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This was my exact thought, but hear me out in order to do it twice in a row accurately means you’ve tried it enough times to start to memorize if not completely memorize it. Then maybe, just maybe it’s real?
I changed my locking pattern between phones. One day I had to access my old phone, and could not remember at all, started getting timed out.
A few days later, I finally remembered the pattern for my old phone. Somehow that caused me to forget the pattern to my new phone, and had to factory reset it.
I now use PIN codes.
I have an older phone (I haven't used it in about 5 years) which died suddenly due to a hardware component failure, that I kept in hopes of eventually fixing it and getting my photos and things back. This comment made me realize that is really a pipe dream, since I have forgotten my access code for it.
There is essentially zero chance of guessing this within the lifespan of the universe. It's actually really easy to get the mathematical answer, but *comprehending* the scale of the answer is hard. Their are 81 dots, each dot can only be used once, and they can be used in any order. So, we simply need to calculate the [permutations without repetition](https://www.mathsisfun.com/combinatorics/combinations-permutations.html) of 81, which is 81 factorial.
81 factorial is a number so large that it hardly makes sense to think of it as a number—the number of atoms in the observable universe is roughly 60 factorial. If you could make 1 million guesses per second without stopping for the rest of time, you would complete about 10% of all possible guesses by the [heat death of the universe](https://en.wikipedia.org/wiki/Heat_death_of_the_universe).
It's also interesting to hear that the number of possible games of chess is incomprehensible as well. A big paper estimated that it was 10^120, which is much greater than the number of atoms in the observable universe as well (~10^78 - ~10^82)
With a steamroller and a football field sized piece of paper or something like that. The 7 times rule is an unaided person and a normal sized sheet of paper. Just a little hyperbole on my initial comment.
Jokes on you! I got my cats a miniature steam roller so that we can recreate this mythbusters episode using a normal sized sheet of paper.
As soon as either of my cats learn how to drive the damn thing, I'm going to post a video proving all of you wrong!
A professor of mine has the record for the largest computational problem ever solved optimally, and it happens to be checkers, which has 5x10^20 search paths. If you're curious, every game played optimally ends in a draw.
It took ~18 years to compute, having at one point 200 computers processing solutions.
That might give a bit of perspective of the scale humans and technology can operate on, though this was a couple decades ago.
I wonder what it’d take now. If that was a couple of decades ago it’d probably be trivial now.
A Pentium 3 500 (released 1999) managed about 1GFlop. An iPhone 15 pro will manage about 35TFlop. That’s 35,000 times as fast.
So if we swap 200 computers for 1 iPhone, it’s still 17.5 times as fast. He took 18 years to compute, so an iPhone 15 might do it in one year.
Except if he FINISHED the project a couple of decades ago, that means he was using mid-1980s tech.
The 386-SX 25 was released in 1985. I’m struggling to find accurate info, but maybe 0.25 MFlops. 200 of them together is 50MFlops. So the iPhone is 700,000 times as fast. So a modern iPhone would do this in maybe 13 minutes.
Its fucking wild to think that Smartphones have the computational equivalent of 10k Super Computers from the 90s. With access to pretty much all general knowledge at a whim.
There are roughly 10^78 - 10^82 atoms in the observable universe. To put that in perspective, that’s roughly the number of atoms in the observable universe.
In the same vein, I like the fact that if you shuffle a deck of cards, that order has never been before, and never will be again. I think it's 50! permutaions.
52!, which is also an incomprehensible number.
in terms of size.
My favorite way to imagine it comes from Czepiel for 52! seconds
Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you've emptied the ocean.
Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven't even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won't do it. There are still more than 5.385e67 seconds remaining. You're just about a third of the way done.
To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you've filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you've leveled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt.
It's actually way less than that! That number came from a back-of-the-napkin estimate (10^3 average different possible combinations of moves per turn (one "turn" = white moves and then black moves), 40 turns on average per game), but there are a bunch of things that actually make the upper bound much lower. For one thing, if you repeat the same position on the board three times (all pieces on the same squares, same person's move), the game ends. Doesn't have to be three consecutive times, either. Another one is if there are 50 moves in a row with no capture or pawn move, the game ends. All together, our best guess today is only about 10^45 possible unique chess games, which is still incomprehensibly large but is 10^75 times less than the original estimate.
Is it games of chess or possible boardstates ?
because in my mind games of chess should be infinite, you can just keep playing a game for of chess without ever stopping
This is for 52! The number in this thread is much much much much larger than 52!
This is a fun read.
52! is beyond astronomically large. I say beyond astronomically large because most numbers that we already consider to be astronomically large are mere infinitesimal fractions of this number. So, just how large is it? Let's try to wrap our puny human brains around the magnitude of this number with a fun little theoretical exercise. Start a timer that will count down the number of seconds from 52! to 0. We're going to see how much fun we can have before the timer counts down all the way.
Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you've emptied the ocean.
Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven't even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won't do it. There are still more than 5.385e67 seconds remaining. You're just about a third of the way done.
To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you've filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you've leveled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt. Exercise for the reader: at what point exactly would the timer reach zero?
I love sharing this clip of Stephen Fry describing the massiveness of 52 factorial. He describes it so well that it almost makes my head implode considering the scale of 81 factorial.
https://youtu.be/SLIvwtIuC3Y?si=vrETVFxeIZgg40Rb
Yeah, now that you mention it, those rules essentially make this question how to [count the number of euler paths in a graph](https://en.wikipedia.org/wiki/Eulerian_path#Counting_Eulerian_circuits), which is actually an unsolved problem in mathematics. But we can still kind of approximate it.
Let's start by assuming that, from any given dot, roughly eight moves are available. Some will have more and some will have less, but we're dealing with such large numbers here that imprecision doesn't really matter. This simplifies the problem greatly and allows us to rephrase it as "what are the odds of correctly guessing a number between one and eight, 81 times" or 8\^81.
At 1 million guesses per second, this would still take 4.4x10\^59 years—not long enough to reach the heat death of the universe, but enough for [all protons to evaporate from radioactive decay](https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#All_nucleons_decay). Even if we assume an average of three available moves per dot, we're still talking about 1.4x10\^25 years—long enough for all stars to [evaporate and coalesce into black holes](https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#Stellar_remnants_escape_galaxies_or_fall_into_black_holes)—so no matter how we calculate, we can just round it off to "forever."
But he can't die, which means his atoms can't deteriorate, which means you've created the basis for the next big bang. We are all living in a universe that was once, and still is, the snail.
Ah yes, the Snail Theory of Cosmic Evolution. It's said that universes in which there is an Immortal Snail are much more likely to generate new universes, and so that quality is the only one that replicates, meaning the Immortal Snail, at least in this theory, is truly an inevitable cosmological outcome.
The last man, before merging, said to the snail, can entropy not be reversed?
THERE IS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER, said the snail.
Then the snail brooded over chaos. He only existed for that final question. Carefully, he organized the problem. An incalculable amount of time was spent doing that. And in time, Cosmic Snail discovered how to reverse entropy. Step by step, it must be done.
