Representing the Integer Factorization Problem Using Ordered Binary Decision Diagrams

Abstract

A method is given to reduce the problem of finding a nontrivial factorization of a positive integer $\alpha$, greater than one, to the problem of finding a solution to a system of Boolean equations, that is, a system of equations such that each equation is of the form $f=g$ where f and g are Boolean functions, meaning $\{0,1\}$-valued functions in zero or more Boolean ($\{0,1\}$-valued) variables. Our system is obtained by applying a sequence of reductions to an initial system of equations of the form

$$\begin{aligned} \{f_i(\vec {x},\vec {y}) = \alpha _i \mid i \in \{0, \dots , 2n\} \} \end{aligned}$$

where for each i $f_i(\vec {x},\vec {y})=f_i(x_0,\dots ,x_n,y_0,\dots ,y_n)$ gives the coefficient of $2^i$ in the binary expansion of

$$\begin{aligned} (x_0+2x_1+\dots +2^{n}x_n)(y_0+2y_1+\dots +2^{n}y_n), \end{aligned}$$

$\alpha _i$ gives the coefficient of $2^i$ in the binary expansion of $\alpha$, and $x_i$ and $y_i$ are $\{0,1\}$-valued variables. That is, the initial system represents a binary multiplier whose output bits have been set equal to the bits of $\alpha$. It is shown that each Boolean function in our reduced system, that is, each Boolean function

$$\begin{aligned} (g-h)~ {\text {mod}} 2 = {\left\{ \begin{array}{ll} 0 &{} \iff g=h \\ 1 &{} \iff g \ne h \end{array}\right. } \end{aligned}$$

such that $g=h$ is an equation in the reduced system, can be represented by a type of graph called an ordered binary decision diagram (OBDD) with an upper bound on its number of vertices of $\mathcal {O} \left( \log _2(\alpha /\log _2(\alpha )))^3 \right)$. Previous work has shown that the initial system has at least one Boolean function with an OBDD representation with number of vertices exponential in $\log _2(\alpha )$.

Appendix: Proofs for Section 2 (Preliminaries)

Proposition 1 For all $N \in \mathbb {Z}_{+}$, $\alpha ,\beta \in \mathbb {N}$, and $x,y \in \mathbb {Z}$

• (1) $(x \pm y) ~\text {mod}~ N = (x \pm (y ~\text {mod}~ N))~\text {mod}~ N$

• (2) $(xy) ~\text {mod}~ N = ((x ~\text {mod}~ N)y) ~\text {mod}~ N$

• (3) $(x ~\text {mod}~ (N^{\beta }) ) ~\text {mod}~ (N^{\alpha }) = x ~\text {mod}~ (N^{\min (\alpha ,\beta )})$

• (4) $(x \pm y) ~\text {div}~ N = ( (x \pm (y ~\text {mod}~ N)) ~\text {div}~ N ) \pm (y ~\text {div}~ N)$

• (5) $(xy) ~\text {div}~ N = (x ~\text {div}~ N)y+(((x ~\text {mod}~ N)y) ~\text {div}~ N)$

• (6) $( x ~\text {div}~ (N^{\beta }) ) ~\text {div}~ (N^{\alpha }) = x ~\text {div}~ (N^{\alpha +\beta })$

• (7) $x = N^{\beta }(x ~\text {div}~ (N^{\beta }))+\sum _{k=0}^{\beta - 1}((x ~\text {div}~ (N^k)) ~\text {mod}~ N)N^k$

• (8) If $x,y \in \mathbb {Z}$ and $x \le y$ then $((x ~\text {div}~ N) \le (y ~\text {div}~ N))$.

• (9) $(-1-x) ~\text {mod}~ N = N-1-(x ~\text {mod}~ N)$

Additionally, if $\alpha \le \beta$ then

$$ (x ~\text {mod}~ (N^{\beta })) ~\text {div}~ (N^{\alpha }) = (x ~\text {div}~ (N^{\alpha })) ~\text {mod}~ (N^{\beta -\alpha }). $$

Proof

For all $w \in \mathbb {Z}$ $w ~\text {mod}~ 1 = 0$ and $w ~\text {div}~ 1 = w$. Thus if $N=1$ then 1 through 9 hold.

