Modular multiplicative inverse
Last Updated : 24 Jul, 2024
Given two integers A and M, find the modular multiplicative inverse of A under modulo M.
The modular multiplicative inverse is an integer X such that:
A X ≡ 1 (mod M)
Note: The value of X should be in the range {1, 2, ... M-1}, i.e., in the range of integer modulo M. ( Note that X cannot be 0 as A*0 mod M will never be 1). The multiplicative inverse of "A modulo M" exists if and only if A and M are relatively prime (i.e. if gcd(A, M) = 1)
Examples:
Input: A = 3, M = 11
Output: 4
Explanation: Since (4*3) mod 11 = 1, 4 is modulo inverse of 3(under 11).
One might think, 15 also as a valid output as "(15*3) mod 11"
is also 1, but 15 is not in range {1, 2, ... 10}, so not valid.
Input: A = 10, M = 17
Output: 12
Explamation: Since (10*12) mod 17 = 1, 12 is modulo inverse of 10(under 17).
Naive Approach: To solve the problem, follow the below idea:
A naive method is to try all numbers from 1 to m. For every number x, check if (A * X) % M is 1
Below is the implementation of the above approach:
C++ // C++ program to find modular // inverse of A under modulo M #include <bits/stdc++.h> using namespace std; // A naive method to find modular // multiplicative inverse of 'A' // under modulo 'M' int modInverse(int A, int M) { for (int X = 1; X < M; X++) if (((A % M) * (X % M)) % M == 1) return X; } // Driver code int main() { int A = 3, M = 11; // Function call cout << modInverse(A, M); return 0; } // Java program to find modular inverse // of A under modulo M import java.io.*; class GFG { // A naive method to find modulor // multiplicative inverse of A // under modulo M static int modInverse(int A, int M) { for (int X = 1; X < M; X++) if (((A % M) * (X % M)) % M == 1) return X; return 1; } // Driver Code public static void main(String args[]) { int A = 3, M = 11; // Function call System.out.println(modInverse(A, M)); } } /*This code is contributed by Nikita Tiwari.*/ # Python3 program to find modular # inverse of A under modulo M # A naive method to find modulor # multiplicative inverse of A # under modulo M def modInverse(A, M): for X in range(1, M): if (((A % M) * (X % M)) % M == 1): return X return -1 # Driver Code if __name__ == "__main__": A = 3 M = 11 # Function call print(modInverse(A, M)) # This code is contributed by Nikita Tiwari.
// C# program to find modular inverse // of A under modulo M using System; class GFG { // A naive method to find modulor // multiplicative inverse of A // under modulo M static int modInverse(int A, int M) { for (int X = 1; X < M; X++) if (((A % M) * (X % M)) % M == 1) return X; return 1; } // Driver Code public static void Main() { int A = 3, M = 11; // Function call Console.WriteLine(modInverse(A, M)); } } // This code is contributed by anuj_67. <script> // Javascript program to find modular // inverse of a under modulo m // A naive method to find modulor // multiplicative inverse of // 'a' under modulo 'm' function modInverse(a, m) { for(let x = 1; x < m; x++) if (((a % m) * (x % m)) % m == 1) return x; } // Driver Code let a = 3; let m = 11; // Function call document.write(modInverse(a, m)); // This code is contributed by _saurabh_jaiswal. </script> <?php // PHP program to find modular // inverse of A under modulo M // A naive method to find modulor // multiplicative inverse of // A under modulo M function modInverse( $A, $M) { for ($X = 1; $X < $M; $X++) if ((($A%$M) * ($X%$M)) % $M == 1) return $X; } // Driver Code $A = 3; $M = 11; // Function call echo modInverse($A, $M); // This code is contributed by anuj_67. ?> Time Complexity: O(M)
Auxiliary Space: O(1)
Modular multiplicative inverse when M and A are coprime or gcd(A, M)=1:
The idea is to use Extended Euclidean algorithms that take two integers 'a' and 'b', then find their gcd, and also find 'x' and 'y' such that
ax + by = gcd(a, b)
To find the multiplicative inverse of 'A' under 'M', we put b = M in the above formula. Since we know that A and M are relatively prime, we can put the value of gcd as 1.
Ax + My = 1
If we take modulo M on both sides, we get
Ax + My ≡ 1 (mod M)
We can remove the second term on left side as 'My (mod M)' would always be 0 for an integer y.