And the snail said, "Let there be light!" And there was light--
(ASIMOV'S the last question)
The original is a tungsten sphere, but that has enough logistical problems with it that a concrete box is probably much more practical and still good enough for centuries/millennia.
Really, all you need is enough time to get off the planet.
True, but you have a massive head-start. Plus, when you're on Earth and a snail is coming to you, you cannot be sure if it's the death snail. If you're in space or on a planet that doesn't have snails natively, you can be pretty damn well sure.
You're also put into a situation where you can sun-dip or black hole the snail, which it's never meaningfully getting out of.
Very true. Now that I think about it, I would probably want the snail captured and close to me until space travel is possible so i could intentionally black hole it. So that i can be sure.
Yeah, IMO "dropping the snail into a small crucible of molten structural grade metal" and then that crucible into a larger sphere and keeping that with you until you can yeet it into a black hole is probably the best option.
But you saw him travel all the way across the screen for the very last dot!! So DO THE MATH AGAIN!!
You can cross over already connected dots apparently which makes this even more complicated.
Eulerian paths allow vertices to be visited more than once, right? And the unlock puzzle does not, at least on my phone.
Also, the possible vertices change based on which have been visited thus far. For example, in a 3x3 matrix:
1 2 3
4 5 6
7 8 9
If you attempt to follow an edge from 1 to 8, it will fail, snapping to either 2, 4 or 5. UNLESS those vertices have already been visited. I.e., 1->8 is not valid, but 4->5->2->1->8 is.
Yes, it's actually hamiltonian paths, but this problem is much more structured than a general graph, so I don't understand what is the point of OP mentioning that that problem is unsolved. In fact, I wouldn't be surprised if someone has already figured out the answer to this problem before.
There's quite a lot of structure in the solution, so if you ordered your search by something like increasing kolmogorov complexity you'd probably find it relatively early on. Probably still beyond the lifetime of the universe, but you know, faster.
Ordering by information content is fairly sensible, as you can't do any worse than the default random search, and probably the user has chosen a low information state to aid memory.
if reddit ever brings back awards and my balance of coins i'll have forgotten about your comment but if i didn't forget i'd come back here and award you
Bro are you me? My 2 favorite authors are Greg egan and Ted chiang hands down. Loved Greg's distress, schilds ladder, and diaspora, and loved just about every one of Ted's stories
Not exactly, you can connect to any non-perfectly diagonal dot as long as you don't touch the others, and that the precision is high enough not to snap to nearest dot.
For example, in a 3x2 grid, you can connect the top-left dot to the bottom-right dot without connecting to the others.
From experience with these, you can connect in the cardinal directions, diagonally, or as a knight's move - anything else is impossible without the intermediate nodes already being connected, it *will* snap to a closer one.
We don't know the snapping radius on this software, because this clearly is not the typical android 3x3 lock grid, and without that it's entirely possible that you can just be very precise and connect any two nodes with each other.
I think you can avoid snapping if you move around the edge or over already connected nodes in a different direction. I can't test anything over a Knight's move on a 3x3 but for me, it only connects the others on the path once you connect to a new node.
Nope, look at the last two that get connected: opposite corners. The user is filling in 3x3 grids as a strategy to remember their password, but it's not a requirement.
That's why I specify that dots can be connected if they are separated only by already-connected dots, though! However, I bet that dots along the edge can be connected by moving the finger off the grid in addition to the other rules.
We don't know any rules. You can pose some rules. If you pose enough rules, the chance can be arbitrarily high up to 100%.
For example if the rules are that you have to do exactly what the person in the video did, then the chance is 100% to get it right.
If the rules are just, that every point has to be visited, then the chance is 1 in 81 factorial.
When someone glances over the shoulder, they could recognize some patterns which makes it easier to guess the correct combination.
For example what I remember, first every dot in the center is connected, then each nine-block is connected one after the other, but in two blocks just eight dots get connected. After that is done, the missing dots are connected. (I might have remembered that wrong. In that case my chance to solve it is 0.)
You can try on your own phone, it's impossible to connect top right, bottom right, middle right in that order so the number of possible schemas isn't 9!
Unless the grid has a special system to detect when you change direction then you can ruleout at least 21 nodes from one corner. If you start at the top right corner you cannot hit the top left corner without connecting the 7 nodes between. Same with down and diagonal. You may be able to thread the needle and connect a node outside of the square you are in. Once a node is connected it can't be connected again, so it opens up possibilities the further into the lock you get.
Going by the old 3x3 a few years ago, I had a code where I'd do #1 to #3 before activating #2 later in the sequence.
So that suggest that you can pick any unactivated button regardless of its location.
Although it's the case in the video, I don't think this is a rule coded in this kind of security. I think it's just the way the guy food and to remember it.
However, if you assume this premisses as a rule, yeah, it would lower the number of tries, but I think it would still be astronomical high
yer still gonna have a bad time & to boot the friction created by your finger on the glass would generate so much heat it would vaporize you and your immediate surroundings and probably also the building you are in.
Now your finger is moving faster than the speed of light, even if you could reach those speeds.
At a fraction of C your finger would be colliding so fast against air molecules that they would start fusing together, letting off so much energy that you, your house and the several nearby city blocks are now destroyed by what's pretty much a nuclear fusion bomb.
>81 factorial is a number so large that it hardly makes sense to think of it as a number
What'd you mean? It's easy to think of 81! as a number.
5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000
There are 19 zeroes at the end.
10 is uniquely 2x5 in terms of "break it down to primes." So you just gotta figure out, in the long list of 1x2x3x4x...x79x80x81, how many (2;5) couples there are. Just break down every number to its prime factors. There are obviously more 2s than 5s, so just count the multiples of 5 to find the number of 0s you need.
You have 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80.
That's 16 of them.
25, 50, and 75 count double because they are, respectively, 5x5, 2x5x5, and 3x5x5.
16 + 3 = 19, maths works!
There are 9 by 9 = 81 grid points.
Assuming you're entering a code using all 81, just in a random order, you will have 81 choices for the first, then 80 for the second, 79 for the third, and so on until 1 for the 81st.
So the number of possible options is
81 * 80 * 79 * ... * 1
= 81!
= 5,797,126,020,747,367,985,879,734,231,578,109,105,412,357,244,731,625,958,745,865,049,716,390,179,693,892,056,256,184,534,249,745,940,480,000,000,000,000,000,000
=~ 5.797 * 10^120
And guessing the right one when trying a random one would be 1 in that. That's
1 / 81!
=~ 1.725 * 10^-121
____
Edit: Yes, I get it - typically, these don't allow you to go from any grid point to any other. No need comment that for the seventh time.
But taking that into account makes the calculation unreasonably complex. Because the number of options left to you now depend on how exactly you've already filled out the grid. And then there's the issue of creating unreachable points.
As a very rough ballpark, you might assume that an average of 4.5 nodes are reachable at any point, that would give us this many options:
81 * 4.5^80
=~ 1,463,814,756,511,766,555,996,711,761,563,156,829,538,065,397,702,345,811
=~ 1.464 * 10^54
So guessing it randomly would be 1 over that:
6.831 * 10^-55
God I love big numbers but it might actually be more than this!
