effect an extension to the RNG state size (since they're running around soaking up RNG rolls in a chaotic manner), but I still seriously doubt there are enough effective bits there to matter. *TL;DR:* There are only around 1ᴇ10 (give or take a few orders of magnitude) possible sequences of guest direction choices that the game is capable of producing. That's a modestly large number, but it's so small in comparison that "go through the maze without taking a single detour" almost certainly _isn't_ one of the possible sequences.
Every time I watch one of your videos I think "This MUST be the last one. He CAN'T find anything else in this game to talk about" And then BAM! "The Impossible Maze"
5:55 I'm mad that I am very likely not going to live to see the dawn of September 9th, 2420. The train has already left on the actual year 420, so that's my next best chance, lol.
Been playing this game somewhat obsessively since I was 10 in the year 2000, and have been happily binging your videos for the past few days. I am playing RCTC on my iPad right now, but I definitely need to step up my game and get OpenRCT. Thanks for the videos!
Wow. Just wow. It starts so mild, I was thinking, "oh I know why they do that, this video may not be so interesting," and it just literally ramps up exponentially from there.
@@MarcelVos I suppose the file-format of exported rides is known, so I guess it would technically possible to program a thing that would generate such a file for a X size maze like this
Here's my prediction for how long it takes a guest to solve the park-sized Impossible Maze: precisely as long as it would take a guest to ride The Century Coaster from start to finish - 323,599,680 RCT days, or 1,320,815 RCT years Update from seeing how long it takes: Oh. My. God. RCT guests are really stupid.
they have maps in hand and still cant get to bathroom 1 on the next path over... im surprised you had any hope for them. also: why do staff get lost in their own park?
@@jackradzelovage6961 I believe it's called "Malingering" They'll get paid so long as it looks like they're working, so they're always going somewhere, but really they're just wandering around.
@@aprinnyonbreak1290 malingering is more faking a medical problem to get out of work, but yeah theyve gotta be doing it deliberately. i mean i can definitely find my way when there are four ferris wheels stacked on top of each other XD
I don't know how I ended up here, but I'm glad I have. I got this game in a cereal box about 20 years ago and used to play it on my grandmother's Windows XP computer when I was barely old enough to read. Completely forgot this game existed or that I had a cd copy in an old jewelcase for decades, and now it's interesting to suddenly see it again with some mathematical equations and NPC AI breakdowns all these years later.
Guest 292189: "This ride is good value for money."
Is that his age?
@@UnknownCleric2420 it's his killstreak
It's amazing value. You ride for the rest of your life!
@@carpetman9191 and many others
Guest 9929399 at indent 4, Looks like I am close to solving the maze I can see the exit
Marcel was actually stuck in this maze, it’s why he’s been gone for a while. Glad you found your way out, good to see you back.
Marcel has been building his parks from the inside all along. Soon enough he will free himself from his video game prison.
I want to get off 10 tile maze!
I want to get off 10 tile maze!
I want to get off 10 tile maze!
I want to get off 10 tile maze!
TH-cam 01 looks too intense for me!
I want to get off 10 tile maze!
I want to get off 10 tile maze!
I want to get off 10 tile maze!
I'm not going on TH-cam 01 - It isn't safe!
He thoroughly tests for quality no matter what.... even if you must get personally involved
I guessed 5.9 million years. I was a few digits off or so...
He had to test the mazes out himself.
“Don’t go into that maze. Your chances of going through are only 9.3 * 10^-55,000%!”
“So you’re saying there’s a chance.”
Every gacha player
chances of getting through first time*
dream stan when someone said their god cheated :
Gacha players : "So thats roughly fifty fifty."
"never tell me the odds!"
I remember as a kid that I would make very easy mazes and they would always walk in weird patterns and I'm like: "Wtf is wrong with these people" hahaha.
The fact that he made people run his mazes for 2000 years is some Black Mirror shit that gives me existential dread
you live in a black mirror episode, its called "the West"
I want to get off Mr Bone’s Wild Ride
@@ValaAssistant China literally has a social credit system...they've got the West beat
@@camhenchy2511 tiananmen square
Good thing these NPC's aren't conscience. I think. 0_0
"I want to get off Mr. Marcel's Wild Labyrinth!"
to make it ultimate mr bones, change the exit of the maze a path right back to the entrance.
*The labyrinth never ends!*
Ludicrous labyrinth*
Illusionism Ay I know you from brick hill! We talked a while ago my name was Danton
"My gut feeling says I should head towards the entrance again." - In-game figure probably
My favourite part about this maze is the bias against moving forward exists right to the end. Which means a guest that makes it all the way to the last couple tiles of the park-filling maze will likely turn around and slowly walk all the way back to the start.
Statistically for every one that did solve it, countless more would have likely done that very thing
@@coolkidsclub9943 Statistically for everyone that did solve it, the Earth formed and evolved for billions of years and we existed a second time as exact down to the atom clones of ourselves countless times. For every time that that maze is solved, I write this comment billions of times.
Imagine how frustrating it would be to witness this.
It's very, very unlikely they'd make it all the way back to the beginning. For every indent they pass, the chances increase that they'd enter it, at which point they have a 50-50 chance of turning around. Although it is pretty likely that a guest who made it almost to the end could still take several more days of real time to solve the maze, or even just get infinitely stuck despite literally being able to see the end of the maze.