Ax ≡ 1 (mod M)
So the 'x' that we can find using Extended Euclid Algorithm is the multiplicative inverse of 'A'
Below is the implementation of the above approach:
C++ // C++ program to find multiplicative modulo // inverse using Extended Euclid algorithm. #include <bits/stdc++.h> using namespace std; // Function for extended Euclidean Algorithm int gcdExtended(int a, int b, int* x, int* y); // Function to find modulo inverse of a void modInverse(int A, int M) { int x, y; int g = gcdExtended(A, M, &x, &y); if (g != 1) cout << "Inverse doesn't exist"; else { // m is added to handle negative x int res = (x % M + M) % M; cout << "Modular multiplicative inverse is " << res; } } // Function for extended Euclidean Algorithm int gcdExtended(int a, int b, int* x, int* y) { // Base Case if (a == 0) { *x = 0, *y = 1; return b; } // To store results of recursive call int x1, y1; int gcd = gcdExtended(b % a, a, &x1, &y1); // Update x and y using results of recursive // call *x = y1 - (b / a) * x1; *y = x1; return gcd; } // Driver Code int main() { int A = 3, M = 11; // Function call modInverse(A, M); return 0; } // This code is contributed by khushboogoyal499 // C program to find multiplicative modulo inverse using // Extended Euclid algorithm. #include <stdio.h> // C function for extended Euclidean Algorithm int gcdExtended(int a, int b, int* x, int* y); // Function to find modulo inverse of a void modInverse(int A, int M) { int x, y; int g = gcdExtended(A, M, &x, &y); if (g != 1) printf("Inverse doesn't exist"); else { // m is added to handle negative x int res = (x % M + M) % M; printf("Modular multiplicative inverse is %d", res); } } // C function for extended Euclidean Algorithm int gcdExtended(int a, int b, int* x, int* y) { // Base Case if (a == 0) { *x = 0, *y = 1; return b; } int x1, y1; // To store results of recursive call int gcd = gcdExtended(b % a, a, &x1, &y1); // Update x and y using results of recursive // call *x = y1 - (b / a) * x1; *y = x1; return gcd; } // Driver Code int main() { int A = 3, M = 11; // Function call modInverse(A, M); return 0; } // java program to find multiplicative modulo // inverse using Extended Euclid algorithm. public class GFG { // Global Variables public static int x; public static int y; // Function for extended Euclidean Algorithm static int gcdExtended(int a, int b) { // Base Case if (a == 0) { x = 0; y = 1; return b; } // To store results of recursive call int gcd = gcdExtended(b % a, a); int x1 = x; int y1 = y; // Update x and y using results of recursive // call int tmp = b / a; x = y1 - tmp * x1; y = x1; return gcd; } static void modInverse(int A, int M) { int g = gcdExtended(A, M); if (g != 1) { System.out.println("Inverse doesn't exist"); } else { // m is added to handle negative x int res = (x % M + M) % M; System.out.println( "Modular multiplicative inverse is " + res); } } // Driver code public static void main(String[] args) { int A = 3, M = 11; // Function Call modInverse(A, M); } } // The code is contributed by Gautam goel (gautamgoel962) # Python3 program to find multiplicative modulo # inverse using Extended Euclid algorithm. # Global Variables x, y = 0, 1 # Function for extended Euclidean Algorithm def gcdExtended(a, b): global x, y # Base Case if (a == 0): x = 0 y = 1 return b # To store results of recursive call gcd = gcdExtended(b % a, a) x1 = x y1 = y # Update x and y using results of recursive # call x = y1 - (b // a) * x1 y = x1 return gcd def modInverse(A, M): g = gcdExtended(A, M) if (g != 1): print("Inverse doesn't exist") else: # m is added to handle negative x res = (x % M + M) % M print("Modular multiplicative inverse is ", res) # Driver Code if __name__ == "__main__": A = 3 M = 11 # Function call modInverse(A, M) # This code is contributed by phasing17 // C# program to find multiplicative modulo // inverse using Extended Euclid algorithm. using System; public class GFG { public static int x, y; // Function for extended Euclidean Algorithm static int gcdExtended(int a, int