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Modular exponentiation can be performed with a negative exponent e by finding the modular multiplicative inverse d of b modulo m using the extended Euclidean algorithm. That is: c = b e mod m = d −e mod m, where e < 0 and b ⋅ d ≡ 1 (mod m). Modular exponentiation is efficient to compute, even for very large integers.
The modular inverse of aR mod N is REDC((aR mod N) −1 (R 3 mod N)). Modular exponentiation can be done using exponentiation by squaring by initializing the initial product to the Montgomery representation of 1, that is, to R mod N, and by replacing the multiply and square steps by Montgomery multiplies.
Here, complexity refers to the time complexity of performing computations on a multitape Turing machine. [1] See big O notation for an explanation of the notation used. Note: Due to the variety of multiplication algorithms, () below stands in for the complexity of the chosen multiplication algorithm.
This can be accomplished via modular exponentiation, which is the slowest part of the algorithm. The gate thus defined satisfies U r = I {\displaystyle U^{r}=I} , which immediately implies that its eigenvalues are the r {\displaystyle r} -th roots of unity ω r k = e 2 π i k / r {\displaystyle \omega _{r}^{k}=e^{2\pi ik/r}} .
Many algorithms for exponentiation do not provide defence against side-channel attacks. Namely, an attacker observing the sequence of squarings and multiplications can (partially) recover the exponent involved in the computation. This is a problem if the exponent should remain secret, as with many public-key cryptosystems.
The first step is relatively slow but only needs to be done once. Modular multiplicative inverses are used to obtain a solution of a system of linear congruences that is guaranteed by the Chinese Remainder Theorem. For example, the system X ≡ 4 (mod 5) X ≡ 4 (mod 7) X ≡ 6 (mod 11) has common solutions since 5,7 and 11 are pairwise coprime ...
Using fast algorithms for modular exponentiation and multiprecision multiplication, the running time of this algorithm is O(k log 2 n log log n) = Õ(k log 2 n), where k is the number of times we test a random a, and n is the value we want to test for primality; see Miller–Rabin primality test for details.
Since the additions, subtractions, and digit shifts (multiplications by powers of B) in Karatsuba's basic step take time proportional to n, their cost becomes negligible as n increases. More precisely, if T(n) denotes the total number of elementary operations that the algorithm performs when multiplying two n-digit numbers, then