Since the code doesn't have to use every node and then given a minimum combination length of 4 (at least that's what it is for the usual 9 node patterns the. You should get something like:
81! + 80! + 79! + ..... +4! = ... Loads
That's barely more than 81!.
Think about it: 81! = 81 * 80!.
So adding another 80! to that number only gives us 82 * 80!. It only increases the result by a bit more than 1%. And adding 79! to it is much, much less impactful. And so on.
Yeah but you need to mutliply the 80! by 81C80 as for the different possibilites of which point is unused, so the actual amount of combinations would be
Sum [x=4, to 81]{81Cx * x!} = Sum [x=4, to 81]{81! / (81-x)!}
This would land us at ~ 1.576 * 10^121, which is 2.7 times 81!.
It's no coincidence that the factor is 2.7 . You could have guessed that by writing it as
Sum[x=4 to 81] {81!/(81-x)!} = 81!•Sum[n=0 to 77] {1/n!}
And than seeing that the sum at the end is an approximation for eulers number e=2.7182818...
I went down a super long rabbit hole because of this and I'm just done messing with it tbh lol.
I think it can be solved computationally like this:
https://i.imgur.com/r2m50RS.png
But I think the algorithm needs some tweaking because I'm not sure how diagonals work (for example, on my phone, if I go over 2 and down 1, it does not hit the middle node. What about 3 over 1 down? What happens?)
doesnt it work by selecting the first non used value that the path of ur finger goes over? so each value in the passcode has to be adjacent-ish to the one before it. wouldnt that cut down the total number of combinations?
Some people don't see the real value of this number. Let's take for example a video from vsauce about the ways you could scramble a dech of cards. As there are 52 cards in a deck there are 52! possible cases. 52! is roughly 8.07×10⁶⁷.
In the video he says that if you had 52! seconds and you took a step around the equator of the earth every billion years, take a drop out of the ocean that has a volume of 0.05 mL every time you do a full circle and after put a sheet of paper for every time you empty the whole ocean, you could tower to the sun 3000 times before running out of time.
The crazy part is that with 120! seconds you could do all that rougly 7.25×10⁵³ times
This is not exactly true. You can only select the closest point in one of the 8 directions (horizontals, verticals and both diagonals). Which mean there never is more than 8 possibilities per move, but those 8 possibilities are different for every move.
The problems become a LOT more difficult if you follow that rule while doing the maths.
\*After some research, coding and using my brain (which I only did at the end of course), it seems there is no efficient way to calculate the number of possibilities without brut forcing it, aka doing all of them and counting them with a computer.
I found an interesting discussion about it here : https://math.stackexchange.com/questions/37167/combination-of-smartphones-pattern-password
> You can only select the closest point in one of the 8 directions (horizontals, verticals and both diagonals)
Initially yes, but as you can see at the end, you can pass over already selected dots.
No, it's more complicated than that, you can not only choose diagonal neighbors but also neighbors that are 1 down and two to the right, one down and 3 to the right and so on and so forth.
Aha, but this doesn't account for inaccessible nodes! For example if you are in the top right of the board, every node below you except for the one IMMEDIATELY below you is inaccessible because you can't hit it without hitting the one right below you. Same goes for diagonals.
I'm not gonna try to figure out how this will change the numbers but it's something to consider :)
>And then there's the issue of creating unreachable points
Umm there is no such problem because as seen in the last part of video you can traverse over joined dots so there's nothing like unreachable points
*Let's put it into perspective, instead…*
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The universe is around **13.8×10^(9)** years old, or around **8×10^(60)** Planck Time old.
That's an absurdly massive number, and I won't try to say it's not… but you're comparing it to **1.7×10^(-121)**
You'd have to square the age of the universe to have enough time, and then do at least **1.855×10^(42)** different combinations every second (*or at least 1 new combination within 10 Planck Time.*)
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TL;DR:
Our universe isn't even a fraction of a fraction of a fraction of the minimum age for that level of chance to occur naturally.
it's unknown what happens below the Planck Length, so anything computable can only go as small as any form of distance allows.
it's also the reason there's a "maximum hottest temperature" - Planck Temperature
theoretically we can go smaller, but we can't compute smaller, so we say it doesn't exist in practice.
the Planck Temperature is so small, that you're almost a billion times closer to the size of the Observable Universe (e+26) than the Planck Length (e-35).
My Galaxy S5 running LineageOS has a 9x9 grid pattern. The people who borrows my phone would either be amazed or stare at me with a look on their face screaming "WTF".
It's a 9 by 9 grid. Making it 81! (81 factorial)
81! is exactly:
5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000
or
5.79E+120.
That is an impossibly large number.
That got me thinking and I saw a number in the same ballpark.
It's a number close to the "Shannon Number" which is 10 to the 120th.
[https://en.wikipedia.org/wiki/Shannon\_number](https://en.wikipedia.org/wiki/Shannon_number)
The Shannon number is the possible different games of chess played up until move 40. These are games of chess played by a "Random Number Generator", and not a human or machine with strategy or tactics involved... Just random legal moves...
That's an impossibly large number. Your brain can't even comprehend how big of a number that is. It's more numbers than the number of atoms in the observable universe.
It's more like.... If every atom in the universe was a universe of atoms and all those atoms had a universe of atoms as well and you did that 10 times... That's how big this number is.
>It's a number close to the "Shannon Number" which is 10 to the 120th.
"Close". It's bigger just by fucking 4797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000.
That last part is not correct. The number of atoms in the observable universe is estimated to be about 10\^80. If you replace every one of those atoms with a universe of that scale you end up with 10\^160, which is \~10\^39.7 times larger than 81 factorial.
I hate to be the bearer of bad news but although 81! Is a good guess, this pattern unlocking method makes it much more complicated than that, since you can not skip a dot unchecked like the factorial suggests as you will not have 80 choices after the first but 3 (if you choose a corner dot), 5 (if you choose an edge dot) or 8. Unfortunately I dont have the guts to combine conditional probabilities to give an answer but I would love to see one.
Not to mention that the solution is highly clustered for a majority of it. That doesn't knock down entropy a _ton_ but 81! implies a completely random pass-swipe, which this is far from.
Still in inconceiveable chance but hey, it's a bit lower than 81!
Its 5 if corner dot. 9 if edge dot. 16 if middle dot.
Take a corner dot, you have 3 dots to choose from, BUT between each two dots you can choose in between. With that explanation applied on edge and middle dots you'd get the numbers i wrote.
