Problem 38 - Discrete Mathematics with Applications

Question

Let \(f(n)\) denote the number of bits in the binary representation of a positive integer \(n\). Show that \(f(n)=\mathbf{O}(\lg n)\)

Step-by-Step Solution

Answer

The function f(n), which represents the number of bits in the binary representation of a positive integer n, is upper-bounded by some constant multiple of the logarithm base 2 of n (denoted as lg(n)). We can prove this by showing that there exists a positive constant c such that f(n) <= c * lg(n) for all positive integers n. Since we can observe that f(n) is always greater than or equal to lg(n) + 1 for any positive integer n, we can rewrite the inequality as f(n) <= lg(n) + 1. Let c = 1, then we have f(n) <= 1 * (lg(n) + 1), which holds true for all positive integers n. Therefore, we have proven that f(n) = O(lg(n)).

Step 1: Understanding binary representation and logarithm

In binary representation, each digit position represents a power of 2. So, for a positive integer n, the maximum number of bits required to represent a n can be obtained by finding the highest power of 2 that can be represented by n (i.e., 2^k <= n, k is the highest exponent of 2 that can be represented). A logarithm function, particularly log base 2, can help determine the power of 2 needed to represent a given number n. In fact, the logarithm base 2 of n (denoted as log_2(n) or lg(n)) indicates at which power we need to place 2 in order to get n.

Step 2: Identifying the relationship between f(n) and lg(n)

To understand the relationship between f(n) and lg(n), let's consider a few examples: - n = 1: The binary representation is 1. One bit is needed, so f(1) = 1. Also, lg(1) = 0. - n = 4: The binary representation is 100. Three bits are needed, so f(4) = 3. Also, lg(4) = 2. In both examples, we can observe that f(n) is always greater than or equal to lg(n) + 1 for any positive integer n (i.e., f(n) >= lg(n) + 1).

Step 3: Proving f(n) = O(lg(n))

Now, we need to prove that f(n) is upper-bounded by some constant multiple of lg(n). Let's show that there exists a positive constant c such that f(n) <= c * lg(n) for all positive integers n: Since f(n) >= lg(n) + 1, we can rewrite the inequality as f(n) <= lg(n) + 1. Let c = 1, then we have: f(n) <= 1 * (lg(n) + 1) This inequality holds true for all positive integers n, and therefore, we have proven that f(n) = O(lg(n)).

Key Concepts

Binary Representation

Binary representation is a way of expressing numbers using only two digits: 0 and 1. This system is called "binary" because it relies on powers of 2 for each digit place. For example, the binary number 101 represents:

• \(2^2 = 4\) for the first digit '1',
• \(2^1 = 2\) for the second digit '0',
• \(2^0 = 1\) for the third digit '1'.

Thus, 101 in binary equals 5 in decimal because \(4 + 1 = 5\). This system is efficient for computers because they operate using on/off signals, which correspond to 1s and 0s.

The number of bits needed to represent any positive integer \(n\) is denoted by \(f(n)\). Each bit signifies a power of 2, and the largest bit in the binary form of \(n\) points to the highest power of 2 that fits in \(n\). Understanding this concept helps in determining how compactly information can be stored or transmitted.

Logarithmic Function

A logarithmic function is the inverse of an exponential function. Specifically, the logarithm base 2, written as \(\log_2(n)\) or \(\lg(n)\), asks: "To what power must 2 be raised to yield \(n\)?"

For example:

• \(\lg(8) = 3\) because \(2^3 = 8\).
• \(\lg(16) = 4\) since \(2^4 = 16\).

Logarithmic functions are important in computing because they help explain how complex processes grow. When we talk about the number of bits needed to represent a number, the logarithm gives a concise way to describe growth because it reflects the number of times you must multiply by 2 to reach \(n\).

This understanding paves the way for connecting binary representation \(f(n)\) with \(\lg(n)\). Because \(\lg(n)\) denotes the power of 2 that reaches \(n\), the bits needed are naturally linked. With logarithms, we compactly express storage or calculation requirements, a crucial feature in algorithm analysis.

Upper Bound

The concept of an upper bound is vital in mathematics and computing. It describes a threshold above which a particular function does not grow. When we say \(f(n) = O(\lg(n))\), we're defining an asymptotic upper bound on the number of bits to represent \(n\).

Let's break this down:

• \(O(\lg(n))\) notation captures a growth rate or scaling factor.
• It implies that as \(n\) increases, the function \(f(n)\) will not exceed a constant multiple of \(\lg(n)\).

The exercise demonstrates that there exists a constant \(c\) such that \(f(n) \leq c \cdot \lg(n)\) for all positive integers \(n\).

This is crucial in algorithm analysis and complexity theory because it gives a ceiling on how resource demands, like storage, grow. By proving an upper bound, we ensure efficiency and predictability in computations. This helps in comparing algorithms and deciding on the most practical approach when dealing with large data sets or developing software solutions.