(adsbygoogle = window.adsbygoogle || []).push({}); i have some cryptography scholars working on finding the chipper i really want to reverse engineer this game
rsa is the most complex form of encrypting i have seen yet your brain will hurt with the algorithms

ill keep you post waiting for them to find chipper

don't flame i'm trying

i know how to cryptography but these guys are better
im trying to form some sort of team to reverse engineering this game now we need people who are experienced to help us i thought of this because if no ones helping we have to help are selves post here if you in

HERE SOME INFO ON RSA:
[multipage=Key Generation Algorithm ]
Key Generation Algorithm:

1.Generate two large random primes, p and q, of approximately equal size such that their product n = pq is of the required bit length, e.g. 1024 bits. [See note 1].
2.Compute n = pq and (φ phi = (p-1)(q-1).
3.Choose an integer e, 1 < e < phi, such that gcd(e, phi) = 1. [See note 2].
4.Compute the secret exponent d, 1 < d < phi, such that ed ≡ 1 (mod phi). [See note 3].
5.The public key is (n, e) and the private key is (n, d). Keep all the values d, p, q and phi secret.
•n is known as the modulus.
•e is known as the public exponent or encryption exponent or just the exponent.
•d is known as the secret exponent or decryption exponent.

[multipage=Encryption ]
Encryption:

Sender A does the following:-
1.Obtains the recipient B's public key (n, e).
2.Represents the plaintext message as a positive integer m [see note 4].
3.Computes the ciphertext c = me mod n.
4.Sends the ciphertext c to B.

[multipage=Decryption ]

Decryption:

Recipient B does the following:-
1.Uses his private key (n, d) to compute m = c^d mod n.
2.Extracts the plaintext from the message representative m.

[multipage=Digital signing ]
Digital signing:
Sender A does the following:-
1.Creates a message digest of the information to be sent.
2.Represents this digest as an integer m between 0 and n-1. [See note 5].
3.Uses her private key (n, d) to compute the signature s = md mod n.
4.Sends this signature s to the recipient, B.
[multipage=Signature verification ]

Signature verification:
Recipient B does the following:-
1.Uses sender A's public key (n, e) to compute integer v = s^e mod n.
2.Extracts the message digest from this integer.
3.Independently computes the message digest of the information that has been signed.
4.If both message digests are identical, the signature is valid.


[multipage=Summary of RSA ]
Summary of RSA:

•n = pq, where p and q are distinct primes.
•phi, φ = (p-1)(q-1)
•e < n such that gcd(e, phi)=1
•d = e^-1 mod phi.
•c = m^e mod n, 1<m<n.
•m = c^d mod n.

[multipage=Key length ]
Key length:

When we talk about the key length of an RSA key, we are referring to the length of the modulus, n, in bits. The minimum recommended key length for a secure RSA transmission is currently 1024 bits. A key length of 512 bits is now no longer considered secure, although cracking it is still not a trivial task for the likes of you and me. The longer your information is needed to be kept secure, the longer the key you should use. Keep up to date with the latest recommendations in the security journals.

There is small one area of confusion in defining the key length. One convention is that the key length is the position of the most significant bit in n that has value '1', where the least significant bit is at position 1. Equivalently, key length = ceiling(log2(n+1)). The other convention, sometimes used, is that the key length is the number of bytes needed to store n multiplied by eight, i.e. ceiling(log256(n+1)).
The key used in the RSA Example paper [KALI93] is an example. In hex form the modulus is


The key used in the RSA Example paper [KALI93] is an example. In hex form the modulus is 0A 66 79 1D C6 98 81 68 DE 7A B7 74 19 BB 7F B0
C0 01 C6 27 10 27 00 75 14 29 42 E1 9A 8D 8C 51
D0 53 B3 E3 78 2A 1D E5 DC 5A F4 EB E9 94 68 17
01 14 A1 DF E6 7C DC 9A 9A F5 5D 65 56 20 BB AB