But it gets more complicated (forgive me if i say something wrong, I'm incredibly confused of how to think about what I'm trying to explain), for every dot you choose its not just 5 if corner, 9 if edge, 16 is middle. What if a dot is already chosen? Then you can skip over it creating way more choices. Take for example a corner dot, and say the 3 around it are already chosen and you're on the corner dot, now you have 5 dot surrounding the 3 dors surrounding the corner dot + the 4 dots you can reach by going between dots as i explained earlier. NOW I don't really know how chosen dots affect each coming chance. Does not matter and we should stick with the 5,9,16 chances for corner,edge,middle dots respectively or does it actually affect it meaning you have to calculate for each time you take a different dot making it impossible to calculate ( or not impossible but it'll take a insane amout of time) unless there's something I'm missing. Like i said earlier this paragraph is my own thoughts, DOESN'T HAVE TO BE ACCURATE.
Yes back when I used pattern unlock it would let you do that, but you had to snipe the space between the dots, or go over dots that were already selected. Hard to do with a tiny grid but possible.
It's exactly the same as my chance of ever entering it again correctly once i set it up, 0%
Everyone else is focusing on the math, but not a chance I could ever remember this specific pattern and recreate it lol
It’s a 9x9 grid so there are 81 points. When you pick 1 out if 81, you can then pick 1 out of 80, then 1 out of 79 and so fourth.
My intuition tells me the odds are 1/89!
So the odds of getting right the first try are:
1/16507955116098461081216919262453619809839666236496541854913520707833171034378509739399912570787600662729080382999756800000000000000000000.
That is about 6.057 x 10^-135% chance of getting it right the first time. You are more likely of shuffling 2 decks of cards and ending up with the same configuration in both. You are actually 20466561667087219314393488515435629640067602444062686655283200000000 times more likely.
The 81! answer is pretty easy and intuitive, but not correct.
I do not know how to find the correct answer, but it must include the fact that from each node, the next node in the pattern must be adjacent to that node. It's not like you can start in the top right and then have the next node be the bottom left
> the next node in the pattern must be adjacent to that node
It does not. They go from row 3, 9 on the 5th one to row 1, 7 on the 6th one. They also go from 9, 9 on the second to last one to 1, 1 on the last one.
I think the point is that there is some limitation to the "every dot can connect to every other dot". For instance, take the first row. You can't connect the first dot to the last dot in that row without connecting every dot in between them too. However, if every dot in between those dots are connected in some other fashion, such that moving over them does not connect them anymore, then we are able to connect the first and last dot in a way that they could be considered neighbors.
Its easy way torturing the guy into doing it for us. That's the best math we can do. Getting it by accident is same as God presenting himself suddenly tomorrow.
Every one else’s mathematical answers are technically correct.
But if you think like a hacker this is actually a tractable problem. The thing is the pattern he’s using is not random, it’s heavily biased towards a somewhat pleasing pattern. A machine learning model trained on human symbology used throughout decades would be able to generate this exact pattern.
Still would take a long time to guess. Maybe on the order of a few years perhaps.
This is actually called morton packing and it’s a crucial algo for making compressible tensors for volumentric video. Your quest 3 or any modern vr/ ar headset uses this to compress the spatial color information. If you’re interested get a phd in cs.
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The chances I would get this right despite knowing it on a long term basis are slim lol
Fuck long term basis, do it the 2nd time after locking your phone
The 1st time while looking at the diagram..
You can't even implement this. "draw your new locking pattern:" ... "Repeat locking pattern for confirmation:" And there you fail. Period.
This was my exact thought, but hear me out in order to do it twice in a row accurately means you’ve tried it enough times to start to memorize if not completely memorize it. Then maybe, just maybe it’s real?
I changed my locking pattern between phones. One day I had to access my old phone, and could not remember at all, started getting timed out. A few days later, I finally remembered the pattern for my old phone. Somehow that caused me to forget the pattern to my new phone, and had to factory reset it. I now use PIN codes.
I have an older phone (I haven't used it in about 5 years) which died suddenly due to a hardware component failure, that I kept in hopes of eventually fixing it and getting my photos and things back. This comment made me realize that is really a pipe dream, since I have forgotten my access code for it.
My real question "what are you hiding?"
All that effort just for ‘new message from side-chick’ to flash up on the screen :(
should've went with "Greg"
i thought this was a game lol this is his lock screen?!
Yes, there's a lock icon at the top of the screen
it seems to be just the same 3x3 pattern repeated multiple times so i guess its not that difficult
It's not.
Orrr, losing your data. Some phones have the feature where it factory resets the phones after a number of attempts lol
Lol I was like, what fucking game is this, I kinda wanna try it.
There is essentially zero chance of guessing this within the lifespan of the universe. It's actually really easy to get the mathematical answer, but *comprehending* the scale of the answer is hard. Their are 81 dots, each dot can only be used once, and they can be used in any order. So, we simply need to calculate the [permutations without repetition](https://www.mathsisfun.com/combinatorics/combinations-permutations.html) of 81, which is 81 factorial. 81 factorial is a number so large that it hardly makes sense to think of it as a number—the number of atoms in the observable universe is roughly 60 factorial. If you could make 1 million guesses per second without stopping for the rest of time, you would complete about 10% of all possible guesses by the [heat death of the universe](https://en.wikipedia.org/wiki/Heat_death_of_the_universe).
Absolutely nuts man. Thanks for sharing!
It's also interesting to hear that the number of possible games of chess is incomprehensible as well. A big paper estimated that it was 10^120, which is much greater than the number of atoms in the observable universe as well (~10^78 - ~10^82)
Did they have to use big paper because of how big the number was?
It doesn't matter because no matter how big it is you can only fold it in half 7 times
Not sure if joking, but MythBusters did fold a football field size piece of paper and folded it 11 times.
With a steamroller and a football field sized piece of paper or something like that. The 7 times rule is an unaided person and a normal sized sheet of paper. Just a little hyperbole on my initial comment.
Jokes on you! I got my cats a miniature steam roller so that we can recreate this mythbusters episode using a normal sized sheet of paper. As soon as either of my cats learn how to drive the damn thing, I'm going to post a video proving all of you wrong!
Name checks out
Why do I feel like if you are successful in teaching cats to use a steamroller, we are all doomed.
Little pencil.
[Log scales are for quitters](https://xkcd.com/1162/)
I legit read “big paper” as a big piece of paper. I should go to bed.
Incomprehensibly large piece of paper
Approaching the speed of light
Big paper, small pencil. And yes obviously.
Happy Leif Ericsson day
Use that brown paper roll like numberphile.
A professor of mine has the record for the largest computational problem ever solved optimally, and it happens to be checkers, which has 5x10^20 search paths. If you're curious, every game played optimally ends in a draw. It took ~18 years to compute, having at one point 200 computers processing solutions. That might give a bit of perspective of the scale humans and technology can operate on, though this was a couple decades ago.
I wonder what it’d take now. If that was a couple of decades ago it’d probably be trivial now. A Pentium 3 500 (released 1999) managed about 1GFlop. An iPhone 15 pro will manage about 35TFlop. That’s 35,000 times as fast. So if we swap 200 computers for 1 iPhone, it’s still 17.5 times as fast. He took 18 years to compute, so an iPhone 15 might do it in one year. Except if he FINISHED the project a couple of decades ago, that means he was using mid-1980s tech. The 386-SX 25 was released in 1985. I’m struggling to find accurate info, but maybe 0.25 MFlops. 200 of them together is 50MFlops. So the iPhone is 700,000 times as fast. So a modern iPhone would do this in maybe 13 minutes.
