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Chapter 8: Problem 10

Find \(f(n)\) when \(n=2^{k},\) where \(f\) satisfies the recurrence relation \(f(n)=f(n / 2)+1\) with \(f(1)=1\)

Short Answer

Expert verified
The function is \(f(2^k) = k + 1\).

Step by step solution

01

- Understand the Given Recurrence Relation

The given recurrence relation is \(f(n) = f(n / 2) + 1\) with the initial condition \(f(1) = 1\). Our goal is to find the function \(f(n)\) specifically when \(n = 2^k\).
02

- Substitute n

Substitute \(n = 2^k\) into the recurrence relation. Thus, we get \(f(2^k) = f(2^{k-1}) + 1\).
03

- Apply Recursively

Apply the recurrence relation recursively: \[ f(2^k) = f(2^{k-1}) + 1 \] \[ f(2^{k-1}) = f(2^{k-2}) + 1 \] Continue this process until the base case is reached.
04

- Reach the Base Case

Notice that this process continues until reaching the base case where \(f(2^0) = f(1) = 1\).
05

- Sum the Increments

Summing up the recursive increments, we get: \[ f(2^k) = 1 + (k-1) + (k-2) + \ldots + 1 + 1 \] Which simplifies to: \[ f(2^k) = k + 1\]
06

- Summarize the Result

Therefore, the final form of the function is \(f(2^k) = k + 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Conditions
Initial conditions are a fundamental part of solving recurrence relations. They provide the specific starting point or the base case for a recursive function. In our problem, the initial condition is given as \(f(1) = 1\). This condition means that when the input \(n\) is 1, the value of the function \(f(n)\) is also 1. Initial conditions are crucial because:
  • They ensure the recurrence relation has a well-defined starting value.
  • They help in uniquely determining the function values for all other cases.
For example, in our problem, if we don't have the initial condition \(f(1) = 1\), it would be impossible to evaluate \(f(n)\) for any other \(n\). Without initial conditions, the solutions to the recurrence could be ambiguous or undefined.
Recursive Functions
A recursive function is a function that refers to itself within its own definition to break down the problem into smaller and more manageable sub-problems. In the given exercise, the function \(f(n)\) is defined recursively:
\(f(n) = f(n / 2) + 1\)
Here, the function \(f\) at a larger value of \(n\) is defined in terms of the function at a smaller value (specifically \(n/2\)). Key points to understand about recursive functions are:
  • They make use of smaller instances of the same problem, which makes them efficient for problems that can be broken down into subproblems of the same type.
  • They rely significantly on initial conditions to provide a base case.
In our problem, for any value of \(n\) that is a power of 2 (like \(2^k\)), the recurrence relation helps in forming a sequence of smaller and simpler evaluations until the base case is met. Hence, it simplifies the overall computation.
Mathematical Induction
Mathematical induction is a powerful proof technique used to establish the truth of an infinite sequence of propositions. In the context of recurrence relations, induction helps in proving that the function obtained truly satisfies the given recurrence relation and initial conditions. To use mathematical induction:
  1. **Base Case:** Verify the proposition for the initial value (e.g., \(k = 1\)). This involves checking that \(f(2^0) = f(1) = 1\).
  2. **Inductive Step:** Assume the proposition holds for some arbitrary \(k = m\). This is called the inductive hypothesis. For our example, assume that \(f(2^m) = m + 1\).
  3. **Prove for \(k = m+1\):** Using the hypothesis, show that the proposition holds for \(k = m+1\). This would involve proving \(f(2^{m+1}) = (m+1) + 1\) based on the recurrence \(f(2^{m+1}) = f(2^m) + 1\).
Completing these steps will verify the correctness of the function. For our example, this means proving \(f(2^k) = k + 1\) for all \(k\) by induction.

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Most popular questions from this chapter

a) Find the recurrence relation satisfied by \(R_{n},\) where \(R_{n}\) is the number of regions that a plane is divided into by \(n\) lines, if no two of the lines are parallel and no three of the lines go through the same point. b) Find \(R_{n}\) using iteration.

Generating functions are useful in studying the number of different types of partitions of an integer \(n .\) A partition of a positive integer is a way to write this integer as the sum of positive integers where repetition is allowed and the or- der of the integers in the sum does not matter. For exam- ple, the partitions of 5 (with no restrictions) are \(1+1+1+1+\) \(1+1,1+1+1+2,1+1+3,1+2+2,1+4,2+3,\) and \(5 .\) Exercises \(53-58\) illustrate some of these uses. Show that if \(n\) is a positive integer, then the number of partitions of \(n\) into distinct parts equals the number of partitions of \(n\) into odd parts with repetitions allowed; that is, \(p_{o}(n)=p_{d}(n) .[\text { Hint: Show that the generating }\) functions for \(p_{o}(n)\) and \(p_{d}(n)\) are equal. \(]\)

Exercises 33鈥�37 deal with a variation of the Josephus problem described by Graham, Knuth, and Patashnik in [GrKnPa94]. This problem is based on an account by the historian Flavius Josephus, who was part of a band of 41 Jewish rebels trapped in a cave by the Romans during the Jewish Roman war of the first century. The rebels preferred suicide to capture; they decided to form a circle and to repeatedly count off around the circle, killing every third rebel left alive. However, Josephus and another rebel did not want to be killed this way; they determined the positions where they should stand to be the last two rebels remaining alive. The variation we consider begins with n people, numbered 1 to n, standing around a circle. In each stage, every second person still left alive is eliminated until only one survives. We denote the number of the survivor by J(n). Determine the value of \(J(n)\) for each integer \(n\) with \(1 \leq\) \(n \leq 16 .\)

Suppose that \(p\) and \(q\) are distinct primes. Use the principle of inclusion- exclusion to find \(\phi(p q),\) the number of positive integers not exceeding \(p q\) that are relatively prime to \(p q\) .

Exercises \(38-45\) involve the Reve's puzzle, the variation of the Tower of Hanoi puzzle with four pegs and \(n\) disks. Before presenting these exercises, we describe the Frame-Stewart algorithm for moving the disks from peg 1 to peg 4 so that no disk is ever on top of a smaller one. This algorithm, given the number of disks \(n\) as input, depends on a choice of an integer \(k\) with \(1 \leq k \leq n .\) When there is only one disk, move it from peg 1 to peg 4 and stop. For \(n>1\) , the algorithm proceeds recursively, using these three steps. Recursively move the stack of the \(n-k\) smallest disks from peg 1 to peg \(2,\) using all four pegs. Next move the stack of the \(k\) largest disks from peg 1 to peg \(4,\) using the three-peg algorithm from the Tower of Hanoi puzzle without using the peg holding the \(n-k\) smallest disks. Finally, recursively move the smallest \(n-k\) disks to peg \(4,\) using all four pegs. Frame and Stewart showed that to produce the fewest moves using their algorithm, \(k\) should be chosen to be the smallest integer such that \(n\) does not exceed \(t_{k}=k(k+1) / 2,\) the \(k\) th triangular number, that is, \(t_{k-1}
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