Re: adacrypt encryption algoruthm
- From: "ping pong" <mosescuadro@xxxxxxxx>
- Date: Sun, 20 Mar 2011 17:11:44 +1030
lower c that is intended not capital C. Apologies.
I have a problem with lack of precision on the matters you mentioned also.
The parameter limits
Explicit identification of the key material
that you are asking about can only be made clear by: adacrypt.
Actually this quite good practice.
It's part of the process before analysis can start.
That is good I am learning!
<robertwessel2@xxxxxxxxx> wrote in message news:c27d34f6-3c8d-4348-924f-57ed6a12a768@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Mar 20, 12:32 am, "ping pong" <mosescua...@xxxxxxxx> wrote:adacrypt symmetric stream cipher
2 p is plaintext byte
3 C is ciphertext
5 a[i] are arrays of integers
6 all other labels are integers
7 p1 = p * 10
8 p2 = p1 + a1[i]
9 ds = a2[p2]
10 co = a3[j]
11 N = a4[k]
12 choose n
13 c = co + ds + n*N
14 Encryption Example:
15 a1= (0,1,2,3,4,5,6,7,8,9)
16 a2= (-526 to +487)
17 a3= (+300 to +1300)
18 a4= (+14000 to +28000)
19 P = 80
20 p1 = 80 * 10 = 800
21 p2 = 800 + 7 = 807
22 ds = -428
23 co = 512
24 N = 14802
25 n = 14
26 c = 512 - 428 + (14 * 14802) = 207312
Ping Pong comment: T
The ciphered is y-coordinate of a point on a line c = ax + o + b, where b
is plaintext derived
The algorithm follows according to what I got from a dialogue with adacrypt
Adacrypt to validate, algorithm and example, please.
Plaintext is the secret.
Not sure what else is required to be secret?
In line 3, I assume you meant “c is ciphertext” (not uppercase C)?
Where do i, j and k come from in lines 8, 10 and 11?
What exactly does "choose n" do in line 12?
Where does the key data factor into this?
What is the point of the arrays, when it appears simple addition or
subtraction of a constant from the index value would do the same thing
(or nothing at all in the case of a1)?
p, which I assume is an input plaintext symbol, has a range of
What is the point of having 1014 entries in a2, when the maximum legal
limit of its index apears to be (p*10+9), which would surely imply
that 1009 entries would be enough assuming the 0..100 range limit on
p, and that the extra five entries are superfluous? Or if the
apparent range of p is incorrect, the size of a2 should still be
What is the range of outputs c? The dependency is obviously on the
undefined "choose n", but it appears to be about (1787 + n*28000),
which given that n can apparently range to at least 14, implies at
least 0..393787, or at least 18.6 bits, for an input about 6.7 bits of
Not so much.
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