@@professorracc.9780 No, poincare recurrence time for the universe vastly exceeds the expected solving time of this game. It's not even remotely close - it's not even large enough to be a rounding error in the exponent.
The 2018 Marcel: "You can charge a lot for this coaster, but the troughput is very low, making this a very inefficient ride". The 2020 Marcel: "This is how you make a maze with a 100% participation and 0 throughput"
can't minmax with out the min ;)
Not 0% throughput, but a very very very tiny amount
@@Tomwesstein -- It's throughput is one guest every entire lifetime of a universe where one second is the length of the lifetime of another universe. Practically every guest coming out of the maze makes Xe-124 look like a toddler.
I am the 666th like
@@TlalocTemporal Now I wonder what the half-life of such a maze would be once you let it fill with guests and then close the entrance.
This explains why my mazes seemed to suck for the guests. I'd build a somewhat large, but easily solvable maze by our standards, but I watched guests spend over an ingame year before they got out.
That means your guests are braindead
Just build some food-options in there
Yep, so the strategy to making easily solvable mazes is to branch most of your dead-end paths from the right of the correct path.
"How many seconds in eternity?" The Shepard Boy says.
"This is a labyrinth by Marcel Vos! It takes an hour to get past the first intersection, and an hour to go back where you came from!
Every hundred years, a visitor comes and finishes it.
And when every visitor finished the labyrinth once, the first second of eternity will have passed.
You must think that's a helluva long time.
Personally? I think that's a helluva maze."
Awesome reference! Hats off sir... or should that be a fez?
@@DeetexSeraphine As long as it's not a bowtie
@@SaHaRaSquad of course it is, bowties are cool
@Xx Yy dr. Who.
@Xx Yy Doctor Who. The episode "Heaven Sent" amazing episode btw
After witnessing the fall of civilization, the extinction of the human race, our own Sun blacking out and eventually going supernova, the entirety of the Milky Way Galaxy being consumed by a supermassive black hole, the heat death of the universe, every single particle being attracted to a single point and, eventually, forming an entire new universe, over and over again, Guest #1422 had this to say:
"Impossible Maze was great!"
_Guest __#1422__ can't find entrance to Impossible Maze_
"What a great value!"
That's not how the universe works. Our sun would not "black out" and would never go supernova(it's too small for that). It would instead blow up into a red giant possibly swallowing earth, but definitely mercury and venus. A few milion years later it would shed it's outer layers and become a white dwarf. After a really long time(in human timescales) it would eventually becoma a black dwarf(blacking out finally). The milky way galaxy would not be consumed by a black hole if our current predictions are correct. The heat death of the universe is literally every particle ceasing to exist due to entrophy. So they would not unite if heat death happens. And a new universe would not form. Guest #1422 would however finish before plance reocurrance time or something, which is the largest number that makes sense in out physical universe.
@@The360MlgNoscoper Oh, I'm fairly certain that an astrophysicist could point out every single thing that I got wrong, but I also didn't feel like peer researching my comment for the sake of a joke.
If he observed the heat death of the universe a trillion trillion times, it wouldn't even be equivalent to a single atom in the entire universe with regards to the [average] amount of time it would take to solve the maze.
“Daddy, what’s the biggest number?”
“The number of years it takes a guest to finish Mr. Marcel’s Hilariously Huge Hedge Maze, son.”
tree(3)
@@OriginalPiMan Rayo's Number >>>>>>>>>>>>>>>>>>>>>>> tree(3)
tree(tree(tree(3)))
@@nhilario7638 still way smaller than Rayo's Number. Its not even in the same scale of comparability.
@@OriginalPiMan pretty sure that tree(4) is bigger than tree(3)
And of course the game is paused at 6/9/420 during the 10 maze testing results screen
nice!
Yes! Someone noticed it!
@@MarcelVos you are an amazing human being
😂 💯
sorry, can u explane?
The Spiffing Brit: "Marcel Vos is a guy who's trying to find the most optimal way of playing RC2."
Marcel Vos: "Here's how to trap a guest in a maze until the heat death of the universe."
Edit: Look. It's been three years. It's painfully obvious that this is an unserious joke comment. If you feel the need to make an 'um acksually' reply, I promise you that you're not as smart as you think you are.
I believe this number is so big that universe would create and destroy itself multiple times before first guest gets through.
Though the same thing.
@@thicccrusade2302 But what if that number is also the number of universes that each have a guest trying to solve this exact maze?
@@Gigusx Then one of them would complete the maze in one third of the time.
Yeah, actually, when the heat death of the universe is happening, there will still be some park quests in the maze who hasn't even reached the 100th tile.
Whoa, the guest's strangely horrible "maze solving algorithm" might be similar to a gambling game stacked against you! It's like, "heads you win $1, tails you lose $1", but the coin is three times as likely to land tails than heads. And "beating the maze" means you've profitted $n overall, where n is the number of indents in the maze. I guess the benefit of this maze game is that you can't go into the negatives, so that's a nice plus...
hello cary koaster heckler
I see the youtube algorythm led you here as well
The newer algorithm is slightly better
Oh hey I know you.