b) { // Base Case if (a == 0) { x = 0; y = 1; return b; } // To store results of recursive call int gcd = gcdExtended(b % a, a); int x1 = x; int y1 = y; // Update x and y using results of recursive // call x = y1 - (b / a) * x1; y = x1; return gcd; } // Function to find modulo inverse of a static void modInverse(int A, int M) { int g = gcdExtended(A, M); if (g != 1) Console.Write("Inverse doesn't exist"); else { // M is added to handle negative x int res = (x % M + M) % M; Console.Write( "Modular multiplicative inverse is " + res); } } // Driver Code public static void Main(string[] args) { int A = 3, M = 11; // Function call modInverse(A, M); } } // this code is contributed by phasing17 <script> // JavaScript program to find multiplicative modulo // inverse using Extended Euclid algorithm. // Global Variables let x, y; // Function for extended Euclidean Algorithm function gcdExtended(a, b){ // Base Case if (a == 0) { x = 0; y = 1; return b; } // To store results of recursive call let gcd = gcdExtended(b % a, a); let x1 = x; let y1 = y; // Update x and y using results of recursive // call x = y1 - Math.floor(b / a) * x1; y = x1; return gcd; } function modInverse(a, m) { let g = gcdExtended(a, m); if (g != 1){ document.write("Inverse doesn't exist"); } else{ // m is added to handle negative x let res = (x % m + m) % m; document.write("Modular multiplicative inverse is ", res); } } // Driver Code { let a = 3, m = 11; // Function call modInverse(a, m); } // This code is contributed by Gautam goel (gautamgoel962) </script> <?php // PHP program to find multiplicative modulo // inverse using Extended Euclid algorithm. // Function to find modulo inverse of a function modInverse($A, $M) { $x = 0; $y = 0; $g = gcdExtended($A, $M, $x, $y); if ($g != 1) echo "Inverse doesn't exist"; else { // m is added to handle negative x $res = ($x % $M + $M) % $M; echo "Modular multiplicative " . "inverse is " . $res; } } // function for extended Euclidean Algorithm function gcdExtended($a, $b, &$x, &$y) { // Base Case if ($a == 0) { $x = 0; $y = 1; return $b; } $x1; $y1; // To store results of recursive call $gcd = gcdExtended($b%$a, $a, $x1, $y1); // Update x and y using results of // recursive call $x = $y1 - (int)($b/$a) * $x1; $y = $x1; return $gcd; } // Driver Code $A = 3; $M = 11; // Function call modInverse($A, $M); // This code is contributed by chandan_jnu ?> OutputModular multiplicative inverse is 4
Time Complexity: O(log M)
Auxiliary Space: O(log M), because of the internal recursion stack.
Iterative Implementation of the above approach:
C++ // Iterative C++ program to find modular // inverse using extended Euclid algorithm #include <bits/stdc++.h> using namespace std; // Returns modulo inverse of a with respect // to m using extended Euclid Algorithm // Assumption: a and m are coprimes, i.e., // gcd(A, M) = 1 int modInverse(int A, int M) { int m0 = M; int y = 0, x = 1; if (M == 1) return 0; while (A > 1) { // q is quotient int q = A / M; int t = M; // m is remainder now, process same as // Euclid's algo M = A % M, A = t; t = y; // Update y and x y = x - q * y; x = t; } // Make x positive if (x < 0) x += m0; return x; } // Driver Code int main() { int A = 3, M = 11; // Function call cout << "Modular multiplicative inverse is " << modInverse(A, M); return 0; } // this code is contributed by shivanisinghss2110 // Iterative C program to find modular // inverse using extended Euclid algorithm #include <stdio.h> // Returns modulo inverse of a with respect // to m using extended Euclid Algorithm // Assumption: a and m are coprimes, i.e., // gcd(A, M) = 1 int modInverse(int A, int M) { int m0 = M; int y = 0, x = 1; if (M == 1) return 0; while (A > 1) { // q is quotient int q = A / M; int t = M; // m is remainder now, process same as // Euclid's algo M = A % M, A = t; t = y; // Update y and x y = x - q * y; x = t; } // Make x positive if (x < 0) x += m0; return x; } // Driver Code int main() { int A = 3, M = 11; // Function call printf("Modular multiplicative