Search results
Results From The WOW.Com Content Network
The multiplicative inverse x ≡ a −1 (mod m) may be efficiently computed by solving Bézout's equation a x + m y = 1 for x, y, by using the Extended Euclidean algorithm. In particular, if p is a prime number, then a is coprime with p for every a such that 0 < a < p ; thus a multiplicative inverse exists for all a that is not congruent to ...
The constants R mod N and R 3 mod N can be generated as REDC(R 2 mod N) and as REDC((R 2 mod N)(R 2 mod N)). The fundamental operation is to compute REDC of a product. When standalone REDC is needed, it can be computed as REDC of a product with 1 mod N. The only place where a direct reduction modulo N is necessary is in the precomputation of R ...
The pattern shown by 8 and 16 holds [6] for higher powers 2 k, k > 2: {,}, is the 2-torsion subgroup, so (/) cannot be cyclic, and the powers of 3 are a cyclic subgroup of order 2 k − 2, so: ( Z / 2 k Z ) × ≅ C 2 × C 2 k − 2 . {\displaystyle (\mathbb {Z} /2^{k}\mathbb {Z} )^{\times }\cong \mathrm {C} _{2}\times \mathrm {C} _{2^{k-2}}.}
For 8-bit integers the table of quarter squares will have 2 9 −1=511 entries (one entry for the full range 0..510 of possible sums, the differences using only the first 256 entries in range 0..255) or 2 9 −1=511 entries (using for negative differences the technique of 2-complements and 9-bit masking, which avoids testing the sign of ...
The linear congruence 4x ≡ 5 (mod 10) has no solutions since the integers that are congruent to 5 (i.e., those in ¯) are all odd while 4x is always even. However, the linear congruence 4x ≡ 6 (mod 10) has two solutions, namely, x = 4 and x = 9. The gcd(4, 10) = 2 and 2 does not divide 5, but does divide 6.
In computing, the modulo operation returns the remainder or signed remainder of a division, after one number is divided by another, called the modulus of the operation.. Given two positive numbers a and n, a modulo n (often abbreviated as a mod n) is the remainder of the Euclidean division of a by n, where a is the dividend and n is the divisor.
This formula can be used to generate sets of equations, that can be used in CRT (Chinese remainder theorem): [12] g ( M p , n − 1 ) 2 ≡ − 1 ( mod M p , n ) {\displaystyle g^{\frac {(M_{p,n}-1)}{2}}\equiv -1{\pmod {M_{p,n}}}} , where g is a number such that there exists an x where x 2 ≡ g ( mod M p , n ) {\displaystyle x^{2}\equiv g ...
The values X i are always odd (bit 0 never changes), bits 2 and 1 alternate (the lower 3 bits repeat with a period of 2), the lower 4 bits repeat with a period of 4, and so on. Therefore, the application using these random numbers must use the most significant bits; reducing to a smaller range using a modulo operation with an even modulus will ...