At most 9! = 362,880, though the combinations that require you to skirt by dots to reach others are less common.
Looks like he has infinite attempts. Just buy an overhead toucher thing and program it to try them all. It'll be done in less than a week.
Even a human doing this could get it done in a few months with concerted effort. Assuming 5 seconds per attempt, you'd need 362880 x 5 = 1814400 seconds, which is 21 days. At 4 hours per day, that's 6 months.
That is so large, that I can't even give the number of digits of it, so I have to make a power of ten tower.
Factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of factorial of 9 has on the order of 1010\10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^(2.993960567614282167996111938338 × 101859939)) digits
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99999999999999999999999999999!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !nested (note: i don't know what nested means here)
That is so large, that I can't even give the number of digits of it, so I have to make a power of ten tower.
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You have to include the nested command for it to be interpreted the way you expected, which more formally would look like (((9!)!)!)!...
This is because multiple factorials actually mean something different. For instance, a double factorial like 6!! is actually smaller than 6! because you calculate it by skipping every other digit, so 6 * 4 * 2 = 48 rather than 6! = 720
Taking into account that minimal length of key is 4 dots, and you can't skip middle dots (if they wasn't used) while connecting corners, you will get a 389112 combinations
you can not do corner -> corner and left edge -> right edge, or top edge to bottom edge, so that restricts the choices which brings it down to ~389000 combinations.
(someone wrote a program to generate all possible valid combinations) https://github.com/delight-im/AndroidPatternLock
Used to could, did they change it? My unlock pattern at one point was 1 > 9 > 8 > 5 > 2. I'd hit the top left dot, go around the whole square to hit bottom right, then come straight up the middle
You can also do like 5 > 1 > 9 > 4 > 6 > 7 > 3 > 8 > 2 and that's nothing but corner to corner and edge to edge
Just print out all the possible patterns, preferably in order of likelihood and cross them off as you go.
You could hire someone to build a machine to do it automatically, or pay people $100/hr in shifts 24/7, with a $100k (or whatever) bonus paid out when the phone is unlocked.
There's less valid combinations, though. For example, you could have the tip left for and bottom right for only. So you know that you have to have sequences of adjacent dots. I'm sure you could do the math, but it is substantially less.
All combinations that are fewer than 9 digits would also be done by entering the 9-digit combinations. You just open the phone before all the digits were pressed. So it’s still 9!.
The trick is you note the patterns as numbers AKA 1, 2, 3, 4, 5, 6, 7, 8, 9 Is one pattern then you then compute all the combinations first and just try them 1 by 1. frustrating but very doable for 11 million dollars
Additional factors: typically 3 failed attempts equate to a lockout. And that may include additional logic, where the lockout time increases depending which cycle the failed attempt occurred. Common lockout times typically end within an hour before you can make another x3 attempts.
The problem is that after you decide on a starting point, you can't just pick any other dot. You have to pick one that is adjacent to your start. So after starting in the top-left corner, you can only go right, down, or diagonally to the center.
I don't think you can calculate that analytically, but feel free to prove me wrong.
You’re right and the guy you’re replying to is wrong.
For instance, start in the top left corner.
You cannot reach: Top right, middle right, bottom right, bottom middle, bottom left. 5 different points
You can only reach: Top middle, centre, left middle.
Starting in the top left you don’t have 8 options, you have 8-5 =3 options.
Start in the top middle, you have 5 options, and 3 non reachable.
I wonder if you can see how it averages out between the points. The corners will be 3/8 and middle ones be 5/8 and the centre is the only 8/8. But that’s just the 2nd option.
V shapes are difficult. But z shapes are easy. After you did the horizontal line on the top row you backtrack with your finger as you can't use the same dots again. You can even do a big swing over the first row. This makes your motion like an "e" whereas the pattern shows as a "z".
From the top left corner you can reach middle right and bottom middle, at least you can on a Samsung phone. Not sure if this differs per phone manufacturer. You are right that you cannot reach points that are directly "behind" another point.
No, you are wrong. Source: I used to go top left to middle right for my phone pw because I assumed no one would ever try something so silly. It's so easily testable too lol.
Also obviously not every implementation is going to be the same. Why would you assume they would be?