Yeah as soon as he started trying to calculate probabilities I got worried that we'd have to start working with Markov chains...
Imagine going through the enormous maze, and then at the end someone raised the land up twice and you fall into the void
lmao that feels like a metaphor for life
Guest 3452 has drowned!
That would be the longest ride in RCT. Time? Infinity
@@bensullivan5562 Ben, despite the incredibly long timespan it would take guest to complete the maze it's nowhere near infinity. Number marcel proposed is very large, but nothing compared to infinity
r/foundsatan
it is impressive that they managed to make maze pathfinding worse than RNG
If you think about it, many things are terrible RNGs. If you hit a light switch, the light will turn on most of the time. If it doesn't, it usually won't no matter how often you hit the switch.
Anything not RNG, you can set a strategy though
I am talking about pure RNG, if there was an equal chance to try any direction
@@SuperAWaC
That would take multiple RNG calls, which would have been computationally expensive and by an unpredictable amount as well. What if the guests all keep randomly selecting directions they can't go and have to pick again? This was a pretty reasonable solution IMO.
@@Graknorke things like xorshift are so fast (a few clock cycles) its not really different from simply adding 1 to the direction (1 clock cycle) considering every other aspect of the game needs more computational power than that.
A chilling documentary of what a videogame looks like through the eyes of a mathematician
You should see what the loop prevention algorithm looks like in large corporate networks. It's an impressive body of work under the hood - and far more reasonable to deal with at the configuration level. www.diva-portal.org/smash/get/diva2:1004743/FULLTEXT01.pdf
I'm just imagining some brave hero visitor trudging for an eternity and finally reaching the exit gate, just 1 tile away, and the RNG decides to send him into a backwards spiral that takes him all the way back to the beginning. I imagine all the children asking him what it was like out there at the end.
So I've had a nice scroll through the comments and I'm fairly certain no one picked up on the date being 6/9/420 at 6:24. Well played Marcel, well played.
That´s only because you´r a junkie.
4:47 aw man, that one guy was doing so well too
Basically, just like what happened to Odysseus after his shipmates opened the wind bag of Aeolus.
A great illustration of what he was talking about at that moment, with the bias to returning to the entrance
He looked over the hedge, saw what was on the other side, and said "Fuck it, I'm leaving."
That's how I'm feeling about "Viviene S"!!!
So what you're saying is, roller coaster tycoon god has made a maze so long even he can't solve it before the heat death of the universe.
How big would the maze have to be have the average solve time be the death of the sun, the heat death of the universe? I think this would be a neat way to represent galactic timescales.
The heat death of the universe is around 10^100 years according to wikipedia. To get there we would only need around 680 indents, which is less than 1% of the park filling maze.
Marcel Vos 🤯😂
@@MarcelVos WOW YOU CALCULATED THAT
@Mathew DeCoste To add, I don't know if this helps, or makes it even worse to visualize, but taking the avg solve time Marcel came up with divided by the heat death of the universe to find how many heat deaths would occur gives us 6.6 x 10^19,658 individual heat deaths added together, which is, extremely close to the avg solve time, showing off just the raw difference in scale between the two.
@@MarcelVos but wait. Theres coal power
Actually, if you map out the random number generator function that the game uses, it's incredibly likely that there's no achievable sequence of RNG calls that makes the maze possible to complete at all.
I wonder what they use? It might even be trivial to walk the entire period and calculate.
I don't know the first thing about how RCT assigns RNG, but my initial reaction is that the large number of events in the park pulling RNG is going to create an amount of noise that means it should be possible.
Yeah, other events make rng calls too. It's not only the maze
I think you're right, but it is hard to say with certainty. Chris Sawyer did most of his programming in either C or Assembly, as I recall, and the rand() function in C has a maximum value of 32767.
If the random number generator could produce a list of every possible permutation of numbers from 1 to 32767 (which it can't), then it could finish even much larger mazes. But because I don't know how random rand() is, I can't say what the largest possible solvable maze could be.
OriginalPiMan. It wasn't C. It was pure assembly. He did help to translate it to C for RollerCoaster Tycoon Classic, though
For some reason, I was laughing so hard when he was explaining just how large those numbers were. Imagining a guest being unable to complete that maze before the heat death of the universe is just too good. lmao
"You are born in the maze. You will die in the maze. Because in maze 101: no one ever re-enters, and no one ever leaves.
Mazes. Mazes never changes
In Maze 101, you meet the worst thing in the world.
To follow the code of honour
Die with a sword in your hand
That is the way of the warrior
Your forefathers left on the land
To live by the sword
you must die by the sword
Haha "leaves"
Just like the green hedge walls
Welcome to the hotel California
This is literally like a problem out of an introductory statistics textbook.
If, when he's been showing calculations/conversions, he explicitly showed units, these kind of videos would be a good educational tool.
what does it say about me that i want to solve it
@@deadlykitten4471 I also have this urge. It's definitely possible to solve this without the testing that Marcel did.
I think it requires Markov Chains, which are not introductory stats
@@BlueRaja Introductory statistics definitely covers things like calculating the minimum/maximum amount of time spent in such a maze, it's just some probability multiplied with itself, in this case "slightly" more often than with the typical examples with dice rolling etc. but the principle stays the same and a calculator won't care.
