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Give a positive integer n, find modular multiplicative inverse of all integer from 1 to n with respect to a big prime number, say, 'prime'.
The modular multiplicative inverse of a is an integer 'x' such that.
a x ? 1 (mod prime)
Examples:
Input : n = 10, prime = 17
Output : 1 9 6 13 7 3 5 15 2 12
Explanation :
For 1, modular inverse is 1 as (1 * 1)%17 is 1
For 2, modular inverse is 9 as (2 * 9)%17 is 1
For 3, modular inverse is 6 as (3 * 6)%17 is 1
.......
Input : n = 5, prime = 7
Output : 1 4 5 2 3
A simple solution is to one by one find modular inverse for every number.
Output:
1 9 6 13 7 3 5 15 2 12
Time Complexity: O(n*prime)
Auxiliary Space: O(1)
An efficient solution is based on extended Euclid algorithm.
Extended Euclidean algorithm finds integer coefficients x and y such that:
ax + by = gcd(a, b) Let us put b = prime, we get ax + prime * y = gcd(a, prime) We know gcd(a, prime) = 1 because one of the numbers is prime. So we know ax + prime * y = 1 ---- (i) Since prime * y is a multiple of prime, x is modular multiplicative inverse of a. ax ? 1 (mod prime)
We can recursively find x using below expression (see extended Euclid algorithm for details).
if we take for gcd(prime%a,prime) it'll be 1
so (prime%a)*x1+prime*y1 = gcd(prime%a, prime)
=> (prime%a)*x1+prime*y1 = 1 -----(ii)
=>(prime - (prime/a)*a)x1 + prime*y1 = 1
=>-(prime/a)*x1*a+(x1+y1)*prime
using eq(i) and eq(ii) comparing the co-eeficient of a & prime we get
x = -(prime/a)*x1, & y = (x1+y1)
x = inv(a) & x1 = inv(prime%a)We use above relation to compute inverse using previously computed values.
=> inverse(a) = -(prime/a)* inverse(prime % a) % prime => inverse(a) = (prime - (prime/a)) * inverse(prime % a) % prime
(-x % m = (m-x) % m)
We use Dynamic Programming approach that uses above recursive structure.
dp[1] = 1,
dp[2] = dp[17%2]*(17-17/2)%17 = 9
dp[3] = dp[17%3]*(17-17/3)%17 = 6
and so on..
Output:
1 9 6 13 7 3 5 15 2 12
Time Complexity: O(n)
Auxiliary Space: O(n)