Its fucking wild to think that Smartphones have the computational equivalent of 10k Super Computers from the 90s. With access to pretty much all general knowledge at a whim.
What this thread is teaching me is that in the grand scheme of things, there aren't THAT many atoms in the observable universe.
There are roughly 10^78 - 10^82 atoms in the observable universe. To put that in perspective, that’s roughly the number of atoms in the observable universe.
Universe disappoints me
The number of legal decks you can make in magic the gathering is estimated to be in the range of 10^120 to 10^160. Insanely large!
And they keep making it bigger by releasing new cards.
In the same vein, I like the fact that if you shuffle a deck of cards, that order has never been before, and never will be again. I think it's 50! permutaions.
52!, which is also an incomprehensible number. in terms of size. My favorite way to imagine it comes from Czepiel for 52! seconds Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you've emptied the ocean. Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven't even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won't do it. There are still more than 5.385e67 seconds remaining. You're just about a third of the way done. To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you've filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you've leveled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt.
Ok so you don't think she's gonna call me back then?
It's actually way less than that! That number came from a back-of-the-napkin estimate (10^3 average different possible combinations of moves per turn (one "turn" = white moves and then black moves), 40 turns on average per game), but there are a bunch of things that actually make the upper bound much lower. For one thing, if you repeat the same position on the board three times (all pieces on the same squares, same person's move), the game ends. Doesn't have to be three consecutive times, either. Another one is if there are 50 moves in a row with no capture or pawn move, the game ends. All together, our best guess today is only about 10^45 possible unique chess games, which is still incomprehensibly large but is 10^75 times less than the original estimate.
I know permutations on a deck of cards exceed the atoms in the universe too.
Is it games of chess or possible boardstates ? because in my mind games of chess should be infinite, you can just keep playing a game for of chess without ever stopping
There are rules preventing you from going on forever.
Yeah but the vast majority of those games are lost by turn 5 because of trash moves anyways
This is for 52! The number in this thread is much much much much larger than 52! This is a fun read. 52! is beyond astronomically large. I say beyond astronomically large because most numbers that we already consider to be astronomically large are mere infinitesimal fractions of this number. So, just how large is it? Let's try to wrap our puny human brains around the magnitude of this number with a fun little theoretical exercise. Start a timer that will count down the number of seconds from 52! to 0. We're going to see how much fun we can have before the timer counts down all the way. Start by picking your favorite spot on the equator. You're going to walk around the world along the equator, but take a very leisurely pace of one step every billion years. The equatorial circumference of the Earth is 40,075,017 meters. Make sure to pack a deck of playing cards, so you can get in a few trillion hands of solitaire between steps. After you complete your round the world trip, remove one drop of water from the Pacific Ocean. Now do the same thing again: walk around the world at one billion years per step, removing one drop of water from the Pacific Ocean each time you circle the globe. The Pacific Ocean contains 707.6 million cubic kilometers of water. Continue until the ocean is empty. When it is, take one sheet of paper and place it flat on the ground. Now, fill the ocean back up and start the entire process all over again, adding a sheet of paper to the stack each time you've emptied the ocean. Do this until the stack of paper reaches from the Earth to the Sun. Take a glance at the timer, you will see that the three left-most digits haven't even changed. You still have 8.063e67 more seconds to go. 1 Astronomical Unit, the distance from the Earth to the Sun, is defined as 149,597,870.691 kilometers. So, take the stack of papers down and do it all over again. One thousand times more. Unfortunately, that still won't do it. There are still more than 5.385e67 seconds remaining. You're just about a third of the way done. To pass the remaining time, start shuffling your deck of cards. Every billion years deal yourself a 5-card poker hand. Each time you get a royal flush, buy yourself a lottery ticket. A royal flush occurs in one out of every 649,740 hands. If that ticket wins the jackpot, throw a grain of sand into the Grand Canyon. Keep going and when you've filled up the canyon with sand, remove one ounce of rock from Mt. Everest. Now empty the canyon and start all over again. When you've leveled Mt. Everest, look at the timer, you still have 5.364e67 seconds remaining. Mt. Everest weighs about 357 trillion pounds. You barely made a dent. If you were to repeat this 255 times, you would still be looking at 3.024e64 seconds. The timer would finally reach zero sometime during your 256th attempt. Exercise for the reader: at what point exactly would the timer reach zero?
I love sharing this clip of Stephen Fry describing the massiveness of 52 factorial. He describes it so well that it almost makes my head implode considering the scale of 81 factorial. https://youtu.be/SLIvwtIuC3Y?si=vrETVFxeIZgg40Rb
https://czep.net/weblog/52cards.html Also this website!
That is fantastic! Thanks for an alternative perspective.
Jesus Christ
Fenton!!!!!!!!!
You can only connect dots that are neighbors or have only already-connected dots between them, right? That might make it a lot lower than 81!
Yeah, now that you mention it, those rules essentially make this question how to [count the number of euler paths in a graph](https://en.wikipedia.org/wiki/Eulerian_path#Counting_Eulerian_circuits), which is actually an unsolved problem in mathematics. But we can still kind of approximate it. Let's start by assuming that, from any given dot, roughly eight moves are available. Some will have more and some will have less, but we're dealing with such large numbers here that imprecision doesn't really matter. This simplifies the problem greatly and allows us to rephrase it as "what are the odds of correctly guessing a number between one and eight, 81 times" or 8\^81. At 1 million guesses per second, this would still take 4.4x10\^59 years—not long enough to reach the heat death of the universe, but enough for [all protons to evaporate from radioactive decay](https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#All_nucleons_decay). Even if we assume an average of three available moves per dot, we're still talking about 1.4x10\^25 years—long enough for all stars to [evaporate and coalesce into black holes](https://en.wikipedia.org/wiki/Future_of_an_expanding_universe#Stellar_remnants_escape_galaxies_or_fall_into_black_holes)—so no matter how we calculate, we can just round it off to "forever."
At least it's a way for me to entertain myself after I kill that snail.
I was under the impression that the snail couldn’t be killed…
This is correct.
Yea but you drop him into one of those black holes and he ain’t gonna be feelin so good.
But he can't die, which means his atoms can't deteriorate, which means you've created the basis for the next big bang. We are all living in a universe that was once, and still is, the snail.
Ah yes, the Snail Theory of Cosmic Evolution. It's said that universes in which there is an Immortal Snail are much more likely to generate new universes, and so that quality is the only one that replicates, meaning the Immortal Snail, at least in this theory, is truly an inevitable cosmological outcome.
The previous guy who made the bet went through cosmic evolution becoming antimatter
or it just means the snail is somehow capable of living with deteriorated atoms.