inverse is %d\n", modInverse(A, M)); return 0; } // Iterative Java program to find modular // inverse using extended Euclid algorithm class GFG { // Returns modulo inverse of a with // respect to m using extended Euclid // Algorithm Assumption: a and m are // coprimes, i.e., gcd(A, M) = 1 static int modInverse(int A, int M) { int m0 = M; int y = 0, x = 1; if (M == 1) return 0; while (A > 1) { // q is quotient int q = A / M; int t = M; // m is remainder now, process // same as Euclid's algo M = A % M; A = t; t = y; // Update x and y y = x - q * y; x = t; } // Make x positive if (x < 0) x += m0; return x; } // Driver code public static void main(String args[]) { int A = 3, M = 11; // Function call System.out.println("Modular multiplicative " + "inverse is " + modInverse(A, M)); } } /*This code is contributed by Nikita Tiwari.*/ # Iterative Python 3 program to find # modular inverse using extended # Euclid algorithm # Returns modulo inverse of a with # respect to m using extended Euclid # Algorithm Assumption: a and m are # coprimes, i.e., gcd(A, M) = 1 def modInverse(A, M): m0 = M y = 0 x = 1 if (M == 1): return 0 while (A > 1): # q is quotient q = A // M t = M # m is remainder now, process # same as Euclid's algo M = A % M A = t t = y # Update x and y y = x - q * y x = t # Make x positive if (x < 0): x = x + m0 return x # Driver code if __name__ == "__main__": A = 3 M = 11 # Function call print("Modular multiplicative inverse is", modInverse(A, M)) # This code is contributed by Nikita tiwari. // Iterative C# program to find modular // inverse using extended Euclid algorithm using System; class GFG { // Returns modulo inverse of a with // respect to m using extended Euclid // Algorithm Assumption: a and m are // coprimes, i.e., gcd(A, M) = 1 static int modInverse(int A, int M) { int m0 = M; int y = 0, x = 1; if (M == 1) return 0; while (A > 1) { // q is quotient int q = A / M; int t = M; // m is remainder now, process // same as Euclid's algo M = A % M; A = t; t = y; // Update x and y y = x - q * y; x = t; } // Make x positive if (x < 0) x += m0; return x; } // Driver Code public static void Main() { int A = 3, M = 11; // Function call Console.WriteLine("Modular multiplicative " + "inverse is " + modInverse(A, M)); } } // This code is contributed by anuj_67. <script> // Iterative Javascript program to find modular // inverse using extended Euclid algorithm // Returns modulo inverse of a with respect // to m using extended Euclid Algorithm // Assumption: a and m are coprimes, i.e., // gcd(a, m) = 1 function modInverse(a, m) { let m0 = m; let y = 0; let x = 1; if (m == 1) return 0; while (a > 1) { // q is quotient let q = parseInt(a / m); let t = m; // m is remainder now, // process same as // Euclid's algo m = a % m; a = t; t = y; // Update y and x y = x - q * y; x = t; } // Make x positive if (x < 0) x += m0; return x; } // Driver Code let a = 3; let m = 11; // Function call document.write(`Modular multiplicative inverse is ${modInverse(a, m)}`); // This code is contributed by _saurabh_jaiswal </script> <?php // Iterative PHP program to find modular // inverse using extended Euclid algorithm // Returns modulo inverse of a with respect // to m using extended Euclid Algorithm // Assumption: a and m are coprimes, i.e., // gcd(a, m) = 1 function modInverse($A, $M) { $m0 = $M; $y = 0; $x = 1; if ($m == 1) return 0; while ($A > 1) { // q is quotient $q = (int) ($A / $M); $t = $M; // m is remainder now, // process same as // Euclid's algo $M = $A % $M; $A = $t; $t = $y; // Update y and x $y = $x - $q * $y; $x = $t; } // Make x positive if ($x < 0) $x += $m0; return $x; } // Driver Code $A = 3; $M = 11; // Function call echo "Modular multiplicative inverse is: ", modInverse($A, $M); // This code is contributed by Anuj_67 ?> OutputModular multiplicative inverse is 4
Time Complexity: O(log m)
Auxiliary Space: O(1)
Modular multiplicative inverse when M is prime:
If we know M is prime, then we can also use Fermat's little theorem to find the inverse.
