SOT-177: Add Montgomery term #81
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zk-steve
reviewed
Aug 25, 2024
| This technique replaces division by $n$ with division by a power of $2$, , which is easily performed on a computer since numbers are | ||
| represented in binary form. | ||
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| To achieve this, we need to convert all the numbers we use into Montgomery space. The $n$-residue of $a$ in Montgomery space is given |
zk-steve
reviewed
Aug 25, 2024
| To compute this product, we need to determine $r^{-1}$, which satisfies the property: | ||
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| $r \cdot r^{-1} = 1 \mod n$ | ||
| $\iff r \cdot r^{-1} - n \cdot n' = 1$ |
zk-steve
reviewed
Aug 25, 2024
| We show how to compute $9 * 11 \mod 13$ ($= 8$). We have: | ||
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| - $r = 2^4 = 16$ | ||
| - Using the Extended Euclid algorithm, we determine that $16 ** 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ |
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Suggested change
| - Using the Extended Euclid algorithm, we determine that $16 ** 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ | |
| - Using the Extended Euclid algorithm, we determine that $16 * 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ |
zk-steve
reviewed
Aug 25, 2024
| We show how to compute $x = 7^{10} \mod 13$ ($=4$). We have: | ||
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| - $r = 2^4 = 16$ | ||
| - Using the Extended Euclid algorithm, we determine that $16 ** 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ |
There was a problem hiding this comment.
Suggested change
| - Using the Extended Euclid algorithm, we determine that $16 ** 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ | |
| - Using the Extended Euclid algorithm, we determine that $16 * 9 − 13 * 11 = 1$, thus, $r^{-1} = 9, n' = 11$ |
zk-steve
reviewed
Aug 25, 2024
| Therefore, we can find $r^{-1}$ and $n'$ using the | ||
| [Extended Euclid algorithm](https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm). We can express $\bar{R}$ as follows: | ||
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| $\bar R = \bar a \cdot \bar b \cdot r^{-1} \mod n$ |
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use array to align equations
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