Using Chinese Remainder Theorem to Combine Modular equations
Given N modular equations: A ≅ x1mod(m1) . . A ≅ xnmod(mn) Find x in the equation A ≅ xmod(m1*m2*m3..*mn) where mi is prime, or a power of a prime, and i takes values from 1 to n. The input is given as two arrays, the first being an array containing values of each xi, and the second array containing the set of values of each prime. mi Output an integer for the value of x in the final equation.
Consider the two equations A ≅ 2mod(3) A ≅ 3mod(5) Input : 2 3 3 5 Output : 8 Consider the four equations, A ≅ 3mod(4) A ≅ 4mod(7) A ≅ 1mod(9) (32) A ≅ 0mod(11) Input : 3 4 1 0 4 7 9 11 Output : 1243
Explanation : We aim to solve these equations two at a time. We take the first two equations, combine it, and use that result to combine with the third equation, and so on. The process of combining two equations is explained as follows, by taking example 2 for reference:
- A ≅ 3mod(4) and A ≅ 4mod(7) are the two equations that we are provided with at first. Let the resulting equation be some A0 ≅ x0 mod(m1 * m2).
- A0 is given by m1‘ * m1 * x0 + m0‘ * m0 * x1 where m1‘ = modular inverse of m1 modulo m0 and m0‘ = modular inverse of m0 modulo m1
- We can calculate modular inverse using extended euclidean algorithm.
- We find x0 to be A0 mod (m1 * m2)
- We get our new equation to be A ≅ 11mod(28), where A is 95
- We now try to combine this with equation 3, and by a similar method, we get A ≅ 235mod(252), where A = 2503
- And finally, on combining this with equation 4, we get A ≅ 1243mod(2772) where A = 59455 and x = 1243
We observe that 2772 is rightly equal to 4 * 7 * 9 * 11. We have thus found the value of x for the final equation. You can refer to Extended Euclidean Algorithm and Modular multiplicative inverse for extra information on these topics.
Time Complexity : O(l) ,where l is the size of remainder list.
Space Complexity : O(1) ,as we are not using any extra space.
This theorem and algorithm has excellent applications. One very useful application is in calculating nCr % m where m is not a prime number, and Lucas Theorem cannot be directly applied. In such a case, we can calculate the prime factors of m, and use the prime factors one by one as a modulus in our nCr % m equation which we can calculate using Lucas Theorem, and then combine the resulting equations together using the above shown Chinese Remainder Theorem. This article is contributed by Deepak Srivatsav. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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