Legend say that the first guest the enters the maze gives up, and become a janitor and stay in the park forever
"Another Visitor! Stay awhile! Stay FOREVER"
That's what happened to the hundreds of guests walking around in circles outside my park entrance.
Yeeees, I shall maintain this maze so that it shall never break down. I cannot leave. No one can leeeeeaaaave!
@@PlittHD destroy him my robots.
It's better than the second guest. He became an entertainer.
*Marcel* "You will die ten trillion deaths before you finish this maze."
*George* "Hold my lemonade."
this is still sooner than the day my dad comes back
Oh great. Another very original "Hold My Beer" comment.
George: "great maze was great"
Yes george, that was the purpose😂
Due to almost all games using pseudorandom number generation, and since there is a finite set of seeds for said number-generation, this maze *may* be genuinely impossible. For example, the Mersner Twister PRNG has a period of just 10^6001. I have no idea how one might formulate a proof that this particular maze is impossible, as specifically testing every possible seed across every step of its period would be almost as impossible as the maze itself.
It IS impossible. A chance that low = equal to it being 0%, but it seems like too many people don't understand that.
@@badidea6034 Err, no. Outside of what I discussed (i.e. the fact that this is pseudorandomness, and not randomness); Even if the chance of it occuring is on the order of 1/10^1000, mathematically speaking, it is not 0. Thus, it is only *nearly* impossible, unless it can be proven that no outcome possible from the PRNG would result in a completion.
@@tales9476 Great point, but RCT running on a different computer - say a NASA supercomputer may have a different implementation of RNG that has a longer period or is "true random" (based on quantum events)
@@kw4093-v3p Er, no. The algorithm RCT uses is deterministic. Start with the same seed, and get the same results. The period of the RNG algorithm is a property of the algorithm itself, not the hardware it is run on. If a computer were using true randomness (i.e. particle-detectors, etc.), instead of the PNRG algorithm, then that isn't vanilla RCT.
@@tales9476 Good to know thanks for clarifying
The park filling maze might actually be unsolvable by the guests' AI due to how random number generators aren't actually random. When you have to create that many random numbers the algorithm will eventually loop back on itself.
I guessed 65 millions years. Good to know that it would take longer than the heat death of the universe to solve this maze. Praise be!
I guessed the lifetime of the universe (in years) squared. I was not even close.
At that time, Mr. Bones Wild Ride will be halfway through its course.
I guessed "the game crashes or the year counter stack overflows before the guest finishes it."
When he asked that, I simply said "Years....". At least I was partially right.
11:39 *insert gif of Jim Carrey* “So you’re telling me there’s a chance?”
Imagine being stuck in a maze for over 57 days just because you're too incompetent to walk forward.
Whaddaya mean "walk forward"??!1?
I CAN'T use "just walk forward algorithm" when "in maze"-flag is TRUE!
Ugh, stupid malfunction people don't even know their algos...
57 days? Try 2000+ years like he showed in the video lol.
"If you want to solve a maze, just keep going left"
"Uhh are you sure? This one just looks like a straight path with a bunch of tiny dead ends."
"Shhh just trust me it works"
Always taking left or always taking right will work 100% of the time, but it might no be the fastest way
you'd be dizzy as shit but youll get there. better hire more janitors...
In a maze like this, you'd actually be done relatively quickly, as you'd move towards the exit every single time. You'd end up stepping on every single tile of the world, but only once (in the fingers) or twice (in the main pathway) each.
@Stale Bagelz i Think the only ones that that can happen would be with the end in the center
@@zgthor No, all you need is at least one single loop within the maze. Think a simple straight wall that can be circled. Or in other words, a maze that consists of not only dead ends.
I'm surprised no one has done this yet - I'll give you the exact math for how much each left indentation gives.
Because walking backwards has different behavior depending on where you start walking backwards, let's track the current indentation with a variable: x. For the first indentation, x=0. For the second, x=1, etc. We have the first indentation at x=0 because, if the NPC enters the indentation and exits to the right (so they are facing the entrance), they make zero choices before getting back to the first indentation again. Right now, our goal is to calculate the expected number of units an NPC moves to reach position x+1 from position x.
Let's say the total number of indentations is n.
Now we are ready to begin: let b be a NPC, and let b start at position x, facing forward. As mentioned in the video, there is a 3/4*1/2=3/8 chance they end up at position x-1, facing backwards, when they make their next choice not at position x. Similarly, there is a 1/4+3/4*1/2=5/8 chance they end up at position x+1, facing forwards, when they make their next choice not at position x. B moves 4 units when they end up facing backwards at position x-1 (1 into the indentation, one out of the indentation, and two to get to x-1), and has a 1/4 chance of reaching x+1 in just 2 moves, and a 3/4*1/2=3/8 chance of reaching x+1 in 4 moves.
Now let's say b is at position x, facing backwards. As mentioned in the video, there is a 1/4*1/2=1/8 chance they end up at position x+1 facing forwards when they make their next choice not at position x, and a 3/4+1/4*1/2=7/8 chance they end up at position x-1 facing backwards. Upon ending at position x+1 facing forwards, b has moved 4 units. There is a 3/4 chance b ends at position x-1 facing backwards in 2 moves, and a 1/4*1/2=1/8 chance b ends at position x-1 facing backwards in 4 moves.
