Did anyone know...

[PureStr]

theres only 9 numbers used... 10 is two digits.

0.1.2.3.4.5.6.7.8.9

10 digits

True, but the 0 isnt used.

0 isint used where ?
in the verification image
or in my example ?

[Lachlan]

Even if there are only 50 letters and numbers (A-Z, a-z, 2-9 with O, o, 0, 1, l, I, and maybe a few others excluded) I do not think there would actually be 50^6 (50x50x50x50x50x50) or 15,625,000,000 (15.6 billion) actual codes. The issue I see with this is that the work to create a combination, filter it for invalid combinations (like rude words), create an image of the code with the various dots and stripes and then SEND the image to the client seems like a lots of excess CPU and bandwith usage, especially since a new image pops up almost immediately when you enter an invalid code.

More likely there is a group of images in the SRO Client and the server simply sends a random pointer to the image to display. For example, if there are 10,000 images, the server might send a value to the client between 1 and 10,000 (N) and the Nth image in the file would be displayed. When the player enters the code, the 6 characters are sent back to central to be confirmed that those 6 characters are the correct characters for image N.

Like a few other people mentioned, I am sure I have seen the same code appear more than once, and even in a relatively short time. Of course, it is just possible that I saw a SIMILAR code, and only assume it was the same.


BTW: Someone in the thread mentioned using Factorial Equations to get the number of possible codes. This would only be true if the characters could only be used once. For example, if there are 50 characters available, then the formula for all permutaions of 50 characters where 6 are selected is P(50, 6) or 50! / (50 - 6)!. It is the bottom portion that is important, as the resulting equation becomes 50x49x48x47x46x45 which is only 11,441,304,000 or 11.4 billion ... about 4 billion less than the power equation. Even so, this formula is not the correct one in this case, because reusing the characters is allowed. At least, I have seen several codes with characters repeated, such as 466DQA (the last code I had to enter).


Another way to look at this, is say that the formula for calculating the number of ways to create a 6 character string from 50 characters is F(50, 6). More generally, the formula is F(N, L) where N is the number of characters and L is the length of the code.

Since there are 50 characters that can go in the first position, the result could also be calculated as 50 x the number of ways to make 5 character codes.

F(50, 6) = 50 x F(N, 5)

The question then is, "What is N?"

If N is reduced by 1 to represent the first character, then no code could have 2 characters the same. But we have seen plenty of repeats, so this cannot be the case.

Therefore, N must still be 50, so the equation is:

F(50, 6) = 50 x F(50, 5)

Continue the logic, and you will see that:

F(50, 6) = 50 x F(50, 5) = 50 x 50 x F(50, 4) = 50 x 50 x 50 x F(50, 3) ... 50 x 50 x 50 x 50 x 50 x 50 x F(50, 0)

Presumably there is only 1 way to select zero items from a group (a group of 0 items is 1 way), so the result is 50^6.


If however, N had been reduced by 1 to represent the character that had been used in the previous spot, the formula would become:

F(50, 6) = 50 x F(49, 5)

And again continuing this logic, will give:

F(50, 6) = 50 x F(49, 5) = 50 x 49 x F(48, 4) = 50 x 49 x 48 x F(47, 3) ... 50 x 49 x 48 x 47 x 46 x 45 x F(44, 0)

Even though there are 44 character remaining at the end, there is no place to put them, so the "Factorial"-like equation ends at 45.


Yeah I know ... chalk this one up to insomnia and not wanting to get online bad enough to enter 50 validation codes.


Good Luck!

[moley]

and after all that math you all have failed to realise that even though you have several well educated guesses as to the maximum amount of combinations of the characters that appear on the verification image, you haven't calculated the full number of verification images that there are because of the random noise added to the image.

[PureStr]

random noise added to the image

?

[Nitro]

NuclearSilo is PRO at Math ... he knows this xD ...
As far as I know its - number of letters ^ number of spaces ...