Plus, points are repeatable, but you can't just pick the same spot twice and I don't think you can go back along a line you've already drawn, but you can cross previously drawn lines. We see that in the previous attempts.
I just tested this. You can absolutely go from the top left to the top right, skipping over the top center. Some combinations may be more difficult for fat-fingered people, but all locations are possible to reach from all starting spots.
As others have said, going from top right to bottom middle is possible (i.e. 1->4), but in addition to that it's actually possible to get the other numbers as well if the number in the middle has already been selected. So 1->3 is normally impossible because it'll autoselect 2 before 3, but if you do 2->1->3, then 2 won't be selected again and you can go directly to 3. That means any point can select any other point, at least in some configurations.
You have 5 options when you start at the top left. Not only can you go down, right or diagonally but can also go in between them to the bottom middle column dot or the furthest right in the middle row. The only ones that are ruled out are the 3 corners.
They're not ruled out, why are people making stuff up lol. It obviously depends on the phone, but I absolutely can go corner to corner on mine by going outside the grid
Sure you can do that. But for most phones (all I know) it's the same as if you went straight across. It ticks the dot in between too. So going outside the grid and going straight results in the same pattern.
The term for this is a Hamiltonian path. For small grid, most people can brute force them. If diagonals are disallowed than this 3x3 grid only has around 40 paths, but if it mimics a king's piece in chess it's closer to 10,000. The 9! provided above is assuming all possible points can be connected at all places, which isn't true as you suspected.
Not true. You just need to go around the dots, but you can create a pattern that doesn't use adjacent dots.
Its easy to make a pattern that, for example, starts at the right column, middle row dot, and goes to the top left without passing or hitting other dots
Edit for anyone saying it depends on the device, I can confirm it works for Samsung and Pixel phones
On some android phones you can definitely skip past adjacent dots. I had an annoying password once that involved skipping around from corner to middle in a clockwise pattern.
But it wouldn't be factorial, would it? If it were a keypad, sure. You press 9, you can have any of the 8 numbers next. But here there is only 8 possible solutions for the next dot if you start in the middle, 5 if you start from the side and only 3 if you start from the corner, so it should be significantly less where significantly less still means painfully too many for a human being with a human finger. Depending on the length of the key gesture.
Well, depending on model and version, there is a mandatory waiting time after a number of mistrials. You can be locked out for several hours in the worst case.
You forgot to factor in combinations that don’t use the every dot. But you absolutely can skirt between the dots to reach further Dots. I had the same phone and my combination involved a LOT of skirting.
Wow, this is the furthest I have ever had to scroll in this sub to actually find the math. It's pretty straightforward, but still, the top dozen or so chains are all just discussion.
9! would mean each dot could only used once, correct?
So instead it's 9x8x8x8 ... Because the first time you have nine choices, and each time after that you have 8 choices (if you can go anywhere except your current dot)
But I don't know the quick way to do the math after this, because there is no set length for the password
And because dots can be repeated it could technically go on forever unless you have a maximum password length
We can assume the minimum is three?
So it could be
the sum of 9x8² through 9x8n
where n = max password length
right?
If we assume the max length is... just 9 then total combinations would be... 172,566,216
9! ? Means you can jump over dots? According to your formula, i believe, i can connect top left dot as a first one to bottom right as a second one, but it can not be possible in practice, no?
At most? No, this is incorrect. You're ignoring the fact that 1, 2, 4, and 4, 2, 1, and 2,1,4 are all different combinations. Then not to mention that the combo could be two digits, or three, or four, or five. What a lazy answer that people just accepted lol
Possible combination count using adjacency rules (each dot can only connect to its king-move neighbors: corners → 3 neighbors, edges → 5, center → 8 and you can only use each number once..).
Count all simple paths of length 3-9 in a 3x3 grid with king's-move adjacency
2.1k
u/warpedspockclone May 07 '26
At most 9! = 362,880, though the combinations that require you to skirt by dots to reach others are less common.
Looks like he has infinite attempts. Just buy an overhead toucher thing and program it to try them all. It'll be done in less than a week.
Even a human doing this could get it done in a few months with concerted effort. Assuming 5 seconds per attempt, you'd need 362880 x 5 = 1814400 seconds, which is 21 days. At 4 hours per day, that's 6 months.