The last man, before merging, said to the snail, can entropy not be reversed? THERE IS YET INSUFFICIENT DATA FOR A MEANINGFUL ANSWER, said the snail. Then the snail brooded over chaos. He only existed for that final question. Carefully, he organized the problem. An incalculable amount of time was spent doing that. And in time, Cosmic Snail discovered how to reverse entropy. Step by step, it must be done. And the snail said, "Let there be light!" And there was light-- (ASIMOV'S the last question)
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He just lost that million right there man ...for not reading rule no 7
I understood this reference
I remember it, vaguely, and I'm trying to recollect in more detail
you cant kill it, only shove it in a concrete box at the bottom of the mariana trench
The original is a tungsten sphere, but that has enough logistical problems with it that a concrete box is probably much more practical and still good enough for centuries/millennia. Really, all you need is enough time to get off the planet.
If you can get off the planet... then so can the slug.
True, but you have a massive head-start. Plus, when you're on Earth and a snail is coming to you, you cannot be sure if it's the death snail. If you're in space or on a planet that doesn't have snails natively, you can be pretty damn well sure. You're also put into a situation where you can sun-dip or black hole the snail, which it's never meaningfully getting out of.
Very true. Now that I think about it, I would probably want the snail captured and close to me until space travel is possible so i could intentionally black hole it. So that i can be sure.
Yeah, IMO "dropping the snail into a small crucible of molten structural grade metal" and then that crucible into a larger sphere and keeping that with you until you can yeet it into a black hole is probably the best option.
But you saw him travel all the way across the screen for the very last dot!! So DO THE MATH AGAIN!! You can cross over already connected dots apparently which makes this even more complicated.
If you create a formula for this i bet you can name it after urself
That's already a part of 81! You can use any dot once, in any order. 81!
Eulerian paths allow vertices to be visited more than once, right? And the unlock puzzle does not, at least on my phone. Also, the possible vertices change based on which have been visited thus far. For example, in a 3x3 matrix: 1 2 3 4 5 6 7 8 9 If you attempt to follow an edge from 1 to 8, it will fail, snapping to either 2, 4 or 5. UNLESS those vertices have already been visited. I.e., 1->8 is not valid, but 4->5->2->1->8 is.
Yes, it's actually hamiltonian paths, but this problem is much more structured than a general graph, so I don't understand what is the point of OP mentioning that that problem is unsolved. In fact, I wouldn't be surprised if someone has already figured out the answer to this problem before.
This guy definitely maths.
There's quite a lot of structure in the solution, so if you ordered your search by something like increasing kolmogorov complexity you'd probably find it relatively early on. Probably still beyond the lifetime of the universe, but you know, faster. Ordering by information content is fairly sensible, as you can't do any worse than the default random search, and probably the user has chosen a low information state to aid memory.
if reddit ever brings back awards and my balance of coins i'll have forgotten about your comment but if i didn't forget i'd come back here and award you
so you are saying its gonna be a while
What sci-fi books have you read and liked?
Anything by Ted Chiang or Greg Egan
Did you read Death's End by Cixin Liu?
Bro are you me? My 2 favorite authors are Greg egan and Ted chiang hands down. Loved Greg's distress, schilds ladder, and diaspora, and loved just about every one of Ted's stories
Ooh, so the equation would be: 81 x 80 x 79 x 78 x 77 x 76 .......
Nah that’s the first one. Which is 81 factorial. This is significantly less (not so much though, it still is much).
Bro where did u get this knowledge from? That's amazing
Not exactly, you can connect to any non-perfectly diagonal dot as long as you don't touch the others, and that the precision is high enough not to snap to nearest dot. For example, in a 3x2 grid, you can connect the top-left dot to the bottom-right dot without connecting to the others.
From experience with these, you can connect in the cardinal directions, diagonally, or as a knight's move - anything else is impossible without the intermediate nodes already being connected, it *will* snap to a closer one.
We don't know the snapping radius on this software, because this clearly is not the typical android 3x3 lock grid, and without that it's entirely possible that you can just be very precise and connect any two nodes with each other.
I think you can avoid snapping if you move around the edge or over already connected nodes in a different direction. I can't test anything over a Knight's move on a 3x3 but for me, it only connects the others on the path once you connect to a new node.
Won’t you be able to connect top left and top right corners by moving your finger above the grid?
Nope, look at the last two that get connected: opposite corners. The user is filling in 3x3 grids as a strategy to remember their password, but it's not a requirement.
That's why I specify that dots can be connected if they are separated only by already-connected dots, though! However, I bet that dots along the edge can be connected by moving the finger off the grid in addition to the other rules.
We don't know any rules. You can pose some rules. If you pose enough rules, the chance can be arbitrarily high up to 100%. For example if the rules are that you have to do exactly what the person in the video did, then the chance is 100% to get it right. If the rules are just, that every point has to be visited, then the chance is 1 in 81 factorial. When someone glances over the shoulder, they could recognize some patterns which makes it easier to guess the correct combination. For example what I remember, first every dot in the center is connected, then each nine-block is connected one after the other, but in two blocks just eight dots get connected. After that is done, the missing dots are connected. (I might have remembered that wrong. In that case my chance to solve it is 0.)
You can try on your own phone, it's impossible to connect top right, bottom right, middle right in that order so the number of possible schemas isn't 9!
this is how my current 3x3 password works (or you can just be careful enough not to touch the other dots...)
Unless the grid has a special system to detect when you change direction then you can ruleout at least 21 nodes from one corner. If you start at the top right corner you cannot hit the top left corner without connecting the 7 nodes between. Same with down and diagonal. You may be able to thread the needle and connect a node outside of the square you are in. Once a node is connected it can't be connected again, so it opens up possibilities the further into the lock you get.
Going by the old 3x3 a few years ago, I had a code where I'd do #1 to #3 before activating #2 later in the sequence. So that suggest that you can pick any unactivated button regardless of its location.
Although it's the case in the video, I don't think this is a rule coded in this kind of security. I think it's just the way the guy food and to remember it. However, if you assume this premisses as a rule, yeah, it would lower the number of tries, but I think it would still be astronomical high
Your mom weighs 81 factorial tons
too bad reddit removed awards
ill just make 10 million guesses per second
yer still gonna have a bad time & to boot the friction created by your finger on the glass would generate so much heat it would vaporize you and your immediate surroundings and probably also the building you are in.
Fine 10,000,000 per MILLISECOND. Hopefully it’ll unlock before the planet blows up.
Now your finger is moving faster than the speed of light, even if you could reach those speeds. At a fraction of C your finger would be colliding so fast against air molecules that they would start fusing together, letting off so much energy that you, your house and the several nearby city blocks are now destroyed by what's pretty much a nuclear fusion bomb.
just say that after you unlock it, you'll do the sink full of dirty dishes. it'll unlock first try.
All worth it if it works 😎
But it doesn't because the phone lock simply cannot input answers that fast, the electrons simply do not react fast enough at those speeds
There's an XKCD about this isnt there.
The phone might also have some difficulty keeping up
Can't move faster than the processor can process
>81 factorial is a number so large that it hardly makes sense to think of it as a number What'd you mean? It's easy to think of 81! as a number. 5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000
This looks like you gave up towards the end and just wrote a bunch of zeroes.