This is all we need to begin the hard part: calculating the expected number of moves (aka units moved) to reach position x+1, facing forwards, from position x, facing forwards.
I will use the notation Y_x to mean the random variable denoting the number of moves to reach position x+1 (facing forwards) from position x (facing forwards), and Z_x to mean the random variable denoting the number of moves to reach position x+1 (facing forwards) from position x (facing backwards). Note that it is impossible to reach position x+1 facing backwards without first reaching it facing forwards, so this sort of notation comes naturally from asking the question.
Let's start with a simple case: starting the maze. In this case, b is at x=0 facing forward, so we are talking about the random variable Y_0. There is a 3/8 chance b enters the first indentation, does a 180, exits the first indentation, turns right toward the start, makes one move, then does a 180 and makes one move back to where they started (4 moves). There is a 3/8 chance b reaches position x=1 in 4 moves, and a 1/4 chance b reaches position x=1 in 2 moves. Therefore we can write the expected number of moves Y_0 as E[Y_0]=3/8*(4+E[Y_0])+3/8*4+1/4*2, which we can rewrite as 5/8*E[Y_0]=28/8, so E[Y_0]=28/5. We can also calculate the expected number of moves Z_0 as E[Z_0]=1/8*4+1/8*(4+E[Y_0])+3/4*(2+E[Y_0]) = 5/2+7/8*E[Y_0]=5/2+49/10=37/5.
We can confirm these values by compiling and running the C code I uploaded to pastebin here: pastebin.com/kcxugAhB. dist is the distance to travel (if, for example, this is set to 2, a round in the maze ends when x=2 is reached for the first time), startdir is 0 for forwards and 1 for backwards, and startpos is the current x value (x as used in this comment).
Now, note that, for x>0, we can calculate E[Y_x]=3/8*4+3/8*(4+E[Z_{x-1}]+E[Y_x])+1/4*2=7/2+3/8*E[Z_{x-1}]+3/8*E[Y_x] so E[Y_x]=56/10+3/5*E[Z_{x-1}],
and E[Z_x]=1/8*4+1/8*(4+E[Z_{x-1}]+E[Y_x])+3/4*(2+E[Z_{x-1}]+E[Y_x])=5/2+7/8*E[Z_{x-1}]+7/8*E[Y_x].
This is a complex recurrence relation, but using the base cases and Wolfram Alpha, we can solve it, giving E[Y_x]=
1/10*(-55 + 111*(7/5)^x*UnitStep(x-1) + 111*UnitStep(-x)). The UnitStep(n) function is 0 when n0. I think it's easier to see what's going on if we instead write it casewise: for x>1, we have E[Y_x]=11.1*(7/5)^x-5.5, and E[Y_1]=10.04 and E[Y_0]=5.6.
We don't really care about E[Z_x] but Wolfram Alpha solves it as well, giving E[Z_x]=37/10*(-5+7*(7/5)^x).
We are almost done now: we have calculated the expected number of moves between each value of x, so we just need to sum up the expected values from E[Y_0] to E[Y_n]. Assuming n>0, this gives the expected number of moves required to reach the nth position in the maze as f(n)=38.85*(7/5)^n-5.5n-33.25. In asymptotic notation, this is in Θ((7/5)^n), which means that, as n gets arbitrarily large, adding an additional indentation multiplies the time required by 7/5=1.4, which is pretty close to what you got in the video! Neat! I think I may be slightly inaccurate by assuming the hedge is as wide as the indentation is deep (i.e. hedges are placed in a unit grid and each takes up a 1x1 space).
Even Mr. Bones would look at this and be like "dude, that takes too much time"
I laughed way to hard at this.
This is just mind blowing Marcel! awesome found
When the world needed him most, he returned
it would be cool if someone set up a 24/7 livestream of the 64k tile maze, and we could watch guests fill up more and more of the maze
No it wouldn´t
"I guess I'm just lucky" - Dream after walking through the entire maze with no errors
Dreams luck is significantly easier to achieve than this lol. What would take longer, getting struck by lightning twice in a row, or picking up every atom in the universe and waiting one Google amount of years in-between? The universe hasn't even been alive for one TRILLIONTH of the time that each of those atoms would take to pick up XD
@@vulblhotdiessfi I think it was a joke, and your tone implies you did not at all get that.
@@dig8634 ( ͡° ͜ʖ ͡°)
Yeah, except dreams luck compared to this would seem more like a coin flip.
Imagine being so petty you'd lie to your nine-year-old fans about cheating and still get fifth place in a speedrun. I just don't understand people like Dream.
God gives 1 human immortality
God: How did you spend your immortal life?
Human: I finished Marcel's maze
God: ....
cuh-LEARLY i wasted that gift on you...
@@jackradzelovage6961 'Gift', you say?
sadly, even this maze's number is incredibly tiny compared to infinity
@@NoNameAtAll2 Naw, this number just makes infinity even bigger then previously thought. Lol
That's a good lesson in exponential growth. It will scale up like this to heights that the century coaster could never even dream of reaching.
*giggles* I love absurd calculations and huge numbers, they are always fun. I also love how people argue that however unlikely, it's still possible
Probability: The chance that one of the NPCs finishing the maze is so vanishingly close to 0 that any system that simulates it will invariably break down and decay completely into radiation before any of them can complete the maze.
