# Has anyone heard of matrix-based “Russian Doll” encryption techniques?

I've been told that prime-number factoring is based on a "russian-doll" sequence of matrices, where a seed matrix of height Y and width X exists consisting of all zeros except for a single 1 at row Y column X. For every x columns added, the matrix grows by y, and two closely related formulas sum the column vectors and the row vectors.

Put another way, given a matrix of height $Y+ny$ and width $X+nx$, the matrix consists of all zeros and ones such that the sum across the rows is less than $f((Y+ny)/j,Y+ny)$, and the sum across the columns is less than $g(X+nx,(X+nx)/k)$. $j$ and $k$ provide the noise or entropy of the encryption algorithm.

The distribution of products of numbers is an example of one method of filling out such a matrix. Has anyone ever studied the relationship between $f$ and $g$ to come up with a family of "roman-doll" encryption techniques with difficulty related to prime-number factoring, but security closer to a "one-time pad" method of encryption?

### Reference

I don't have a good online reference for this, but the methods are readily understandable by anyone with a junior high education level. A similar approach is described at http://www.gbbservices.com/math/squarediff.html#Cracking_RSA

### Postscript

The algorithm I'm familiar with has the following theorems:

$$\begin{eqnarray} \forall j \in \mathcal{N}\cdot h_{0,j}&=&1\\ \sum_{i\in \mathcal{N}} h_{i,0}&=&1\\ \forall J\times k \in \mathcal{N}^2 &\cdot& (\exists s\cdot (s^2=2*J+1)) \vee \sum_{j\in \mathcal{N}}{h_{2^k(2J+1),j}}=(k+1)\sum_{i\in \mathcal{N}} h_{i,J}\\ \forall J \times k \in \mathcal{N}^2 &\cdot& \exists s\cdot (s^2=2*J+1) \implies \sum_{j\in \mathcal{N}}{h_{2^k(2J+1),j}}=(k+1)\sum_{i\in \mathcal{N}} h_{i,J}-\lceil\frac{k+1}{2}\rceil \end{eqnarray}$$

An additional constraint that results in the "russian doll" effect are: $$\forall I \in \mathcal{N} \cdot I\ge X\implies \sum_{j\in \mathcal{N}} h_{I,j}=\sum_{j=Y+(y/x-m)(I-X)}^{Y+y(I-X)/x} h_{I,j}$$

where x, y, X, and Y are as introduced in the original question.

Private keys are those rows whose sum is one. Public keys are those whose sum is 2. The higher rows are more cryptographically secure.

Also, if the matrix is populated according to standard number system, then:

$$\forall I \in \mathcal{N} \cdot \sum_{j\in \mathcal{N}}\sum_{i=1}^{I} h_{i,j}=\sum_{n=1}^{\lfloor\sqrt{I}\rfloor} (\lfloor\frac{I}{n}\rfloor-n)=(D(x)-u)/2$$

where D(x) is the Divisor Summatory Function

### Examples

Matrix 0:

1    (1)


Matrix 1:

1   0   1   (9)
1   0   0   (7)
1   0   0   (5)
1   0   0   (3)
1   0   0   (1)


Matrix 2:

1   0   0   0   0   0   1   ?   ?    (33)
1   0   0   0   0   0   0   ?   0    (31)
1   0   0   0   0   0   0   ?        (29)
1   0   0   0   0   1   0   0        (27)
1   0   0   0   0   0   1   (25)
1   0   0   0   0   0   0   (23)
1   0   0   0   1   0   0   (21)
1   0   0   0   0   0   0   (19)
1   0   0   0   0   0   0   (17)
1   0   0   1   0   0   0   (15)
1   0   0   0   0   0   0   (13)
1   0   0   0   0   0   0   (11)
1   0   1   0   0   0   0   (9)
1   0   0   0   0   0   0   (7)
1   0   0   0   0   0   0   (5)
1   0   0   0   0   0   0   (3)
1   0   0   0   0   0   0   (1)

• Who told you this? Do you have a reference? Where do those theorems come from? If you won't cite sources, it is going to be impossible for anyone to help you. – mikeazo Jul 21 '15 at 1:37
• I've never heard of this, but a statement like "security closer to a one-time-pad" means (most likely) the author either doesn't understand perfect secrecy or doesn't have a proof. A reference would help, but right now this looks like snake oil. – tylo Jul 28 '15 at 18:38
• The author didn't provide a proof. That's why I'm wondering if anyone else has seen this approach. I don't have a good online reference for this, but the methods are readily understandable by anyone with a junior high education level. A similar approach is described at gbbservices.com/math/squarediff.html#Cracking_RSA – zQAycX Jul 29 '15 at 9:55
• I´ve added that link to the question for your convenience, cleaned up some obsolete comments, and reopened the question accordingly. – e-sushi Jul 29 '15 at 21:44