Modular exponentiation

Modular exponentiation is a type of exponentiation performed over a modulus. It is particularly useful in computer science, especially in the field of cryptography.

Generally, modular exponentiation problems take the form where given base b, exponent e, and modulus m, one wishes to calculate c such that:

$c\equiv b^{e}{\pmod {m}}$

If b, e, and m are non-negative and b < m, then a unique solution c exists and has the property 0 ≤ c < m. For example, given b = 5, e = 3, and m = 13, the solution c works out to be 8.

Modular exponentiation problems similar to the one described above are considered easy to do, even if the numbers involved are enormous. On the contrary, computing the discrete logarithm (finding b given c, e, and m) is believed to be difficult. This one way function behavior makes modular exponentiation a good candidate for use in cryptographic algorithms.

Straightforward method

The most straightforward method to calculating a modular exponent is to calculate b^e directly, then to take this number modulo m. Consider trying to compute c, given b = 4, e = 13, and m = 497:

$c\equiv 4^{13}{\pmod {497}}$

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 b^e is 10 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 b^e is 1304 decimal digits in length. Such calculations are possible on modern computers, but the sheer enormity of such numbers causes the speed of calculations to slow considerably. As b and e increase even further to provide better security, the value b^e becomes unwieldy.

The time required to perform the exponentiation depends on the operating environment and the processor. If exponentiation is performed as a series of multiplications, then this requires O(e) time to complete.

Memory-efficient method

A second method to compute modular exponentiation requires more operations than the first method. Because the required memory footprint 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:

$c\equiv (a\cdot b){\pmod {m}}$

$c\equiv (a\ ({\mbox{mod}}\ m))\cdot (b\ ({\mbox{mod}}\ m)){\pmod {m}}$

The algorithm follows:

1. Set c = 1, e' = 0.
2. Increase e' by 1.
3. Set $c\equiv (b\cdot c){\pmod {m}}$.
4. If e' < e, goto step 2. Else, c contains the correct solution to $c\equiv b^{e}{\pmod {m}}$.

Note that in every pass through step 3, the equation $c\equiv b^{e'}{\pmod {m}}$ holds true. When step 3 has been executed e times, then, c contains the answer that was sought.

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) time to complete. However, since the numbers used in these calculations are much smaller than the numbers used in the first algorithm's calculations, the constant factor in this method tends to be smaller.

Most efficient method

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 binary exponentiation (also known as exponentiation by squaring).

First, it is required that the exponent e be converted to binary notation. That is, e can be written as:

$e=\sum _{i=0}^{n-1}a_{i}2^{i}$

In such notation, the length of e is n bits. a_i can take the value 0 or 1 for any i such that 0 ≤ i < n - 1. By definition, a_{n - 1} = 1.

The value b^e can then be written as:

$b^{e}=b^{\left(\sum _{i=0}^{n-1}a_{i}2^{i}\right)}=\prod _{i=0}^{n-1}\left(b^{2^{i}}\right)^{a_{i}}$

The solution c is therefore:

$c\equiv \prod _{i=0}^{n-1}\left(b^{2^{i}}\right)^{a_{i}}{\mbox{m}}$

Such an algorithm can be easily implemented in a programming language that includes operator overloading and garbage collection. The following example is in C#. Let the class Bignum represent an arbitrarily large positive integer. Input base is the base (b), exp is the exponent (e), and m is the modulus.

Bignum modpow(Bignum base, Bignum exp, Bignum m) {

   Bignum result = 1;

   while (exp > 0) {
      if (exp & 1 > 0) result = (result * base) % m;
      exp >>= 1;
      base = (base * base) % m;
   }

   return result;

}

This code, based on that on page 244 of Bruce Schneier's Applied Cryptography, 2e, ISBN 0471117099, uses a single while loop to perform all work necessary to compute the modular exponentiation.

Note that upon entering the loop for the first time, the code variable base is equivalent to b. However, the repeated squaring in the third line of code ensures that at the completion of every loop, the variable base is equivalent to $b^{2^{i}}\ ({\mbox{mod}}\ m)$, where i is the number of times the loop has been iterated. (This makes i the next working bit of the binary exponent exp, where the least-significant bit is exp₀).

The first line of code simply carries out the multiplication in $\prod _{i=0}^{n-1}\left(b^{2^{i}}\right)^{a_{i}}\ ({\mbox{mod}}\ m)$. If a_i is zero, no code executes since this effectively multiplies the running total by one. If a_i instead is one, the variable base (containing the value $b^{2^{i}}\ ({\mbox{mod}}\ m)$ of the original base) is simply multiplied in.

Finally, the example b = 4, e = 13, and m = 497 is put through this test. Note that e is 1101 in binary notation. Because e is four binary digits in length, the loop executes only four times:

• Upon entering the loop for the first time, variables base = 4, exp = 1101 (binary), and result = 1. Because the right-most bit of exp is 1, result is changed to be (1 * 4) % 497, or 4. exp 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 exp is 0, causing result to retain its present value of 4. exp 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 exp is 1. result is changed to be (4 * 256) % 497, or 30. exp 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 exp is 1. result is changed to be (30 * 429) % 497, or 445. exp is right-shifted to become 0, and base is squared to be (429 * 429) % 497, or 151.

The loop then terminates since exp is zero, and the result 445 is returned. This agrees with the previous two algorithms.

The running time of this algorithm is O(log e). When working with large values of e, this offers a substantial speed benefit over both of the previous two algorithms.