Marcel: *So you're saying there's a chance*
"If you picked up an atom and waited a googl years to pick up the next one, you still wouldn't come close"
B R U H
Googol
*Picks up one single atom, and starts stopwatch...*
Love how you can hear the laugh in his voice at 7:35 when he talks about filling up the entire park with a maze.
Good thing my preferred maze strategy is hugging the right wall, I'll be out in a few hours.
The stupid thing is the guests will jump to peek over the wall when they go through a maze, but even then they can't see that they just need to go straight lol
Im a left wall hugger, but they do the same thing. Smart man!
It is a common strategy for mazes, but it no longer works if there is a loop in the maze (unless of course you realize you've been walking in circles. I have actually designed mazes in the past purposefully designed to not make this strategy work regardless if you are a right or left hugger)
@@sashashadowhive6128 Yes, the term is called simply connected en.wikipedia.org/wiki/Simply_connected_space
@@sashashadowhive6128 Well good, but most mazes have terrible design.
To see that you are regularly back on TH-cam really makes me happy. Thank you so much for shedding light on such a legendary game 🙌🏻
6.6 Quinoctogintaquingentillamillinillion years is a very long time.
Yes. Yes it is.
I was wondering what the term for the answer was. I don't care if you made it up, there is now a name for that number and that is all I care about.
I like how you can see the guests going from one maze to the next, thinking "oh that was easy let's check out the other one!"
And then get trapped for the rest of their existence.
“A big number looks too intense for me”
Lmao
My takeaway: if you want a solvable hedge maze that looks nice without just making a linear path to the exit, put the branches on the right.
when you get to this levels of random, you might want to consider looking into how RCT handles random number generation, as you might go long enough to create an RNG loop
As Grey from 'Grey Still Plays' would say: "Those are rookie numbers!" and "More zeros damnit!"
More endless mental torture damnit!
These ain't rookie zeros.
10^19758 goes beyond all physical meaning.
For you see, a plank length is 1.6*10^-35 meters.
A plank cube would be 2.5*10^-103 cubic meters.
The radius of the observable universe is 4.4*10^26 meters.
The volume of the observable universe is 3.6*10^80 cubic meters.
There are about 1.4 * 10^183 plank volumes in the observable universe.
Now let's have a million universes, each a hundred times larger than our own.
We get 1.4*10^191
Hm...still 19,567 orders of magnitude short.
Let's say each of these plank volumes blinks every microsecond, from the dawn of time until our sun burns out.
Across all of our million universes, there will be 2.2*10^214 blinks.
Still 19,544 orders of magnitude short
*We haven't even made a dent*
Now let's say a god is writing a book for his 10^214 friends. To be specific, he is writing the complete dictionary of all 470,000 words, with an extensive definition for each 500 words long. But this god can only write a letter every 500 billion years.When he finishes, he begins translating the book into every language that has ever existed (about 10,000), one character at a time.
By the time he is done creating a complete set of dictionaries for all his friends, 2*10^239 years have passed.
Still not even a dent.
Now don't get me long, it's certainly possible to conjure up a much larger number. But only if you resort to abstract math. 10^19758 goes beyond all physical meaning.
@@tomc.5704,
But see that's when you need to start bringing in a larger scale. Just because the amount of zeros goes beyond all physical meaning to humans, doesn't mean that it has no meaning. Because on the universal scale, humans are as small as, if not smaller, than a higgs boson particle.
@@tomc.5704
R O O K I E
N U M B E R S!!!!!!
MORE ZEROS DAMNIT!!!!!!!!!!!!!
My question is, how does this affect the rating of the maze, if at all? Does increasing the difficulty of it increase it's excitement rating, or does it only count towards the amount of tiles used? How many tiles of left indents do you need for guests to pay $20.00 for the maze?
i don't think it gets stats until a guest completes it
the stats of a maze are determined by the amount of tiles used - each additional tile adds a very small fixed amount to the excitement and intensity. however, there's a cap at one hundred tiles, so any maze bigger than that will have identical stats despite being more expensive and having worse throughput
My guess was "an absolutely meaninglessly, incomprehensibly large number" so I'd say my guess was basically spot on. If anything it was a bit smaller than I was expecting. Also to all the people talking about the finite period of RNGs and so on: While it's true that the PRNG in games like this has a finite (albeit very large) period, just because it has a small period doesn't make it trivial to prove that it's impossible to use that PRNG to generate the necessary sequence to let a guest escape. There are many, many other things in the game that will advance the PRNG by a most likely unpredictable amount, which will lead to an incredibly difficult to predict system. So while the PRNG is easy to predict, the behavior of the system is not.
If you could see state of the RNG and manipulate it through actions in the game, to cause it to be incremented a different number of times (for example, how tool assisted speedruns work), you could probably trivially make a single guest walk all the way to the end of the maze without ever turning into a single alcove.
Six Flags park manager: *builds this maze irl*
Park guests: *enter maze, finish it in less than two minutes*
Park manager: *confused screaming*
Me: But you can't buy things while on a ride.
Six Flags Manager: Mod in food stalls.