The most significant byte 0x0A in binary is 00001010'B. The most significant bit is at position 508, so its key length is 508 bits. On the other hand, this value needs 64 bytes to store it, so the key length could also be referred to by some as 64 x 8 = 512 bits. We prefer the former method. You can get into difficulties with the X9.31 method for signatures if you use the latter convention.
[multipage=A very simple example of RSA encryption ]
A very simple example of RSA encryption:

This is an extremely simple example using numbers you can work out on a pocket calculator (those of you over the age of 35 45 can probably even do it by hand).
1.Select primes p=11, q=3.
2.n = pq = 11.3 = 33
phi = (p-1)(q-1) = 10.2 = 20
3.Choose e=3
Check gcd(e, p-1) = gcd(3, 10) = 1 (i.e. 3 and 10 have no common factors except 1),
and check gcd(e, q-1) = gcd(3, 2) = 1
therefore gcd(e, phi) = gcd(e, (p-1)(q-1)) = gcd(3, 20) = 1
4.Compute d such that ed ≡ 1 (mod phi)
i.e. compute d = e-1 mod phi = 3^-1 mod 20
i.e. find a value for d such that phi divides (ed-1)
i.e. find d such that 20 divides 3d-1.
Simple testing (d = 1, 2, ...) gives d = 7
Check: ed-1 = 3.7 - 1 = 20, which is divisible by phi.
5.Public key = (n, e) = (33, 3)
Private key = (n, d) = (33, 7).
This is actually the smallest possible value for the modulus n for which the RSA algorithm works.
Now say we want to encrypt the message m = 7,
c = m^e mod n = 73 mod 33 = 343 mod 33 = 13.
Hence the ciphertext c = 13.

To check decryption we compute
m' = c^d mod n = 13^7 mod 33 = 7.
Note that we don't have to calculate the full value of 13 to the power 7 here. We can make use of the fact that
a = bc mod n = (b mod n).(c mod n) mod n
so we can break down a potentially large number into its components and combine the results of easier, smaller calculations to calculate the final value.

One way of calculating m' is as follows:-
m' = 13^7 mod 33 = 13^(3+3+1) mod 33 = 133.133.13 mod 33
= (13^3 mod 33).(13^3 mod 33).(13 mod 33) mod 33
= (2197 mod 33).(2197 mod 33).(13 mod 33) mod 33
= 19.19.13 mod 33 = 4693 mod 33
= 7.

Now if we calculate the ciphertext c for all the possible values of m (0 to 32), we get
m 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
c 0 1 8 27 31 26 18 13 17 3 10 11 12 19 5 9 4

m 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32
c 29 24 28 14 21 22 23 30 16 20 15 7 2 6 25 32

Note that all 33 values of m (0 to 32) map to a unique code c in the same range in a sort of random manner. In this case we have nine values of m that map to the same value of c - these are known as unconcealed messages. m = 0, 1 and n-1 will always do this for any n, no matter how large. But in practice, higher values shouldn't be a problem when we use large values for n in the order of several hundred bits.

If we wanted to use this system to keep secrets, we could let A=2, B=3, ..., Z=27. (We specifically avoid 0 and 1 here for the reason given above). Thus the plaintext message "HELLOWORLD" would be represented by the set of integers m1, m2, ...

(9,6,13,13,16,24,16,19,13,5)

Using our table above, we obtain ciphertext integers c1, c2, ... (3,18,19,19,4,30,4,28,19,26)


Note that this example is no more secure than using a simple Caesar substitution cipher, but it serves

to illustrate a simple example of the mechanics of RSA encryption.
Remember that calculating me mod n is easy, but calculating the inverse c^-e mod n is very difficult, well, for large n's anyway. However, if we can factor n into its prime factors p and q, the solution becomes easy again, even for large n's. Obviously, if we can get hold of the secret exponent d, the solution is easy, too.

i was asking the same question can we use the debug keys to unlock the retail?....
-If anyone wants to make a crypt tool and try this notify me of the results (pm me).

im trying to form some sort of team to reverse engineering this game now we need people who are experienced to help us i thought of this because if no ones helping we have to help are selves post here if you in

ermm. ile try lol

all right man

im not really experienced at rsa but i can give it a go, i know quite a bit, im prepared to research about it and learn the full way it works...