Open in App
Not now

# Discrete logarithm (Find an integer k such that a^k is congruent modulo b)

• Difficulty Level : Medium
• Last Updated : 29 Dec, 2021

Given three integers a, b and m. Find an integer k such that where a and m are relatively prime. If it is not possible for any k to satisfy this relation, print -1.
Examples:

Input: 2 3 5
Output: 3
Explanation:
a = 2, b = 3, m = 5
The value which satisfies the above equation
is 3, because
=> 23 = 2 * 2 * 2 = 8
=> 23 (mod 5) = 8 (mod 5)
=> 3
which is equal to b i.e., 3.

Input: 3 7 11
Output: -1

A Naive approach is to run a loop from 0 to m to cover all possible values of k and check for which value of k, the above relation satisfies. If all the values of k exhausted, print -1. Time complexity of this approach is O(m)
An efficient approach is to use baby-step, giant-step algorithm by using meet in the middle trick.

Baby-step giant-step algorithm

Given a cyclic group G of order ‘m’, a generator ‘a’ of the group, and a group element ‘b’, the problem is to find an integer ‘k’ such that
So what we are going to do(according to Meet in the middle trick) is to split the problem in two parts of each and solve them individually and then find the collision.

Now according to the baby-step giant-step
algorithm, we can write 'k' as
with
and  and .
Therefore, we have:

Therefore in order to solve, we precompute
for different values of 'i'.
Then fix 'b' and tries values of 'j'
In RHS of the congruence relation above. It
tests to see if congruence is satisfied for
any value of 'j', using precomputed
values of LHS.

Let’s see how to use above algorithm for our question:-
First of all we have to write , where Obviously, any value of k in the interval [0, m) can be represented in this form, where and
Replace the ‘k’ in above equality, we get:-

1. The term left and right can take only n distinct values as . Therefore we need to generate all these terms for either left or right part of equality and store them in an array or data structure like map/unordered_map in C/C++ or Hashmap in java.
2. Suppose we have stored all values of LHS. Now iterate over all possible terms on the RHS for different values of j and check which value satisfies the LHS equality.
3. If no value satisfies in above step for any candidate of j, print -1.

## C++

 // C++ program to calculate discrete logarithm #include using namespace std;   /* Iterative Function to calculate (x ^ y)%p in    O(log y) */ int powmod(int x, int y, int p) {     int res = 1;  // Initialize result       x = x % p;  // Update x if it is more than or                 // equal to p       while (y > 0)     {         // If y is odd, multiply x with result         if (y & 1)             res = (res*x) % p;           // y must be even now         y = y>>1; // y = y/2         x = (x*x) % p;     }     return res; }   // Function to calculate k for given a, b, m int discreteLogarithm(int a, int b, int m) {       int n = (int) sqrt (m) + 1;       unordered_map value;       // Store all values of a^(n*i) of LHS     for (int i = n; i >= 1; --i)         value[ powmod (a, i * n, m) ] = i;       for (int j = 0; j < n; ++j)     {         // Calculate (a ^ j) * b and check         // for collision         int cur = (powmod (a, j, m) * b) % m;           // If collision occurs i.e., LHS = RHS         if (value[cur])         {             int ans = value[cur] * n - j;             // Check whether ans lies below m or not             if (ans < m)                 return ans;         }     }     return -1; }   // Driver code int main() {     int a = 2, b = 3, m = 5;     cout << discreteLogarithm(a, b, m) << endl;       a = 3, b = 7, m = 11;     cout << discreteLogarithm(a, b, m); }

## Java

 // Java program to calculate discrete logarithm   class GFG{ /* Iterative Function to calculate (x ^ y)%p in O(log y) */ static int powmod(int x, int y, int p) {     int res = 1; // Initialize result       x = x % p; // Update x if it is more than or                 // equal to p       while (y > 0)     {         // If y is odd, multiply x with result         if ((y & 1)>0)             res = (res*x) % p;           // y must be even now         y = y>>1; // y = y/2         x = (x*x) % p;     }     return res; }   // Function to calculate k for given a, b, m static int discreteLogarithm(int a, int b, int m) {       int n = (int) (Math.sqrt (m) + 1);       int[] value=new int[m];       // Store all values of a^(n*i) of LHS     for (int i = n; i >= 1; --i)         value[ powmod (a, i * n, m) ] = i;       for (int j = 0; j < n; ++j)     {         // Calculate (a ^ j) * b and check         // for collision         int cur = (powmod (a, j, m) * b) % m;           // If collision occurs i.e., LHS = RHS         if (value[cur]>0)         {             int ans = value[cur] * n - j;             // Check whether ans lies below m or not             if (ans < m)                 return ans;         }     }     return -1; }   // Driver code public static void main(String[] args) {     int a = 2, b = 3, m = 5;     System.out.println(discreteLogarithm(a, b, m));       a = 3;     b = 7;     m = 11;     System.out.println(discreteLogarithm(a, b, m)); } } // This code is contributed by mits

