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In Homomorphic Encryption from Learning with Errors: Conceptually-Simpler, Asymptotically-Faster, Attribute-Based, Gentry et. al defined Flattening as follows;

Let $\vec{a},\vec{b}$ be vectors of some dimention $k$ over $\mathbb{Z}_q$. Let $\ell = \lfloor \log_2q \rfloor +1$ and $N = k \cdot \ell$

Define $\operatorname{BitDecomp}(\vec{a})$ be the $N$-dimentional vector $ = (a_{1,0},\ldots,a_{1,\ell-1} \ldots,a_{k,0},\ldots,a_{k,\ell-1})$, where $a_{i,j}$ is the $j$-th bit in the $a_i$'s binary representation, bit ordered least significant to most significant.

For $\vec{a}' = (a_{1,0},\ldots,a_{1,\ell-1}, \ldots,a_{k,0},\ldots,a_{k,\ell-1})$, let $$\operatorname{BitDecomp}^{-1}(\vec{a}) = (\sum 2^j\cdot a_{1,j}, \ldots, \sum 2^j\cdot a_{k,j}) $$ be the inverse of $\operatorname{BitDecomp}$, but well defined when the input is not a $0/1$ vector.

This is nothing but usual decomposition into bits and the data has already stored in the decomposed way.

For $N$-dimensional vector $\vec{a}'$, let $\operatorname{Flatten}(\vec{a}') = \operatorname{BitDecomp}(\operatorname{BitDecomp}^{-1} (\vec{a}'))$.

When $A$ is a matrix, let $\operatorname{BitDecomp}(A), \operatorname{BitDecomp}^-1(A)$, or $\operatorname{Flatten}(A)$ be matrix formed by applying the operation to each row of $A$ seperately.

An interesting feature of $\operatorname{Flatten}$ is that it makes the coefficients of a vector matrix $small$ without affecting its product with $\operatorname{Powersof2}(\vec{b})$, and without knowing of $\vec{b}'$

It is not clear for me how the $\operatorname{Flatten}$ makes the coefficients of a vector matrix $\boldsymbol{small}$. What I see that the two operations, $\operatorname{BitDecomp}$ and $\operatorname{BitDecomp}$, are just the inverse of each other. Could someone elaborate on what I'm missing here?

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The important part you missed is the last sentence of the first block:

$BitDecdomp^{-1}$ is still well-defined, if it doesn't get a $N$-dimensional bit vector (all entries from $\{0,1\}$).

In the second quote, $Flatten$ is defined as first using $BitDecdomp^{-1}$, and then use $BitDecomp$. So the output of $Flatten$ will be a bit-vector. This will be the same as the input only if that input was also a vector from $\{0,1\}$ - but it is defined for any vector over $\mathbb{Z}_q$.

The basic idea here is to use a continuation from a smaller domain to a larger one.

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  • $\begingroup$ could you elaborate on the last part? as far as, I can see, the dimensions are still the same. $\endgroup$ – kelalaka Feb 7 at 8:37
  • $\begingroup$ @kelalaka The dimensions are not different. The difference in the domain / range is on one hand $\mathbb{Z}_q^d$ and on the other hand $\mathbb{Z}_2^d$. $\endgroup$ – tylo Feb 7 at 10:12

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