Six Flags Employee: Sir, we don't need mods, we can just cut the hedges.
Six Flags Manager: Even better!
There's a news story about a couple at a corn maze, they called 911 two hours after the maze had closed because they were lost and unable to find the exit... Or think about simply stepping through the corn to go to the building that they could clearly see from where they called 911 from...
Never underestimate human stupidity.
@@veggiet2009 can you please give me a article link? just wanna know more
@@veggiet2009 Protip: Any maze that doesn't have a 4 way intersection can be solved by turning the same way at every intersection, though it might take a while if it's particularly complicated. Even if it does have a 4 way intersection, it can still be solved by picking a different (cardinal) direction every time you reach it. It only starts to get complicated when the maze has multiple 4 way intersections.
Life sometimes feels like a left indented maze
Minotaur: I am the king of mazes.
George: I completed Marcel's Hedge Maze.
Minotaur: Please forgive me, almighty.
Numberphile: I'm about to end this number's whole career
they need to do a video on this number. call it marcel's number
@@lemminjuice They've done videos on far larger numbers.
I want to get off Hedge Maze 1
T H E
H E D G E
N E V E R
E N D S
Welcome to actual hell
mo
mo
@tommyjay_97 "Viviene S." literally just said that with a grumpy face, prompting me to try to figure out how to end her maze misery. 🥺😅...worse case, I'll just delete the maze entirely 🤦🏾♀️😂
Imagine being citizen when Attilla the Hun was starting his campaign in Europe. You escape into a maze and wander around it till you find the exit. When you exit you find out this thing called WWII was happening.
"damn huns at it again"
@@mrjwvd war... war never changes.
@@Hellothere-lk7xi Everything changed when the Fire Nation attacked.
Basically, Daedalus's labyrinth, except the minotaur teabags you for all eternity.
Fantastic
Math
Weird seeing you here :D Are you a RCT fan? Lol
Hey man, when in Rome.
There is a parallel universe out there, where a guest completed this maze in only a few hours.
*mirror universe
-Since the pathfinding is chiral-
Nah the reality is;
There are a thousand mazes trying to navigate a left indented human
Hi, its me, George. It took me a long long while, but I got out of the maze.
*drops into water*
And the exit is blocked by a no entry sign, leading George back into the entrance to the Maza again.
@@SrFreakus George has drowned!
Damn. They murdered you George. F
Cern: Yeah, six sigma is a high enough confidence interval to determine the mass of subatomic particles.
Marcel Vos: Plebs, grahams number sigma or GTFO
My guess is it takes many times longer than the lifespan of the universe.
Ok I guess I'm off by a factor of 10^18,900
I guessed a googol years. I'm also pretty off.
Note that the end of the universe is not really agreed upon and 10^100 (a googol) years is the assumption for heat death with proton decay. Without proton decay the equivalent would be slightly higher with 10^(10^120) years. And it still isn't entirely dead after that. Probably.
What I'm saying is: Start the computers, there is still a chance.
@@sparschlumpf I didn't know about the connection between a googol years and the heat death. I just knew it is a large number.
This is the most amazing thing I've watched on youtube this year, I almost cried laughing at the absurdity of the numbers.
I've never played this game, and it doesn't matter - I literally enjoyed this video more than anything else on TH-cam to date!
A computer game, analysed to the extreme, dripping with snark.... brilliant!
Day 1: This maze looks easy.
Day 57: WILSON!
you mean day 529292518473829572719384726274928263728283726174939172538492726393917374829204750602963318204017274920274829472949282747493010373838263636
This is exactly the kind of high-quality content for which we’re supporting you on Patreon. Keep up the great work and have a nice weekend!
Being a mathematician, I really like your scientific approach. Your materials on longest ride time records on RCT2 (the synchronised coasters with your deterministic analysis and this one with your probabilistic analysis) could probably make a publishable scientific article.
"Don't get lost was great!"
-Joel
It’s like I discovered a brand new world of Tycoon. Incredibly interesting video. I’m hooked
I guessed that it would take 400 million real-life years to solve.
I was wrong by approximately 6.6 x 10^19758 years. 400 million years isn't even noticeable on that timescale. It would be roughly as noticeable as a single atom being removed from our universe. If our universe was multiplied by itself 233 times.
It would be much more than 10^200 times less noticeable than removing an atom from our universe, actually. There's only 10^80 of em.
Potential names:
- The Maze Of A Lifetime!
-Get Lost
-Maze: Hardcore Edition
-Worse Than Hell
"Get Lost" is absolutely genius and you deserve a gold medal for coming up with that name.
@@Shredster7 Thanks! It's quite a miracle, normally I absolutely suck at giving names to things.
Hey, whoa, don't make Hell seem more cruel than it actually is by comparing it to this maze!
@@notmuch_23 you're absolutely right let me make a small modification.
@@n.stephan9848 That's better!
I love these vids. Just a guy torturing virtual people in an ancient game. It would be hilarious if once a person finished the maze, they went back in it.
18 years is hardly ancient.
I mean there's nothing else to do in the parks he shows in the video lol.
tbf this game still has a very active community. It's not like he dug up some obscure title ;p
“Hedge Maze 1 was great!”