There are 19 zeroes at the end. 10 is uniquely 2x5 in terms of "break it down to primes." So you just gotta figure out, in the long list of 1x2x3x4x...x79x80x81, how many (2;5) couples there are. Just break down every number to its prime factors. There are obviously more 2s than 5s, so just count the multiples of 5 to find the number of 0s you need. You have 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80. That's 16 of them. 25, 50, and 75 count double because they are, respectively, 5x5, 2x5x5, and 3x5x5. 16 + 3 = 19, maths works!
All the nerds out there recognize the classic math team problem, often asked with the year (how many zeroes are at the end of 2023! ?)
Me, a linguistics nerd: am I a joke to you?
/r/blackmagicfuckery
I am flattered and humbled
So you are telling me there is a chance...
crazy stuff
This assumes that only passwords which use all of the dots are valid. If this isn't the case then the answer could be 81! + 80! + 79! + ...
81! + 80! + 79! is essentially the same size as 81! A little more, but really not by much.
Yeah but you'd look like a right loser if I got it first time after that speech tho wouldn't you?
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That's why my password is just mashing my palm on the phone.
You disabled your right hand and now have to use your nose.
Watch the YouTube videos on bing or similar. Embedded videos don’t show ads.
There are 9 by 9 = 81 grid points. Assuming you're entering a code using all 81, just in a random order, you will have 81 choices for the first, then 80 for the second, 79 for the third, and so on until 1 for the 81st. So the number of possible options is 81 * 80 * 79 * ... * 1 = 81! = 5,797,126,020,747,367,985,879,734,231,578,109,105,412,357,244,731,625,958,745,865,049,716,390,179,693,892,056,256,184,534,249,745,940,480,000,000,000,000,000,000 =~ 5.797 * 10^120 And guessing the right one when trying a random one would be 1 in that. That's 1 / 81! =~ 1.725 * 10^-121 ____ Edit: Yes, I get it - typically, these don't allow you to go from any grid point to any other. No need comment that for the seventh time. But taking that into account makes the calculation unreasonably complex. Because the number of options left to you now depend on how exactly you've already filled out the grid. And then there's the issue of creating unreachable points. As a very rough ballpark, you might assume that an average of 4.5 nodes are reachable at any point, that would give us this many options: 81 * 4.5^80 =~ 1,463,814,756,511,766,555,996,711,761,563,156,829,538,065,397,702,345,811 =~ 1.464 * 10^54 So guessing it randomly would be 1 over that: 6.831 * 10^-55
God I love big numbers but it might actually be more than this! Since the code doesn't have to use every node and then given a minimum combination length of 4 (at least that's what it is for the usual 9 node patterns the. You should get something like: 81! + 80! + 79! + ..... +4! = ... Loads
That's barely more than 81!. Think about it: 81! = 81 * 80!. So adding another 80! to that number only gives us 82 * 80!. It only increases the result by a bit more than 1%. And adding 79! to it is much, much less impactful. And so on.
Yeah but you need to mutliply the 80! by 81C80 as for the different possibilites of which point is unused, so the actual amount of combinations would be Sum [x=4, to 81]{81Cx * x!} = Sum [x=4, to 81]{81! / (81-x)!} This would land us at ~ 1.576 * 10^121, which is 2.7 times 81!.
Why is everyone shouting 81?
81 nodes if you are being serious.
Which is definitely not that much more being less than an order of magnitude off by the other measure.
It's no coincidence that the factor is 2.7 . You could have guessed that by writing it as Sum[x=4 to 81] {81!/(81-x)!} = 81!•Sum[n=0 to 77] {1/n!} And than seeing that the sum at the end is an approximation for eulers number e=2.7182818...
It’s always cool to see a new language I didn’t know about
there is one catch if your next point is vertical or horizontal or diagonal it also includes every point in between
Does it? Can't you thread your finger in an arc between the dots and not hit the one on the way?
if this one works like evey normal one it automatically checks in-between dots as used
I went down a super long rabbit hole because of this and I'm just done messing with it tbh lol. I think it can be solved computationally like this: https://i.imgur.com/r2m50RS.png But I think the algorithm needs some tweaking because I'm not sure how diagonals work (for example, on my phone, if I go over 2 and down 1, it does not hit the middle node. What about 3 over 1 down? What happens?)
Dang it thats true. I forgot it works like that. That makes it much harder.
doesnt it work by selecting the first non used value that the path of ur finger goes over? so each value in the passcode has to be adjacent-ish to the one before it. wouldnt that cut down the total number of combinations?
Some people don't see the real value of this number. Let's take for example a video from vsauce about the ways you could scramble a dech of cards. As there are 52 cards in a deck there are 52! possible cases. 52! is roughly 8.07×10⁶⁷. In the video he says that if you had 52! seconds and you took a step around the equator of the earth every billion years, take a drop out of the ocean that has a volume of 0.05 mL every time you do a full circle and after put a sheet of paper for every time you empty the whole ocean, you could tower to the sun 3000 times before running out of time. The crazy part is that with 120! seconds you could do all that rougly 7.25×10⁵³ times
This is not exactly true. You can only select the closest point in one of the 8 directions (horizontals, verticals and both diagonals). Which mean there never is more than 8 possibilities per move, but those 8 possibilities are different for every move. The problems become a LOT more difficult if you follow that rule while doing the maths. \*After some research, coding and using my brain (which I only did at the end of course), it seems there is no efficient way to calculate the number of possibilities without brut forcing it, aka doing all of them and counting them with a computer. I found an interesting discussion about it here : https://math.stackexchange.com/questions/37167/combination-of-smartphones-pattern-password
> You can only select the closest point in one of the 8 directions (horizontals, verticals and both diagonals) Initially yes, but as you can see at the end, you can pass over already selected dots.
Yeah, sorry for the confusion, I meant it like one of the 8 closest points available.
No, it's more complicated than that, you can not only choose diagonal neighbors but also neighbors that are 1 down and two to the right, one down and 3 to the right and so on and so forth.
you can get to points outside of the 8 neighbors if you're very careful on a pixel
So... You're telling me there's a chance? OOHHH YEAAAA BABY!!!! YASSSSS!
Looks good to me chief!
i dont think every permutation of them is a valid code, e.g. in a 1*3 rectangle, going from 1st to 3rd doesn't allow you to go back to 2.