[zZh4ku]

MAthematicians

[Lachlan]

and after all that math you all have failed to realise that even though you have several well educated guesses as to the maximum amount of combinations of the characters that appear on the verification image, you haven't calculated the full number of verification images that there are because of the random noise added to the image.

But, the "random noise" is not going to change the number of codes, therefore it does not chagne the probability of "guessing" a code randomly, or of getting two codes the same.


What it would effect, potentially, is a method of looking at the image as a whole and determining from that what the code might be. In this case, the actual code is irrelavent, as many pixel patterns could result in the same code, therefore it would only matter how many pixels are in the image.

As far as I can tell, on my screen the image is in a box defined by upper left 674,637 and 874,702 excluding the boarders. It might be that there is a line or two at the bottom of the image that never chages ... I cannot quite tell. This is an area of 13,266 pixels.

Given that the image appears to be practically a black & white image, I will treat the pixels as binary values, where they can be either Black (0) or White (1). The result is that 2^13266 possible images can be made. This is a big number ... somewhere around 10^3993 ... or a 1 with 3993 zeros behind it.

In contrast, the number of atoms in the universe is estimated to be about 10^78, so this number would represent the number of atoms in about 10^3915 Universes, give or take a few.

When you consider the codes again (say 16 billion of them) you get approximately 10^3983 possible images that map to each code.

Good Luck!

[PureStr]

As far as I know its - number of letters ^ number of spaces ...

Its :

Number of different possiblities per digit ^ number of digits.

Cant beleive people are still arguing on this...
its like 7th grade math.

[Shimrod]

Pfft, it comes down to one fact really.....

either you can login or you can't


So what's the problem?

[Lachlan]

As far as I know its - number of letters ^ number of spaces ...

Its :

Number of different possiblities per digit * number of digits.

Cant beleive people are still arguing on this...
its like 7th grade math.

Who’s arguing? It is a discussion of a simple question with some tangents and a ubiquitous distain for the ICV system.


By "*" do you mean multiplication? If so, you are incorrect. Of course, ** often means power, but I am pretty sure here the symbol ^ has been generally accepted to mean "raised to the power of". The answer is N "raised to the power of" 6. We are still guessing a bit on what N might be, but it cannot be larger than 62.

Assuming there are 50 characters and 6 digits, according to your formula (assuming you meant multiplication) there would only be 300 possible codes. Even with only 10 possible values in each of 6 positions, it should be obvious that there could be 1 million “codes”.


000000
000001
000002
.
.
.
999999


Look familiar? That is 10 characters (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) in 6 positions and it allows for 1 million possible values (10 to the 6th power or 10^6). And that is just using 10 characters, which is much less that the 50 or so characters that can be used in a code. The correct answer is 15,625,000,000 (assuming there are 50 available characters).


BTW: After several attempts to sign on, I have not seen any of the following characters:

Upper Case:
I J K L O P S U V W X Y Z

Lower Case:
j k l o p t u v w x y z

Digits:
0 1 9

If this is accurate (since it is likely close to random, it might take a while for some characters to show up) it would mean the number of available characters is only 34, and 34^6 is just over 1.5 billion (1,544,804,416). Of course, if more characters show up, then the number will get bigger.


Good Luck!

[PureStr]

yup i meant ^ (to the power of), not * (multiplication)

[Lachlan]

Kool ...

BTW: I just had an L show up, so now there are definitely 35 characters in the list. 35^6 = 1,838,265,625

I do wish they would exclude B and/or 8 because it seems I am always getting them wrong.

Good Luck!

[brentonsbbe359]

Not really.

The image code doesn't call for the letters like z, x, and o. Some letters are not included.

Recalculate lol.

it does some times. ive gotten 'z' and 'x' in my image verifacation.



i think its awesome you did that

~Rawr~ *everytime sro lets me on i shall exit so i can try again to get the same image code*

[XMoshe]

Not really.

The image code doesn't call for the letters like z, x, and o. Some letters are not included.