Suppose that $N>1$ is a positive integer, $x,y \in \mathbb {Z}$, and $\alpha ,\beta \in \mathbb {N}$. Statements 1 and 2 follow from the fact that congruence mod N respects addition, subtraction, and multiplication. If $\alpha \le \beta$ then it follows from 1 and 2 that

$$\begin{aligned} (x ~\text {mod}~ (N^{\beta }) ) ~\text {mod}~ (N^{\alpha })&= ((x ~\text {mod}~ (N^{\beta })) + N^{\beta }(x ~\text {div}~ (N^{\beta }))) ~\text {mod}~ (N^{\alpha }) \\&= x ~\text {mod}~ (N^{\alpha }) \\&= x ~\text {mod}~ (N^{\min (\alpha ,\beta )}). \end{aligned}$$

If $\alpha > \beta$ then since $\{0,\dots ,N^{\beta }-1\} \subset \{0,\dots ,N^{\alpha }-1\}$,

$$\begin{aligned} (x ~\text {mod}~ (N^{\beta })) ~\text {mod}~ (N^{\alpha })&= x ~\text {mod}~ (N^{\beta }) \\&= x ~\text {mod}~ (N^{\min (\alpha ,\beta )}). \end{aligned}$$

Therefore 3 holds. To see that 4 holds, observe that

$$\begin{aligned} (x+y) ~\text {div}~ N&= \frac{x+y-((x+y) ~\text {mod}~ N)}{N} \\&= \frac{x+y-(y ~\text {mod}~ N)+(y ~\text {mod}~ N)-((x+(y ~\text {mod}~ N)) ~\text {mod}~ N)}{N} \\&= \frac{x+(y ~\text {mod}~ N)-((x+(y ~\text {mod}~ N)) ~\text {mod}~ N)}{N} + \frac{y-(y ~\text {mod}~ N)}{N} \\&= ((x+(y ~\text {mod}~ N)) ~\text {div}~ N) + (y ~\text {div}~ N) \end{aligned}$$

and

$$\begin{aligned} (x-y) ~\text {div}~ N&= \frac{x-y-((x-y) ~\text {mod}~ N)}{N} \\&= \frac{x-(y-(y ~\text {mod}~ N)+(y ~\text {mod}~ N))-((x-(y ~\text {mod}~ N)) ~\text {mod}~ N)}{N} \\&= \frac{x-(y ~\text {mod}~ N)-((x-(y ~\text {mod}~ N)) ~\text {mod}~ N)}{N}-\frac{y-(y ~\text {mod}~ N)}{N} \\&= ((x-(y ~\text {mod}~ N)) ~\text {div}~ N)-(y ~\text {div}~ N). \end{aligned}$$

The proof of 5 is another derivation similar to that for 4:

$$\begin{aligned} (xy) ~\text {div}~ N&= \frac{xy - ((xy) ~\text {mod}~ N)}{N} \\&= \frac{N(x ~\text {div}~ N)y + (x ~\text {mod}~ N)y - (((x ~\text {mod}~ N)y) ~\text {mod}~ N)}{N} \\&= (x ~\text {div}~ N)y + \frac{(x ~\text {mod}~ N)y - (((x ~\text {mod}~ N)y) ~\text {mod}~ N)}{N} \\&= (x ~\text {div}~ N)y+(((x ~\text {mod}~ N)y) ~\text {div}~ N). \end{aligned}$$

For the proof of 6 first observe that

$$\begin{aligned} x&= (x ~\text {mod}~ (N^{\beta })) + N^{\beta } (x ~\text {div}~ (N^{\beta })) \\&= (x ~\text {mod}~ (N^{\beta })) + N^{\beta } ((x ~\text {div}~ (N^{\beta })) ~\text {div}~ (N^{\alpha })) + N^{\beta + \alpha }((x ~\text {div}~ (N^{\beta })) ~\text {div}~ (N^{\alpha })). \end{aligned}$$