aM-1 ≡ 1 (mod M)
If we multiply both sides with a-1, we get
a-1 ≡ a M-2 (mod M)
Below is the implementation of the above approach:
C++ // C++ program to find modular inverse of A under modulo M // This program works only if M is prime. #include <bits/stdc++.h> using namespace std; // To find GCD of a and b int gcd(int a, int b); // To compute x raised to power y under modulo M int power(int x, unsigned int y, unsigned int M); // Function to find modular inverse of a under modulo M // Assumption: M is prime void modInverse(int A, int M) { int g = gcd(A, M); if (g != 1) cout << "Inverse doesn't exist"; else { // If a and m are relatively prime, then modulo // inverse is a^(m-2) mode m cout << "Modular multiplicative inverse is " << power(A, M - 2, M); } } // To compute x^y under modulo m int power(int x, unsigned int y, unsigned int M) { if (y == 0) return 1; int p = power(x, y / 2, M) % M; p = (p * p) % M; return (y % 2 == 0) ? p : (x * p) % M; } // Function to return gcd of a and b int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Driver code int main() { int A = 3, M = 11; // Function call modInverse(A, M); return 0; } // Java program to find modular // inverse of A under modulo M // This program works only if // M is prime. import java.io.*; class GFG { // Function to find modular inverse of a // under modulo M Assumption: M is prime static void modInverse(int A, int M) { int g = gcd(A, M); if (g != 1) System.out.println("Inverse doesn't exist"); else { // If a and m are relatively prime, then modulo // inverse is a^(m-2) mode m System.out.println( "Modular multiplicative inverse is " + power(A, M - 2, M)); } } static int power(int x, int y, int M) { if (y == 0) return 1; int p = power(x, y / 2, M) % M; p = (int)((p * (long)p) % M); if (y % 2 == 0) return p; else return (int)((x * (long)p) % M); } // Function to return gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Driver Code public static void main(String args[]) { int A = 3, M = 11; // Function call modInverse(A, M); } } // This code is contributed by Nikita Tiwari. # Python3 program to find modular # inverse of A under modulo M # This program works only if M is prime. # Function to find modular # inverse of A under modulo M # Assumption: M is prime def modInverse(A, M): g = gcd(A, M) if (g != 1): print("Inverse doesn't exist") else: # If A and M are relatively prime, # then modulo inverse is A^(M-2) mod M print("Modular multiplicative inverse is ", power(A, M - 2, M)) # To compute x^y under modulo M def power(x, y, M): if (y == 0): return 1 p = power(x, y // 2, M) % M p = (p * p) % M if(y % 2 == 0): return p else: return ((x * p) % M) # Function to return gcd of a and b def gcd(a, b): if (a == 0): return b return gcd(b % a, a) # Driver Code if __name__ == "__main__": A = 3 M = 11 # Function call modInverse(A, M) # This code is contributed by Nikita Tiwari. // C# program to find modular // inverse of a under modulo M // This program works only if // M is prime. using System; class GFG { // Function to find modular // inverse of A under modulo // M Assumption: M is prime static void modInverse(int A, int M) { int g = gcd(A, M); if (g != 1) Console.Write("Inverse doesn't exist"); else { // If A and M are relatively // prime, then modulo inverse // is A^(M-2) mod M Console.Write( "Modular multiplicative inverse is " + power(A, M - 2, M)); } } // To compute x^y under // modulo M static int power(int x, int y, int M) { if (y == 0) return 1; int p = power(x, y / 2, M) % M; p = (p * p) % M; if (y % 2 == 0) return p; else return (x * p) % M; } // Function to return // gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Driver Code public static void Main() { int A = 3, M = 11; // Function call modInverse(A, M); } } // This code is contributed by nitin mittal. <script> // Javascript program to find modular inverse of a under modulo m // This program works only if m is prime. // Function to find modular inverse of a under modulo m // Assumption: m is prime function modInverse(a, m) { let g = gcd(a, m); if (g != 1) document.write("Inverse doesn't exist"); else { // If a and m are relatively prime, then modulo // inverse is a^(m-2) mode m document.write("Modular multiplicative inverse is " + power(a, m - 2, m)); } } // To compute x^y under modulo m function power(x, y, m) { if (y == 0) return 1; let p = power(x, parseInt(y / 2), m) % m; p = (p * p) % m; return (y % 2 == 0) ? p : (x * p) % m; } // Function to return gcd of a and b function gcd(a, b) { if (a == 0) return b; return gcd(b % a, a); } // Driver code let a = 3, m = 11; // Function call modInverse(a, m); // This code is contributed by subham348. </script> <?php // PHP program to find modular // inverse of A under modulo M // This program works only if M // is prime. // Function to find modular // inverse of A under modulo // M Assumption: M is prime function modInverse( $A, $M) { $g = gcd($A, $M); if ($g != 1) echo "Inverse doesn't exist"; else { // If A and M are relatively // prime, then modulo inverse // is A^(M-2) mod M echo "Modular multiplicative inverse is " , power($A, $M - 2, $M); } } // To compute x^y under modulo m function power( $x, $y, $M) { if ($y == 0) return 1; $p = power($x, $y / 2, $M) % $M; $p = ($p * $p) % $M; return ($y % 2 == 0)? $p : ($x * $p) % $M; } // Function to return gcd of a and b function gcd($a, $b) { if ($a == 0) return $b; return gcd($b % $a, $a); } // Driver Code $A = 3; $M = 11; // Function call modInverse($A, $M); // This code is contributed by anuj_67. ?> OutputModular multiplicative inverse is 4
Time Complexity: O(log M)
Auxiliary Space: O(log M), because of the internal recursion stack.
