Question

Give a \(\Theta(n \text { lg } n)\) algorithm that computes the reminder when \(x^{n}\) is divided by \(p .\) For simplicity, you may assume that \(n\) is a power of 2 That is, \(n=2^{k}\) for some positive integer \(k\)

Short answer

By repeatedly squaring the base and halving the exponent, you can compute the remainder when \(x^n\) is divided by \(p\) where \(n = 2^k\) in \(O(\log n)\) time complexity.

Step-by-step solution

Initialization

Start with \(base = x\), \(exponent = n\) and \(result = 1\).

Exponentiation

While \(exponent > 0\), perform the following steps: If \(exponent\) is odd, then multiply \(result\) by \(base\) and compute the remainder when dividing by \(p\). If \(exponent\) is even, square the \(base\) and compute the remainder when dividing by \(p\). Then halve the \(exponent\) .

Final result

After the loop finishes, \(result\) will hold the value of \(x^n \mod p\).

Key concepts

Computational Complexity

Computational complexity is a way of measuring the efficiency of an algorithm with respect to its time and space requirements. It tells us how the resources needed by the algorithm grow with the input size.

An algorithm with a complexity of \(\Theta(n \log n)\) indicates that as the input size \(n\) grows, the time complexity grows at a rate proportional to \(n \log n\). This is common in many efficient algorithms, such as mergesort and certain divide-and-conquer algorithms.

• \(\Theta\) notation gives a tight bound, telling us both the upper and lower limits of the algorithm's performance on large inputs.
• It is crucial to understand this as it helps in predicting the behavior of the algorithm on large data sets.

In the context of the given exercise, the algorithm uses \(\Theta(n \log n)\) time complexity to handle the computation of exponentiation by squaring efficiently.

Modular Arithmetic

Modular arithmetic is like the clock arithmetic where numbers wrap around after reaching a certain value, called modulus. It is a key tool in number theory and cryptography.

The expression \(x^n \mod p\) means we calculate \(x^n\) and then find the remainder when it's divided by \(p\). Modular arithmetic keeps numbers manageable by avoiding excessively large numbers in computations.

• For every multiplication, we apply a modulus operation, which helps to reduce number size within our calculations.
• This operation keeps the intermediate numbers smaller, thus preventing overflow which might occur during calculations with very large numbers.

The exercise exploits modular arithmetic by continuously taking the modulus of the intermediate results, ensuring that the numbers involved remain within a practical range.

Exponentiation by Squaring

Exponentiation by squaring is an efficient method for computing powers of a number. It reduces the number of multiplications required compared to naive methods.

This method involves pre-computing multiple powers of 2 and using these to build up to the desired exponent level. When the exponent is even, the base is squared; when odd, the current result is multiplied by the base.

• This method ensures we perform the multiplication operation only \(\log n\) times, drastically reducing the computational load.
• Each squaring cuts the remaining exponent to half, quickly converging to zero.

The exercise uses this method within a loop that effectively calculates \(x^n\). It skips unnecessary multiplications and efficiently updates the result within mod operation constraints, making it ideal for large power calculations.