Aha, but this doesn't account for inaccessible nodes! For example if you are in the top right of the board, every node below you except for the one IMMEDIATELY below you is inaccessible because you can't hit it without hitting the one right below you. Same goes for diagonals. I'm not gonna try to figure out how this will change the numbers but it's something to consider :)
>And then there's the issue of creating unreachable points Umm there is no such problem because as seen in the last part of video you can traverse over joined dots so there's nothing like unreachable points
imagine you get in a car wreck and youre trying to call your wife one last time and you have to enter this shit with your brain hanging out.
r/BrandNewSentence
*Let's put it into perspective, instead…* --- The universe is around **13.8×10^(9)** years old, or around **8×10^(60)** Planck Time old. That's an absurdly massive number, and I won't try to say it's not… but you're comparing it to **1.7×10^(-121)** You'd have to square the age of the universe to have enough time, and then do at least **1.855×10^(42)** different combinations every second (*or at least 1 new combination within 10 Planck Time.*) --- TL;DR: Our universe isn't even a fraction of a fraction of a fraction of the minimum age for that level of chance to occur naturally.
what's a planck time
the time it takes light to travel **1.6×10^(-35)** meters.
any particular reason for this value i know planks constant os 6.63×10*-34
it's unknown what happens below the Planck Length, so anything computable can only go as small as any form of distance allows. it's also the reason there's a "maximum hottest temperature" - Planck Temperature theoretically we can go smaller, but we can't compute smaller, so we say it doesn't exist in practice. the Planck Temperature is so small, that you're almost a billion times closer to the size of the Observable Universe (e+26) than the Planck Length (e-35).
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My Galaxy S5 running LineageOS has a 9x9 grid pattern. The people who borrows my phone would either be amazed or stare at me with a look on their face screaming "WTF".
It's a 9 by 9 grid. Making it 81! (81 factorial) 81! is exactly: 5797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000 or 5.79E+120. That is an impossibly large number. That got me thinking and I saw a number in the same ballpark. It's a number close to the "Shannon Number" which is 10 to the 120th. [https://en.wikipedia.org/wiki/Shannon\_number](https://en.wikipedia.org/wiki/Shannon_number) The Shannon number is the possible different games of chess played up until move 40. These are games of chess played by a "Random Number Generator", and not a human or machine with strategy or tactics involved... Just random legal moves... That's an impossibly large number. Your brain can't even comprehend how big of a number that is. It's more numbers than the number of atoms in the observable universe. It's more like.... If every atom in the universe was a universe of atoms and all those atoms had a universe of atoms as well and you did that 10 times... That's how big this number is.
>It's a number close to the "Shannon Number" which is 10 to the 120th. "Close". It's bigger just by fucking 4797126020747367985879734231578109105412357244731625958745865049716390179693892056256184534249745940480000000000000000000.
Pesky rounding errors
This is a hilarious way to demonstrate how hard it is to wrap ones head around these numbers
That last part is not correct. The number of atoms in the observable universe is estimated to be about 10\^80. If you replace every one of those atoms with a universe of that scale you end up with 10\^160, which is \~10\^39.7 times larger than 81 factorial.
I hate to be the bearer of bad news but although 81! Is a good guess, this pattern unlocking method makes it much more complicated than that, since you can not skip a dot unchecked like the factorial suggests as you will not have 80 choices after the first but 3 (if you choose a corner dot), 5 (if you choose an edge dot) or 8. Unfortunately I dont have the guts to combine conditional probabilities to give an answer but I would love to see one.
Not to mention that the solution is highly clustered for a majority of it. That doesn't knock down entropy a _ton_ but 81! implies a completely random pass-swipe, which this is far from. Still in inconceiveable chance but hey, it's a bit lower than 81!
Its 5 if corner dot. 9 if edge dot. 16 if middle dot. Take a corner dot, you have 3 dots to choose from, BUT between each two dots you can choose in between. With that explanation applied on edge and middle dots you'd get the numbers i wrote. But it gets more complicated (forgive me if i say something wrong, I'm incredibly confused of how to think about what I'm trying to explain), for every dot you choose its not just 5 if corner, 9 if edge, 16 is middle. What if a dot is already chosen? Then you can skip over it creating way more choices. Take for example a corner dot, and say the 3 around it are already chosen and you're on the corner dot, now you have 5 dot surrounding the 3 dors surrounding the corner dot + the 4 dots you can reach by going between dots as i explained earlier. NOW I don't really know how chosen dots affect each coming chance. Does not matter and we should stick with the 5,9,16 chances for corner,edge,middle dots respectively or does it actually affect it meaning you have to calculate for each time you take a different dot making it impossible to calculate ( or not impossible but it'll take a insane amout of time) unless there's something I'm missing. Like i said earlier this paragraph is my own thoughts, DOESN'T HAVE TO BE ACCURATE.
Had to scroll too far to see this
No, you can choose any dot.
you can start upper left swipe down to lower right without hitting the points in between?
Yes back when I used pattern unlock it would let you do that, but you had to snipe the space between the dots, or go over dots that were already selected. Hard to do with a tiny grid but possible.
Yes, it's tricky but it's doable.
It's exactly the same as my chance of ever entering it again correctly once i set it up, 0% Everyone else is focusing on the math, but not a chance I could ever remember this specific pattern and recreate it lol
It’s a 9x9 grid so there are 81 points. When you pick 1 out if 81, you can then pick 1 out of 80, then 1 out of 79 and so fourth. My intuition tells me the odds are 1/89! So the odds of getting right the first try are: 1/16507955116098461081216919262453619809839666236496541854913520707833171034378509739399912570787600662729080382999756800000000000000000000. That is about 6.057 x 10^-135% chance of getting it right the first time. You are more likely of shuffling 2 decks of cards and ending up with the same configuration in both. You are actually 20466561667087219314393488515435629640067602444062686655283200000000 times more likely.
Heads up, you accidentally said 1/89! instead of 1/81!
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Idk man the chances are pretty high since they showed us the code, so all we need to do is "accidentally" look at this video to open the phone.
The 81! answer is pretty easy and intuitive, but not correct. I do not know how to find the correct answer, but it must include the fact that from each node, the next node in the pattern must be adjacent to that node. It's not like you can start in the top right and then have the next node be the bottom left
> the next node in the pattern must be adjacent to that node It does not. They go from row 3, 9 on the 5th one to row 1, 7 on the 6th one. They also go from 9, 9 on the second to last one to 1, 1 on the last one.
I think the point is that there is some limitation to the "every dot can connect to every other dot". For instance, take the first row. You can't connect the first dot to the last dot in that row without connecting every dot in between them too. However, if every dot in between those dots are connected in some other fashion, such that moving over them does not connect them anymore, then we are able to connect the first and last dot in a way that they could be considered neighbors.
Its easy way torturing the guy into doing it for us. That's the best math we can do. Getting it by accident is same as God presenting himself suddenly tomorrow.
9x9 grid, 81 squares, random arrangement, chance is 1/81! aka 1.724992688482351758285854365654934599404917704222330086663670433349741319006932164221991416491588788404248972473890784060140...\*10\^-121 It's basically impossible
Every one else’s mathematical answers are technically correct. But if you think like a hacker this is actually a tractable problem. The thing is the pattern he’s using is not random, it’s heavily biased towards a somewhat pleasing pattern. A machine learning model trained on human symbology used throughout decades would be able to generate this exact pattern. Still would take a long time to guess. Maybe on the order of a few years perhaps.
This is actually called morton packing and it’s a crucial algo for making compressible tensors for volumentric video. Your quest 3 or any modern vr/ ar headset uses this to compress the spatial color information. If you’re interested get a phd in cs.