Applications:
Computation of the modular multiplicative inverse is an essential step in RSA public-key encryption method.
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How to compute mod of a big number?Given a big number 'num' represented as string and an integer x, find value of "num % a" or "num mod a". Output is expected as an integer. Examples : Input: num = "12316767678678", a = 10 Output: num (mod a) ? 8 The idea is to process all digits one by one and use the property that xy (mod a) ? ((x
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Exponential Squaring (Fast Modulo Multiplication)Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7 Examples: Input : base = 2, exp = 2Output : 4Input : base = 5, exp = 100000Output : 754573817In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the valu
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Trick for modular division ( (x1 * x2 .... xn) / b ) mod (m)Given integers x1, x2, x3......xn, b, and m, we are supposed to find the result of ((x1*x2....xn)/b)mod(m). Example 1: Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007 Naive Method : Simply calculate the product (55*54*53*52*51)= say x,Divide x by 120 a
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Modular MultiplicationGiven three integers a, b, and M, where M is the modulus. Compute the result of the modular multiplication of a and b under modulo M.((a×b) mod M)Examples:Input: a = 5, b = 3, M = 11Output: 4Explanation: a × b = 5 × 3 = 15, 15 % 11 = 4, so the result is 4.Input: a = 12, b = 15, M = 7Output: 5Explana
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Difference between Modulo and ModulusIn the world of Programming and Mathematics we often encounter the two terms "Modulo" and "Modulus". In programming we use the operator "%" to perform modulo of two numbers. It basically finds the remainder when a number x is divided by another number N. It is denoted by : x mod N where x : Dividend
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Modulo Operations in Programming With Negative ResultsIn programming, the modulo operation gives the remainder or signed remainder of a division, after one integer is divided by another integer. It is one of the most used operators in programming. This article discusses when and why the modulo operation yields a negative result. Examples: In C, 3 % 2 r
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Modulo 10^9+7 (1000000007)In most programming competitions, we are required to answer the result in 10^9+7 modulo. The reason behind this is, if problem constraints are large integers, only efficient algorithms can solve them in an allowed limited time.What is modulo operation: The remainder obtained after the division opera
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Fibonacci modulo pThe Fibonacci sequence is defined as F_i = F_{i-1} + F_{i-2} where F_1 = 1 and F_2 = 1 are the seeds. For a given prime number p, consider a new sequence which is (Fibonacci sequence) mod p. For example for p = 5, the new sequence would be 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4 … The minimal zero of the
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Modulo of a large Binary StringGiven a large binary string str and an integer K, the task is to find the value of str % K.Examples: Input: str = "1101", K = 45 Output: 13 decimal(1101) % 45 = 13 % 45 = 13 Input: str = "11010101", K = 112 Output: 101 decimal(11010101) % 112 = 213 % 112 = 101 Approach: It is known that (str % K) wh
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Modular multiplicative inverse from 1 to nGive 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 Ex
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Modular exponentiation (Recursive)Given three numbers a, b and c, we need to find (ab) % cNow why do “% c†after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store suc
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Chinese Remainder Theorem
Introduction to Chinese Remainder TheoremWe are given two arrays num[0..k-1] and rem[0..k-1]. In num[0..k-1], every pair is coprime (gcd for every pair is 1). We need to find minimum positive number x such that: x % num[0] = rem[0], x % num[1] = rem[1], .......................x % num[k-1] = rem[k-1] Basically, we are given k numbers which
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Implementation of Chinese Remainder theorem (Inverse Modulo based implementation)We are given two arrays num[0..k-1] and rem[0..k-1]. In num[0..k-1], every pair is coprime (gcd for every pair is 1). We need to find minimum positive number x such that: x % num[0] = rem[0], x % num[1] = rem[1], ....................... x % num[k-1] = rem[k-1] Example: Input: num[] = {3, 4, 5}, rem[
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Cyclic Redundancy Check and Modulo-2 DivisionCyclic Redundancy Check or CRC is a method of detecting accidental changes/errors in the communication channel. CRC uses Generator Polynomial which is available on both sender and receiver side. An example generator polynomial is of the form like x3 + x + 1. This generator polynomial represents key
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Using Chinese Remainder Theorem to Combine Modular equationsGiven 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
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