Recalculate lol.

it does some times. ive gotten 'z' and 'x' in my image verifacation.

i think its awesome you did that

~Rawr~ *everytime sro lets me on i shall exit so i can try again to get the same image code*

I dare your to do that xD

[brentonsbbe359]

Not really.

The image code doesn't call for the letters like z, x, and o. Some letters are not included.

Recalculate lol.

it does some times. ive gotten 'z' and 'x' in my image verifacation.

i think its awesome you did that

~Rawr~ *everytime sro lets me on i shall exit so i can try again to get the same image code*

I dare your to do that xD

omg i totally will X)

but how much will you give me if it works? X) 90mil? mm?

[Cresta]

Just say that it is a VERY little possibility that you will get the same image code

[SpInKsTaR]

Even if there are only 50 letters and numbers (A-Z, a-z, 2-9 with O, o, 0, 1, l, I, and maybe a few others excluded) I do not think there would actually be 50^6 (50x50x50x50x50x50) or 15,625,000,000 (15.6 billion) actual codes. The issue I see with this is that the work to create a combination, filter it for invalid combinations (like rude words), create an image of the code with the various dots and stripes and then SEND the image to the client seems like a lots of excess CPU and bandwith usage, especially since a new image pops up almost immediately when you enter an invalid code.

More likely there is a group of images in the SRO Client and the server simply sends a random pointer to the image to display. For example, if there are 10,000 images, the server might send a value to the client between 1 and 10,000 (N) and the Nth image in the file would be displayed. When the player enters the code, the 6 characters are sent back to central to be confirmed that those 6 characters are the correct characters for image N.

Like a few other people mentioned, I am sure I have seen the same code appear more than once, and even in a relatively short time. Of course, it is just possible that I saw a SIMILAR code, and only assume it was the same.


BTW: Someone in the thread mentioned using Factorial Equations to get the number of possible codes. This would only be true if the characters could only be used once. For example, if there are 50 characters available, then the formula for all permutaions of 50 characters where 6 are selected is P(50, 6) or 50! / (50 - 6)!. It is the bottom portion that is important, as the resulting equation becomes 50x49x48x47x46x45 which is only 11,441,304,000 or 11.4 billion ... about 4 billion less than the power equation. Even so, this formula is not the correct one in this case, because reusing the characters is allowed. At least, I have seen several codes with characters repeated, such as 466DQA (the last code I had to enter).


Another way to look at this, is say that the formula for calculating the number of ways to create a 6 character string from 50 characters is F(50, 6). More generally, the formula is F(N, L) where N is the number of characters and L is the length of the code.

Since there are 50 characters that can go in the first position, the result could also be calculated as 50 x the number of ways to make 5 character codes.

F(50, 6) = 50 x F(N, 5)

The question then is, "What is N?"

If N is reduced by 1 to represent the first character, then no code could have 2 characters the same. But we have seen plenty of repeats, so this cannot be the case.

Therefore, N must still be 50, so the equation is:

F(50, 6) = 50 x F(50, 5)

Continue the logic, and you will see that:

F(50, 6) = 50 x F(50, 5) = 50 x 50 x F(50, 4) = 50 x 50 x 50 x F(50, 3) ... 50 x 50 x 50 x 50 x 50 x 50 x F(50, 0)

Presumably there is only 1 way to select zero items from a group (a group of 0 items is 1 way), so the result is 50^6.


If however, N had been reduced by 1 to represent the character that had been used in the previous spot, the formula would become:

F(50, 6) = 50 x F(49, 5)

And again continuing this logic, will give:

F(50, 6) = 50 x F(49, 5) = 50 x 49 x F(48, 4) = 50 x 49 x 48 x F(47, 3) ... 50 x 49 x 48 x 47 x 46 x 45 x F(44, 0)

Even though there are 44 character remaining at the end, there is no place to put them, so the "Factorial"-like equation ends at 45.


Yeah I know ... chalk this one up to insomnia and not wanting to get online bad enough to enter 50 validation codes.


Good Luck!

If all posts were this long + informative The world would be a better place