Doing a "modular exponentiation" means calculating the remainder when dividing by a positive integer m (called the modulus) a positive integer b (called the base) raised to the e-th power (e is called the exponent). In other words, problems take the form where given base b, exponent e, and modulus m, one wishes to calculate c such that:
For example, given b = 5, e = 3, and m = 13, the solution c is the remainder of dividing by 13, namely the rest of the division 125 / 13, which works out to be 8.
If b, e, and m are non-negative and b < m, then there exists a unique solution c with the property 0 ≤ c < m.
- where e < 0 and
Modular exponentiation problems similar to the one described above are considered easy to do, even if the numbers involved are enormous. On the other hand, computing the discrete logarithm (finding e given b, c, and m) is believed to be difficult. This one way function behavior makes modular exponentiation a good candidate for use in cryptographic algorithms.
The most straightforward method of calculating a modular exponent is to calculate be directly, then to take this number modulo m. Consider trying to compute c, given b = 4, e = 13, and m = 497:
One could use a calculator to compute 413; this comes out to 67,108,864. Taking this value modulo 497, the answer c is determined to be 445.
Note that b is only one digit in length and that e is only two digits in length, but the value be is 8 digits in length.
In strong cryptography, b is often at least 256 binary digits (77 decimal digits). Consider b = 5 × 1076 and e = 17, both of which are perfectly reasonable values. In this example, b is 77 digits in length and e is 2 digits in length, but the value be is 1309 decimal digits in length. Such calculations are possible on modern computers, but the sheer magnitude of such numbers causes the speed of calculations to slow considerably. As b and e increase even further to provide better security, the value be becomes unwieldy.
The time required to perform the exponentiation depends on the operating environment and the processor. The method described above requires O(e) multiplications to complete.
A second method to compute modular exponentiation requires more operations than the first method. Because the required memory is substantially less, however, operations take less time than before. The end result is that the algorithm is faster.
This algorithm makes use of the fact that, given two integers a and b, the following two equations are equivalent:
The algorithm is as follows:
- Set c = 1, e' = 0.
- Increase e' by 1.
- Set .
- If e' < e, goto step 2. Else, c contains the correct solution to .
Note that in every pass through step 3, the equation holds true. When step 3 has been executed e times, then, c contains the answer that was sought. In summary, this algorithm basically counts up e' by ones until e' reaches e, doing a multiply by b and the modulo operation each time it adds one (to ensure the results stay small).
The example b = 4, e = 13, and m = 497 is presented again. The algorithm passes through step 3 thirteen times:
- e' = 1. c = (1 * 4) mod 497 = 4 mod 497 = 4.
- e' = 2. c = (4 * 4) mod 497 = 16 mod 497 = 16.
- e' = 3. c = (16 * 4) mod 497 = 64 mod 497 = 64.
- e' = 4. c = (64 * 4) mod 497 = 256 mod 497 = 256.
- e' = 5. c = (256 * 4) mod 497 = 1024 mod 497 = 30.
- e' = 6. c = (30 * 4) mod 497 = 120 mod 497 = 120.
- e' = 7. c = (120 * 4) mod 497 = 480 mod 497 = 480.
- e' = 8. c = (480 * 4) mod 497 = 1920 mod 497 = 429.
- e' = 9. c = (429 * 4) mod 497 = 1716 mod 497 = 225.
- e' = 10. c = (225 * 4) mod 497 = 900 mod 497 = 403.
- e' = 11. c = (403 * 4) mod 497 = 1612 mod 497 = 121.
- e' = 12. c = (121 * 4) mod 497 = 484 mod 497 = 484.
- e' = 13. c = (484 * 4) mod 497 = 1936 mod 497 = 445.
The final answer for c is therefore 445, as in the first method.
Like the first method, this requires O(e) multiplications to complete. However, since the numbers used in these calculations are much smaller than the numbers used in the first algorithm's calculations, the computation time decreases by a factor of at least O(e) in this method.
In pseudocode, this method can be performed the following way:
function modular_pow(base, exponent, modulus) c := 1 for e_prime = 1 to exponent c := (c * base) mod modulus return c
Right-to-left binary methodEdit
A third method drastically reduces both the number of operations and the memory footprint required to perform modular exponentiation. It is a combination of the previous method and a more general principle called exponentiation by squaring (also known as binary exponentiation).
First, it is required that the exponent e be converted to binary notation. That is, e can be written as:
In such notation, the length of e is n bits. ai can take the value 0 or 1 for any i such that 0 ≤ i < n - 1. By definition, an - 1 = 1.
The value be can then be written as:
The solution c is therefore:
function modular_pow(base, exponent, modulus) result := 1 while exponent > 0 if (exponent & 1) equals 1: result = (result * base) mod modulus exponent := exponent >> 1 base = (base * base) mod modulus return result
Note that upon entering the loop for the first time, the code variable base is equivalent to . However, the repeated squaring in the third line of code ensures that at the completion of every loop, the variable base is equivalent to , where i is the number of times the loop has been iterated. (This makes i the next working bit of the binary exponent exponent, where the least-significant bit is exponent0).
The first line of code simply carries out the multiplication in . If ai is zero, no code executes since this effectively multiplies the running total by one. If ai instead is one, the variable base (containing the value of the original base) is simply multiplied in.
Example: base = 4, exponent = 13, and modulus = 497. Note that exponent is 1101 in binary notation. Because exponent is four binary digits in length, the loop executes only four times:
- Upon entering the loop for the first time, variables base = 4, exponent = 1101 (binary), and result = 1. Because the right-most bit of exponent is 1, result is changed to be (1 * 4) % 497, or 4. exponent is right-shifted to become 110 (binary), and base is squared to be (4 * 4) % 497, or 16.
- The second time through the loop, the right-most bit of exponent is 0, causing result to retain its present value of 4. exponent is right-shifted to become 11 (binary), and base is squared to be (16 * 16) % 497, or 256.
- The third time through the loop, the right-most bit of e is 1. result is changed to be (4 * 256) % 497, or 30. exponent is right-shifted to become 1, and base is squared to be (256 * 256) % 497, or 429.
- The fourth time through the loop, the right-most bit of exponent is 1. result is changed to be (30 * 429) % 497, or 445. exponent is right-shifted to become 0, and base is squared to be (429 * 429) % 497, or 151.
The loop then terminates since exponent is zero, and the result 445 is returned. This agrees with the previous two algorithms.
The running time of this algorithm is O(log exponent). When working with large values of exponent, this offers a substantial speed benefit over both of the previous two algorithms.
Montgomery reduction, for calculating the modulo when the numbers involved are very large.