# 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 ``=` `[``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``), ````" "````; ` ` `  `\$a` `= 3; ` `\$b` `= 7; ` `\$m` `= 11; ` `echo` `discreteLogarithm(``\$a``, ``\$b``, ``\$m``), ````" "````; ` ` `  `// 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 ``=` `[``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 `

## PHP

 ` `

Output:

```3
-1
```

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