Therefore since

$$ (x ~\text {mod}~ (N^{\beta })) + N^{\beta } ((x ~\text {div}~ (N^{\beta })) ~\text {div}~ (N^{\alpha }) ) \le N^{\beta }-1 + N^{\beta }(N^{\alpha }-1) = N^{\beta + \alpha } - 1 $$

it follows that

$$ x ~\text {mod}~ (N^{\beta +\alpha }) = (x ~\text {mod}~ (N^{\beta })) + N^{\beta } ((x ~\text {div}~ (N^{\beta })) ~\text {mod}~ (N^{\alpha })). $$

And so

$$ (x ~\text {div}~ (N^{\beta })) ~\text {div}~ (N^{\alpha }) = x ~\text {div}~ (N^{\alpha +\beta }). $$

We use statement 6 to prove 7 by induction. For the base case observe

$$ x = (x ~\text {mod}~ N) + N(x ~\text {div}~ N). $$

For the inductive step, suppose that m is a positive integer and

$$ x = N^{m}(x ~\text {div}~ (N^{m}))+\sum _{k=0}^{m - 1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^{k}. $$

It follows that

$$\begin{aligned} x&= N^{m}(x ~\text {div}~ (N^{m}))+\sum _{k=0}^{m - 1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k \\&= N^{m}(N((x ~\text {div}~ (N^{m})) ~\text {div}~ N) + ((x ~\text {div}~ (N^{m}))) ~\text {mod}~ N)+\sum _{k=0}^{m - 1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k \\&= N^{m+1}(x ~\text {div}~ (N^{m+1}))) + N^{m}((x ~\text {div}~ (N^{m})) ~\text {mod}~ N)+\sum _{k=0}^{m - 1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k \\&= N^{m+1}(x ~\text {div}~ (N^{m+1}))) + \sum _{k=0}^{(m+1) - 1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k. \end{aligned}$$

Statement 8 follows immediately from the definition of $~\text {div}~$ and the fact that $w \mapsto \left\lfloor w/N \right\rfloor$ is an increasing function.

For statement 9, observe that for $x \in \mathbb {Z}$ since $(x ~\text {mod}~ N) \in \{0,\dots ,N-1\}$, $N-1-(x ~\text {mod}~ N) \in \{0,\dots ,N-1\}$, and so

$$ N-1-(x ~\text {mod}~ N) = (N-1-(x ~\text {mod}~ N)) ~\text {mod}~ N = (-1-x) ~\text {mod}~ N. $$

Now suppose that $\alpha \le \beta$. It is evident from the preceding that for $x \in \mathbb {Z}$

$$ x = \sum _{k \ge 0}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k, $$

$$ x ~\text {mod}~ (N^{\alpha }) = \sum _{k=0}^{\alpha -1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k, $$

and

$$ x ~\text {div}~ (N^{\alpha }) = \sum _{k \ge \alpha }((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^{k - \alpha }. $$

A direct derivation gives

$$\begin{aligned} (x ~\text {mod}~ (N^{\beta })) ~\text {div}~ (N^{\alpha })&= \left( \sum _{k=0}^{\beta -1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^k \right) ~\text {div}~ (N^{\alpha }) \\&= \sum _{k=\alpha }^{\beta -1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^{k-\alpha } \\&= \sum _{k=0}^{\beta -\alpha -1}((x ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^{k} \\&= \sum _{k=0}^{\beta -\alpha -1}(((x ~\text {div}~ (N^{\alpha })) ~\text {div}~ (N^{k})) ~\text {mod}~ N)N^{k} \\&= (x ~\text {div}~ (N^{\alpha })) ~\text {mod}~ (N^{\beta -\alpha }). \end{aligned}$$

$\square$