@@roddydykes7053 Maze 1 has broken down.
my guess was "heat death of the universe" and it turns out that was actually lowballing i can't-
(had completely forgotten this video until binge watching the whole channel and oh WOW do i not envy you when building that maze. the dedication sure, but the actual process? surprised you didn't go insane during it. well, any more than you already might've been at least ;p)
Marcel: Guess how long.
Me: Oh Jesus, a googol years?
Marcel: A googol TO THE POWER OF 200 years.
Me: ...
HP printer factory worker: 'Sir, we've run out of printer ink!'
manager: 'How is this possible?!'
worker: 'Marcel Vos accidentally printed out his impossible maze calculations'
DJAvren
I understand the joke you’re trying to make but the equation was only about 5 pages long.
@@matthew8153 6, but that was just one of the numbers to be fair.
Still would only be 100 pages or so.
Carbon 12
You are correct with six, sorry about that.
I feel like my guess of 80 billion years kinda undershot the maze's escape time by a bit.
just a little
11:10 "There is a chance a guest will walk from the entrance to the exit without going into a single indent." Maybe with a true random number generator, but some rough math indicates you'd need your RNG to have a state size of on the order of at least 8 kiB for there to be even a 50% chance for this to be possible.
Even a _top-of-the-range_ Mersenne Twister has a state size of only around 2.5 kiB, so even if it used that there'd still only be about a 1 in 1ᴇ48836 chance of there even _being_ a configuration that results in a guest traversing the maze in a single go.
With the (at best) 32 bit state size of the linear congruential generator RCT probably uses, the chances are more like 1 in 1ᴇ54992.
I guess you could argue that the states of the thousands of guests traversing the maze are in effect an extension to the RNG state size (since they're running around soaking up RNG rolls in a chaotic manner), but I still seriously doubt there are enough effective bits there to matter.
*TL;DR:* There are only around 1ᴇ10 (give or take a few orders of magnitude) possible sequences of guest direction choices that the game is capable of producing. That's a modestly large number, but it's so small in comparison that "go through the maze without taking a single detour" almost certainly _isn't_ one of the possible sequences.
This people hanging out at the entrance thing reminds me of that one giant park where the entertainers all hung out in one very happy little corner
Honestly, this is an awesome demonstration about how binomial distributions work, and a neat model of brownian motion with some kind of potential!
That number of years just made my head explode.
Mechanic 1 can't find exit to RetroRobby's head
Every time I watch one of your videos I think "This MUST be the last one. He CAN'T find anything else in this game to talk about" And then BAM! "The Impossible Maze"
5:55 I'm mad that I am very likely not going to live to see the dawn of September 9th, 2420. The train has already left on the actual year 420, so that's my next best chance, lol.
9.3 x 10^-55000% chance? Get Dream on the line he'll have it done in 15 minutes
Been playing this game somewhat obsessively since I was 10 in the year 2000, and have been happily binging your videos for the past few days. I am playing RCTC on my iPad right now, but I definitely need to step up my game and get OpenRCT. Thanks for the videos!
I love how much detail devs put into older games like this. Rather than just have them pass through they literally have AI for navigating the maze.
Plot twist: the guests are actually drawing arrows on the ground with lipstick and aren't noticing the goblins who keep changing their direction.
Someone show this to Brady from Numberphile
This is some fantastic, quality content. I wish this type of stuff would be suggested more often.
Wow. Just wow. It starts so mild, I was thinking, "oh I know why they do that, this video may not be so interesting," and it just literally ramps up exponentially from there.
“I want to get off of Impossible Maze 1.”
Mr. Bones Wild Maze
@@johnfoltz8183 Mr. Marcels Tame Maze
7:55 how long does it take to build that maze? xD
Took me about 6 hours with an autohotkey script. Even with a script you are limited as the game can only process inputs so fast.
@@MarcelVos I suppose the file-format of exported rides is known, so I guess it would technically possible to program a thing that would generate such a file for a X size maze like this
Here's my prediction for how long it takes a guest to solve the park-sized Impossible Maze: precisely as long as it would take a guest to ride The Century Coaster from start to finish - 323,599,680 RCT days, or 1,320,815 RCT years
Update from seeing how long it takes: Oh. My. God. RCT guests are really stupid.
they have maps in hand and still cant get to bathroom 1 on the next path over... im surprised you had any hope for them.
also: why do staff get lost in their own park?
@@jackradzelovage6961 Seems pretty human to me
@@jackradzelovage6961
I believe it's called "Malingering"
They'll get paid so long as it looks like they're working, so they're always going somewhere, but really they're just wandering around.
@@aprinnyonbreak1290 malingering is more faking a medical problem to get out of work, but yeah theyve gotta be doing it deliberately. i mean i can definitely find my way when there are four ferris wheels stacked on top of each other XD
@@jackradzelovage6961 Ah, my mistake. Hopefully my internet is still apparant. I wonder if pretending to be bad at a job had a name.
Literally cackled like a banshee. Quality content here
I don't know how I ended up here, but I'm glad I have.
I got this game in a cereal box about 20 years ago and used to play it on my grandmother's Windows XP computer when I was barely old enough to read. Completely forgot this game existed or that I had a cd copy in an old jewelcase for decades, and now it's interesting to suddenly see it again with some mathematical equations and NPC AI breakdowns all these years later.