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

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


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 thuis 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 of 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 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<int, int> 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 =  * 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.  ?>

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 =  * 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

## PHP

 

Output:

3
-1


Time complexity: O(sqrt(m))
Auxiliary space: O(sqrt(m))

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Improved By : jit_t, Mithun Kumar