## Python3

 # Python3 program to calculate # discrete logarithm import math;   # Iterative Function to calculate # (x ^ y)%p in O(log y) def powmod(x, y, p):       res = 1; # Initialize result       x = x % p; # Update x if it is more                # than or equal to p       while (y > 0):                   # If y is odd, multiply x with result         if (y & 1):             res = (res * x) % p;           # y must be even now         y = y >> 1; # y = y/2         x = (x * x) % p;     return res;   # Function to calculate k for given a, b, m def discreteLogarithm(a, b, m):     n = int(math.sqrt(m) + 1);       value = [0] * m;       # Store all values of a^(n*i) of LHS     for i in range(n, 0, -1):         value[ powmod (a, i * n, m) ] = i;       for j in range(n):                   # Calculate (a ^ j) * b and check         # for collision         cur = (powmod (a, j, m) * b) % m;           # If collision occurs i.e., LHS = RHS         if (value[cur]):             ans = value[cur] * n - j;                           # Check whether ans lies below m or not             if (ans < m):                 return ans;           return -1;   # Driver code a = 2; b = 3; m = 5; print(discreteLogarithm(a, b, m));   a = 3; b = 7; m = 11; print(discreteLogarithm(a, b, m));   # This code is contributed by mits

## C#

 // C# program to calculate discrete logarithm using System; class GFG{ /* Iterative Function to calculate (x ^ y)%p in O(log y) */ static int powmod(int x, int y, int p) {     int res = 1; // Initialize result       x = x % p; // Update x if it is more than or                 // equal to p       while (y > 0)     {         // If y is odd, multiply x with result         if ((y & 1)>0)             res = (res*x) % p;           // y must be even now         y = y>>1; // y = y/2         x = (x*x) % p;     }     return res; }   // Function to calculate k for given a, b, m static int discreteLogarithm(int a, int b, int m) {       int n = (int) (Math.Sqrt (m) + 1);       int[] value=new int[m];       // Store all values of a^(n*i) of LHS     for (int i = n; i >= 1; --i)         value[ powmod (a, i * n, m) ] = i;       for (int j = 0; j < n; ++j)     {         // Calculate (a ^ j) * b and check         // for collision         int cur = (powmod (a, j, m) * b) % m;           // If collision occurs i.e., LHS = RHS         if (value[cur]>0)         {             int ans = value[cur] * n - j;             // Check whether ans lies below m or not             if (ans < m)                 return ans;         }     }     return -1; }   // Driver code static void Main() {     int a = 2, b = 3, m = 5;     Console.WriteLine(discreteLogarithm(a, b, m));       a = 3;     b = 7;     m = 11;     Console.WriteLine(discreteLogarithm(a, b, m)); } } // This code is contributed by mits

## PHP

 0)     {         // If y is odd, multiply x with result         if ($y & 1)$res = ($res *$x) % $p; // y must be even now$y = $y >> 1; // y = y/2$x = ($x *$x) % $p; } return$res; }   // Function to calculate k for given a, b, m function discreteLogarithm($a,$b, $m) {$n = (int)sqrt($m) + 1;$value = array_fill(0, $m, NULL); // Store all values of a^(n*i) of LHS for ($i = $n;$i >= 1; --$i)$value[ powmod ($a,$i * $n,$m) ] = $i; for ($j = 0; $j <$n; ++$j) { // Calculate (a ^ j) * b and check // for collision$cur = (powmod ($a,$j, $m) *$b) % $m; // If collision occurs i.e., LHS = RHS if ($value[$cur]) {$ans = $value[$cur] * $n -$j;                           // Check whether ans lies below m or not             if ($ans <$m)                 return $ans; } } return -1; } // Driver code$a = 2; $b = 3;$m = 5; echo discreteLogarithm($a,$b, $m), "\n";$a = 3; $b = 7;$m = 11; echo discreteLogarithm($a,$b, \$m), "\n";   // This code is contributed by ajit. ?>

## Javascript



Output:

3
-1

Time complexity: O(sqrt(m)*log(b))
Auxiliary space: O(sqrt(m))
A possible improvement is to get rid of binary exponentiation or log(b) factor in the second phase of the algorithm. This can be done by keeping a variable that multiplies by ‘a’ each time as ‘an’. Let’s see the program to understand more.

