Problem 4 - Applied Discrete Structures

Question

Prove by induction that if \(T(n)=1+T(\lfloor n / 2\rfloor), T(0)=0,\) and \(2^{r-1} \leq\) \(n<2^{r}, r \geq 1,\) then \(T(n)=r\)

Step-by-Step Solution

Answer

By induction, \(T(n) = r\) for \(2^{r-1} \leq n < 2^r\).

Step 1: Basis Case

First, verify the base case for the smallest value of \(n\). When \(n = 1\), we have \(n = 2^0\), so \(r = 1\). According to the definition of \(T(n)\), \(T(0)=0\) and \(T(1) = 1 + T(\lfloor 1/2 \rfloor) = 1 + T(0) = 1\). Thus, the base case holds as \(T(1) = 1 = r\).

Step 2: Inductive Hypothesis

Assume that the statement holds for a natural number \(k\), such that if \(2^{r-1} \leq k < 2^r\), then \(T(k) = r\). This is our induction hypothesis.

Step 3: Inductive Step

We need to prove that \(T(k+1) = r\) under the premise \(2^{r-1} \leq k+1 < 2^r\). If \(k+1 = 2^{r-1}\), when divided by 2, \(k+1\) becomes an integer half its size, i.e., \(\lfloor (k+1)/2 \rfloor = 2^{r-2}\). Applying the hypothesis, \(T(2^{r-2}) = r-1\). Now calculate \(T(k+1) = 1 + T(\lfloor (k+1)/2 \rfloor) = 1 + (r-1) = r\).

Step 4: Conclusion

Since both the base case and the inductive step have been verified, by mathematical induction, \(T(n) = r\) for all \(n\) such that \(2^{r-1} \leq n < 2^r\). This completes the proof.

Key Concepts

Recurrence Relations

A recurrence relation is a way to define a sequence of numbers where each term is formulated based on the preceding terms. In our original problem, we deal with a specific recurrence relation:

• \( T(n) = 1 + T(\lfloor n/2 \rfloor) \)
• with \( T(0) = 0 \)

This tells us that the value of \( T(n) \) for any integer \( n \), is determined by the value at half of that integer, rounded down to the closest lesser integer, plus one. Recurrence relations are powerful because they can help break down complex sequences into simpler, more manageable parts. They are often paired with mathematical induction to prove properties about the sequence.

Base Case

In mathematical induction, the base case is the initial step where we verify the statement for the smallest possible value. It's crucial as it anchors the entire inductive proof process.
For our exercise, the base case is when \( n = 1 \). We check whether the function \( T(n) \) behaves as expected at this starting point. We compute:

• \( T(1) = 1 + T(\lfloor 1/2 \rfloor) = 1 + T(0) = 1 \)

Since 1 matches the expected value of \( r = 1 \) when \( n = 1 \) falls under the condition \( 2^0 \leq n < 2^1 \), the base case is proven true which establishes a foundation for the induction to proceed.

Inductive Step

The inductive step is where we assume the statement holds for a particular instance and then prove it for the next. The process can be summarized as follows:

• Assume that for some integer \( k \) satisfying \( 2^{r-1} \leq k < 2^r \), that \( T(k) = r \).
• Next, show that \( T(k+1) = r \) must also hold.

To achieve this:

• Consider \( k+1 \) such that \( 2^{r-1} \leq k+1 < 2^r \).
• Compute \( T(k+1) = 1 + T(\lfloor (k+1)/2 \rfloor) \).
• Since \((k+1)/2\) can be represented as half the power of 2 specified, derive that the floor value leads back to our assumed step \( T(2^{r-2}) = r-1 \), hence it requires simplifying \(1 + (r-1) = r \).

The ability to make the step from \( k \) to \( k+1 \) holds the key to continuing indefinitely, confirming the original statement for all numbers in the specified range.

Floor Function

The floor function, denoted as \( \lfloor x \rfloor \), returns the largest integer less than or equal to a given number \( x \). It plays a significant role in our example as it determines the reduced value of \( n \) in the recurrence relation.

• Given \( \lfloor n/2 \rfloor \), the function effectively shrinks the problem by halving the original value of \( n \) and rounding down.

This concept is crucial as it ensures the sequence approaches a base case quickly and efficiently, allowing the induction process to unfold correctly. Understanding this function helps appreciate the elegance and efficiency of how sequence problems are structured and simplified by recurrence relations.