## C++

 // C++ program to calculate discrete logarithm #include using namespace std;   int discreteLogarithm(int a, int b, int m) {     int n = (int) sqrt (m) + 1;       // Calculate a ^ n     int an = 1;     for (int i = 0; i value;       // Store all values of a^(n*i) of LHS     for (int i = 1, cur = an; i<= n; ++i)     {         if (! value[ cur ])             value[ cur ] = i;         cur = (cur * an) % m;     }       for (int i = 0, cur = b; i<= n; ++i)     {         // Calculate (a ^ j) * b and check         // for collision         if (value[cur])         {             int ans = value[cur] * n - i;             if (ans < m)                 return ans;         }         cur = (cur * a) % m;     }     return -1; }   // Driver code int main() {     int a = 2, b = 3, m = 5;     cout << discreteLogarithm(a, b, m) << endl;       a = 3, b = 7, m = 11;     cout << discreteLogarithm(a, b, m); }

## Java

 // Java program to calculate discrete logarithm   class GFG {       static int discreteLogarithm(int a, int b, int m)     {         int n = (int) (Math.sqrt (m) + 1);           // Calculate a ^ n         int an = 1;         for (int i = 0; i < n; ++i)             an = (an * a) % m;           int[] value=new int[m];           // Store all values of a^(n*i) of LHS         for (int i = 1, cur = an; i <= n; ++i)         {             if (value[ cur ] == 0)                 value[ cur ] = i;             cur = (cur * an) % m;         }           for (int i = 0, cur = b; i <= n; ++i)         {             // Calculate (a ^ j) * b and check             // for collision             if (value[cur] > 0)             {                 int ans = value[cur] * n - i;                 if (ans < m)                     return ans;             }             cur = (cur * a) % m;         }         return -1;     }       // Driver code     public static void main(String[] args)     {         int a = 2, b = 3, m = 5;         System.out.println(discreteLogarithm(a, b, m));         a = 3;         b = 7;         m = 11;         System.out.println(discreteLogarithm(a, b, m));     } }   // This code is contributed by mits

## Python3

 # Python3 program to calculate # discrete logarithm import math;   def discreteLogarithm(a, b, m):       n = int(math.sqrt (m) + 1);       # Calculate a ^ n     an = 1;     for i in range(n):         an = (an * a) % m;       value = [0] * m;       # Store all values of a^(n*i) of LHS     cur = an;     for i in range(1, n + 1):         if (value[ cur ] == 0):             value[ cur ] = i;         cur = (cur * an) % m;           cur = b;     for i in range(n + 1):                   # Calculate (a ^ j) * b and check         # for collision         if (value[cur] > 0):             ans = value[cur] * n - i;             if (ans < m):                 return ans;         cur = (cur * a) % m;       return -1;   # Driver code a = 2; b = 3; m = 5; print(discreteLogarithm(a, b, m));   a = 3; b = 7; m = 11; print(discreteLogarithm(a, b, m));   # This code is contributed by mits

## C#

 // C# program to calculate discrete logarithm using System;   class GFG {       static int discreteLogarithm(int a, int b, int m) {     int n = (int) (Math.Sqrt (m) + 1);       // Calculate a ^ n     int an = 1;     for (int i = 0; i < n; ++i)         an = (an * a) % m;       int[] value = new int[m];       // Store all values of a^(n*i) of LHS     for (int i = 1, cur = an; i<= n; ++i)     {         if (value[ cur ] == 0)             value[ cur ] = i;         cur = (cur * an) % m;     }       for (int i = 0, cur = b; i<= n; ++i)     {         // Calculate (a ^ j) * b and check         // for collision         if (value[cur] > 0)         {             int ans = value[cur] * n - i;             if (ans < m)                 return ans;         }         cur = (cur * a) % m;     }     return -1; }   // Driver code static void Main() {     int a = 2, b = 3, m = 5;     Console.WriteLine(discreteLogarithm(a, b, m));       a = 3;     b = 7;     m = 11;     Console.WriteLine(discreteLogarithm(a, b, m)); } }   // This code is contributed by mits



## Javascript



Output:

3
-1

Time complexity: O(sqrt(m))
Auxiliary space: O(sqrt(m))
Reference:
http://e-maxx-eng.appspot.com/algebra/discrete-log.html
https://en.wikipedia.org/wiki/Baby-step_giant-step
This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

My Personal Notes arrow_drop_up
Related Articles